Page 1 :
Current Electricity 1, , Chapter, , 19, , Current Electricity, (5) For a given conductor current does not change with, , Electric Current, (1) The time rate of flow of charge through any crosssection is called current. i  Lim, , Δt 0, , then i , , change in cross-sectional area. In the following figure i1 = i2 = i3, , ΔQ dQ, . If flow is uniform, , Δt, dt, , i2, , i1, , Q, . Current is a scalar quantity. It's S.I. unit is ampere, t, , i3, Fig. 19.2, , (A) and C.G.S. unit is emu and is called biot (Bi), or ab ampere., 1A = (1/10) Bi (ab amp.), , (6) Current due to translatory motion of charge : If n particle, , (2) Ampere of current means the flow of 6.25  1018, , nq, through a given area in time t then i , +, t, , (3) The conventional direction of current is taken to be the, direction of flow of positive charge, i.e. field and is opposite to, i, , , E, , Fig. 19.1, , +, , +, , electrons/sec through any cross-section of the conductor., , the direction of flow of negative charge as shown below., , +, , +, , each having a charge q, pass, , i, , +, Fig. 19.3, , If n particles each having a charge q pass per second per, unit area, the current associated with cross-sectional area A is, i  nqA, , , E, , If there are n particle per unit volume each having a charge, , q and moving with velocity v, the current thorough, cross section, (4) The net charge in a current carrying conductor is zero., , A is i  nqvA, , Table : 19.1 Types of current, Alternating current (ac), (i), , Direct current (dc), (i) (Pulsating dc), , i, , (Constant dc), , i, , i, , +, –, , t, t, , t

Page 2 :
2 Current Electricity, Magnitude and direction, both varies with time, , dc  Inverter  ac, , ac  Rectifier  dc, (ii) Shows heating effect only, , (ii), , Shows heating effect, chemical effect and magnetic effect of, , current, (iii) It’s symbol is, , +, , (iii) It’s symbol is, , ~, , (7) Current due to rotatory motion of charge : If a point, , –, , (1) Current density at point P is given by J , , charge q is moving in a circle of radius r with speed v (frequency, , , angular speed  and time period T) then corresponding, q, qv, qω, current i  q ν  , , T 2 πr 2 π, , r, , q, , ˆ, dA, , dA, i, , di, n, dA, , P, , n, , J, , dA cos, , Fig. 19.5, , Fig. 19.4, , , , i, , J, , , , (2) If the cross-sectional area is not normal to the current,, (8) Current carriers : The charged particles whose flow in a, definite direction constitutes the electric current are called, , but makes an angle  with the direction of current then, J, , current carriers. In different situation current carriers are, different., (i) Solids : In solid conductors like metals current carriers, are free electrons., (ii) Liquids : In liquids current carriers are positive and, negative ions., (iii) Gases : In gases current carriers are positive ions and, free electrons., (iv) Semi conductor : In semi conductors current carriers, are holes and free electrons., , Current Density (J ), , di,  di  JdA cos  J .dA  i   J  dA, dA cos, , (3) If current density J is uniform for a normal crosssection A then J , , i, A, , (4) Current density J is a vector quantity. It's direction is, same as that of E . It's S.I. unit is amp/m2 and dimension [L–2A]., (5) In case of uniform flow of charge through a crosssection normal to it as i  nqvA  J , (6), J  E , , Current, E, , , , i,  nqv ., A, , density relates with electric field as, , ; where  = conductivity and  = resistivity or, , specific resistance of substance., , Drift Velocity, , Current density at any point inside a conductor is defined, , Drift velocity is the average uniform velocity acquired by, , as a vector having magnitude equal to current per unit area, , free electrons inside a metal by the application of an electric, , surrounding that point. Remember area is normal to the, , field which is responsible for current through it. Drift velocity is, , direction of charge flow (or current passes) through that point., , very small it is of the order of 10–4 m/s as compared to thermal, l, , speed (~– 10 5 m / s) of electrons at room temperature., A, , vd, E, +, , V, , –, , Fig. 19.6

Page 3 :
Current Electricity 1039, metallic, , , , lattice, , is, , defined, , as, , relaxation, , time, , mean free path, , . With rise in temperature, , r.m.s. velocityof electrons v rms, , vrms increases consequently  decreases., (2) Mobility : Drift velocity per unit electric field is called, mobility of electron i.e.  , If suppose for a conductor, , vd, m2, . It’s unit is, ., volt  sec, E, , Ohm's Law, , n = Number of electron per unit volume of the conductor, , If the physical conditions of the conductor (length,, , A = Area of cross-section, , temperature, mechanical strain etc.) remains some, then the, , V = potential difference across the conductor, , current flowing through the conductor is directly proportional to, , E = electric field inside the conductor, , the potential difference across it’s two ends i.e. i  V, , i = current, J = current density,  = specific resistance,  =, , 1, conductivity     then current relates with drift velocity as, , , i  neAv, , d, , we can also write, , vd , , , , V  iR where R is a proportionality constant, known as electric, , resistance., (1) Ohm’s law is not a universal law, the substances, which, , i, J, E, E, V, , , , , ., neA ne ne, ne  l n e, , obey ohm’s law are known as ohmic substance., , (1) The direction of drift velocity for electron in a metal is, , opposite to that of applied electric field (i.e. current density J )., v d  E i.e., greater the electric field, larger will be the drift, , (2) Graph between V and i for a metallic conductor is a, straight line as shown. At different temperatures V-i curves are, different., V, , V, , T1, , velocity., , 1, , (2) When a steady current flows through a conductor of, non-uniform cross-section drift velocity varies inversely with, 1, , area of cross-section  v d  , A, , vd1, , i, , A1 < A2, so vd1  vd 2, , i, A1, , , , 1 2, i, , vd 2, , T2, , 2, , i, , (A) Slope of the line, = tan  , , V, , (B) Here tan1 > tan2, So R1 > R2, ,  R, , i, , A2, , i.e. T1 > T2, , Fig. 19.9, , Fig. 19.7, , (3) If diameter (d) of a conductor is doubled, then drift, velocity of electrons inside it will not change. +, +, , V, , –, , V, , –, , Less – d, , More – d, , Same – vd, , Some – vd, Fig. 19.8, , (3) The device or substances which don’t obey ohm’s law, , e.g. gases, crystal rectifiers, thermoionic valve, transistors etc., are known as non-ohmic or non-linear conductors., For these V-i, Crystal, i, , Static resistance R st , (1) Relaxation time () : The time interval between two, successive collisions of electrons with the positive ions in the, , rectifier, , curve is not linear., V, 1, , i, tan , , , , , V, , Fig. 19.10

Page 4 :
1040 Current Electricity, Dynamic resistance R dyn , , V, 1, , I, tan , , gives R2 = R1 [1 +  (t2 – t1)]. This formula gives an approximate, value., , Resistance, , Table 19.2 : Variation of resistance of some electrical material, , (1) The property of substance by virtue of which it, , with temperature, , opposes the flow of current through it, is known as the, Material, , resistance., , Temp. coefficient of, , Variation of resistance, , resistance (), , with temperature rise, , Metals, , Positive, , Increases, , Solid non-metal, , Zero, , Independent, , Semi-conductor, , Negative, , Decreases, , Electrolyte, , Negative, , Decreases, , Ionised gases, , Negative, , Decreases, , Alloys, , Small positive value, , Almost constant, , (2) Formula of resistance : For a conductor if l = length, of a conductor A = Area of cross-section of conductor, n =, No. of free electrons per unit volume in conductor,  =, relaxation, R, , time, , then, , resistance, , of, , conductor, , l, m, l, , . ; where  = resistivity of the material of, 2, A ne  A, , conductor, (3) Unit and dimension : It’s S.I. unit is Volt/Amp. or Ohm ()., Also 1 ohm , , 8, , 1volt 10 emu of potential, , = 109 emu of, 1 Amp 10 1 emu of current, , resistance. It’s dimension is [ML2 T 3 A 2 ] ., (4) Dependence of resistance : Resistance of a conductor, , Resistivity (), Conductivity () and Conductance (C), , depends upon the following factors., , (1) Resistivity : From R  , , (i) Length of the conductor : Resistance of a conductor is, directly proportional to it’s length i.e. R  l and inversely, , R   i.e. resistivity is numerically equal to the resistance of a, , substance having unit area of cross-section and unit length., , 1, proportional to it’s area of cross-section i.e. R , A, , (i) Unit and dimension : It’s S.I. unit is ohm  m and, dimension is [ML3 T 3 A 2 ], , (ii) Temperature : For a conductor, Resistance  temperatur e ., , If, , (ii) It’s formula :  , , R0 = resistance of conductor at 0oC, Rt = resistance of conductor at toC, , and ,  = temperature co-efficient of resistance, then Rt  R0 (1   t   t 2 ) for t > 300oC and, Rt  R0 (1  t ) for t  300oC, , or  , , Rt  R0, R0  t, , If R1 and R2 are the resistances at t1oC and t2oC, R, 1   t1, respectively then 1 , ., R2 1   t2, The value of  is different at different temperature., Temperature, , coefficient, , of, , resistance, , l, ; If l = 1m, A = 1 m2 then, A, , averaged over the, R 2  R1, temperature range t1oC to t2oC is given by  , which, R1 (t 2  t1 ), , m, ne 2, , (iii) Resistivity is the intrinsic property of the substance. It, is independent of shape and size of the body (i.e. l and A)., (iv) For different substances their resistivity is also, different e.g. silver = minimum = 1.6  10–8 -m and fused, quartz, , = maximum  1016 -m, insulator, (Maximum for fused quartz), ,   alloy   semi - conductor , ,  conductor, (Minimum for silver ), , (v) Resistivity depends on the temperature. For metals, , t  0 (1  t) i.e. resitivity increases with temperature., (vi) Resistivity increases with impurity and mechanical, stress.

Page 5 :
Current Electricity 1041, (vii) Magnetic field increases the resistivity of all metals, except iron, cobalt and nickel., (viii) Resistivity of certain substances like selenium,, cadmium, sulphides is inversely proportional to intensity of light, falling upon them., (2) Conductivity : Reciprocal of resistivity is called, conductivity () i.e.  , , 1, , of, , resistances, , before, , l, R1, l, A,  1  2  1, R2 l 2 A1  l 2, , , A,   2, , A, ,  1, , 2, , and, , (2) If radius is given then R , , 2, , , r,   2, , r, ,  1, , (1) If length is given then R  l 2 , , with unit mho/m and dimensions, , , , after, 4, , stretching, , , d,   2, , d, ,  1, , R1  l1, , R 2  l 2, , , , , , , , , , , , 4, , 2, , r , R, 1,  1   2 , R 2  r1 , r4, , 4, , Electrical Conducting Materials For Specific Use, , [M 1 L3 T 3 A 2 ] ., (3) Conductance : Reciprocal of resistance is known as, conductance. C , , Ratio, , 1, 1, It’s unit is, or –1 or “Siemen”., R, , i, , (1) Filament of electric bulb : Is made up of tungsten which, has high resistivity, high melting point., (2) Element of heating devices (such as heater, geyser or, press) : Is made up of nichrome which has high resistivity and, high melting point., , , , (3) Resistances of resistance boxes (standard resistances), V, , : Are made up of alloys (manganin, constantan or nichrome), , Fig. 19.11, , these materials have moderate resistivity which is practically, independent of temperature so that the specified value of, , Stretching of Wire, , resistance does not alter with minor changes in temperature., , If a conducting wire stretches, it’s length increases, area of, cross-section decreases so resistance increases but volume, remain constant., Suppose for a conducting wire before stretching it’s length =, , l1, area of cross-section = A1, radius = r1, diameter = d1, and, resistance R1  , , l1, A1, , (4) Fuse-wire : Is made up of tin-lead alloy (63% tin + 37%, lead). It should have low melting point and high resistivity. It is, used in series as a safety device in an electric circuit and is, designed so as to melt and thereby open the circuit if the current, exceeds a predetermined value due to some fault. The function, of a fuse is independent of its length., , Before stretching, , After stretching, , l1, , i  r 3/2 ., , l2, , , , , Safe current of fuse wire relates with it’s radius as, , (5) Thermistors : A thermistor is a heat sensitive resistor, , Volume remains constant i.e. A1l1 = A2l2, Fig. 19.12, , usually prepared from oxides of various metals such as nickel,, copper, cobalt, iron etc. These compounds are also semiconductor. For thermistors  is very high which may be positive, , After stretching length = l2, area of cross-section = A2,, radius = r2, diameter = d2 and resistance  R2  , , l2, A2, , or negative. The resistance of thermistors changes very rapidly, with change of temperature., i, , V, Fig. 19.13

Page 6 :
1042 Current Electricity, , Thermistors are used to detect small temperature change, and to measure very low temperature., , O, , Orange, , 3, , 103, , Y, , Yellow, , 4, , 104, , G, , Green, , 5, , 105, , B, , Blue, , 6, , 106, , V, , Violet, , 7, , 107, , G, , Grey, , 8, , 108, , W, , White, , 9, , 109, , Colour Coding of Resistance, To remember the sequence of colour code following, , To know the value of resistance colour code is used. These, code are printed in form of set of rings or strips. By reading the, , sentence should kept in memory., , B B R O Y Great Britain Very Good Wife., , values of colour bands, we can estimate the value of resistance., The carbon resistance has normally four coloured rings or, , Grouping of Resistance, (1) Series grouping, , bands say A, B, C and D as shown in following figure., A B, , C, , D, , (i) Same current flows through each resistance but potential, difference distributes in the ratio of resistance i.e. V  R, , Fig. 19.14, , R1, , R2, , R3, , V1, , V2, , V3, , i, +, , Colour band A and B : Indicate the first two significant, , V, , –, , Fig. 19.15, , figures of resistance in ohm., Band C : Indicates the decimal multiplier i.e. the number of, (ii), , zeros that follows the two significant figures A and B., Band D : Indicates the tolerance in percent about the, , than the maximum value of resistance in the combination., (iii) If n identical resistance are connected in series, , indicated value or in other words it represents the percentage, accuracy of the indicated value., , R eq  nR, , The tolerance in the case of gold is  5% and in silver is , 10%. If only three bands are marked on carbon resistance, then, it indicate a tolerance of 20%., Table 19.3 : Colour code for carbon resistance, Letters as an, , Colour, , aid to memory, , Req  R1  R 2  R 3 equivalent resistance is greater, , V' , , and potential difference across each resistance, , V, n, , (2) Parallel grouping, , appeared across each resistance, , Multiplier, , but current distributes in the, , (A, B), , (C), , reverse ratio of their resistance, , Black, , 0, , 10o, , B, , Brown, , 1, , 101, , R, , Red, , 2, , 102, , i2, , (i) Same potential difference, , Figure, , B, , i1, , i.e. i , , 1, R, , i3, i, , R1, R2, R3, V, , Fig. 19.16

Page 7 :
Current Electricity 1043, (ii) Equivalent resistance is given by, or R eq  (R11  R 21  R 31 )1 or R eq , , 1, 1, 1, 1, , , , R eq R1 R 2 R 3, , R1 R 2 R 3, R1 R 2  R 2 R 3  R 2 R1, , Equivalent resistance is smaller than the minimum value of, resistance in the combination., , R eq, , terminals of a cell when it is not supplying any current is called, , (iv) If two resistance in parallel, R1 R 2, Multiplication, , , R1  R 2, Addition, , it’s emf., (2) Potential difference (V) : The voltage across the, , (v) Current through any resistance, , terminals of a cell when it is supplying current to external, ,  Resistance of oppositebranch , i'  i  , , Total resistance, , , , Where i = required current (branch current),, i1, , i = main current, , resistance is called potential difference or terminal voltage., Potential difference is equal to the product of current and, resistance of that given part i.e. V = iR., , R1, , (3) Internal resistance (r) : In case of a cell the opposition, ,  R2, i1  i ,  R1  R 2, , , , , , ,  R1, and i2  i ,  R1  R 2, , , , , , , i, , of electrolyte to the flow of current through it is called internal, i2, , R2, , resistance of the cell. The internal resistance of a cell depends, , Fig. 19.17, , on the distance between electrodes (r  d), area of electrodes [r,  (1/A)] and nature, concentration (r  C) and temperature of, , (vi) In n identical resistance are connected in parallel, R eq , , (1) Emf of cell (E) : The potential difference across the, , i, R, and current through each resistance i' , n, n, , electrolyte [r  (1/ temp.)]., A cell is said to be ideal, if it has zero internal resistance., , Cell in Various Positions, , Cell, , (1) Closed circuit : Cell supplies a constant current in the, R, , circuit., i, , V = iR, , E, r, Fig. 19.19, , The device which converts chemical energy into electrical, energy is known as electric cell. Cell is a source of constant emf, but not constant current., Anode Cathode, , A, , E, Rr, , +, , (ii) Potential difference across the resistance V  iR, , –, , (iii) Potential drop inside the cell = ir, , –, , +, , (i) Current given by the cell i , , +, , –, , –, +, , Symbol of cell, , Electrolyte, Fig. 19.18, , (iv) Equation of cell E  V  ir (E > V), E, , (v) Internal resistance of the cell r    1   R, V, , , (vi) Power dissipated in external resistance (load)

Page 8 :
1044 Current Electricity, (i), , 2, , P  Vi  i 2 R , , V2  E , ,  .R, R, R r, , Power delivered will be maximum when, Pmax , , Rr, , so, , 2, , Maximum current (called short circuit current) flows, , momentarily isc , , E, r, , (ii) Potential difference V = 0, , E, ., 4r, , This statement in generalised from is called “maximum, , power transfer theorem”., , Grouping of Cells, , Pmax = E2/4r, , P, , R=r, R, , Fig. 19.20, , Group of cell is called a battery., (vii) When the cell is being charged i.e. current is given to, the cell then E = V – ir and E < V., , subtractive while their internal resistances are always additive. If, , (2) Open circuit : When no current is taken from the cell it, , dissimilar plates of cells are connected together their emf’s are, added to each other while if their similar plates are connected, , is said to be in open circuit, R, , C, , In series grouping of cell’s their emf’s are additive or, , A, , D, , B, , together their emf’s are subtractive., E1, , E, r, Fig. 19.21, , (i), , Current through the circuit i = 0, , (ii) Potential difference between A and B, VAB = E, , E2, , E1, , E2, , Eeq = E1 – E2 (E1 > E2), , Eeq = E1 + E2, , req = r1 + r2, , req = r1 + r2, Fig. 19.23, , (1) Series grouping : In series grouping anode of one cell is, connected to cathode of other cell and so on. If n identical cells, E, r, , are connected in series, , E, r, , E, r, , E, r, , (iii) Potential difference between C and D, VCD = 0, (3) Short circuit : If two terminals of cell are join together by, a thick conducting wire, R=0, , E, r, Fig. 19.22, , i, R, Fig. 19.24, , (i) Equivalent emf of the combination Eeq  nE, (ii) Equivalent internal resistance req  nr, (iii) Main current = Current from each cell  i , , nE, R  nr, , (iv) Potential difference across external resistance V  iR

Page 9 :
Current Electricity 1045, (v) Potential difference across each cell V ' , , V, n, 2, ,  nE , (vi) Power dissipated in the external circuit  ,  .R,  R  nr , , (vii), , Condition, , for, , maximum, , power, , R  nr, , and, ,  E2 , , Pmax  n , ,  4r , , (viii) This type of combination is used when nr << R., (i) Equivalent emf of the combination Eeq  nE, , (2) Parallel grouping : In parallel grouping all anodes are, , (ii) Equivalent internal resistance of the combination, , connected at one point and all cathode are connected together, at other point. If n identical cells areEconnected, ,r, in parallel, , req, , E, r, , nr, , m, , (iii), , Main, , current, , flowing, , E, r, i, R, , (v) Potential difference across each cell V ' , (vi) Current from each cell i ' , (vii), , Condition, , (i) Equivalent emf Eeq = E, and Pmax  (mn ), , (ii) Equivalent internal resistance Req  r / n, , (v) Current from each cell i' , , Pmax, , maximum, , power, , R, , nr, m, , Kirchoff's Laws, rule or current law (KCL). According to it the algebraic sum of, , i, n, , currents meeting at a junction is zero i.e. i = 0., ,  E , (vi) Power dissipated in the circuit P  ,  .R,  R r/n , , E,  n ,  4r, , i, n, , E2, 4r, , 2, , Condition, , for, , V, n, , (1) Kirchoff’s first law : This law is also known as junction, , across each cell = V = iR, , 2, , load, , (viii) Total number of cell = mn, , E, R r/n, , (iv) potential difference across external resistance = p.d., , (vii), , the, , (iv) Potential difference across load V  iR, , Fig. 19.25, , (iii) Main current i , , through, , nE, mnE, i, , nr mR  nr, R, m, , for, , max., , power, , is, , R  r/n, , i1, i4, , and, , i2, , , , , , , i3, , Fig. 19.27, , (viii) This type of combination is used when nr >> R, (3) Mixed Grouping : If n identical cell’s are connected in, a row and such m row’s are connected in parallel as shown., E, r E, r, 1, , 1, , E, r, n, , 2, , entering the junction must equal the sum of the currents leaving, the junction. i1  i3  i2  i4, (ii) This law is simply a statement of “conservation of, , charge”., , 2, , i, , In a circuit, at any junction the sum of the currents, , m, , V, R, Fig. 19.26

Page 10 :
1046 Current Electricity, (2) Kirchoff’s second law : This law is also known as loop, rule or voltage law (KVL) and according to it “the algebraic sum, of the changes in potential in complete traversal of a mesh, , (iv) The change in voltage in traversing an inductor in the, direction of current is  L, Adi, , L, , (closed loop) is zero”, i.e. V = 0, , dt, , ., , L, , i, , B, L, , (A), , (i) This law represents “conservation of energy”., , di, while in opposite direction it is, dt, A, , di, , L, , i, L, , dt, , di, dt, , B, (B), , Fig. 19.31, , (ii) If there are n meshes in a circuit, the number of, independent equations in accordance with loop rule will be (n – 1)., (3) Sign convention for the application of Kirchoff’s law :, For the application of Kirchoff’s laws following sign convention, are to be considered, , Different Measuring Instruments, , (i) The change in potential in traversing a resistance in, the direction of current is – iR while in the opposite direction +iR, A, , R, , i, , B, , – iR, , A, , R, , i, , B, , + iR, , Fig. 19.28, , (ii) The change in potential in traversing an emf source, from negative to positive terminal is +E while in the opposite, direction – E irrespective of the direction of current in the circuit., E, , A, , B, , E, , A, , (1) Galvanometer : It is an instrument used to detect, small current passing through it by showing deflection., Galvanometers are of different types, , B, , galvanometer,, –E, , +E, , Fig. 19.29, , moving, , magnet, , e.g. moving coil, , galvanometer,, , hot, , wire, , galvanometer. In dc circuit usually moving coil galvanometer are, used., , (iii) The change in potential in traversing a capacitor from, the negative terminal to the positive terminal is , opposite direction , A, , –, , C, , q, ., C, B, , +, , q, , (A), , , , A, , q, , –, , , , C, , q, C, , Fig. 19.30, , C, , q, while in, C, , (i) It’s symbol :, , ; where G is the, , G, , total internal resistance of the galvanometer., (ii) Full scale deflection current : The current required for, full scale deflection in a galvanometer is called full scale, , +, , q, , (B), , B, , deflection current and is represented by ig., (iii) Shunt : The small resistance connected in parallel to, galvanometer coil, in order to control current flowing through the, galvanometer is known as shunt., Table 19.4 : Merits and demerits of shunt, Merits of shunt, , Demerits of shunt

Page 11 :
Current Electricity 1047, To protect the galvanometer, , Shunt, , resistance, , coil from burning, , the sensitivity of galvanometer., , (c) To pass nth part of main current (i.e. ig , , decreases, , the galvanometer, required shunt S , , It can be used to convert any, , i, ) through, n, , G, ., (n  1), , (3) Voltmeter : It is a device used to measure potential, , galvanometer into ammeter of, , difference and is always put in parallel with the ‘circuit element’, , desired range., , (2) Ammeter : It is a device used to measure current and, , across which potential difference, is to be measured., V, , is always connected in series with the ‘element’ through which, current is to be measured., , R, , i, , R, , +, , i, , A, +, , V, , V, , –, , Fig. 19.34, , –, , Fig. 19.32, , (i) The reading of a voltmeter is always lesser than true, value., (i) The reading of an ammeter is always lesser than, actual current in the circuit., , (ii) Greater the resistance of voltmeter, more accurate will, be its reading. A voltmeter is said to be ideal if its resistance is, infinite, i.e., it draws no current from the circuit element for its, , (ii) Smaller the resistance of an ammeter more accurate, will be its reading. An ammeter is said to be ideal if its, resistance r is zero., , operation., (iii) Conversion of galvanometer into voltmeter : A, galvanometer may be converted into a voltmeter by connecting, , (iii) Conversion of galvanometer into ammeter : A, galvanometer may be converted into an ammeter by connecting, , R, a large resistance R in series with the galvanometer, as shown, , a low resistance (called shunt S) in parallel to the galvanometer, , Vg = igG, , V, , i – ig, ig, , (V – Vg), , ig, , S, , G as shown in figure., i, , G, , in the figure., , Fig. 19.35, , G, , Ammeter, Fig. 19.33, , (a) Equivalent resistance of the combination = G + R, (b) According to ohm’s law, (a) Equivalent resistance of the combination , , GS, GS, , (b) G and S are parallel to each other hence both will, have equal potential difference i.e. ig G  (i  ig )S ; which gives, Required shunt S , , ig, (i  i g ), , G, , V = ig (G + R); which gives, , Required series resistance R , ,  V, , V, G , 1G, , , ig,  Vg, , , (c) If nth part of applied voltage appeared across, galvanometer (i.e. Vg , R  (n  1) G ., , V, ) then required series resistance, n

Page 12 :
1048 Current Electricity, (4) Wheatstone bridge, , Potentiometer is a device mainly used to measure emf of a, B, , : Wheatstone bridge is an, arrangement, , of, , P, , four, , resistance which can be, , A, , used to measure one of, , K1, , them in terms of rest. Here, , G, , (1) Circuit diagram : Potentiometer consists of a long, , S, , –, , arms AB and BC are called, , measure internal resistance of a given cell., C, , R, +, , given cell and to compare emf’s of cells. It is also used to, Q, , D, , resistive wire AB of length L (about 6m to 10 m long) made up, K2, , of mangnine or constantan and a battery of known voltage e, , Fig. 19.36, , and internal resistance r called supplier battery or driver cell., , ratio arm and arms AC and BD are called conjugate arms, , Connection of these two forms primary circuit., One terminal of another cell (whose emf E is to be, , (i) Balanced bridge : The bridge is said to be balanced, , measured) is connected at one end of the main circuit and the, , when deflection in galvanometer is zero i.e. no current flows, , other terminal at any point on the resistive wire through a, , through the galvanometer or in other words VB = VD. In the, P, R, balanced condition, , on mutually changing the position, , Q, S, , R, , h, ,r, galvanometer G. This eforms, theKsecondary circuit., Other details, , Primary, are as follows, circuit, , of cell and galvanometer this condition will not change., , Secondary, , (ii) Unbalanced bridge : If the bridge is not balanced current, , circuit, , will flow from D to B if VD > VB i.e. (VA  VD )  (VA  VB ) which, , J, , A, E, , B, , G, Fig. 19.38, , gives PS > RQ., (iii) Applications of wheatstone bridge : Meter bridge, post, office box and Carey Foster bridge are instruments based on, the principle of wheatstone bridge and are used to measure, unknown resistance., , J = Jockey, , (5) Meter bridge : In case of meter bridge, the resistance, , K = Key, , wire AC is 100 cm long. Varying the position of tapping point B,, , R = Resistance of potentiometer wire,, , bridge is balanced. If in balanced position of bridge AB = l, BC, P, R, Q (100  l), (100  l), (100 – l) so that, . Also,  S , , , R, Q SS, l, P, l, R, , P, , = Specific resistance of potentiometer wire., , Rh = Variable resistance which controls the current, through the wire AB, , G, A, , , , R.B., , Q, , B, , l cm, E, , be high but its temperature coefficient of resistance () must be, K, , Fig. 19.37, , (i) The specific resistance () of potentiometer wire must, , C, , (100 – l) cm, , low., (ii) All higher potential points (terminals) of primary and, secondary circuits must be connected together at point A and all, lower potential points must be connected to point B or jockey., , Potentiometer

Page 13 :
Current Electricity 1049, (iii) The value of known potential difference must be, greater than the value of unknown potential difference to be, measured., (iv) The potential gradient must remain constant. For this, the current in the primary circuit must remain constant and the, jockey must not be slided in contact with the wire., (v) The diameter of potentiometer wire must be uniform, everywhere., (2) Potential gradient (x) : Potential difference (or fall in, potential) per unit length of wire is called potential gradient i.e., x, , x, , one direction, If V < E then current will flow in galvanometer circuit in, opposite direction, If V = E then no current will flow in galvanometer circuit, , , , e, V volt, .R ., where V  iR  , , R, , R, , r, L m, h, , , , So, , If V > E then current will flow in galvanometer circuit in, , this condition to known as null deflection position, length l is, known as balancing length., , V iR iρ, e, R, , , , ., L, L, A (R  Rh  r) L, , In balanced condition E  xl, , (i) Potential gradient directly depends upon, , or E  xl , , (a) The resistance per unit length (R/L) of potentiometer, , ,  R, V, iR, e, .  l, l, l  , , L, L,  R  Rh  r  L, , If V is constant then L  l , , wire., (b) The radius of potentiometer wire (i.e. Area of crosssection), (c), , x1, L, l,  1  1, x 2 L2 l2, , (6) Standardization of potentiometer : The process of, determining potential gradient experimentally is known as, , The, , specific, , resistance, , of, , the, , material, , of, , standardization of potentiometer., , potentiometer wire (i.e. ), , e, r, , Rh, , K, , (d) The current flowing through potentiometer wire ( i), J, , (ii) potential gradient indirectly depends upon, , A, , (a) The emf of battery in the primary circuit ( i.e. e), (b) The resistance of rheostat in the primary circuit ( i.e., , Rh), , E0, , B, , G, , E, Fig. 19.40, , (3) Working : Suppose jocky is made to touch a point J on, wire then potential difference between A and J will be, , V  xl, , At this length (l) two potential difference are obtained, Let the balancing length for the standard emf E0 is l0 then, , (i) V due to battery e and, , by the principle of potentiometer E0 = xl0  x , , (ii) E due to unknown cell, e, r, l, J1, , A, , Rh, , K, , G, , J, , J2, , G, , G, , E, Fig. 19.39, , B, , E0, l0

Page 14 :
1050 Current Electricity, (7) Sensitivity of potentiometer : A potentiometer is said to, be more sensitive, if it measures a small potential difference, more accurately., , (i) Initially in secondary circuit key K' remains open and, balancing length (l1) is obtained. Since cell E is in open circuit, , (i) The sensitivity of potentiometer is assessed by its, , so it’s emf balances on length l1 i.e. E = xl1, , potential gradient. The sensitivity is inversely proportional to the, potential gradient., , …. (i), , (ii) Now key K is closed so cell E comes in closed circuit. If, the process of balancing repeated again then potential, , (ii) In order to increase the sensitivity of potentiometer, , difference V balances on length l2 i.e. V = xl2, , (a) The resistance in primary circuit will have to be, , …. (ii), E, , (iii) By using formula internal resistance r    1  . R ', V, , , decreased., (b) The length of potentiometer wire will have to be increased, , l l, r   1 2,  l2, , so that the length may be measured more accuracy., Table 19.5 : Difference between voltmeter and potentiometer, Voltmeter, , Potentiometer, , It’s resistance is high but finite, , Its resistance is infinite, , It draws some current from, , It does not draw any current from, , source of emf, , the source of unknown emf, , The potential difference, , The potential difference, , measured by it is lesser than, , measured by it is equal to, , the actual potential difference, , actual potential difference, , Its sensitivity is low, , Its sensitivity is high, , It is a versatile instrument, , It measures only emf or, , ,  . R', , , , (2) Comparison of emf’s of two cell : Let l1 and l2 be the, balancing lengths with the cells E1 and E2 respectively then E1 =, E1 l1, , E 2 l2 K, e, r, , xl1 and E2 = xl2 , , Rh, , J, B, , A, E1, , G, , 1, , E2, , 2, Fig. 19.42, , potential difference, It is based on deflection, , It is based on zero deflection, , method, , method, , Let E1 > E2 and both are connected in series. If balancing, , Application of Potentiometer, , length is l1 when cell assist each other and it is l2 when they, , (1) To determine the internal resistance of a primary cell, Rh, , K, , e, r, , oppose each other as shown then :, +, , J, , A, , R, , K, Fig. 19.41, , –, , +, , E2, , +, , –, , E1, , –, , –, , E2, , B, , G, , E, , E1, , (E1  E 2 )  xl 1, , , , E1  E 2, l,  1, E1  E 2 l2, , (E1  E 2 )  xl 2, , or, , E1, l l,  1 2, E 2 l1  l2, , +

Page 15 :
Current Electricity 1051, (3) Comparison of resistances : Let the balancing length for, resistance R1 (when XY is connected) is l1 and let balancing, length for resistance R1 + R2 (when YZ is connected) is l2., Rh, , K, , X, , Y, , R1, , i, , G, , iR, L, ,  e, , where L = length of, , potentiometer wire,  = resistance per unit length, l = balancing, length for e, , ammeter readings with the help of potentiometer is called, B, Z, , calibration of ammeter., (i) In the process of calibration of an ammeter the current, flowing in a circuit is measured by an ammeter and the same, , R2, , current is also measured with the help of potentiometer. By, Rh, , K1, , comparing both the values, the errors in the ammeter readings, , 1, , Fig. 19.43, , e, , are determined., A, , K1, , + –, +, , B, , E1, , + –, , Then iR1 = xl1 and i(R1 + R2) = xl2, , R 2 l2  l1, , , R1, l1, , +, , –, , R, , G, , 3, +, , +, , Rh, , K, , 1, 2, , 1, , (4) To determine thermo emf, , A, , iRl, L, , (5) Calibration of ammeter : Checking the correctness of, , J, , A, , (iv) x  i , , –, , A, , –, , K2, Fig. 19.45, , A, , HRB, , B, G, , +, , E0, , –, , G, , 1 2 3, , (ii) For the calibration of an ammeter, 1  standard, Cold ice Hot sand, , Fig. 19.44, , resistance coil is specifically used in the secondary circuit of the, potentiometer, because the potential difference across 1  is, equal to the current flowing through it i.e. V = i., (iii) If the balancing length for the emf E0 is l0 then E0 = xl0, , (i) The value of thermo-emf in a thermocouple for ordinary, temperature difference is very low (10–6 volt). For this the, , x , , E0, l0, , (Process of standardisation), , potential gradient x must be also very low (10–4 V/m). Hence a, , (iv) Let i ' current flows through 1 resistance giving, , high resistance (R) is connected in series with the potentiometer, , potential difference as V '  i' (1)  xl1 where l1 is the balancing, , wire in order to reduce current., (ii) The potential difference across R must be equal to the, emf of standard cell i.e. iR = E0  i , , E0, R, , (iii) The small thermo emf produced in the thermocouple e =, , xl, , length. So error can be found as i  i  i'  i  xl1  i , , E0,  l1, l0, , (6) Calibration of voltmeter, (i) Practical voltmeters are not ideal, because these do not, have infinite resistance. The error of such practical voltmeter

Page 16 :
1052 Current Electricity, can be found by comparing the voltmeter reading with, , conductor is zero., , calculated value of p.d. by potentiometer., (ii) If l0 is balancing length for E0 the emf of standard cell by, connecting 1 and 2 of bi-directional key, then x = E0/l0., (iii) The balancing length l1 for unknown potential difference, , V is given by (by closing 2 and 3) V '  xl1  (E0 / l0 )l1 ., e, ,  For a given conductor JA = i = constant so that J , , Rh, , K1, , + –, +, , A, , +, , V, , 2, , –, , G, , 3, , RB, +, , i.e., J1 A1 = J2 A2 ; this is called equation of continuity, , B, , C, , E0 1, , + –, , 1, A, , K2, , –, , Rh, Fig. 19.46, ,  The drift velocity of electrons is small because of the, frequent collisions suffered by electrons., ,  The small value of drift velocity produces a large amount, If the voltmeter reading is V then the error will be ( V – V), which may be + ve, – ve or zero., , of electric current, due to the presence of extremely large, number of free electrons in a conductor., The propagation of current is almost at the speed of light, and involves electromagnetic process. It is due to this reason, that the electric bulb glows immediately when switch is on., ,  In the absence of electric field, the paths of electrons, between successive collisions are straight line while in, ,  Human body, though has a large resistance of the order of, , presence of electric field the paths are generally curved., , k (say 10 k), is very sensitive to minute currents even as, low as a few mA. Electrocution, excites and disorders the, ,  Free electron density in a metal is given by n , , N Ax d, A, , nervous system of the body and hence one fails to control the, , where NA = Avogadro number, x = number of free electrons, , activity of the body., , per atom, d = density of metal and A = Atomic weight of, ,  dc flows uniformly throughout the cross-section of, , metal., , conductor while ac mainly flows through the outer surface, ,  In the absence of radiation loss, the time in which a fuse will, , area of the conductor. This is known as skin effect., , melt does not depends on it’s length but varies with radius as, ,  It is worth noting that electric field inside a charged, , t  r4 ., , conductor is zero, but it is non zero inside a current carrying, ,  If length (l) and mass (m) of a conducting wire is given, , conductor and is given by E , , V, l, , where V = potential, , difference across the conductor and l = length of the, conductor., + +, +, +Electric, + + field out side the l current carrying, Ein = 0, +, , +, , +, , +, , J1 +, i, +, , A+1, , i, , J2, Ein = V/l, A2, , –, , then R , , l2, ., m, ,  Macroscopic form of Ohm’s law is R , , V, , while it’s, i

Page 17 :
Current Electricity 1053, microscopic form is J =  E., ,  After stretching if length increases by n times then, resistance will increase by n2 times i.e. R 2  n 2 R1 . Similarly, 1, if radius be reduced to, times then area of cross-section, n, , decreases, , 1, n2, , times so the resistance becomes n times i.e., 4, , The longest diagonal (EC or AG) , , The diagonal of face (e.g. AC, ED, ....) , A side (e.g. AB, BC.....) , , R 2  n 4 R1 ., ,  After stretching if length of a conductor increases by x%, , 5, R, 6, 3, R, 4, , 7, R, 12, ,  Resistance of a conducting body is not unique but, , then resistance will increases by 2x % (valid only if x < 10%), , depends on it’s length and area of cross-section i.e. how the, ,  Decoration of lightning in festivals is an example of, , potential difference is applied. See the following figures, , series grouping whereas all household appliances connected, , c, , c, , in parallel grouping., , , , , b, ,  Using n conductors of equal resistance, the number of, , a, , a, , possible combinations is 2n – 1., , b, ,  If the resistance of n conductors are totally different, then, the number of possible combinations will be 2n., ,  If n identical resistances are first connected in series and, then in parallel, the ratio of the equivalent resistance is given, by, , Rp, Rs, , , , n2, ., 1, , Length = a, , Length = b, , Area of cross-section = b  c, , Area of cross-section = a  c, ,  a , , bc, ,  If a wire of resistance R, cut in n equal parts and then, these parts are collected to form a bundle then equivalent, resistance of combination will be, , R, n2, , be, , Rs, , and, , Rp, , R1, , ., , A, , respectively, , Resistance R   , ,  Some standard results for equivalent resistance, , then, , R2, B, , R5, ,  If equivalent resistance of R1 and R2 in series and, parallel, ,  b , , ac, , Resistance R   , , R3, , R4, , 1, 1, R1   Rs  Rs2  4 Rs R p  and R2   Rs  Rs2  4 Rs R p  ., , , 2 , 2 , ,  If a skeleton cube is made with 12 equal resistance each, having resistance R then the net resistance across, H, E, , R1 R2 (R3  R4 )  (R1  R2 )R3 R4  R5 (R1  R2 ) (R3  R4 ), R5 (R1  R2  R3  R4 )  (R1  R3 )(R2  R4 ), , G, F, , D, A, , R AB , , R1, A, , B, , R3, , C, B, , R2, , R2, , R1

Page 18 :
1054 Current Electricity, present in that branch. In practical situation it always happen, because we can never have an ideal cell or battery with zero, R AB , , resistance., , 2 R1 R2  R3 (R1  R2 ), 2 R3  R1  R2, , A, , R1, , R1, , R1, , connected then it will cancel out the effect of two cells e.g. If, , R1, , in the combination of n identical cells (each having emf E and, R3, , R3, B, ,  In series grouping of identical cells. If one cell is wrongly, , R3, , R3, , , , internal resistance r) if x cell are wrongly connected then, equivalent emf, , R2, , R2, , R2, , R2, , Eeq  (n  2 x ) E, , and equivalent internal, , resistance req  nr ., ,  Graphical view of open circuit and closed circuit of a cell., R AB , , , , , , V, , 1/2, 1, 1, (R1  R2 )  (R1  R2 )2  4 R3 (R1  R2 ), 2, 2, , A, , R1, , R1, , R1, R2, , R2, , Vmax =E; i = 0, , R1, imax =E/r ; V = 0, , R2  , , i, , B, ,  If n identical cells are connected in a loop in order, then, R AB , , R, 1 , R1 1  1  4  2, 2 ,  R1, , , , , , , , emf between any two points is zero., E, r, , E, r, , E, r, ,  It is a common misconception that “current in the circuit, , Close, , E, r loop, , will be maximum when power consumed by the load is, maximum.”, , E, r, , n cell, , Actually current i  E /(R  r) is maximum (= E/r) when R =, min = 0 with PL  (E / r)2  0  0 min . while power consumed, by the load E2R/(R + r)2 is maximum (= E2/4r) when R = r and, i  (E / 2r)  max(  E / r)., ,  In parallel grouping of two identical cell having no, internal resistance, R, ,  Emf is independent of the resistance of the circuit and, , R, , E, , depends upon the nature of electrolyte of the cell while, , E, , E, , potential difference depends upon the resistance between the, , E, , two points of the circuit and current flowing through the, circuit., ,  Whenever a cell or battery is present in a branch there, must be some resistance (internal or external or both), , E eq  E, , , , E eq  0, , When two cell’s of different emf and no internal, , resistance are connected in parallel then equivalent emf is, , R, E1

Page 19 :
Current Electricity 1055, indeterminate, note that connecting a wire with a cell with no, resistance is equivalent to short circuiting. Therefore the total, , 1., , Current of 4.8 amperes is flowing through a conductor., The number of electrons per second will be, , current that will be flowing will be infinity., , 2., , (a) 3  1019, , (b) 7.68  10 21, , (c) 7.68  10 20, , (d) 3  10 20, , [CPMT 1986], , When the current i is flowing through a conductor, the, drift velocity is v . If, , 2i, , current is flowed through the, , same metal but having double the area of cross-section,, ,  In the parallel combination of non-identical cell's if they, , then the drift velocity will be, , are connected with reversed polarity as shown then, , (a) v / 4, , (b) v / 2, , (c) v, , (d) 4 v, , equivalent emf, , Eeq , , E1r2  E2r1, r1  r2, , i, , i1, , E1,r1, , i2, , E2, r2, , 3., , drift velocity of electrons will be, , R, ,  Wheatstone bridge is most sensitive if all the arms of, , 4., , [CPMT 1986], , (a) 1010 m / sec, , (b) 10 2 cm / sec, , (c) 10 4 cm / sec, , (d) 10 1 cm / sec, , Every atom makes one free electron in copper. If 1.1, ampere current is flowing in the wire of copper having 1, , bridge have equal resistances i.e. P = Q = R = S, , mm diameter, then the drift velocity (approx.) will be, ,  If the temperature of the conductor placed in the right, , (Density of copper  9  10 3 kg m 3 and atomic weight =, , gap of metre bridge is increased, then the balancing length, , 63), , decreases and the jockey moves towards left., , [CPMT 1989], ,  In Wheatstone bridge to avoid inductive effects the, battery key should be pressed first and the galvanometer key, afterwards., ,  The measurement of resistance by Wheatstone bridge is, , When current flows through a conductor, then the order of, , 5., , (a) 0.3 mm / sec, , (b) 0.1 mm / sec, , (c) 0.2 mm / sec, , (d) 0.2 cm / sec, , Which one is not the correct statement, , not affected by the internal resistance of the cell., , (a) 1 volt  1 coulomb  1 joule, ,  In case of zero deflection in the galvanometer current, , (b) 1 volt  1 ampere  1 joule / second, , flows in the primary circuit of the potentiometer, not in the, , [NCERT 1978], , (c) 1 volt  1 watt  1 H .P., , galvanometer circuit., , (d) Watt-hour can be expressed in eV, ,  A potentiometer can act as an ideal voltmeter., 6., , If a 0.1 % increase in length due to stretching, the, percentage increase in its resistance will be, [MNR 1990; MP PMT 1996; UPSEAT 1999; MP PMT 2000], , Electric Conduction, Ohm's Law and Resistance, , (a) 0.2 %, , (b) 2 %, , (c) 1 %, , (d) 0.1 %

Page 20 :
1056 Current Electricity, 7., , The specific resistance of manganin is 50  10 8 ohm  m ., The resistance of a cube of length 50 cm will be, (a) 10 6 ohm, , (b) 2.5  10 5 ohm, , (c) 10 8 ohm, , (d) 5  10 4 ohm

Page 21 :
Current Electricity 1051, 8., , The resistivity of iron is 1  10 7 ohm  m . The resistance of a iron, wire of particular length and thickness is 1 ohm. If the length and, the diameter of wire both are doubled, then the resistivity in, [CPMT 1983; DPMT 1999], ohm  m will be, (a) 1  10 7, , 9., , (b), , 18., , [CPMT 1984], , 2  10 7, , (c), 19., , 10., , 11., , 12., , 13., , (a) For metallic conductors at low temperature, , 20., , 21., , 22., , 15., , 23., , 16., , 24., , Drift velocity v d varies with the intensity of electric field as per the, relation, [CPMT 1981; BVP 2003], 1, E, , vd  E, , (b) v d , , (c), , v d  constant, , (d) v d  E 2, , 25., , (d) None of the above, , (a) 4 volts, , (b) 4 amperes, , (c) 2 amperes, , (d) 2 volts, , (b) 2.2  10 6 ohm  m, (d) 0.22  10 6 ohm  m, , A certain wire has a resistance R . The resistance of another wire, identical with the first except having twice its diameter is, (a), , 2R, , (b) 0.25 R, , (c), , 4R, , (d) 0 .5 R, , In hydrogen atom, the electron makes 6.6  10 15 revolutions per, , A wire of length, What length of, temperature and, (a) 1.25 m, , 26., , In a conductor 4 coulombs of charge flows for 2 seconds. The value, of electric current will be, [CPMT 1984], , 4.4  10 6 ohm  m, , (d) 1.6  10 19 A, 5 m and radius 1 mm has a resistance of 1 ohm., the wire of the same material at the same, of radius 2 mm will also have a resistance of 1, , ohm, , (b) Mass of the electrons increases, (c) Electron density decreases, 17., , (a) Two times, (b) Four times, (c) Eight times, (d) Sixteen times, A wire 100 cm long and 2.0 mm diameter has a resistance of 0.7, ohm, the electrical resistivity of the material is, , (c) 1 A, , On increasing the temperature of a conductor, its resistance, increases because, [CPMT 1982], (a) Relaxation time decreases, , 1, 3, , second around the nucleus in an orbit of radius 0.5  10 10 m . It, is equivalent to a current nearly, (a) 1 A, (b) 1 mA, , (d) Tungston wire, , (a), , , , A piece of wire of resistance 4 ohms is bent through 180 at its, mid point and the two halves are twisted together, then the, resistance is, [CPMT 1971], (a) 8 ohms, (b) 1 ohm, (c) 2 ohms, (d) 5 ohms, When a[MP, piece, of, aluminium, wire, of, finite length is drawn through a, PET 1989], series of dies to reduce its diameter to half its original value, its, resistance will become, , (c) 1.1  10 6 ohm  m, , (d) For diode when current flows, , (c) Diode, , , , 62.5  10 18 electrons per second are flowing through a wire of, , (a), , (c) For electrolytes when current passes through them, The example for non-ohmic resistance is, [MP PMT 1978], (a) Copper wire, (b) Carbon resistance, , (d), , 3, , 3, , [NCERT 1974; AIIMS 1997; MH CET 2000; UPSEAT 2001;, CBSE PMT 2002], , (b) For metallic conductors at high temperature, , 14., , 1, , , , (b), , area of cross-section 0.1 m 2 , the value of current flowing will be, (a) 1 A, (b) 0.1 A, (c) 10 A, (d) 0.11 A, , MP PMT 2001; Orissa JEE 2002], , (a) 1154 K, (b) 1100 K, (c) 1400 K, (d) 1127 K, When the length and area of cross-section both are doubled, then its, resistance, [MP PET 1989], (a) Will become half, (b) Will be doubled, (c) Will remain the same, (d) Will become four times, The resistance of a wire is 20 ohms. It is so stretched that the length, becomes three times, then the new resistance of the wire will be, (a) 6.67 ohms, (b) 60.0 ohms, (c) 120 ohms, (d) 180.0 ohms, The resistivity of a wire, [MP PMT 1984; DPMT 1982], (a) Increases with the length of the wire, (b) Decreases with the area of cross-section, (c) Decreases with the length and increases with the cross-section, of wire, (d) None of the above statement is correct, Ohm's law is true, , 1, , , (a), , (c) 4  10 7, (d) 8  10 7, The temperature coefficient of resistance for a wire is, 0.00125 / C . At 300K its resistance is 1 ohm. The temperature at, which the resistance becomes 2 ohm is, [IIT 1980; MP PET 2002; KCET 2003;, , The specific resistance of a wire is  , its volume is 3 m 3 and its, resistance is 3 ohms, then its length will be, , 27., , (b) 2.5 m, , (c) 10 m, (d) 20 m, When there is an electric current through a conducting wire along, its length, then an electric field must exist, (a) Outside the wire but normal to it, (b) Outside the wire but parallel to it, (c) Inside the wire but parallel to it, (d) Inside the wire but normal to it, Through a semiconductor, an electric current is due to drift of, (a) Free electrons, (b) Free electrons and holes, (c) Positive and negative ions, (d) Protons

Page 22 :
1052 Current Electricity, 28., , 29., , In an electrolyte 3.2  10 18 bivalent positive ions drift to the right, , (a) 1.0 mm / sec, , per second while 3.6  10 18 monovalent negative ions drift to the, left per second. Then the current is, (a) 1.6 amp to the left, (b) 1.6 amp to the right, (c) 0.45 amp to the right, (d) 0.45 amp to the left, A metallic block has no potential difference applied across it, then, the mean velocity of free electrons is T = absolute temperature of, the block), , (c) 0.1 mm / sec, (d) 0.01 mm / sec, It is easier to start a car engine on a hot day than on a cold day., This is because the internal resistance of the car battery, (a) Decreases with rise in temperature, (b) Increases with rise in temperature, (c) Decreases with a fall in temperature, (d) Does not change with a change in temperature, 5 amperes of current is passed through a metallic conductor. The, charge flowing in one minute in coulombs will be, , (a) Proportional to T, , 30., , 31., , 32., , 33., , 39., , 40., , (b) Proportional to T, (c) Zero, (d) Finite but independent of temperature, The specific resistance of all metals is most affected by, (a) Temperature, (b) Pressure, (c) Degree of illumination, (d) Applied magnetic field, The positive temperature coefficient of resistance is for, (a) Carbon, (b) Germanium, (c) Copper, (d) An electrolyte, The fact that the conductance of some metals rises to infinity at, some temperature below a few Kelvin is called, (a) Thermal conductivity, (b) Optical conductivity, (c) Magnetic conductivity, (d) Superconductivity, Dimensions of a block are 1 cm  1 cm  100 cm . If specific, resistance of its material is 3  10 7 ohm  m , then the resistance, between the opposite rectangular faces is, , [MP PET 1984], , 41., , 34., , 35., , 3  10 9 ohm, , [MP PMT 1993], , 42., , 43., , (b) 3  10 7 ohm, , (c) 3  10 5 ohm, (d) 3  10 3 ohm, In the above question, the resistance between the square faces is, (a), , 3  10 9 ohm, , (b) 3  10 7 ohm, , (c), , 3  10 5 ohm, , (d) 3  10 3 ohm, , There is a current of 20 amperes in a copper wire of 10 6 square, metre area of cross-section. If the number of free electrons per cubic, , 44., , 45., , metre is 10 29 , then the drift velocity is, (a) 125  10 3 m / sec, 3, , (c) 1.25  10 m / sec, 36., , (b) 12.5  10 3 m / sec, (d) 1.25  10, , 4, , 46., , m / sec, , The electric intensity E , current density j and specific resistance, k are related to each other by the relation, , 37., , E  j/k, , (b), , (c), , Ek/j, , (d) k  jE, , 38., , 2R, , (b), , 47., , R, , (c) R / 2, (d) R / 4, There is a current of 1.344 amp in a copper wire whose area of, cross-section normal to the length of the wire is 1 mm 2 . If the, number of free electrons per cm 3 is 8.4  10 22 , then the drift, velocity would be, [CPMT 1990], , (a), , E  I2R, , (b), , E  IR, , (c), , E  R/I, , (d), , E  I/R, , The resistances of a wire at temperatures tC and 0C are, related by, [MP PMT 1993], (a), , R t  R 0 (1   t), , (b), , R t  R 0 (1   t), , (c), , R t  R 02 (1   t), , (d), , R t  R 02 (1   t), , An electric wire of length ‘I’ and area of cross-section a has a, resistance R ohms. Another wire of the same material having same, [MP area, PET 1993], length and, of cross-section 4a has a resistance of, (a) 4R, (b) R/4, (c) R/16, (d) 16R, For which of the following the resistance decreases on increasing the, temperature, [MP PET 1993], (a) Copper, (b) Tungsten, (c) Germanium, (d) Aluminium, If n, e,  and m respectively represent the density, charge relaxation, time and mass of the electron, then the resistance of a wire of, length l and area of cross-section A will be, , (c), , [CPMT 1984; MP PET 2002], , (a), , An electric wire is connected across a cell of e.m.f. E. The current I, is measured by an ammeter of resistance R. According to ohm's law, , (a), , E  jk, , The resistance of a wire of uniform diameter d and length L is, R . The resistance of another wire of the same material but, diameter 2d and length 4 L will be, , (b) Half of the second, (d) Four times of the second, , [CPMT 1992], , [DPMT 2001], , (a), , (a) 5, (b) 12, (c) 1/12, (d) 300, Two wires of the same material are given. The first wire is twice as, long as the second and has twice the diameter of the second. The, resistance of the first will be, (a) Twice of the second, (c) Equal to the second, , [MP PET 1993], , (a), , (b) 1.0 m / sec, , 48., , ml, ne A, 2, , ne 2A, 2ml, , (b), , (d), , m 2 A, ne 2 l, ne 2 A, 2m l, , The relaxation time in conductors, (a) Increases with the increase of temperature, (b) Decreases with the increase of temperature, (c) It does not depend on temperature, (d) All of sudden changes at 400 K, Which of the following statement is correct, (a) Liquids obey fully the ohm's law, (b) Liquids obey partially the ohm's law, , [DPMT 2003], , (c) There is no relation between current and p.d. for liquids

Page 23 :
Current Electricity 1053, (d) None of the above, 49., , (c) Voltage, , (d) None of the above, , A certain piece of silver of given mass is to be made like a wire., 58., A solenoid is at potential difference 60 V and current flows through, Which of the following combination of length (L) and the area of, it is 15 ampere, then the resistance of coil will be, cross-sectional (A) will lead to the smallest resistance [MP PMT 1995; CBSE PMT 1997], [AFMC 1995], (a) L and A, , (a), , 4, , (b) 8 , , (b) 2L and A/2, , (c), , 0.25 , , (d) 2 , , (c) L/2 and 2 A, , 59., , All of the following statements are true except, , (d) Any of the above, because volume of silver remains same, 50., , [Manipal MEE 1995], , The resistance of a wire is 10  . Its length is increased by 10% by, stretching. The new resistance will now be, , (a) Conductance is the reciprocal of resistance and is measured in, , Siemens, (b) Ohm's law is not applicable at very low and very high, temperatures, , [CPMT 2000; Pb PET 2004], , (a) 12 , (c), 51., , (b) 1 .2 , , 13 , , (c) Ohm's law is applicable to semiconductors, , (d) 11 , , Resistance of tungsten wire at 150C is 133  . Its resistance, temperature coefficient is 0.0045 / C . The resistance of this wire, at 500C will be, [DPMT 2004], (a) 180 , (c), , 52., , (b), , 258 , , 61., , A metal wire of specific resistance 64  10 6 ohm  cm and, length 198 cm has a resistance of 7 ohm, the radius of the wire will, be, [MP PET 1994], (b) 0.24 cm, , (a) 18 : 1, (c) 6 : 1, , 63., , (d) Becomes one fourth, , If the resistance of a conductor is 5  at 50 C and 7  at 100 C then, the mean temperature coefficient of resistance of the material is, o, , o, , (a) 0.008/ C, , (b) 0.006/ C, , (c) 0.004/ C, , (d) 0.001/ C, , o, , o, , The resistance of a discharge tube is, (a) Ohmic, , (b) Non-ohmic, , (c) Both (a) and (b), , (d) Zero, , We are able to obtain fairly large currents in a conductor because, (b) The number density of free electrons is very high and this can, compensate for the low values of the electron drift speed and, the very small magnitude of the electron charge, , For a metallic wire, the ratio V / i (V  the applied potential, difference, i = current flowing) is [MP PMT 1994; BVP 2003], , (c) The number density of free electrons as well as the electron, drift speeds are very large and these compensate for the very, small magnitude of the electron charge, (d) The very small magnitude of the electron charge has to be, divided by the still smaller product of the number density and, drift speed to get the electric current, , (d) Increases or decreases as temperature rises, depending upon, the metal, The resistance of a wire is R. If the length of the wire is doubled by, stretching, then the new resistance will be, , 64., , A platinum resistance thermometer has a resistance of 50  at, , [Roorkee 1992; AFMC 1995; KCET 1993; AMU (Med.) 1999;, , 20C . When dipped in a liquid the resistance becomes 76.8  ., , CBSE PMT 1999; MP PET 2001; UPSEAT 2001], , The temperature coefficient of resistance for platinum is, , (a) 2R, , (b) 4R, , (c) R, , (d), ,   3.92  10 3 / C . The temperature of the liquid is, , R, 4, , Which of the following has a negative temperature coefficient, (a) C, , (b) Fe, , (c) Mn, , (d) Ag, , The reciprocal of resistance is, (a) Conductance, , (c) Does not change, , (a) The electron drift speed is usually very large, , (b) 9 : 1, (d) 2 : 3, , (c) Decreases as the temperature rises, , 57., , (b) Becomes half, , [AFMC 1996; CBSE PMT 1999], , (b) Increases as the temperature rises, , 56., , (a) Becomes two times, , o, , (a) Independent of temperature, , 55., , A potential difference of V is applied at the ends of a copper wire of, length l and diameter d. On doubling only d, drift velocity, , o, , 62., , (c) 0.024 cm, (d) 24 cm, A copper wire of length 1 m and radius 1 mm is joined in series with, an iron wire of length 2 m and radius 3 mm and a current is passed, through the wires. The ratio of the current density in the copper, and iron wires is, [MP PMT 1994], , 54., , 60., , 225 , , (d) 317 , , (a) 2.4 cm, 53., , (d) Ohm's law is not applicable to electron tubes, discharge tubes, and electrolytes, , [AFMC 1995], , (b) Resistivity, , 65., , (a) 100C, , (b) 137C, , (c) 167C, , (d) 200C, , [AFMC 1995], , In a wire of circular cross-section with radius r, free electrons travel, with a drift velocity V when a current I flows through the wire., What is the current in another wire of half the radius and of the, same material when the drift velocity is 2V, [MP PET 1997]

Page 24 :
1054 Current Electricity, , 66., , 67., , (a) 2I, , (b) I, , (c), , (d), , I/2, , 74., , I/4, , The resistivity of a wire depends on its, , (a) Both  1 and  2 increase, , [MP PMT/PET 1998], , (a) Length, , (b) Area of cross-section, , (b)  1 increases and  2 decreases, , (c) Shape, , (d) Material, , (c)  1 decreases and  2 increases, , The conductivity of a superconductor is, , (d) Both  1 and  2 decrease, , [Similar to KCET 1993; MP PMT/PET 1998], , 68., , (a) Infinite, , (b) Very large, , 1.6 mA current is flowing in conducting wire then the number of, electrons flowing per second is [RPMT 1999], , (c) Very small, , (d) Zero, , (a) 10, , 11, , (b) 10, , 16, , (c) 10, , 19, , (d) 10, , 15, , 75., , In a neon discharge tube 2.9  1018 Ne  ions move to the right, each second while 1.2  1018 electrons move to the left per second., , 76., , Electron charge is 1.6  10 19 C . The current in the discharge tube, , 69., , (a) 1 A towards right, , (b) 0.66 A towards right, , (c) 0.66 A towards left, , (d) Zero, , P, , P, , A steady current flows in a metallic conductor of non-uniform crosssection. The quantity/ quantities constant along the length of the, conductor is/are, , (c) v =, P, , (b), , Ralloy  Rmetal, , 71., , 72., , [KCET 1994], , o, , (a) 6.5 , 79., , Ralloy  Rmetal, , Two wires A and B of same material and same mass have radius, 2rand r. If resistance of wire A is 34  , then resistance of B will be, (a), , 544 , , (b), , (c), , 68 , , (d) 17 , , 80., , 272 , , 81., , Two rods of same material and length have their electric resistance, in ratio 1 : 2 . When both rods are dipped in water, the correct, statement will be, [RPMT 1997], , [KCET 2000], , (c) Both have same loss of weight, 20 A current flows for 30 seconds in a wire, transfer of charge, will be, [RPMT 1997], , (a), , 2  10 4 C, , (b) 4  10 4 C, , (c), , 6  10 4 C, , (d) 8  10 4 C, , (b) 5 , , (c) 3 , (d) 4 , Masses of three wires of copper are in the ratio of 1 : 3 : 5 and their, lengths are in the ratio of 5 : 3 : 1. The ratio of their electrical, resistances are, [AFMC 2000], (a) 1 : 3 : 5, (b) 5 : 3 : 1, (c) 1 : 15 : 125, (d) 125 : 15 : 1, Conductivity, increases, in, the, order, of [AFMC 2000], [RPET 1997], (a) Al, Ag, Cu, (b) Al, Cu, Ag, (c) Cu, Al, Ag, (d) Ag, Cu, Al, A uniform wire of resistance R is uniformly compressed along its, length, until its radius becomes n times the original radius. Now, resistance of the wire becomes, , (a), , R, n4, , (b), , (c), , R, n, , (d) nR, , (b) B has more loss of weight, (d) Loss of weight will be in the ratio 1 : 2, , Q, , o, , (a) A has more loss of weight, , 73., , P, , o, , (c) There is no simple relation between Ralloy and R metal, (d), , (d) v = 2 v, , Q, , The resistance of a coil is 4.2  at 100 C and the temperature, coefficient of resistance of its material is 0.004/ C. Its resistance at, 0 C is, [KCET 1999], , (d) Current only, , Ralloy  Rmetal, , 1, v, 4, , Q, , 78., , (c) Current and drift speed, , (a), , P, , If an electric current is passed through a nerve of a man, then man, (a) Begins to laugh, (b) Begins to weep, (c) Is excited, (d) Becomes insensitive to pain, , (b) Drift speed only, , The resistivity of alloys  Ralloy ; the resistivity of constituent metals, , 1, v, 2, , (b) v =, , Q, , Q, , 77., , (a) Current, electric field and drift speed, , R metal . Then, usually, , A current I is passing through a wire having two sections P and Q, of uniform diameters d and d/2 respectively. If the mean drift, velocity[MP, of PET, electrons, 1999] in sections P and Q is denoted by v and v, respectively, then, [Roorkee 1999], (a) v = v, , [KCET 1994, IIT 1997 Cancelled; CBSE PMT 2001], , 70., ,  1 and  2 are the electrical conductivities of Ge and Na, respectively. If these substances are heated, then, , 82., , R, n2, , The resistance of a conductor is 5 ohm at 50 C and 6 ohm at 100 C., Its resistance at 0 C is, [KCET 2000], (a) 1 ohm, (b) 2 ohm, (c) 3 ohm, (d) 4 ohm, If an electron revolves in the path of a circle of radius of 0.5 × 10, m at frequency of 5 × 10 cycles/s the electric current in the circle is, (Charge of an electron = 1.6 × 10 C ), o, , o, , o, , 83., , –1 0, , 15, , –19, , [EAMCET 2000]

Page 25 :
Current Electricity 1055, , 84., , (a) 0.4 mA, (b) 0.8 mA, (c) 1.2 mA, (d) 1.6 mA, Equal potentials are applied on an iron and copper wire of same, length. In order to have the same current flow in the two wires, the, ratio r (iron)/r (copper) of their radii must be (Given that specific, , 93., , resistance of iron = 1.0  10 7 ohm–m and specific resistance of, , 94., , (a) 120 C, , copper = 1.7  10 8 ohm-m), , (d) About 4.8, , An electron (charge = 1.6 × 10 coulomb) is moving in a circle of, radius 5.1 × 10 m at a frequency of 6.8 × 10 revolutions/sec. The, equivalent current is approximately, , 96., , There is a current of 40 ampere in a wire of 10 6 m 2 area of, , –19, , –11, , 15, , [MP PET 2000], , (a), 86., , 95., , (a) R/4, (b) R/2, (c) R, (d) 2R, The drift velocity does not depend upon, [BHU 2001], (a) Cross-section of the wire, (b) Length of the wire, (c) Number of free electrons, (d) Magnitude of the current, , (b) About 2.4, , (c) About 3.6, 85., , 5.1  10, , 3, , (b) 6.8  10, , amp, , 3, , amp, , cross-section. If the number of free electron per m 3 is 10 29 ,, then the drift velocity will be, [Pb. PMT 2001], , (c) 1.1  10 3 amp, (d) 2.2  10 3 amp, A rod of a certain metal is 1.0 m long and 0.6 cm in diameter. Its, , 87., , (a) 1.35 × 10 8 ohm, , (b) 2.70 × 10 7 ohm, , (c) 4.05 × 10 6 ohm, , (d) 8.10 × 10 5 ohm, , At what temperature will the resistance of a copper wire become, three times its value at 0 C (Temperature coefficient of resistance, for copper = 4 × 10 per C ), , (b) 2.50 × 10 3 m/s, , (a) 1.25 × 10 3 m/s, , 3, , resistance is 3.0 × 10, ohm. Another disc made of the same, metal is 2.0 cm in diameter and 1.0 mm thick. What is the, resistance between the round faces of the disc, , (b) 240 C, , (c) 20 C, (d) 4 C, If a wire of resistance R is melted and recasted to half of its length,, then the new resistance of the wire will be, [KCET (Med.) 2001], , [MP PMT 2000], , (a) About 1.2, , Calculate the amount of charge flowing in 2 minutes in a wire of, resistance 10  when a potential difference of 20 V is applied, between its ends, [Kerala (Engg.) 2001], , 97., , (c) 25.0 × 10 3 m/s, (d) 250 × 10 3 m/s, At room, temperature, copper has free electron density of, [MP PET 2000], 28, 8.4  10 per m 3 . The copper conductor has a cross-section of, 10 m and carries a current of 5.4 A. The electron drift velocity in, copper is, [UPSEAT 2002], (a) 400 m/s, (b) 0.4 m/s, –6, , 2, , (c) 0.4 mm/s, , (d) 72 m/s, , o, , –3, , o, , 98., , [MP PET 2000], , (a) 400 C, , (b) 450 C, , (a) 160 , , (b) 80 , , (c) 500 C, , (d) 550 C, , (c) 40 , , (d) 20 , , o, , o, , o, , 88., , 89., , 90., , o, , An electron revolves 6 × 10 times/sec in circular loop. The current, in the loop is, [MNR 1995; UPSEAT 2000], 15, , (a) 0.96 mA, (b) 0.96  A, (c) 28.8 A, (d) None of these, The charge of an electron is 1.6 × 10 C. How many electrons strike, the screen of a cathode ray tube each second when the beam, current is 16 mA, [AMU (Med.) 2000], (a) 10, (b) 10, (c) 10, (d) 10, , 99., , [KCET 2002], , (a) Greater when switched off, (b) Smaller when switched on, (c) Greater when switched on, , 17, , 19, , –19, , –17, , (d) The same whether it is switched off or switched on, 100., , 0.1  0.2ohm, , (a) 2.2 k , [RPET 2001], , (b) 3.3 k , , (b) 5  2 ohm, , (c) 5.6 k , , (d) None of these, , (d) 9.1 k , , A nichrome wire 50 cm long and one square millimetre crosssection carries a current of 4A when connected to a 2V battery., The resistivity of nichrome wire in ohm metre is, , 101., , [EAMCET 2001], , (a) 1 × 10, , 6, , (c) 3 × 10 7, , (b) 4 × 10, , In the figure a carbon resistor has bands of different colours on its, body as mentioned in the figure. The value of the resistance is, Silver, , If potential V  100  0.5 Volt and current I  10  0.2 amp, are given to us. Then what will be the value of resistance, , (c), , 92., , The resistance of an incandescent lamp is, , –19, , (a) 10  0.7 ohm, , 91., , The resistance of a 5 cm long wire is 10 . It is uniformly stretched, so that its length becomes 20 cm. The resistance of the wire is, , 7, , (d) 2 × 10 7, , If an observer is moving with respect to a stationary electron, then, he observes, [DCE 2001], (a) Only magnetic field, , (b) Only electric field, , (c) Both (a) and (b), , (d) None of the above, , 102., , White, , Brown, , Red, , By increasing the temperature, the specific resistance of a conductor, and a semiconductor, [AIEEE 2002], (a) Increases for both, (b) Decreases for both, (c) Increases, decreases, (d) Decreases, increases, Which of the following is vector quantity, [AFMC 2002], (a) Current density, (b) Current, (c) Wattless current, (d) Power

Page 26 :
1056 Current Electricity, 103., , 104., , Masses of 3 wires of same metal are in the ratio 1 : 2 : 3 and their, lengths are in the ratio 3 : 2 : 1. The electrical resistances are in, ratio, [CPMT 2002], (a) 1 : 4 : 9, (b) 9 : 4 : 1, (c) 1 : 2 : 3, (d) 27 : 6 : 1, A current of 1 mA is flowing through a copper wire. How many, electrons will pass a given point in one second, [e = 1.6 × 10 Coulomb], [RPMT 2000; MP PMT 2002], –19, , (a), 105., , 6.25 × 10, , 19, , (b) 6.25 × 10, , (b), , A strip of copper and another of germanium are cooled from room, temperature to 80 K. The resistance of [AIEEE 2003], (a) Each of these increases, (b) Each of these decreases, (c) Copper strip increases and that of germanium decreases, (d) Copper strip decreases and that of germanium increases, , 113., , 15, , (c) 6.25 × 10 31, (d) 6.25 × 10 8, The drift velocity of free electrons in a conductor is ‘ v’ when a, current ‘i’ is flowing in it. If both the radius and current are, doubled, then drift velocity will be [BHU 2002], (a) v, , 112., , The length of a given cylindrical wire is increased by 100 %. Due to, the consequent decrease in diameter the change in the resistance of, the wire will be, [AIEEE 2003], (a) 300 %, (b) 200 %, (c) 100 %, , 114., , v, 2, , Express which of the following setups can be used to verify Ohm’s, law, [IIT-JEE (Screening) 2003], A, , (a), , v, v, (d), 4, 8, A wire of radius r has resistance R. If it is stretched to a radius of, 3r, , its resistance becomes, [BHU 2002], 4, , (b), , (c), 106., , (a), , 9R, 16, , (b), , V, , 81R, 256R, (d), 81, 256, The resistance of a conductor increases with, , 115., , [CBSE PMT 2002], , 108., , (a) Increase in length, (b) Increase in temperature, (c) Decrease in cross–sectional area, (d) All of these, A copper wire has a square cross-section, 2.0 mm on a side. It, carries a current of 8 A and the density of free electrons is, 8  10 28 m 3 . The drift speed of electrons is equal to, , (a) 0.156 × 10, , 3, , 109., , 110., , m.s, , –1, , (b) 0.156 × 10, , 2, , 2, , m.s, , –1, , (c) 3.12 × 10 m.s, (d) 3.12 × 10 m.s, Two wires of same material have length L and 2L and cross–, sectional areas 4A and A respectively. The ratio of their specific, resistance would be, [MHCET 2002], (a) 1 : 2, (b) 8 : 1, (c) 1 : 8, (d) 1 : 1, When a current flows through a conductor its temperature, , 116., , –1, , 111., , 117., , 118., , [CBSE PMT 2000; Pb. PMT 2002], , (a) 4.1 m, , (b) 3.1 m, , (c) 2.1 m, , (d) 1.1 m, , In a hydrogen discharge tube it is observed that through a given, , (a) 1 mA towards right, , (b) 1mA towards left, , (c), , (d) 2 mA towards right, , 2mA towards left, , A steady current i is flowing through a conductor of uniform crosssection. Any segment of the conductor has, (a) Zero charge, (b) Only positive charge, (c) Only negative charge, (d) Charge proportional to current i, The length of the wire is doubled. Its conductance will be, [Kerala PMT 2004], , 119., , 4.2  (diameter of wire = 0.4, , (d) None of the above, , [MP PET 1996], , What length of the wire of specific resistance 48  10 8  m is, needed to make a resistance of, mm), , A, , and 3.12  10 15 protons are moving from left to right. What is the, electric current in the discharge tube and what is its direction, , –1, , May increase or decrease, Remains same, Decreases, Increases, , V, , cross-section 3.13  10 15 electrons are moving from right to left, , [MHCET 2002], , (a), (b), (c), (d), , V, , (d), , We have two wires A and B of same mass and same material. The, diameter of the wire A is half of that B. If the resistance of wire A, is 24 ohm then the resistance of wire B will be, (a) 12 Ohm, (b) 3.0 Ohm, (c) 1.5 Ohm, , [AMU (Med.) 2002], 3, , A, , (c), , V, , A, , 16R, 9, , (c), 107., , (d) 50 %, , 120., , 121., , (a) Unchanged, (b) Halved, (c) Quadrupled, (d) 1/4 of the original value, A source of e.m.f. E = 15 V and having negligible internal resistance, is connected to a variable resistance so that the current in the, circuit increases with time as i = 1.2 t + 3. Then, the total charge that, will flow in first five second will be, (a) 10 C, (b) 20 C, (c) 30 C, (d) 40 C, The new resistance of wire of R , whose radius is reduced half, is [J & K CET, (a) 16 R, (b) 3 R, (c) 2R, (d) R, A resistance R is stretched to four times its length. Its new, resistance will be, [ISM Dhanbad 1994; UPSEAT 2003], (a) 4 R, (b) 64 R, , [

Page 27 :
Current Electricity 1057, , 122., , 123., , 124., , (c) R / 4, (d) 16 R, What is the resistance of a carbon resistance which has bands of, colours brown, black and brown, [DCE 1999], (a) 100 , (b) 1000 , (c) 10 , (d) 1 , The lead wires should have, [Pb. PMT 2000], (a) Larger diameter and low resistance, (b) Smaller diameter and high resistance, (c) Smaller diameter and low resistance, (d) Larger diameter and high resistance, The alloys constantan and manganin are used to make standard, resistance due to they have, , (c) 21  10  5%, 130., , 131., , 125., , 126., , 132., , 133., , 1., , (b) 5  10 m/s, , (c) 2  10 m/s, , (d) 8 10 m/s, , –4, , 128., , (d) 1 : 4, , The length of the resistance wire is increased by 10%. What is the, corresponding change in the resistance of wire, (a) 10%, , (b) 25%, , (c) 21%, , (d) 9%, , The electric field E, current density J and conductivity  of a, conductor are related as, [Kerala PMT 2005], (a)   E / j, , (b)   j / E, , (c)   jE, , (d)   1 / jE, , Two wires that are made up of two different materials whose, specific resistance are in the ratio 2 : 3, length 3 : 4 and area 4 : 5., The ratio of their resistances is [Kerala PMT 2005], (a) 6 : 5, , (b) 6 : 8, , (c) 5 : 8, , (d) 1 : 2, , The potential difference between points A and B of adjoining, figure is, [CPMT 1991], , 2., , 3., , The following four wires are made of the same material and are at, the same temperature. Which one of them has highest electrical, resistance, [UPSEAT 2004], (a) Length = 50 cm, diameter = 0.5 mm, , 4., , 5, , B, , 8, V, 9, , 4, V, 3, , 2V, 5, , 5, 5, , D, , R2  R  (R1  R2 ), , (d), , (d) Length = 300 cm, diameter = 3 mm, , (a), , 1, ampere, 45, , The colour sequence in a carbon resistor is red, brown, orange and, silver. The resistance of the resistor is, , (b), , 1, ampere, 15, , (c), , 1, ampere, 10, , [DCE 2004], 3, , C, , R  R1, , [IIT 1983; CPMT 1991, 92; MH CET 2002;, Pb. PMT 2001; Kerala PMT 2004], , (c) Length = 200 cm, diameter = 2 mm, , (a) 21  10  10%, , 5, , A wire of resistance R is divided in 10 equal parts. These parts are, connected in parallel, the equivalent resistance of such connection, will be, [CPMT 1973, 91], (a) 0.01 R, (b) 0.1 R, (c) 10 R, (d) 100 R, The current in the adjoining circuit will be, , (b) Length = 100 cm, diameter = 1 mm, , 129., , 5, , A, , Two resistors of resistance R1 and R 2 having R1  R 2 are, connected in parallel. For equivalent resistance R , the correct, statement is, [CPMT 1978; KCET (Med.) 2000], (a) R  R1  R 2, (b) R1  R  R 2, (c), , –10, , 3, , 2, V, 3, , (d) 2 V, , [UPSEAT 2004], –19, , (c) 3 : 1, , (c), , (d) The resistance and the specific resistance, will both remain, unchanged, , (a) 1  10 m/s, , (b) 4 : 1, , (b), , (c) The resistance will be halved and the specific resistance will be, doubled, , 28, , (a) 2 : 1, , (a), , (b) The resistance will be halved and the specific resistance will, remain unchanged, , A wire of diameter 0.02 metre contains 10 free electrons per cubic, metre. For an electrical current of 100 A, the drift velocity of the free, electrons in the wire is nearly, , A thick wire is stretched so that its length become two times., Assuming that there is no change in its density, then what is the, ratio of change in resistance of wire to the initial resistance of wire, , Grouping of Resistances, , (a) The resistance will be doubled and the specific resistance will, be halved, , 127., , 3, , [MH CET 2004], , [MH CET 2000; NCERT 1990], , (a) Low resistivity, (b) High resistivity, (c) Low temperature coefficient of resistance, (d) Both (b) and (c), When a potential difference is applied across the ends of a linear, metallic conductor, [MP PET 1997], (a) The free electrons are accelerated continuously from the lower, potential end to the higher potential end of the conductor, (b) The free electrons are accelerated continuously from the higher, potential end to the lower potential end of the conductor, (c) The free electrons acquire a constant drift velocity from the, lower potential end to the higher potential end of the, conductor, (d) The free electrons are set in motion from their position of rest, The electric resistance of a certain wire of iron is R. If its length and, radius are both doubled, then, [CBSE PMT 2004], , (d) 12  10  5%, , 3, , (b) 23  10  10, 1, , i, 2V, , 30, , 30, 30

Page 28 :
1058 Current Electricity, 1, ampere, 5, There are 8 equal resistances R. Two are connected in parallel, such, four groups are connected in series, the total resistance of the, system will be, [MP PMT 1987], (a) R / 2, (b) 2 R, (c) 4 R, (d) 8 R, Three resistances of one ohm each are connected in parallel. Such, connection is again connected with 2 / 3  resistor in series. The, resultant resistance will be, [MP PMT 1985], , (d), 5., , 6., , 7., , (a), , 5, , 3, , (b), , 3, , 2, , (c), , 1, , (d), , 2, , 3, , The lowest resistance which can be obtained by connecting 10, resistors each of 1/10 ohm is, , 13., , 3, , (a), , (c), , 8., , (b) 1 / 200 , , (c) 1 / 100 , , (d) 1 / 10 , , 15., , (b), , 9., , 16., , Three resistors each of 2 ohm are connected together in a triangular, shape. The resistance between any two vertices will be, , 11., , There are n similar conductors each of resistance R . The resultant, resistance comes out to be x when connected in parallel. If they are, connected in series, the resistance comes out to be, (a), , x /n, , (c), , x /n, , (b) n x, (d) nx, , 18., , Equivalent resistance between A and B will be, , [CPMT 1981], 3, , (a) 2 ohm, , 19., , (b) 18 ohm, (c) 6 ohm, , 3, , 3, , 3, , 3, , 20., , (d) 3.6 ohm, A, , 12., , 3, , 3, , B, , A wire has a resistance of 12 ohm. It is bent in the form of, equilateral triangle. The effective resistance between any two corners, of the triangle is, (a) 9 ohms, , (b) 12 ohms, , (c) 6 ohms, , (d) 8/3 ohms, , 2 V, , 2, , 21., , 3, , 2A, , D, , 1 V, , C, 3, , 2, , B, , Referring to the figure below, the effective resistance of the network, is, [NCERT 1973, 75], r, , r, , r, , r, , r, , r, , r, , 6, ohm., 8, One of the resistance wire is broken and the effective resistance, becomes 2  . Then the resistance in ohm of the wire that got, broken was, , Two resistances are joined in parallel whose resultant is, , [DPMT 2004], , 2, , 3, , A current of 2 A flows in a system of conductors as shown. The, potential difference (V A  VB ) will be [CPMT 1975, 76], , (d) 5r / 2, 17., , (b) 3/4 ohm, (d) 6 ohm, , 2, , (d) 10 volts, , (a) 2 r, (b) 4 r, (c) 10 r, , [CPMT 1983; MP PET 1990; MP PMT 1993; DCE 2004], , 10., , (b) 3 volts, , (c) 5 volts, , (d) 2 V, , 2, , (a) 4/3 ohm, (c) 3 ohm, , B, , (a) 2 volts, , (c), , 2, , 1, A, 2, (d) 2 A, , 3, , 3, , (b) 1 V, , A, , (c), , 3, , Three resistances of magnitude 2, 3 and 5 ohm are connected in, parallel to a battery of 10 volts and of negligible resistance. The, potential difference across 3  resistance will be, , (a), , 2V, , 2, , C, , A, , The reading of the ammeter as per figure shown is, 1, (a), A, 2, 8, 3, A, 4, , 6, , A, , (d) 4 , 14., , 3, , 5, , (b) 2 , , [MP PMT 1984; EAMCET 1994], , (a) 1 / 250 , , The effective resistance between the points A and B in the figure, is, [MPDPET 1994], , [CPMT 1976; DPMT 1982], , (a) 3/5, (b) 2, (c) 6/5, (d) 3, Given three equal resistors, how many different combination of all, the three resistors can be made [NCERT 1970], (a) Six, (b) Five, (c) Four, (d) Three, Lamps used for household lighting are connected in, (a) Series, (b) Parallel, (c) Mixed circuit, (d) None of the above, The equivalent resistance of resistors connected in series is always [CPMT 1984;, (a) Equal to the mean of component resistors, (b) Less than the lowest of component resistors, (c) In between the lowest and the highest of component resistors, (d) Equal to sum of component resistors, A cell of negligible resistance and e.m.f. 2 volts is connected to series, combination of 2, 3 and 5 ohm. The potential difference in volts, between the terminals of 3 ohm resistance will be, (a) 0.6, (b) 2/3

Page 29 :
Current Electricity 1059, , 22., , (c) 3, (d) 6, Four wires of equal length and of resistances 10 ohms each are, connected in the form of a square. The equivalent resistance, between two opposite corners of the square is, , (c) One eight, 29., , [NCERT 1977], , 23., , 24., , 25., , (a) 10 ohm, , (b) 40 ohm, , (c) 20 ohm, , (d) 10/4 ohm, , (a) 2 ohm and 7 ohm, , (b) 3 ohm and 6 ohm, , (c) 3 ohm and 9 ohm, , (d) 5 ohm and 4 ohm, , (b) 0.5 amp, , (c) 1.0 amp, , (d) 1.5 amp, , 6, , 30., , and R 2  2 ohm, , R2, , (c), , [CPMT 1989], , 6, , R2, , R1, , R2, , R2, , R2, , (b) 1 , , 2, , (d) 1 .5 , , In the figure, the value of resistors to be connected between C and, D so that the resistance of the entire circuit between A and B does, not change with the number of elementary sets used is, , (b) 6.0 V, (d) 7.2 V, , R, , (a) 6 ohm, , R, , (a) R B, , 20, , 16, , (b) 8 ohm, , 32., , 16, , C, , R, , (b), , R( 3R  1) R, , D, , (d), , R( 3  1), , R, , R, , R, , R, , R, , In the figure shown, the total resistance between A and B is, 2 C 1, , 6, 18, , 8, , B, 3 3, , 3, , (c), , (a) 54 ohm, , (b) 18 ohm, , (c) 36 ohm, , (d) 9 ohm, , 3, , 33., , 2 D 1, , [MP PET 1985; AFMC 2005], , 1, , 1, , 1, , 1, 4, , 1, , 1, , (b) 4 , , 6, , (d) 8 , , The current from the battery in circuit diagram shown is, 2, , (a) 1 A, (c) 1.5 A, , [IIT 1989], , 7, , A, , 15V, , (b) 2 A, , A wire is broken in four equal parts. A packet is formed by keeping, the four wires together. The resistance of the packet in comparison, to the resistance of the wire will be, (b) One fourth, , 1, 8, , (a) 12 , B, , A, 3 3, , 1, , A, , In the network of resistors shown in the adjoining figure, the, equivalent resistance between A and B is, , 3 3, , R, , B, 9, , 3 3, , R, , (c) 3 R, , A, , (d) 24 ohm, , R, , A, , [CPMT 1990; BVP 2003], , (a) Equal, , R1, , B, , 8, , 28., , R1, , (a) Infinity, , The equivalent resistance of the arrangement of resistances shown in, adjoining figure between the points A and B is, , 3, , R1, , 6, , 6, , 3, , [MP PET 1993], , [CPMT 1984], R1, , Q, , (a) 3.6 V, (c) 3.0 V, , 27., , 20V, , An infinite sequence of resistance is shown in the figure. The, resultant resistance between A and B will be, when R1  1 ohm, , A, , 6, , (c) 16 ohm, , 6, , [CPMT 1984], , 0.5 A, , 26., , 4, , (d) 2 amp and 2 amp, , 31., P, , 6, , (c) 1 amp and 1 amp, , Resistances of 6 ohm each are connected in the manner shown in, adjoining figure. With the current 0.5 ampere as shown in figure,, the potential difference VP  VQ is, 6, , 4, , 2 amp and 4 amp, , (b) 1 amp and 2 amp, , Resistors of 1, 2, 3 ohm are connected in the form of a triangle. If a, 1.5 volt cell of negligible internal resistance is connected across 3, ohm resistor, the current flowing through this resistance will be, (a) 0.25 amp, , Four resistances are connected in a circuit in the given figure. The, electric current flowing through 4 ohm and 6 ohm resistance is, respectively, [MP PET 1993], (a), , Two resistors are connected (a) in series (b) in parallel. The, equivalent resistance in the two cases are 9 ohm and 2 ohm, respectively. Then the resistances of the component resistors are, , 1, th, 16, , (d), , 1, , 6, , 0.5, , (d) 3 A, 34., , B, 8, In the given figure, when key K is opened, 10, the reading of the, ammeter A will be, 10V, (a) 50 A, + –, E, , 5, , A, 4, , D, A, , B, , C, K

Page 30 :
1060 Current Electricity, (b) 2 A, , resistance of the thicker wire is 10  . The total resistance of the, combination will be, [CBSE PMT 1995], , (c) 0.5 A, (d), 35., , 10, A, 9, , In the given circuit, the potential of the point E is, , 40 , , (b), , (c), , 5, , 2, , (d) 100 , , [MP PMT 2003], , (a) Zero, , A, , +, , D, , 42., , 2, , C, , B, , (a), , I, ,R, 3, , 2, , (b), , I, 2 R, , I, , 2R, 3, , (a) Less than 4 , , 38., , 43., , I, , 39., , 4, ohm, 3, , (b), , 2V, , (c), , 5V, , (d), , 20, V, 11, , 40., , (d), , 41., , r, 100, , 6, 1.2 A, , (d) 0.80 ampere, 44., , 4, , Three equal resistances each of value R are joined as shown in the, figure. The equivalent resistance between M and N is, [MP PET 1995], , (a) R, (b) 2R, , 1, , A, , 3, , 3, , B, , 1, , (c), , R, [MP PMT 1994], 2, , R, , M, L, , R, , R, , N, , Z, , R, 3, The equivalent resistance between points A and B of an infinite, network of resistances each of 1  connected as shown, is, , (d), , 3, , 45., , 10V, 1, , 1, , 1, , A, , (d), , (b), (d), , [MP PMT 1994], 1, , (a) Infinite, B, , 1, ohm, 3, , (c), , [AFMC 1995], , (c), , 2, , In the figure given below, the current passing through 6  resistor, is, [Manipal MEE 1995], , (c) 0.72 ampere, , 8, ohm, 3, , A student has 10 resistors of resistance ‘r’. The minimum resistance, made by him from given resistors is, , (a) 10 r, , 2, , (b) 0.48 ampere, , Three resistances, each of 1 ohm, are joined in parallel. Three such, combinations are put in series, then the resultant resistance will be, (a) 9 ohm, (b) 3 ohm, (c) 1 ohm, , 2, , (a) 0.40 ampere, , A battery of e.m.f. 10 V is connected to resistance as shown in figure., The potential difference VA  VB between the points A and B is, 2V, , 2, , (d) 12 , , I, A, + –, ,R, 2, Four wires AB, BC, CD, DA of resistance 4 ohm each and a fifth, wire BD of resistance 8 ohm are joined to form a rectangle ABCD of, which BD is a diagonal. The effective resistance between the points, A and B is, [MP PMT 1994], (a) 24 ohm, (b) 16 ohm, , (a), , 2, , (c) More than 4  but less than 12 , , (d), , (c), , 2, , (b) 4 , , R2, , R, , 37., , 2, , 5, , If a resistance R 2 is connected in parallel with the resistance R in, the circuit shown, then possible value of current through R and the, possible value of R 2 will be, , (c), , The equivalent resistance of the following infinite network of, resistances is, [AIIMS 1995], , 4 /3V, , (d) 4/3 V, 36., , 1, , 8V, , (b)  8 V, (c), , E, , –, , 40, , 3, , (a), , r, 10, r, 5, , Two wires of same metal have the same length but their crosssections are in the ratio 3 : 1 . They are joined in series. The, , 46., , 1 5, , 2, , 1, , 1, , (b) 2 , (d) Zero, , A copper wire of resistance R is cut into ten parts of equal length., Two pieces each are joined in series and then five such combinations, are joined in parallel. The new combination will have a resistance, (a) R, (c), , R, 5, , (b), , R, 4, , (d), , R, 25, , [

Page 31 :
Current Electricity 1061, , effective resistance between the two points on any diameter is equal, to, [JIPMER 1999], (a) 12 , (c), 48., , (d) 2 A, 54., , What is the current (i) in the circuit as shown in figure, , (b) 6 , , 3, , (d), , [AIIMS 1998], , (a) 2 A, , In the circuit shown, the point ‘B’ is earthed. The potential at the, point ‘A’ is, 7, , (b) 1.2 A, , R1 = 2, , (d) 0.5 A, 10, , (b) 24 V, , 55., , C, , 50V, , (c) 26 V, , 3V, , (c) 1 A, , B, , A, , (a) 14 V, , R2 = 2, , i, , 24 , , 5, , E, , R4 = 2, , n equal resistors are first connected in series and then connected in, parallel. What is the ratio of the maximum to the minimum, resistance, [KCET 1994], , 3, , (d) 50 V, 49., , (c) 4.92 A, , A wire has resistance 12  . It is bent in the form of a circle. The, , D, , Three resistors each of 4  are connected together to form a, , (a) n, , (b), , 1, n2, , (d), , 1, n, , network. The equivalent resistance of the network cannot be, (a) 1.33 , (c), 50., , (b) 3 .0 , , 6 .0 , , (c), , (d) 12.0 , , 56., , In the circuit shown below, the cell has an e.m.f. of 10 V and internal, resistance of 1 ohm. The other resistances are shown in the figure., The potential difference VA  VB is, [MP PMT 1997], , E=10V, r=1, , (a) 6 V, (b) 4 V, (c) 2 V, (d), 51., , 4, , A, , 2, , 2, , B, , 4, , 57., , A uniform wire of 16  is made into the form of a square. Two, opposite corners of the square are connected by a wire of resistance, 16  . The effective resistance between the other two opposite, corners is, [EAMCET (Med.) 1995], (a), , 32 , , (b) 20 , , (c), , 8, , (d) 4 , , For what value of R the net resistance of the circuit will be 18 ohms, R, , 2 V, , (a), , 8, , (b) 10 , , 10, , then connected in parallel. The equivalent resistance of the, combination will be, [MP PMT/PET 1998; BHU 2005], , (c) 16 , , 10, , (a) nR, , 52., , 1, , n2, , A wire of resistance R is cut into ‘n’ equal parts. These parts are, , (c), , n, R, , (b), , R, n, , (d), , R, n2, , (d) 24 , 58., , 1, , 1, , A, , 10, , 10, , 10, , 10, , B, , In the figure, current through the 3  resistor is 0.8 ampere, then, potential drop through 4  resistor is, [CBSE PMT 1993; AFMC 1999; MP PMT 2004], 3, , The resistance between the terminal points A and B of the given, infinitely long circuit will be, [MP PMT/PET 1998], A, , R3 = 2, , 47., , (a) 9.6 V, 4, , (b) 2.6 V, 6, , (c) 4.8 V, , 1, , (d) 1.2 V, 1, , Upto, infinity, , 1, , 59., , B, 1, , 53., , 1, , 1, , (a), , ( 3  1), , (b) (1  3 ), , (c), , (1  3 ), , (d) (2  3 ), , The current in the given circuit is, , +, , 3, 8, (d), , , 8, 3, What will be the equivalent resistance between the two points A and, , (c), 60., , D, , [CBSE PMT 1999], , [CBSE PMT 1996], , A, , (a) 8.31 A, (b) 6.82 A, 4.8V, , RA = 3, , –, , Three resistances 4  each of are connected in the form of an, equilateral triangle. The effective resistance between two corners is, (a) 8 , (b) 12 , , 10, , 10, , 10, , RB = 6, , 10, , 10, , C, RC = 6, , B, , D, 10, , 10, , 10

Page 32 :
1062 Current Electricity, 67., (a) 10 , (c), 61., , (b), , 30 , , 20 , , A, , (c), 62., , B, , (b) 6 , , R1 = 2, , (c) 4 , , A, , R, , R, , 68., , R, , 15 , , B, R3 = 4 , , An infinite ladder network is arranged with resistances R and 2 R as, shown. The effective resistance between terminals A and B is, R, , R, (d) None of these, R, What is the equivalent resistance between A and B, , R, , 63., , (a), (b), , 3, R, 2, , 2R, , C, A, , 2R, , 2R, , D, , B, , R, , R, (c), 2, (d) 2 R, The current in the following circuit is [CBSE PMT 1997], 1, (a), A, 8, 2, (b), A, 9, 3, 3, 2V, 2, (c), A, 3, 3, , What is the equivalent resistance of the circuit, , (a), , 6, , 69., , 4V, 1 , , +, , –, , 70., , 2, , A battery of emf 10 V and internal resistance 3  is connected to a, resistor as shown in the figure. If the current in the circuit is 0.5 A., then the resistance of the resistor will be, , (c) 10 , , R, , (d) 12 , The potential drop across the 3 resistor is, (a) 1 V, , 4, , (b) 1.5 V, , 6, , (c) 2 V, (d) 3 V, , (b) 1 , 72., , (d) 0.001 , , V, In the given figure, potential difference3between, A and B is, [RPMT 2000], 10K, , (a) 0, , [CPMT 1999], , (b) 5 volt, , 2, 2, , [CPMT 2000], 3, , The equivalent resistance of the circuit shown in the figure is, , (c) 5 , , 2, ohm, 3, , (b) 17 , , A, , 71., , (c) 0.1 , , (b) 6 , , (d) 2, , (a) 19 , , 10 wires (same length, same area, same material) are connected in, parallel and each has 1 resistance, then the equivalent resistance, will be, [RPMT 1999], , (a) 8 , , 2, ohm, 3, , B, , V, , (a) 10 , , 66., , (c) 1, , A, , [KCET 1998], , 2, 2, , (a) , (b) R, (c) 2 R, (d) 3 R, If all the resistors shown have the value 2 ohm each, the equivalent, resistance over AB is, [JIPMER 1999], (b) 4 ohm, , 4, , 8, , (d) 9 , 65., , 2R, , (a) 2 ohm, , (b) 7 , (c), , 2R, , B, , (d) 1 A, 64., , R, , A, , [BHU 1997; MP PET 2001], , 2, R, 3, , R4 = 2 , , (d) 2 , , 9, , (b) 12 , , R2 = 4 , , (a) 8 , , (d) 40 , , What is the equivalent resistance between A and B in the figure, below if R  3 , [SCRA 1996], (a), , In the given figure, the equivalent resistance between the points A, and B is, [AIIMS 1999], , 30 V, 10K, , (c) 10 volt, , 2, , D, , A, 10K, , (d) 15 volt, , 2, , 73., , B, , If each resistance in the figure is of 9  then reading of ammeter is, , (d) 4 , , +, 9V, , –, A, , [

Page 33 :
Current Electricity 1063, (d) 10 ohms, 80., , 74., , 75., , (a) 5 A, , (b) 8 A, , (c) 2 A, , (d) 9 A, , Four resistances 10 , 5 , 7  and 3  are connected so that they, form the sides of a rectangle AB, BC, CD and DA respectively., Another resistance of 10  is connected across the diagonal AC. The, equivalent resistance between A and B is, (a) 2 , , (b) 5 , , (c) 7 , , (d) 10 , , (a), (c), , 77., ,  1 l1   2 l 2, , (b), , l1  l 2, ,  1 l 2   2 l1, , (d), , l1  l 2, ,  1 l 2   2 l1, , 82., ,  1 l1   2 l 2, l1  l 2, , 1, , (d) 0.5 V, , (a), , 8, , 7, , (b), , 7, , 8, , (c), , 7, , 4, , (d), , 4, , 9, , Effective resistance between A and B is [UPSEAT 2001], , (c), , 1, , 1, , 78., , 1, 5, , (c), , 2, , 1, 3, , (b) 1, , 1, 4, , (d) 3, , 1, 2, , 83., , 84., , (a), , 5, ohms, 8, , (b), , 8, ohms, 5, , (c), , 3, ohms, 8, , (d), , 8, ohms, 3, , 85., , 39 ohms, , (c), , 69 ohms, , 12,  . If one, 7, of the resistors is disconnected the resistance becomes 4 . The, resistance of the other resistor is [MH CET 2002], , The effective resistance of two resistors in parallel is, , 12, , 7, , (d), , 7, , 12, , Two resistance wires on joining in parallel the resultant resistance is, 6, ohms . One of the wire breaks, the effective resistance is 2, 5, ohms. The resistance of the broken wire is, , (a), , 3, ohm, 5, , (b) 2 ohm, , (c), , 6, ohm, 5, , (d) 3 ohm, , In the circuit, the potential difference across PQ will be nearest to, , (b) 6.6 V, (c) 4.8 V, (d) 3.2 V, 86., , 3, 3, , (b) 3 , , 100 , , 10, , P, , 5, , (a) 9.6 V, , In the circuit shown here, what is the value of the unknown resistor, R so that the total resistance of the circuit between points P and Q, is also equal to R, [MP PET 2001], , (b), , B, , [MP PET 2001, 2002], , Two wires of the same material and equal length are joined in, parallel combination. If one of them has half the thickness of the, other and the thinner wire has a resistance of 8 ohms, the resistance, of the combination is equal to, , (a) 3 ohms, , 5, , 5, , 5, , A, , (a) 4 , , B, , [AMU (Engg.) 2000], , 79., , 5, , 2, , [MH CET 2000], , (c), , (a), , 5, , (d) 20 , , [AMU (Engg.) 2000], , 1, , (c) 3 V, , (b) 5 , , Equivalent resistance between the points A and B is (in ), , 1, , (b) 2 V, , (a) 15 , , l1  l 2, , Four resistances of 100  each are connected in the form of square., Then, the effective resistance along the diagonal points is, (a) 200 , (b) 400 , (c) 100 , (d) 150 , , A, , (a) 1 V, , (Med.) 2000], 81. [EAMCET, The resistors, of resistances 2 , 4  and 8  are connected in, parallel, then the equivalent resistance of the combination will be[KCET 2001], , 2, , [EAMCET (Engg.) 2000], , 76., , [Kerala (Engg.) 2001], , Two wires of equal diameters, of resistivities 1 and  2 and, lengths l and l , respectively, are joined in series. The equivalent, resistivity of the combination is, 1, , A uniform wire of resistance 9  is cut into 3 equal parts. They are, connected in the form of equilateral triangle ABC. A cell of e.m.f. 2, V and negligible internal resistance is connected across B and C., Potential difference across AB is, , R, , Q, , 48 V, , 80 , , 100 , 20 , , Q, , Three resistors are connected to form the sides of a triangleP ABC,, the resistance of the sides AB, BC and CA are 40 ohms, 60 ohms, and 100 ohms respectively. The effective resistance between the, points A and B in ohms will be

Page 34 :
1064 Current Electricity, [JIPMER 2002], , 87., , (a) 32, , (b) 64, , (c) 50, , (d) 200, , Find the equivalent resistance across AB, , (a), , 20, V, 7, , (b), , 40, V, 7, , (c), , 10, V, 7, , [Orissa JEE 2002], , A, , (a) 1 , , 88., , 2, , (b) 2 , , 2, , (c) 3 , , 2, , 2, 2, , (d) 4 , B, The equivalent resistance between x and y in the circuit shown is, (a) 10 , (c) 20 , (d), , 89., , 5, , 2, , 95., , x, , 10 , , R, 4, , (b), , R, 3, , 10 , , 91., , 92., , 96., , 97., R, , R, , R, , Q, , Two wires of the same dimensions but resistivities 1 and  2 are, connected in series. The equivalent resistivity of the combination is, (a), , 1   2, , (b), , 1   2, , (c), , 1  2, , (d), , 2( 1   2 ), , 98., , (d) 8/3 ohms, , A series combination of two resistors 1  each is connected to a 12, V battery of internal resistance 0.4 . The current flowing through, it will be, [MH CET (Med.) 1999], (a) 3.5 A, , (b) 5 A, , (c) 6 A, , (d) 10 A, , In the circuit shown in the adjoining figure, the current between B, B, , 12, 1, , A, , (a), , X, C, 1, , 4, [EAMCET 2003], , (d) 12, , (c), , 1, , 3, , D, , A 3volt battery with negligible internal resistance is connected in a, circuit as shown in the figure. The current I, in the circuit will be, , 3, , (d) em.f. of a cell is required to find the value of X, [AIEEE 2003], , 99., , (a) 1/3 A, (b) 1 A, 3, 3V, 3, (c) 1.5 A, (d) 2 A, 3, Find the equivalent resistance between the points a and b, , (d) 16 , , [CPMT 1986], , 4, , In the circuit shown in the figure, the current flowing in 2 , resistance, , [CPMT 1989; MP PMT 2004], , (b) 1.2 A, , 1.4A, , 25, , (d) 1.0 A, 100., 8, , b, , G, , (c) 0.4 A, , 4, , 10, , 2, , 10, , (a) 1.4 A, , [BHU 2003; CPMT 2004], , 94., , (c) 6 ohms, , (c) 8, , a, , 4, , 16, , (b) 12 ohms, , (b) 6, , (c) 8 , , V, , (a) 9 ohms, , (a) 4, , 2, , 2A, , A wire has a resistance of 12 ohm. It is bent in the form of, equilateral triangle. The effective resistance between any two corners, of the triangle is, , (b) 2 , , (b) 4 , , 16, , [KCET, and2003], D is zero, the unknown resistance is of, , Three unequal resistors in parallel are equivalent to a resistance 1, ohm. If two of them are in the ratio 1 : 2 and if no resistance value, is fractional, the largest of the three resistances in ohms is, , (a) 2 , , 4, , 2, , I, , 93., , In the circuit shown below, The reading of the voltmeter V is, , (d) 16 V, y, , (c) 4 R, (d) 2 R, 90., , 3, , (c) 20 V, , 10 , , P, , 4, , (a) 12 [MP, V PMT 2002], , The equivalent resistance between the points P and Q of the circuit, given is, [Pb. PMT 2002], (a), , A, , 10 V, , (b) 8 V, , 10 , , 6, , B, , (d) 0, , 10 , , (b) 40 , , 8, , 5, , Five resistors are connected as shown in the diagram. The equivalent, resistance between A and B is, [MP PMT 1996], , C, 4, , The potential difference between point A & B is, [BHU 2003; CPMT 2004; MP PMT 2005], , (a), , 6 ohm, , 4, , 5, , (b) 9 ohm, , 9, , A, 10, , B, 8, , D

Page 35 :
Current Electricity 1065, (c) 12 ohm, , (c), 106., , (d) 15 ohm, 101., , In the figure given the value of X resistance will be, when the p.d., between B and D is zero, [MP PET 1993], , [MP PMT 1999; KCET 2001; BHU 2001, 05], , X, , 6, , (b), , 3, , 8, 15, , A, 15, , 107., , 4, , 102., , 4, , (a) 4 ohm, (b) 6 ohm, D, (c) 8 ohm, (d) 9 ohm, The effective resistance between points A and B is, , A, , 103., , 10, , 108., , 2, , 8, 2.1A, , G, 20, , 5, , In the Wheatstone's bridge shown, P  2 , Q  3 , R  6 , , Q, , P, , 2, , (b) 3 , , R, , (c), , C, , 4R, , In the given figure, when galvanometer shows no deflection, the, current (in ampere) flowing through 5  resistance will be, , (a), , [MP PMT 1995], , B, , R, , S, , 6, , (d) 8 , R, , R, , 109., , D, , 104., , 6, , and S  8  . In order to obtain balance, shunt resistance across 'S', must be, [SCRA 1998], , (d) None of the above three values, Five resistors of given values are connected together as shown in the, figure. The current in the arm BD will be, , A, , 4, , (c) 0.9, , 10, , R, , Five equal resistances each of value R are connected in a form, shown alongside. The equivalent resistance of the network, (a) Between the points B and D is R, , (a) Half the current in the arm ABC, (b) Zero, (c) Twice the current in the arm ABC, (d) Four times the current in the arm ABC, In the network shown in the figure, each of the resistance is equal to, 2  . The resistance between the points A and B is, , (b) Between the points B and D is, , (a) 1 , , (d) Between the points A and C is, 110., , (b), , 4, , (c), , 3, , A, , (d), , 2, , B, , 20 , , 2,, , 1, , (c) i, i i , i, (d) i, i i , i, , 10, , 20, , R, D, , 2,, , 1,, , B, , g, , g, , 30, , (b) 30 , , ig, , 100, , 1, , 2, , G, , i2, , g, , [MP PMT 1997; RPET 2001], , 2, , 20, , i, , Potential difference between the points P and Q in the electric, 2V 0, circuit shown is, [KCET 1999], P, , (a) 4.5 V, 10, , R, , C, , g, , 15, , Q, , (a), , 2, , (b) i, i i , i, , 111., 10, , R, , A, , R, 2, , 10, , (a) i, i i , i, , P, A, , R, , In the circuit shown below the resistance of the galvanometer is 20, . In which case of the following alternatives are the currents, arranged strictly in the decreasing order, , 1,, , 10, , R, , R, 2, , i1, , In the arrangement of resistances shown below, the effective, resistance between points A and B is, 5, , B, , (c) Between the points A and C is R, , [CBSE PMT 1995], , 105., , B, , (d) 1.5, B, , 10, , 40 , , 7, , A, , (b) 0.6, , 10, , 10, , (b) 20 , (c), , 20, , 3, , (a) 0.5, , [NCERT 1974; MP PMT 2000], , (a) 10 , , 3, , 2, , (d) 6 , , 6, 6, , 10, , 3, , (c) 15 , , C, , 4, , (d) 110 , , Five resistances are connected as shown in the figure. The effective, resistance between the points A and B is, , (a), , B, , 90 , , i = 1.5 A, , RA = 2, , RB = 4, , (b) 1.2 V, , 3, , (c) 2.4 V, RD = 6, , RC = 12, Q

Page 36 :
1066 Current Electricity, (d) 2.88 V, 112., , (b) 40 , (c) 30 , , The current between B and D in the given figure is, B, , [RPET 2000; DCE 2001], 30, , 30, , 118., , (a) 1 amp, A, , (b) 2 amp, , [KCET 2002], , (a) 5 R, 30, , (d) 0.5 amp, , 114., , C, , 60, , l, , (c) Zero, 113., , 30, , (b) 3 R, , D, 30V, , (a), (b), , 3, , 14, , (c), , 9, , 14, , (d), , 14, , 9, , 119., , 6, , B, 8, , 120., , In a typical Wheatstone network, the resistances in cyclic order are, A = 10 , B = 5 , C = 4  and D = 4  for the bridge to be, balanced, [KCET 2000], , D=4, , 121., , (a) 10  should be connected in parallel with A, (b) 10  should be connected in series with A, (c) 5  should be connected in series with B, 115., , (d) 5  should be connected in parallel with B, In the circuit shown in figure, the current drawn from the battery is, 4A. If 10  resistor is replaced by 20  resistor, then current, drawn from the circuit will be, , 122., , [KCET 2000; CBSE PMT 2001], 1, , (c) 3 A, , 7, , 4A, , 21, , +, , –, , (b) 3 , , 2R, , 3, , 3, , 3, , 3, , (c) 6 , , 123., , (a), , 5, , 3, 3, 3, 3, The equivalent resistance between P and Q in the given figure, is, , (b), , (d), 117., , 20 , , (a) 50 , , 20 , , 20 , , [MH CET (Med.) 2001], , (c), , 20 , 20 , , P, , R, , A, R, , R, B, , R, , R, , R, , 1, , 12, 1, , A, , C, 1, , X, , 1, , 6, , D, , Which arrangement of four identical resistances should be used to, draw maximum energy from a cell of voltage V, , 3, , 3, , R, , [MP PMT 2004], , B, , A, , R, , R, , (d) R , R equal resistance R., R arms have, In a Wheatstone’s bridge all the four, If the resistance of the galvanometer arm is also R, the equivalent, resistance of the combination as seen by the battery is, R, (a), (b) R, 2, R, (c) 2 R, (d), 4, For what value of unknown resistance X, the potential difference, between B and D will be zero in the circuit shown in the figure, , (d) 5 , [UPSEAT 2001], , (a), , (c), , (c) 2 , , Calculate the equivalent resistance between A and B, 9, , 2, , 4R, , 3, , (b) 6 , , (d) 0 A, 116., , (b), , R, , R, , (a) 4 , , 10, 10, , (b) 2 A, , B, , B, , 3, , (a) 1 A, , 3, , A, , (a) 2R , , C=4, , 3, , 3, , (c) 6 , 3, 3, (d) None of these, Thirteen resistances each of resistance R ohm are connected in the, circuit as shown in the figure below. The effective resistance, between A and B is, [KCET 2003], , B=5, , A = 10 , , B, , The equivalent resistance of the following diagram A and B is, (a), , 7, , A, R, , 2, , 3, (b) 9 , , 4, , A, , R, , R, , (d) R/2, , 3, , R, , R, , (c) R [CBSE PMT 2000], , In the given figure, equivalent resistance between A and B will be, 14, , 3, , (d) 20 , If each of the resistance of the network shown in the figure is R, the, equivalent resistance between A and B is, , Q

Page 37 :
Current Electricity 1067, 132., (d), 124., , An unknown resistance R is connected in series with a resistance of, 10 . This combinations is connected to one gap of a metre bridge, while a resistance R is connected in the other gap. The balance, point is at 50 cm. Now, when the 10  resistance is removed the, balance point shifts to 40 cm. The value of R is (in ohm), (a) 60, (b) 40, (c) 20, (d) 10, 1, , 2, , 1, , 125., , 126., , 2, , (d) None of these, , 6, , The equivalent resistance and potential difference between A and B, for the circuit is respectively, [Pb. PMT 2003], 6, , (a) 4 , 8 V, , (c) P and R, , (b) 8 , 4 V, , (d) Any two points, Q, R, The total current supplied to the circuit by the battery is, , (c) 2 , 2 V, 135., , (a) 1 A, 6V, 6, (b) 2 A, 3, (c) 4 A, 1.5, (d) 6 A, An electric current is passed through a circuit containing two wires, of the same material, connected in parallel. If the lengths and radii, of the wires are in the ratio of 4/3 and 2/3, then the ratio of the, currents passing through the wire will be, , (a) 1.56 , , (b) 2.44 , , (c) 4 , , (d) 2 , , (a), , 3V, R, , (b), , V, R, , (d), 136., , 19, , 20, , 19, 10, (c), (d), , , 10, 19, In circuit shown below, the resistances are given in ohms and the, battery is assumed ideal with emf equal to 3 volt. The voltage across, the resistance R is, , R, , (a), , 9V, 35, , (b), , 5V, 18, , (c), , 5V, 9, , (c) 1.2 V, (d) 1.5 V, , 50 , , R3, R2, , 60  R4, , R5, , 30 , , B, , D, , E, , 2, 4, , 4, , 3, 6, , i, , 18V, V, 5, When a wire of uniform cross-section a, length l and resistance R is, bent into a complete circle, resistance between any two of, diametrically opposite points will be, , 137., , [CBSE PMT 2005], , (a), , R1, , A, , R, , (d), , 50 , +, 3V, –, , R, , F, , V, 2R, , 2V, R, , R, , R, , [UPSEAT 2004; Kerala PMT 2004], , (b) 0.6 V, , B, , C, , 4, , (a) 0.4 V, , 2.5, , D, , For the network shown in the figure the value of the current i is, , [Pb. PMT 2004], , (b), , A, , C, , 3, (d) 16 , 8 V, Five equal resistances each of resistance R are connected as shown, in the figure. A battery of V volts is connected between A and B., The current flowing in AFCEB will be, , (c), , If three resistors of resistance 2, 4 and 5  are connected in, parallel then the total resistance of the combination will be, 20, , 19, , 6, , 2A, , [CBSE PMT 2004], , (b) 1/3, (d) 2, , If a rod has resistance 4  and if rod is turned as half cycle then, the resistance along diameter, [BCECE 2004], , (a), , 2, , 3, , 6, , (c), , (b) Q and R, , (a) 3, (c) 8/9, , 131., , (b), , 3, , [AIEEE 2004], , 130., , 6, , [KCET 2004], 2, , 134., , [AIEEE 2004], , 129., , (d) 0.4 A, , (a), , 2, , 128., , (b) 20 A, , (c) 0.2 A, , 3, , (a) P and Q, , 127., , (a) 5 A, , 133. [KCET, If you, are provided three resistances 2 , 3  and 6 . How will, 2004], you connect them so as to obtain the equivalent resistance of 4 , , A wire has a resistance of 6 . It is cut into two parts and both half, values are connected in parallel. The new resistance is ...., (a) 12 , (b) 1.5 , (c) 3 , (d) 6 , Six equal resistances are connected between points P, Q and R as, shown in the figure. Then the net resistance will be maximum, between, [IIT-JEE (Screening) 2004], P, , A parallel combination of two resistors, of 1  each, is connected in, series with a 1.5  resistor. The total combination is connected, across a 10 V battery. The current flowing in the circuit is, , 30 , , (b), , R, 8, , R, 2, The current in a simple series circuit is 5.0 amp. When an additional, resistance of 2.0 ohms is inserted, the current drops to 4.0 amp. The, original resistance of the circuit in ohms was, , (c), , 138., , R, 4, , 4R, , (d)

Page 38 :
1068 Current Electricity, (a) 1.25, (c) 10, 139., , (b) 8, (d) 20, , 4., , In the circuit given E = 6.0 V, R = 100 ohms, R = R = 50 ohms, R =, 75 ohms. The equivalent resistance of the circuit, in ohms, is, 1, , 2, , (a) 11.875, , 3, , 4, , (c) 1 .[KCET, 2  2005], 5., , R1, , (b) 26.31, i, , (c) 118.75, (d) None of these, 140., , R4, , R2, , E, , R3, , By using only two resistance coils-singly, in series, or in parallel one, should be able to obtain resistances of 3, 4, 12 and 16 ohms. The, separate resistances of the coil are, , By a cell a current of 0.9 A flows through 2 ohm resistor and 0.3 A, through 7 ohm resistor. The internal resistance of the cell is [KCET 2003], (a) 0 .5 , (b) 1 .0 , , (c) 4E, 6., , (a) 3 and 4, (b) 4 and 12, (c) 12 and 16, (d) 16 and 3, In the given circuit, the voltmeter records 5 volts. The resistance of, the voltmeter in ohms is, [KCET 2005], 7., , V, , (a) 200, 100, , (b) 100, , 50, , 10 V, , (d) 50, , Kirchhoff's Law, Cells, , 8., , (b), , (E  V )R, V, , (c), , (V  E)R, V, , (d), , (V  E)R, E, , Two cells, e.m.f. of each is E and internal resistance r are, connected in parallel between the resistance R . The maximum, energy given to the resistor will be, only when, (a), , R r/2, , (b), , Rr, , (c), , R  2r, , (d), , R0, , Kirchhoff's first law i.e . i  0 at a junction is based on the law, of conservation of, [CBSE PMT 1997; AIIMS 2000;, MP PMT 2002; RPMT 2001; DPMT 2005], , and zero internal resistance while the battery E has an e.m. f . of, , (a) Charge, , (b) Energy, , 2 volt . If the galvanometer G reads zero, then the value of the, , (c) Momentum, , (d) Angular momentum, , 9., G, , A, , [RPET 2003; MH CET 2001], , B, , (b) 100, (c) 500, , X, , E1, , 10., , E, , (d) 200, , 1, , 2, , e, 10V, , (c) 1.3 amp, , b, , 3, , c, , 2amp, , i, , The terminal potential difference of a cell is greater than its e.m.f., when it is, (a) Being discharged, (b) In open circuit, , 7, (b), A from b to a through e, 3, (c) 1A from b to a through e, (d) 1A from a to b through e, A cell of e.m. f . 1 .5 V having a finite internal resistance is, , connected to a load resistance of 2  . For maximum power, transfer the internal resistance of the cell should be, [BIT 1988], , (a) 4 ohm, (c) 2 ohm, , 1.3amp, , (d) 1 amp, 11., , (a), , 1amp, , 2amp, , (b) 3.7 amp, , 4V, , d, 7, A from a to b through e, 3, , (a) Charge, (b) Energy, (c) Momentum, (d) Sum of mass and energy, The figure below shows currents in a part of electric circuit. The, current i is, [CPMT 1981; RPET 1999], (a) 1.7 amp, , C, , D, , The magnitude and direction of the current in the circuit shown will, be, [CPMT 1986, 88], a, , Kirchhoff's second law is based on the law of conservation of, , [NCERT 1990; AIEEE 2005], , 500 , , (a) 10, , 3., , (E  V )R, E, , In the adjoining circuit, the battery E1 has an e.m. f . of 12 volt, , resistance X in ohm is, , 2., , (a), , [MNR 1988; MP PET 2000; UPSEAT 2001], , (c) 10, , 1., , (d) E/4, , A cell of e.m.f. E is connected with an external resistance R , then, p.d. across cell is V . The internal resistance of cell will be [MNR 1987; Kerala P, , [KCET 2005], , 141., , (d) 2.0 , , The e.m.f. of a cell is E volts and internal resistance is r ohm. The, resistance in external circuit is also r ohm. The p.d. across the cell, will be, [CPMT 1985; NCERT 1973], (a) E/2, (b) 2E, , (b) 0.5 ohm, (d) None of these, , (c) Being charged, (d) Being either charged or discharged, 12., , In the circuit shown, potential difference between X and Y will be, (a) Zero, , 40, , X, , Y, , (b) 20 V, (c) 60 V, (d) 120 V, , 20, 120V

Page 39 :
Current Electricity 1069, 13., , 14., , In the above question, potential difference across the 40 , resistance will be, (a) Zero, (b) 80 V, (c) 40 V, (d) 120 V, In the circuit shown, A and V are ideal ammeter and voltmeter, respectively. Reading of the voltmeter will be, , [MNR 1983], , 24., , 2V, , (a) 2 V, 25., , (b) 1 V, A, , (c) 0.5 V, 15., , (d) Zero, 1, 1, When a resistance of 2ohm is connected across the terminals of a cell,, the current is 0.5 amperes. When the resistance is increased to 5 ohm,, the current is 0.25 amperes. The internal resistance of the cell is, , (a), , (a), , (c), , 0.5 ohm, , (c) 1.5 ohm, 16., , 17., , 18., , 19., , V, , [DPMT 2002], , (d), , 2.0 ohm, , The terminal potential difference of a cell when short-circuited is, ( E = E.M.F. of the cell), (a) E, (b) E / 2, (c) Zero, (d) E / 3, A primary cell has an e.m.f. of 1.5 volts, when short-circuited it gives, a current of 3 amperes. The internal resistance of the cell is, (a) 4.5 ohm, (b) 2 ohm, (c) 0.5 ohm, (d) 1/4.5 ohm, A 50V battery is connected across a 10 ohm resistor. The current is, 4.5 amperes. The internal resistance of the battery is, (a) Zero, (b) 0.5 ohm, (c) 1.1 ohm, (d) 5.0 ohm, The potential difference in open circuit for a cell is 2.2 volts. When a, 4 ohm resistor is connected between its two electrodes the potential, difference becomes 2 volts. The internal resistance of the cell will be, , 21., , (a) 1 ohm, (b) 0.2 ohm, (c) 2.5 ohm, (d) 0.4 ohm, A new flashlight cell of e.m.f. 1.5 volts gives a current of 15 amps,, when connected directly to an ammeter of resistance 0.04  . The, internal resistance of cell is, [MP PET 1994], (a), , 0.04 , , (b) 0.06 , , (c), , 0.10 , , (d) 10 , , A cell whose e.m.f. is 2 V and internal resistance is 0 .1  , is, connected with a resistance of 3 .9  . The voltage across the cell, terminal will be, [CPMT 1990; MP PET 1993; CBSE PMT 1999;, AFMC 1999; Pb. PMT 2000; AIIMS 2001], , (a), , 0.50 V, , (c) 1.95 V, 22., , (d), , 2.00 V, , The reading of a high resistance voltmeter when a cell is connected, across it is 2.2 V. When the terminals of the cell are also connected, to a resistance of 5  the voltmeter reading drops to 1.8 V. Find, the internal resistance of the cell, (a) 1 .2 , , (b) 1 .3 , , 1 .1 , , (d) 1 .4 , , (c), 23., , (b) 1.90 V, , When cells are connected in parallel, then, , nE, R  nr, , (b), , [MP PMT 1996], , (b) 1.0 ohm, 26., , nE, nR  r, , E, nE, (d), R  nr, R r, A cell of internal resistance r is connected to an external resistance, R. The current will be maximum in R, if, [CPMT 1982], , (a) R  r, (b) R  r, (c) R  r, (d) R  r / 2, 27., To get the maximum current from a parallel combination of n, identical cells each of internal resistance r in an external resistance, R, when[CPMT 1976, 83], [DPMT 1999], (a) R  r, (b) R  r, (c) R  r, (d) None of these, 28., Two identical cells send the same current in 2  resistance,, [CPMT 1985; BHU 1997; Pb. PMT 2001], whether connected in series or in parallel. The internal resistance of, the cell should be, [NCERT 1982; Kerala PMT 2002], , (a) 1 , (c), , [MP PMT 1984; SCRA 1994; CBSE PMT 2002], , 20., , (a) The current decreases, (b) The current increases, (c) The e.m.f. increases, (d) The e.m.f. decreases, The internal resistance of a cell depends on, (a) The distance between the plates, (b) The area of the plates immersed, (c) The concentration of the electrolyte, (d) All the above, n identical cells each of e.m.f. E and internal resistance r are, connected in series. An external resistance R is connected in series, to this combination. The current through R is, , 29., , (b) 2 , , 1, , 2, , (d) 2.5 , , The internal resistances of two cells shown are 0 .1  and 0 .3  ., If R  0.2  , the potential difference across the cell, , (a) B will be zero, (b) A will be zero, (c) A and B will be 2V, , 2V, 0.1, , 2V, 0.3, , A, , B, , 0.2, , (d) A will be  2V and B will be  2V, 30., A torch battery consisting of two cells of 1.45 volts and an internal, resistance 0.15  , each cell sending currents through the filament, of the lamps having resistance 1.5ohms. The value of current will be[MP PET 199, (a) 16.11 amp, (b) 1.611 amp, (c) 0.1611 amp, (d) 2.6 amp, 31., The electromotive force of a primary cell is 2 volts. When it is shortcircuited it gives a current of 4 amperes. Its internal resistance in, ohms is, [MP PET 1995], (a) 0.5, (b) 5.0, (c) 2.0, (d) 8.0, [KCET 2003; MP PMT 2003], 32., The figure shows a network of currents. The magnitude of currents, is shown here. The current i will be, 15A, , 3A, , (a) 3 A, 8A, , 5A, , i, , [MP PMT 1995]

Page 40 :
1070 Current Electricity, (b) 13 A, , (d) i1  i2  i3, , (c) 23 A, , 39., , (d) – 3 A, 33., , [MP PET 1999, 2000; Pb. PMT 2002; MP PMT 2000], , A battery of e.m.f. E and internal resistance r is connected to a, variable resistor R as shown here. Which one of the following is true, E, , r, , 40., , R, , (a) Potential difference across the terminals of the battery is, maximum when R = r, (b) Power delivered to the resistor is maximum when R = r, (c) Current in the circuit is maximum when R = r, 34., , 35., , (d) Current in the circuit is maximum when R  r, A dry cell has an e.m.f. of 1.5 V and an internal resistance of, 0.05  . The maximum current obtainable from this cell for a very, short time interval is, [Haryana CEE 1996], (a) 30 A, (b) 300 A, (c) 3 A, (d) 0.3 A, Consider the circuit given here with the following parameters, , 41., , (a) 1.0[MP, V PMT 1995], (b) 1.5 V, (c) 2.0 V, (d) 2.5 V, Two non-ideal identical batteries are connected in parallel. Consider, the following statements, [MP PMT 1999], (i) The equivalent e.m.f. is smaller than either of the two e.m.f.s, (ii) The equivalent internal resistance is smaller than either of the, two internal resistances, (a) Both (i) and (ii) are correct, (b) (i) is correct but (ii) is wrong, (c) (ii) is correct but (i) is wrong, (d) Both (i) and (ii) are wrong, If six identical cells each having an e.m.f. of 6V are connected in, parallel, the e.m.f. of the combination is, [EAMCET (Med.) 1995; Pb. PMT 1999; CPMT 2000], , (a) 1 V, , (b) 36 V, , 1, (d) 6 V, V, 6, Consider the circuit shown in the figure. The current I3 is equal to, , (c), 42., , E.M.F. of the cell = 12 V. Internal resistance of the cell  2  ., Resistance R  4 , , When a resistance of 2 ohm is connected across the terminals of a, cell, the current is 0.5 A. When the resistance is increased to 5 ohm,, the current is 0.25 A. The e.m.f. of the cell is, , E, , (c), , 54, , 28, , (a) 5 amp, (b) 3 amp, 3 amp, , 6V, , I3, , (d) 5 / 6 amp, R, , 43., , Which one of the following statements in true, , (a), , (b) Rate of energy conversion in the source is 16 W, (d) Potential drop across R is = 16 V, A current of two amperes is flowing through a cell of e.m.f. 5 volts, and internal resistance 0.5 ohm from negative to positive electrode., If the potential of negative electrode is 10V, the potential of positive, electrode will be, , 38., , (a), , i1  i2, , 44., , (b) i2  i3, (c), , O, , A, , 45., , i3, , 47., C, D, , Two resistances R 1 and R 2 are joined as shown in the figure to, current through R 1 is, , 46., , i1  i3, , X, , two batteries of e.m.f. E1 and E 2 . If E 2 is short-circuited, the, , i2, , i1, , 2V, , (d) 20 , , [MP PMT/PET 1998; MP PMT 2000; DPMT 2000], , B, , B, , A, , (c) 15 , , [MP PMT 1997], , (a) 5 V, (b) 14 V, (c) 15 V, (d) 16 V, 100 cells each of e.m.f. 5 V and internal resistance 1 ohm are to be, arranged so as to produce maximum current in a 25 ohms, resistance. Each row is to contain equal number of cells. The, number of rows should be, [MP PMT 1997], (a) 2, (b) 4, (c) 5, (d) 10, The current in the arm CD of the circuit will be, , [RPET 1997], , 5, , (b) 10 , , (c) Power output in is = 8 W, , 37., , 5V, , 10, , (a) Rate of energy loss in the source is = 8 W, , 36., , If VAB  4 V in the given8 figure,, then resistance, V, 12 V X will be, , (a), , E1 / R 1, , (b), , E 2 / R1, , (c), , E2 / R2, , (d), , E1 /(R 2  R1 ), , [NDA 1995], , R1, , E1, , R2, , E2, , A storage battery has e.m.f. 15 volts and internal resistance 0.05, ohm. Its terminal voltage when it is delivering 10 ampere is, (a) 30 volts, (b) 1.00 volts, (c) 14.5 volts, (d) 15.5 volts, The number of dry cells, each of e.m.f. 1.5 volt and internal, resistance 0.5 ohm that must be joined in series with a resistance of, 20 ohm so as to send a current of 0.6 ampere through the circuit is, (a) 2, (b) 8, (c) 10, (d) 12, Emf is most closely related to, [DCE 1999], (a) Mechanical force, (b) Potential difference

Page 41 :
Current Electricity 1071, , 48., , 49., , (c) Electric field, (d) Magnetic field, For driving a current of 2 A for 6 minutes in a circuit, 1000 J of, work is to be done. The e.m.f. of the source in the circuit is, (a) 1.38 V, (b) 1.68 V, (c) 2.04 V, (d) 3.10 V, Two batteries of e.m.f. 4V and 8 V with internal resistances 1  and, 2  are connected in a circuit with a resistance of 9  as shown in, figure. The current and potential difference between the points P, and Q are, [AFMC 1999], 1, A and 3 V, 3, 4V, 8V, 1, 2, 1, Q, (b), A and 4 V P, r1, r2, 6, 1, (c), A and 9 V, 9, 9, 1, (d), A and 12V, 2, In the shown circuit, what is the potential difference across A and B, 20 V, (a) 50 V, (b) 45 V, (c) 30 V, (d) 20 V, Four identical cells each having an electromotive, force (e.m.f.) of, A, B, 12V, are connected in parallel. The resultant electromotive force, (e.m.f.) of the combination is, , 58., , [AIIMS 2000; MH CET 2003], 10 , , (a) 0.1 A, (b) 0.2 A, , 51., , 52., , 53., , 54., , (c) 90 AH, , A, , B, 20 , , (d) 0.4 A, 59., , 60., , 2V, , A current of 2.0 ampere passes through a cell of e.m.f. 1.5 volts, having internal resistance of 0.15 ohm. The potential difference, measured, in volts, across both the ends of the cell will be, (a) 1.35, , (b) 1.50, , (c) 1.00, , (d) 1.20, , A battery has e.m.f. 4 V and internal resistance r. When this battery, is connected to an external resistance of 2 ohms, a current of 1 amp., flows in[AIIMS, the circuit., 1999] How much current will flow if the terminals of, the battery are connected directly, [MP PET 2001], , 61., , [CPMT 1999], , (a) 48 V, (b) 12 V, (c) 4 V, (d) 3 V, Electromotive force is the force, which is able to maintain a constant, (a) Current, (b) Resistance, (c) Power, (d) Potential difference, A cell of emf 6 V and resistance 0.5 ohm is short circuited. The, current in the cell is, [JIPMER 1999], (a) 3 amp, (b) 12 amp, (c) 24 amp, (d) 6 amp, A storage cell is charged by 5 amp D.C. for 18 hours. Its strength, after charging will be, [JIPMER 1999], (a) 18 AH, (b) 5 AH, , 5V, , (c) 0.3 A, , (a), , 50., , (a) 15 V, (b) 12 V, (c) 9 V, (d) 6 V, [CPMT, 1999]is, The current in the given, circuit, , (a) 1 amp, , (b) 2 amp, , (c) 4 amp, , (d) Infinite, , Two batteries A and B each of e.m.f. 2 V are connected in series to, an external resistance R = 1 ohm. If the internal resistance of battery, A is 1.9 ohms and that of B is 0.9 ohm, what is the potential, difference between the terminals of battery A, A, , (a), (b), (c), (d), 62., , R, , When a resistor of 11  is connected in series with an electric cell,, the current flowing in it is 0.5 A. Instead, when a resistor of 5  is, connected to the same electric cell in series, the current increases by, 0.4 A. The internal resistance of the cell is, (a), , 63., , B, , 2V, 3.8[Pb., V PMT 1999], Zero, None of the above, , 1.5 , , (b) 2 , , (c) 2.5 , (d) 3.5 , The internal resistance of a cell is the resistance of, , (d) 15 AH, , [BHU 1999, 2000; AIIMS 2001], , 55., , A battery having e.m.f. 5 V and internal resistance 0.5  is, , 56., , connected with a resistance of 4.5  then the voltage at the, terminals of battery is, [RPMT 2000], (a) 4.5 V, (b) 4 V, (c) 0 V, (d) 2 V, In the given circuit the current I is, [DCE 2000], , (a), (b), (c), (d), 64., , Electrodes of the cell, Vessel of the cell, Electrolyte used in the cell, Material used in the cell, , How much work is required to carry a 6 C charge from the, negative terminal to the positive terminal of a 9 V battery, , 1, , 30 , , (a) 0.4 A, , I1, , (b) – 0.4 A, (c) 0.8 A, (d) – 0.8 A, 57., , I2, , [KCET (Med.) 2001], , (a) 54 × 10, 40 , 40 , , 40V, , I3, , 80V, , The internal resistance of a cell of e.m.f. 12 V is 5  10 2  . It is, connected across an unknown resistance. Voltage across the cell,, when a current of 60 A is drawn from it, is, [CBSE PMT 2000], , 3, , J, , (b) 54 × 10, , (c) 54 × 10 9 J, 65., , 6, , J, , (d) 54 × 10 12 J, , Consider four circuits shown in the figure below. In which circuit, power dissipated is greatest (Neglect the internal resistance of the, power supply), [Orissa JEE 2002], (a), E, , (b), R, , R, , R, E, R

Page 42 :
1072 Current Electricity, , 72., (c), , (d), , R, , R, , R, R, , E, , E, , R, , R, , 66., , 67., , The emf of a battery is 2 V and its internal resistance is 0.5 ., The maximum power which it can deliver to any external circuit will, be, [AMU (Med.) 2002], (a) 8 Watt, (b) 4 Watt, (c) 2 Watt, (d) None of the above, Kirchoff’s I law and II law of current, proves the, , 73., , 68., , Conservation of charge and energy, Conservation of current and energy, Conservation of mass and charge, None of these, , 70., , (c) 10/9 A, (d) 1 A, Eels are able to generate current with biological cells called, electroplaques. The electroplaques in an eel are arranged in 100, rows, each row stretching horizontally along the body of the fish, containing 5000 electroplaques. The arrangement is suggestively, shown below. Each electroplaques has an emf of 0.15 V and internal, resistance of 0.25 , [AIIMS 2004], 0.15 V, , + –, , (d) 87.5%, , In the given current distribution what is the value of I, , 2A, , I, , (c) 2A, , 3A, , (d) 5A, , 5A, , A capacitor is connected to a cell of emf E having some internal, resistance r. The potential difference across the, , +, , 75., , 76., , 77., , (a) Cell is < E, , (b) Cell is E, , (c) Capacitor is > E, , (d) Capacitor is < E, , When the resistance of 9  is connected at the ends of a battery, its, potential difference decreases from 40 volt to 30 volt. The internal, resistance of the battery is, [DPMT 2003], (a) 6 , , (b) 3 , , (c) 9 , , (d) 15 , , The maximum power drawn out of the cell from a source is given, by (where r is internal resistance), [DCE 2002], (a), , E 2 / 2r, , (b), , E2 / 4r, , (c), , E2 / r, , (d), , E 2 / 3r, , Find out the value of current through 2 resistance for the given, circuit, [IIT-JEE (Screening) 2005], (a) 5 A, (b) 2 A, , – 0.25 , , (c) Zero, + –, , 4A, , [CPMT 2004; MP PMT 2005], , 40, A, 29, A, 10, (b), A, 9, 10V, 4, 5, 5, (c), A, 3, (d) 2 A, In the above question, if the internal resistance of the battery is 1, ohm, then what is the reading of ammeter, (a) 5/3 A, (b) 40/29 A, , + –, , (c) 90%, , (b) 8 A, , In the circuit, the reading of the ammeter is (assume internal, resistance of the battery be zero), , + –, , (b) 80 %, , (a) 3A, , 74., , (a), , 69., , (a) 82.5%, , [Orissa PMT 2004], , [CBSE PMT 1993; BHU 2002; AFMC 2003], , (a), (b), (c), (d), , (c) Internal resistance is less than external resistance, (d) None of these, A battery is charged at a potential of 15 V for 8 hours when the, current flowing is 10 A. The battery on discharge supplies a current, of 5 A for 15 hours. The mean terminal voltage during discharge is, 14 V. The "Watt-hour" efficiency of the battery is, , +, , –, , 5, , 10, , 10V, , (d) 4 A, , 5000 electroplaques per row, , 78., , 100 rows, , 20V, , 2, , Two batteries, one of emf 18 volts and internal resistance 2 and, the other of emf 12 volt and internal resistance 1 , are connected, , +, , –, , +, , –, , +, , as shown. The voltmeter V will record a reading of, , –, , (a) 15 volt, , 71., , The water surrounding the eel completes a circuit between the head, 500 , and its tail. If the water surrounding it has a resistance of 500 ,, the current an eel can produce in water is about, (a) 1.5 A, (b) 3.0 A, (c) 15 A, (d) 30 A, Current provided by a battery is maximum when, [AFMC 2004], , (a) Internal resistance equal to external resistance, (b) Internal resistance is greater than external resistance, , V, , (b) 30 volt, (c) 14 volt, , 18V, , 2, , (d) 18 volt, 79., , 12V, Two sources of equal emf are connected, to1an external resistance R., The internal resistances of the two sources are R 1 and

Page 43 :
Current Electricity 1073, R 2 (R 2  R1 ) . If the potential difference across the source having, , (c), , internal resistance R 2 is zero, then, [AIEEE 2005], , (a), , R  R1 R 2 /(R1  R 2 ), , (b), , R  R1 R 2 /(R 2  R1 ), , (c), , R  R 2  (R1  R 2 ) /(R 2  R1 ), , (d), 80., , R1  R 2, 2, , In meter bridge or Wheatstone bridge for measurement of, resistance, the known and the unknown resistances are interchanged., The error so removed is, [MNR 1988; MP PET 1995], , R  R 2  R1, , An energy source will supply a constant current into the load if its, internal resistance is, [AIEEE 2005], (a) Zero, , 2., , (a) End correction, (b) Index error, (c) Due to temperature effect, (d) Random error, A galvanometer can be converted into an ammeter by connecting, [MP PMT 1987, 93; CPMT 1973, 75, 96, 2000;, MP PET 1994; AFMC 1993, 95; RPET 2000; DCE 2000], , (c) Equal to the resistance of the load, (d) Very large as compared to the load resistance, , (a), (b), (c), (d), , The magnitude of i in ampere unit is [KCET 2005], 60, , (a) 0.1, , i, , (b) 0.3, (c) 0.6, , 3., 15, , 5, , 1A, , 10, To draw maximum current from a combination, of cells, how should, the cells be grouped, [AFMC 2005], , (b) Parallel, (c) Mixed, , (a), , (d) Depends upon the relative values of external and internal, resistance, , 5., , The figure shows a network of currents. The magnitude of currents, is shown here. The current I will be [BCECE 2005], 1A, , (b) 9 A, , 10 A, , 6., , I, , (c) 13 A, (d) 19 A, , 85., , 0, , (c) 1000 , , (a) 3 A, , 84., , A cell of internal resistance 1 .5  and of e.m.f. 1.5 volt balances, , 10 3 amp is flowing through a resistance of 1000  . To measure, the correct potential difference, the voltmeter is to be used of which, the resistance should be, [MP PMT 1985], , 4., , (a) Series, , 83., , Low resistance in series, High resistance in parallel, Low resistance in parallel, High resistance in series, , 500 cm on a potentiometer wire. If a wire of 15  is connected, between the balance point and the cell, then the balance point will, shift, [MP PMT 1985], (a) To zero, (b) By 500 cm, (c) By 750 cm, (d) None of the above, , 1A, , (d) None of these, 82., , R1  R2, 2, , Different Measuring Instruments, 1., , (b) Non-zero but less than the resistance of the load, , 81., , (d), , 6A, , (b) 500 , (d)  1000 , , A galvanometer of 100  resistance gives full scale deflection when, 10 mA of current is passed. To convert it into 10 A range ammeter,, the resistance of the shunt required will be, (a), , 10 , , (b) 1 , , (c), , 0 .1 , , (d) 0.01 , , 50  and 100  resistors are connected in series. This, connection is connected with a battery of 2.4 volts. When a, voltmeter of 100  resistance is connected across 100  resistor,, then the reading of the voltmeter will be, , 2A, , The n rows each containing m cells in series are joined in parallel., Maximum current is taken from this combination across an external, resistance of 3 resistance. If the total number of cells used are 24, and internal resistance of each cell is 0.5  then, (a), , m  8, n  3, , (b) m  6, n  4, , (c), , m  12, n  2, , (d) m  2, n  12, , [MP PMT 1985], , (a) 1.6 V, (c) 1.2 V, 7., , A cell of constant e.m.f. first connected to a resistance R1 and then, connected to a resistance R 2 . If power delivered in both cases is, then the internal resistance of the cell is, [Orissa JEE 2005], , (a), , R1 R2, , (b), , R1, R2, , 8., , (b) 1.0 V, (d) 2.0 V, , A 2 volt, a 15  resistor and a potentiometer of 100 cm, [J &battery,, K CET 2005], length, all are connected in series. If the resistance of potentiometer, wire is 5  , then the potential gradient of the potentiometer wire, is, [AIIMS 1982], (a) 0.005 V/cm, , (b) 0.05 V/cm, , (c) 0.02 V/cm, , (d) 0.2 V/cm, , An ammeter gives full scale deflection when current of 1.0 A is, passed in it. To convert it into 10 A range ammeter, the ratio of its, resistance and the shunt resistance will be, [MP PMT 1985], , (a) 1 : 9, , (b) 1 : 10

Page 44 :
1074 Current Electricity, 9., , (c) 1 : 11, (d) 9 : 1, By ammeter, which of the following can be measured, , 18., , The tangent galvanometer, when connected in series with a standard, resistance can be used as, [MP PET 1994], (a) An ammeter, (b) A voltmeter, (c) A wattmeter, (d) Both an ammeter and a voltmeter, , 19., , In Wheatstone's bridge P  9 ohm, Q  11 ohm, R  4 ohm and, S  6 ohm. How much resistance must be put in parallel to the, resistance S to balance the bridge, , [MP PET 1981; DPMT 2001], , (a) Electric potential, (c) Current, 10., , (b) Potential difference, (d) Resistance, , The resistance of 1 A ammeter is 0.018  . To convert it into 10 A, ammeter, the shunt resistance required will be, [MP PET 1982], , (a), (c), 11., , 0.18 , , (b) 0.0018 , , 0.002 , , (d) 0.12 , , [DPMT 1999], , (a) 24 ohm, , For measurement of potential difference, potentiometer is preferred, in comparison to voltmeter because, [MP PET 1983], , 20., , (a) Potentiometer is more sensitive than voltmeter, (b) The resistance of potentiometer is less than voltmeter, (c) Potentiometer is cheaper than voltmeter, (d) Potentiometer does not take current from the circuit, 12., , 13., , 14., , 15., , (a), , 3 volt, , (b) 0.4 volt, , (c), , 7 volt, , (d) 4 volt, , 100 mA current gives a full scale deflection in a galvanometer of, , 24., , convert it into a voltmeter to measure 5 V is, (a), , 98 , , (b) 52 , , (c), , 50 , , (d) 48 , , When a 12  resistor is connected with a moving coil, galvanometer then its deflection reduces from 50 divisions to 10, divisions. The resistance of the galvanometer is, (a), , 24 , , (b) 36 , , (c), , 48 , , (d) 60 , , A galvanometer can be used as a voltmeter by connecting a, [AFMC 1993; MP PMT 1993, 95; CBSE PMT 2004], , (a) High resistance in series, (c) High resistance in parallel, , (b) Low resistance in series, (d) Low resistance in parallel, , (a) 1 V, (b) 1.5 V, (c) 2 V, (d) 3 V, A galvanometer of 10 ohm resistance gives full scale deflection with, 0.01 ampere of current. It is to be converted into an ammeter for, measuring 10 ampere current. The value of shunt resistance required, will be, [MP PET 1984], , [MNR 1994; UPSEAT 2000], , 10, ohm, (b) 0.1 ohm, 999, (c) 0.5 ohm, (d) 1.0 ohm, A potentiometer is used for the comparison of e.m.f. of two cells, E1 and E 2 . For cell E1 the no deflection point is obtained at, , (a), , 25., , 20 cm and for E 2 the no deflection point is obtained at 30 cm ., The ratio of their e.m.f.'s will be, , [CPMT 2002; DPMT 2003], , 17., , (c) 26.4 ohm, (d) 18.7 ohm, A Daniel cell is balanced on 125 cm length of a potentiometer wire., Now the cell is short-circuited by a resistance 2 ohm and the, balance is obtained at 100 cm . The internal resistance of the Daniel, cell is, [UPSEAT 2002], (a) 0.5 ohm, (b) 1.5 ohm, (c) 1.25 ohm, (d) 4/5 ohm, Sensitivity of potentiometer can be increased by, , 21., In order to pass 10% of main current through a moving coil, [MP PET 1994], galvanometer of 99 ohm, the resistance of the required shunt is [MP PET 1990, 99; MP PMT 1994;, (a), Increasing, the, e.m.f., of, the, cell, RPET 2001; KCET 2003, 05], (b) Increasing the length of the potentiometer wire, (a) 9 .9 , (b) 10 , (c) Decreasing the length of the potentiometer wire, (d) None of the above, (c) 11 , (d) 9 , 22., A potentiometer is an ideal device of measuring potential difference, An ammeter of 5 ohm resistance can read 5 mA. If it is to be used, because, to read 100 volts, how much resistance is to be connected in series, (a) It uses a sensitive galvanometer, [MP PET 1991; MP PMT 1996; MP PMT 2000], (b) It does not disturb the potential difference it measures, (a) 19.9995 , (b) 199.995, (c) It is an elaborate arrangement, (c) 1999.95 , (d) 19995 , (d) It has a long wire hence heat developed is quickly radiated, The potential gradient along the length of a uniform wire is, 23., A battery of 6 volts is connected to the terminals of a three metre, 10 volt / metre . B and C are the two points at 30 cm and, long wire of uniform thickness and resistance of the order of 100  ., The difference of potential between two points separated by 50 cm, 60 cm point on a meter scale fitted along the wire. The potential, on the wire will be, difference between B and C will be [CPMT 1986], [CPMT 1984; CBSE PMT 2004], , 2  resistance. The resistance connected with the galvanometer to, , 16., , (b), , 44, ohm, 9, , [MP PET 1984], , 26., , (a) 2/3, (b) 1/2, (c) 1, (d) 2, Potential gradient is defined as, [MP PET 1994], (a) Fall of potential per unit length of the wire, (b) Fall of potential per unit area of the wire, (c) Fall of potential between two ends of the wire

Page 45 :
Current Electricity 1075, , 27., , (d) Potential at any one end of the wire, In an experiment of meter bridge, a null point is obtained at the, centre of the bridge wire. When a resistance of 10 ohm is connected, in one gap, the value of resistance in other gap is, (a) 10 , (b) 5 , 1, (d) 500 , , 5, If the length of potentiometer wire is increased, then the length of, the previously obtained balance point will, (a) Increase, (b) Decrease, (c) Remain unchanged, (d) Become two times, In potentiometer a balance point is obtained, when, (a) The e.m.f. of the battery becomes equal to the e.m.f. of the, experimental cell, (b) The p.d. of the wire between the +ve end to, jockey becomes equal to the e.m.f. of the experimental cell, (c) The p.d. of the wire between +ve point and jockey becomes, equal to the e.m.f. of the battery, (d) The p.d. across the potentiometer wire becomes equal to the, e.m.f. of the battery, In the experiment of potentiometer, at balance, there is no current, in the, (a) Main circuit, (b) Galvanometer circuit, (c) Potentiometer circuit, (d) Both main and galvanometer circuits, If in the experiment of Wheatstone's bridge, the positions of cells, and galvanometer are interchanged, then balance points will, , 36., , [MP PET 1994], , 29., , 30., , 31., , 32., , 33., , 34., , (a) Change, (b) Remain unchanged, (c) Depend on the internal resistance of cell and resistance of, galvanometer, (d) None of these, The resistance of a galvanometer is 90 ohms. If only 10 percent of, the main current may flow through the galvanometer, in which way, and of what value, a resistor is to be used, (a) 10 ohms in series, (b) 10 ohms in parallel, (c) 810 ohms in series, (d) 810 ohms in parallel, Two cells when connected in series are balanced on 8 m on a, potentiometer. If the cells are connected with polarities of one of the, cell is reversed, they balance on 2m. The ratio of e.m.f.'s of the two, cells is, (a), , 3:5, , (b) 5 : 3, , (c), , 3:4, , (d) 4 : 3, , A voltmeter has a resistance of G ohms and range V volts. The value, of resistance used in series to convert it into a voltmeter of range, nV volts is, , (c), 35., , G, n, , A, 4A, , (d) Greater or less than 5  depends on the material of R, 37., , A moving coil galvanometer has a resistance of 50  and gives full, scale deflection for 10 mA. How could it be converted into an, ammeter with a full scale deflection for 1A, [MP PMT 1996], , 38., , 39., , (a), , 50 / 99  in series, , (b) 50 / 99  in parallel, , (c), , 0.01  in series, , (d) 0.01  in parallel, , The current flowing through a coil of resistance 900 ohms is to be, reduced by 90%. What value of shunt should be connected across, the coil, [Roorkee 1992], (a), , 90 , , (b) 100 , , (c), , 9, , (d) 10 , , A galvanometer of resistance 25  gives full scale deflection for a, current of 10 milliampere, is to be changed into a voltmeter of range, 100 V by connecting a resistance of ‘R’ in series with galvanometer., The value of resistance R in  is, [MP PET 1994], , 40., , 41., , (a) 10000, (b) 10025, (c) 975, (d) 9975, In a potentiometer circuit there is a cell of e.m.f. 2 volt, a resistance, of 5 ohm and a wire of uniform thickness of length 1000 cm and, resistance 15 ohm. The potential gradient in the wire is, (a), , 1, / cm, [MP VPET, 1996], 500, , (b), , 3, V / cm, 2000, , (c), , 3, V / cm, 5000, , (d), , 1, V / cm, 1000, , The resistance of a galvanometer is 25 ohm and it requires 50 A, for full deflection. The value of the shunt resistance required to, convert it into an ammeter of 5 amp is, [MP PMT 1994; BHU 1997], 4, , 42., , G, (n  1), , Which of the following statement is wrong, [MP PET 1994], (a) Voltmeter should have high resistance, (b) Ammeter should have low resistance, (c) Ammeter is placed in parallel across the conductor in a circuit, (d) Voltmeter is placed in parallel across the conductor in a circuit, , R, , (c) Less than 5 , , (b) (n  1)G, (d), , 20V, , (b) Greater from 5 , , [MP PMT 1999; MP PET 2002; DPMT 2004; MH CET 2004], , (a) nG, , V, , (a) Equal to 5 , , (c), 28., , In the diagram shown, the reading of voltmeter is 20 V and that of, ammeter is 4 A. The value of R should be (Consider given ammeter, and voltmeter are not ideal), [RPMT 1997], , 43., , (a) 2.5  10 ohm, (b) 1.25  10 3 ohm, (c) 0.05 ohm, (d) 2.5 ohm, Which is a wrong statement, [MP PMT 1994], (a) The Wheatstone bridge is most sensitive when all the four, resistances are of the same order, (b) In a balanced Wheatstone bridge, interchanging the positions of, galvanometer and cell affects the balance of the bridge, (c) Kirchhoff's first law (for currents meeting at a junction in an, electric circuit) expresses the conservation of charge, (d) The rheostat can be used as a potential divider, A voltmeter having a resistance of 998 ohms is connected to a cell, of e.m.f. 2 volt and internal resistance 2 ohm. The error in the, measurement of e.m.f. will be, [MP PMT 1994], (a), , 4  10 1 volt, , (b) 2  10 3 volt, , (c), , 4  10 3 volt, , (d) 2  10 1 volt

Page 46 :
1076 Current Electricity, 44., , 45., , 46., , For comparing the e.m.f.'s of two cells with a potentiometer, a, standard cell is used to develop a potential gradient along the wires., Which of the following possibilities would make the experiment, unsuccessful, [MP PMT 1994], (a) The e.m.f. of the standard cell is larger than the E e.m.f.'s of the, two cells, (b) The diameter of the wires is the same and uniform throughout, (c) The number of wires is ten, (d) The e.m.f. of the standard cell is smaller than the e.m.f.'s of the, two cells, Which of the following is correct, [BHU 1995], (a) Ammeter has low resistance and is connected in series, (b) Ammeter has low resistance and is connected in parallel, (c) Voltmeter has low resistance and is connected in parallel, (d) None of the above, , (b) Increase the resistance R, (c) Reverse the terminals of battery V, (d) Reverse the terminals of cell E, 51., , (a) 1.85 A, , 48., , 20 , , (d), , 40 , , 52., , The figure shows a circuit diagram of a ‘Wheatstone Bridge’ to, measure the resistance G of the galvanometer. The relation, P R, will be satisfied only when, , Q G, , S, , G, , R, G, A, , B, , C, , connected to E 2 , the null point will be, , [MP PMT 1995], , 53., , B, , (a) To the left of C, (b) To the right of C, A, , B, G, , E2, In an experiment to measure the internal, resistance of a cell by, potentiometer, it is found that the balance point is at a length of 2 m, when the cell is shunted by a 5  resistance; and is at a length of, , 54., , (b) 3.0 V, , R, , P, , X, E, , G, , Y, , 10V, 1, , (a) 1.5 V, , (d) 1 , , V, , A resistance of 4  and a wire of length 5 metres and resistance, , 4, , (b) 10 , , A potentiometer circuit shown in the figure is set up to measure, e.m.f. of a cell E. As the point P moves from X to Y the galvanometer, G shows deflection always in one direction, but the deflection, decreases continuously until Y is reached. In order to obtain balance, point between X and Y it is necessary to, , (d) 99,950  in series, , internal resistance 1  . A parallel combination of two identical cells, is balanced across 300 cm of the wire. The e.m.f. E of each cell is [MP PMT 199, , [Haryana CEE 1996], , 15 , , (b) 5  10 4  in parallel, , 5  are joined in series and connected to a cell of e.m.f. 10 V and, , 3m when the cell is shunted by a 10  resistance. The internal, resistance of the cell is, then, (a) 1 .5 , , 5  10 3  in parallel, , (c) 10 5  in series, , E1, , (d) Nowhere on AB, , (a) The galvanometer shows a deflection when switch S is closed, (b) The galvanometer shows a deflection when switch S is open, (c) The galvanometer shows no change in deflection whether S is, open or closed, (d) The galvanometer shows no deflection, The resistance of a galvanometer is 50 ohms and the current, required to give full scale deflection is 100 A . In order to convert, it into an ammeter, reading upto 10A, it is necessary to put a, resistance of, [MP PMT 1997; AIIMS 1999], (a), , C, , (c) At C itself, , 50., , b, , P, , The circuit shown here is used to compare the e.m.f. of two cells, E1 and E 2 (E1  E 2 ) . The null point is at C when the, , (c), , Y, , Q, , galvanometer is connected to E1 . When the galvanometer is, , 49., , X, , (d) From a to b through c, , (b) 8 , (c), , d, , (c) From b to a through c, , 80 , , R, , B, , c, , (b) From b to a, , (b) Greater than 1.85 A, , 2, , A, , (a) From a to b, , (c) Less than 1.85 A, (d) None of these, AB is a wire of uniform resistance. The galvanometer G shows no, current when the length AC = 20cm and CB = 80 cm. The resistance R, is equal to, [MP PMT 1995; RPET 2001], (a), , In the Wheatstone's bridge (shown in figure) X  Y and A  B ., The direction of the current between ab will be, a, , An ammeter with internal resistance 90  reads 1.85 A when, connected in a circuit containing a battery and two resistors 700 , and 410  in series. Actual current will be, [Roorkee 1995], , 47., , (a) Decreases the resistance R, , 3m, 5, 5m, , (c) 0.67 V, E, , (d) 1.33 V, 55., , E, , G, , The resistivity of a potentiometer wire is 40  10 8 ohm  m and, its area of cross-section is 8  10 6 m 2 . If 0.2 amp current is, flowing through the wire, the potential gradient will be

Page 47 :
Current Electricity 1077, (a) 10 2 volt / m, (c), 56., , 57., , 3.2  10 2 volt / m, , (b) 10 1 volt / m, (d) 1 volt / m, , If only 2% of the main current is to be passed through a, galvanometer of resistance G, then the resistance of shunt will be, (a), , G, 50, , (b), , G, 49, , (c), , 50 G, , (d) 49 G, , 65., , (a) Zero, (c) Very large, , 59., , 60., , 61., , 62., , IP  IG, , (c), , IQ[MP,  IPMT/PET, 1998], G, , (d), , IQ  IR, , In the following Wheatstone bridge P / Q  R / S . If key K is, closed, then the galvanometer will show deflection, [CPMT 1999], , (b) Very low, (d) Infinite, , 210 k , , (b) 315 k , , (c), , 420 k , , (d) 440 k , , 66., , (a), , 3  in series, , (b) 3  in parallel, , (c), , 17  in series, , (d) 30  in series, , 67., , 68., , (d) 10 V / m, , 5V /m, , 69., , 0 .8 , , (b) 0 .5 , , (c), , 0 .4 , , (d) 0.25 , , [IIT-JEE (Screening) 1999], , Q, , P, S, R, , G, V, , (b) 2.5  10 4 V / cm, , (c), , 0.62  10 4 V / mm, , (d) 1  10 5 V / mm, , An ammeter whose resistance is 180  gives full scale deflection, when current is 2 mA. The shunt required to convert it into an, ammeter reading 20 mA (in ohms) is, (a) 18, (b) 20, (c) 0.1, (d) 10, A galvanometer whose resistance is 120  gives full scale, deflection with a current of 0.05 A so that it can read a maximum, current of 10 A. A shunt resistance is added in parallel with it. The, resistance of the ammeter so formed is, (a), , 0.06 , , (b) 0.006 , , (c), , 0 .6 , , (d) 6  s, , In a potentiometer experiment, the galvanometer shows no, deflection when a cell is connected across 60 cm of the, potentiometer wire. If the cell is shunted by a resistance of 6  ,, the balance is obtained across 50 cm of the wire. The internal, resistance of the cell is, [SCRA 1994], 0 .5 , , (b) 0 .6 , , (c) 1 .2 , , (d) 1 .5 , , (a), , 71., , In the circuit shown P  R , the reading of the galvanometer is, same with switch S open or closed. Then, , 5  10 5 V / mm, , [Bihar MEE 1995], , In a meter bridge, the balancing length from the left end (standard, resistance of one ohm is in the right gap) is found to be 20 cm. The, value of the unknown resistance is, (a), , (a), , [EAMCET (Engg.) 1995], , 70., , 2V /m, , (b) 0.8 amp, (d) 0.5 amp, , A potentiometer wire has length 10 m and resistance 20  . A 2. 5, V battery of negligible internal resistance is connected across the, wire with an 80  series resistance. The potential gradient on the, wire will be, [KCET 1994], [MP PMT 1999], , A potentiometer consists of a wire of length 4 m and resistance, 10  . It is connected to a cell of e.m.f. 2 V. The potential, difference per unit length of the wire will be, , (c), , S, , A galvanometer having a resistance of 8 ohm is shunted by a wire of, resistance 2 ohm. If the total current is 1 amp, the part of it passing, through the shunt will be, (a) 0.25, [MPamp, PET 1999], (c) 0.2 amp, , A galvanometer has resistance of 7  and gives a full scale, deflection for a current of 1.0 A. How will you convert it into a, voltmeter of range 10 V, [MP PMT 1999], , (b), , R, , (d) In either side, , [CBSE PMT 1998], , Constantan wire is used in making standard resistances because its, (a) Specific resistance is low, (b) Density is high, (c) Temperature coefficient of resistance is negligible, (d) Melting point is high, The net resistance of a voltmeter should be large to ensure that, (a) It does not get overheated, (b) It does not draw excessive current, (c) It can measure large potential difference, (d) It does not appreciably change the potential difference to be, measured, , (a) 0.5 V / m, , Q, K, , (c) No deflection, , A 100 V voltmeter of internal resistance 20 k  in series with a, high resistance R is connected to a 110 V line. The voltmeter reads 5, V, the value of R is, [MP PET 1999], (a), , P, , (b) In right side, , [CBSE PMT 1999; Pb PMT 2004], , 64., , (b), , (a) In left side, , [CBSE PMT 1999; AFMC 2001], , 63., , IR  IG, , The resistance of an ideal voltmeter is, [EAMCET (Med.) 1995; MP PMT/PET 1998; Pb. PMT 1999;, CPMT 2000], , 58., , (a), , A voltmeter of resistance 1000  gives full scale deflection when a, current of 100 mA flow through it. The shunt resistance required, across it to enable it to be used as an ammeter reading 1 A at full, scale deflection is, [SCRA 1994], (a) 10000 , (c), , 222 , , (b) 9000 , (d) 111 

Page 48 :
1078 Current Electricity, 72., , 73., , The resistance of 10 metre long potentiometer wire is 1ohm/meter. A, cell of e.m.f. 2.2 volts and a high resistance box are connected in, series to this wire. The value of resistance taken from resistance box, for getting potential gradient of 2.2 millivolt/metre will be[RPET 1997], (a), , 790 , , (b) 810 , , (c), , 990 , , (d) 1000 , , 80., , (b) 19989 , , (c) 10000 , , (d) 9989 , , In a balanced Wheatstone’s network, the resistances in the arms Q, and S are interchanged. As a result of this, [KCET 1999], , We have a galvanometer of resistance 25  . It is shunted by a, , (a) Network is not balanced, , 2.5  wire. The part of total current that flows through the, galvanometer is given as, , (b) Network is still balanced, (c) Galvanometer shows zero deflection, , [AFMC 1998; MH CET 1999; Pb. PMT 2002], , 74., , (a) 20000 , , (a), , I, 1, , I0 11, , (b), , I, 1, , I0 10, , (c), , I, 3, , I0 11, , (d), , I, 4, , I0 11, , (d) Galvanometer and the cell must be interchanged to balance, 81., , (a) Exactly 10 ohm, , In the adjoining circuit, the e.m.f. of the cell is 2 volt and the, internal resistance is negligible. The resistance of the voltmeter is 80, ohm. The reading of the voltmeter will be, (a) 0.80 volt, , (c) More than 10 ohm, (d) We cannot definitely say, 82., , 80 , , (c) 1.33 volt, , V, , (a), , 20 wire be  80and,  area of crossIf the resistivity of a potentiometer, section be A, then what will be potential gradient along the wire, I, I, (a), (b), A, A, , 76., , 83., , 78., , (b) 4000 , , (a) 1 , , (c), , 6000 , , (d) 8000 , , (b) 3 , , 0.25 , , (b) 0.50, , (c), , 0.80 , , (d) 1.00 , , (d) 5 , 85., , [CPMT 1999], , (b) Potential adder, (c) Potential substracter, , 86., Total P.D., , 87., , (d) Potential multiplier, 79., , Variable P.D., , A potentiometer wire of length 1 m and resistance 10  is, connected in series with a cell of emf 2V with internal resistance 1 , and a resistance box including a resistance R. If potential difference, between the ends of the wire is 1 mV, the value of R is [KCET 1999], , 2V, 1m, , (c) 10 , , The arrangement as shown in figure is called as, (a) Potential divider, , (d) All of these, , In the given figure, battery E is balanced on 55 cm length of, potentiometer wire but when a resistance of 10  is connected in, parallel with the battery then it balances on 50 cm length of the, potentiometer wire then internal resistance r of the battery is, , 2000 , , (a), , R, 4, , R, (d) 4 R, 3[RPET 1996], If an ammeter is connected in parallel to a circuit, it is likely to be, damaged due to excess, [BHU 2000; BCECE 2004], (a) Current, (b) Voltage, , (a), , For a cell of e.m.f. 2V, a balance is obtained for 50 cm of the, potentiometer wire. If the cell is shunted by a 2  resistor and the, balance is obtained across 40 cm of the wire, then the internal, resistance of the cell is, [SCRA 1998], , (b), , (c) Resistance, 84., , [CPMT 1997, SCRA 1994], , 77., , R, 5, , (c), , IA, , (d) IA, , A voltmeter has resistance of 2000 ohms and it can measure upto, 2V. If we want to increase its range to 10 V, then the required, resistance in series will be, (c), , V, , The resistance of a galvanometer coil is R. What is the shunt, resistance required to convert it into an ammeter of range 4 times, , (d) 2.00 volt, 75., , R, A, , (b) Less than 10 ohm, , [CPMT 1991], , 2V, + –, , (b) 1.60 volt, , The ammeter A reads 2 A and the voltmeter V reads 20 V. the, value of resistance R is (Assuming finite resistance's of ammeter and, voltmeter), [JIPMER 1999; MP PMT 2004], , B, , A, E, , r, , A galvanometer with a resistance of 12  gives full scale deflection, when a current of 3 mA is passed. It is required to convert it into a, voltmeter which can read up to 18 V. the resistance to be connected, is, [Pb. PMT 2000], (a) 6000 , , (b) 5988 , , (c) 5000 , , (d) 4988 , , The resistance of an ideal ammeter is [KCET 2000], (a) Infinite, , (b) Very high, , (c) Small, , (d) Zero, , A galvanometer of 25  resistance can read a maximum current of, 6mA. It can be used as a voltmeter to measure a maximum of 6 V, by connecting a resistance to the galvanometer. Identify the correct, choice in the given answers, [EAMCET (Med.) 2000], (a) 1025  in series, , (b) 1025  in parallel, , (c) 975  in series, , (d) 975  in parallel

Page 49 :
Current Electricity 1079, 88., , 89., , A galvanometer has a resistance of 25 ohm and a maximum of 0.01, A current can be passed through it. In order to change it into an, ammeter of range 10 A, the shunt resistance required is, (a) 5/999 ohm, (b) 10/999 ohm, (c) 20/999 ohm, (d) 25/999 ohm, In the circuit shown, a meter bridge is in its balanced state. The, meter bridge wire has a resistance 0.1 ohm/cm. The value of, unknown resistance X and the current drawn from the battery of, negligible resistance is, [AMU (Engg.) 2000], X, , 96., , 6, , 97., , 6, , B, , 5V, , A galvanometer has 30 divisions and a sensitivity 16 A / div.It can, be converted into a voltmeter to read 3 V by connecting, , 98., , (a) Resistance nearly 6 k  in series, (b) 6 k  in parallel, (c), 91., , 500  in series, , 99., , (d) It cannot be converted, Voltmeters V and V are connected in series across a D.C. line. V, reads 80 volts and has a per volt resistance of 200 ohms. V has a, total resistance of 32 kilo ohms. The line voltage is, (a) 120 volts, (b) 160 volts, (c) 220 volts, (d) 240 volts, A potentiometer having the potential gradient of 2 mV/cm is used to, measure the difference of potential across a resistance of 10 ohm. If, a length of 50 cm of the potentiometer wire is required to get the, null point, the current passing through the 10 ohm resistor is (in, mA), 1, , 2, , 100., , (a) 1, (b) 2, (c) 5, (d) 10, AB is a potentiometer wire of length 100 cm and its resistance is 10, ohms. It is connected in series with a resistance R = 40 ohms and a, battery of e.m.f. 2 V and negligible internal resistance. If a source of, unknown e.m.f. E is balanced by 40 cm length of the potentiometer, wire, the value of E is, [MP PET 2001], R, , (a) 0.8 V, , 94., , 95., , (b) 99 , , (c) 1000 , , (d) None of these, , A voltmeter has a range 0-V with a series resistance R. With a, series resistance 2R, the range is 0-V. The correct relation between, V and V is, [CPMT 2001], V   2V, , (b) V   2V, , (c), , V   2V, , (d) V '  2V, , [UPSEAT 2000], , The measurement of voltmeter in the following circuit is, , (c) 4.0 V, , 6V, , 60, , 101., , 40, , A 36  galvanometer is shunted by resistance of 4. The, percentage of the total current, which passes through the, galvanometer is, [UPSEAT 2002], (a) 8 %, (b) 9 %, (c) 10 %, (d) 91 %, An ammeter and a voltmeter of resistance R are connected in series, to an electric cell of negligible internal resistance. Their readings are, A and V respectively. If another resistance R is connected in parallel, with the voltmeter, [EAMCET 2000; KCET 2002], , B, , (d) 0.16 V, , (a) Both A and V will increase, , An ammeter gives full deflection when a current of 2 amp. flows, through it. The resistance, E of ammeter is 12 ohms. If the same, ammeter is to be used for measuring a maximum current of 5 amp.,, then the ammeter must be connected with a resistance of, , (b) Both A and V will decrease, , (a) 8 ohms in series, (b) 18 ohms in series, (c) 8 ohms in parallel, (d) 18 ohms in parallel, In a circuit 5 percent of total current passes through a, galvanometer. If resistance of the galvanometer is G then value of, the shunt is, [MP PET 2001], (a) 19 G, (b) 20 G, , [AFMC 2001], , –, , V, , (d) 6.0 V, , 102., , 2V, , (a), , (b) 3.4 V, , 40 cm, , A, , (a) 999 , , (a) 2.4 V, , (b) 1.6 V, (c) 0.08 V, , A milliammeter of range 10 mA has a coil of resistance 1 . To use, it as voltmeter of range 10 volt, the resistance that must be, [Kerala, PMT 2005], connected, in series, with it, will be [KCET 2001], , +, , [AMU (Med.) 2000], , 93., , (b) 2 : 1, (d) 4 : 1, , 1, , 2, , 92., , 2, , [Kerala (Engg.) 2001], , (a) 1 : 1, (c) 3 : 1, , (d) 12 , 0.5 amp, 90., , 60 cm, , A, , (c) 4 , 1.0 amp, , 1, , 2, , G, 40 cm, , 3, , 5, , (a) 6 , 5 amp, (b) 10 , 0.1 amp, , G, G, (d), 19, 20, [MP PET 2000], A voltmeter having resistance of 50 × 10 ohm is used to measure the, voltage in a circuit. To increase the range of measurement 3 times, the additional series resistance required is, (a) 10 ohm, (b) 150 k.ohm, (c) 900 k.ohm, (d) 9 × 10 ohm, In a potentiometer experiment two cells of e.m.f. E and E are used, in series and in conjunction and the balancing length is found to be, 58 cm of the wire. If the polarity of E is reversed, then the, E, balancing length becomes 29 cm. The ratio 1 of the e.m.f. of the, E2, two cells is, , (c), , (c) A will decrease and V will increase, 103., , (d) A [MP, will PET, increase, 2001]and V will decrease, A wire of length 100 cm is connected to a cell of emf 2 V and, negligible internal resistance. The resistance of the wire is 3 ., The additional resistance required to produce a potential drop of 1, milli volt per cm is, [Kerala PET 2002], (a) 60 , , (b) 47 , , (c) 57 , , (d) 35 

Page 50 :
1080 Current Electricity, 104., , A galvanometer of resistance 20  is to be converted into an, ammeter of range 1 A. If a current of 1 mA produces full scale, deflection, the shunt required for the purpose is, , 105., , 106., , 10 , , (c) 1.0, , (b) 0.05 , , (c) 0.02 , (d) 0.04 , There are three voltmeters of the same range but of resistances, The best, 10000  , 8000  and 4000  respectively., voltmeter among these is the one whose resistance is, (a) 10000 , , V, , (b) 0.1, , [Kerala PET 2002], , (a) 0.01 , , 900 , , 10, 9, , (a), , 114., , A cell of internal resistance 3 ohm and emf 10 volt is connected to a, uniform wire of length 500 cm and resistance 3 ohm. The potential, gradient in the wire is, [MP PET 2003], [Kerala PET 2002], (a) 30 mV/cm, (b) 10 mV/cm, (c) 20 mV/cm, (d) 4 mV/cm, , 115., , An ammeter of 100  resistance gives full deflection for the current, of 10 amp. Now the shunt resistance required to convert it into, ammeter of 1 amp. range, will be, , (b) 8000 , , (c) 4000 , (d) All are equally good, If an ammeter is to be used in place of a voltmeter then we must, connect with the ammeter a, , 100 , , (d) 10.0, , –5, , [AIEEE 2002; AFMC 2002], , (a), (b), (c), (d), 107., , 108., , 109., , 110., , 111., , 112., , Low resistance in parallel, High resistance in parallel, High resistance in series, Low resistance in series, , [RPET 2003], , (a) 10, (c) 10, 116., , A 10 m long wire of 20 resistance is connected with a battery of 3, volt e.m.f. (negligible internal resistance) and a 10  resistance is, joined to it is series. Potential gradient along wire in volt per meter, is, [MP PMT 2003], (a) 0.02, , (b) 0.3, , (c) 0.2, , (d) 1.3, , [MP PMT 2002], , 118., , To convert a 800 mV range milli voltmeter of resistance 40  into a, galvanometer of 100 mA range, the resistance to be connected as, shunt is, [CBSE PMT 2002], (d) 40 , , (d) 10, , , , 1, 40, , (b), , 1, 4, , 1, 1, (d), 10, 140, If the ammeter in the given circuit reads 2 A, the resistance R is, , (a) 1 ohm, R, [MP PMT 2002], (b) 2 ohm, 6, (c) 3 ohm, (d) 4 ohm, A 50 ohm galvanometer gets fullA scale deflection when a current of, 6V, 0.01 A passes through the coil. When it is converted, to a 10 A, ammeter, the shunt resistance is, [Orissa JEE 2003], , (a), 119., , 120., , [CBSE PMT 2002], , 0.01 , , (b) 0.05 , , (c) 2000 , (d) 5000 , Resistance in the two gaps of a meter bridge are 10 ohm and 30, ohm respectively. If the resistances are interchanged the balance, point shifts by, [Orissa JEE 2003], (a) 33.3 cm, (b) 66.67cm, (c) 25 cm, (d) 50 cm, A potentiometer has uniform potential gradient. The specific, resistance of the material of the potentiometer wire is 10 ohm–, meter and the current passing through it is 0.1 ampere; cross-section, of the wire is 10 m . The potential gradient along the potentiometer, wire is, [KCET 2003], –7, , –6, , [MP PET 2002], , 113., , , , , , 1, , 3, , A 100 ohm galvanometer gives full scale deflection at 10 mA. How, much shunt is required to read 100 mA, (a) 11.11 ohm, (c) 1.1 ohm, , (b) 10, , (c), , (a) Copper, (b) Steel, (c) Manganin, (d) Aluminium, To convert a galvanometer into a voltmeter, one should connect a, (a) High resistance in series with galvanometer, (b) Low resistance in series with galvanometer, (c) High resistance in parallel with galvanometer, (d) Low resistance in parallel with galvanometer, , (c) 30 , , , , A galvanometer of resistance 36  is changed into an ammeter by, using a shunt of 4 . The fraction f of total current passing, through the galvanometer is, [BCECE 2003], (a), , 117., , (b) 20 , , 3, , 5, , 0, , A potentiometer has uniform potential gradient across it. Two cells, connected in series (i) to support each other and (ii) to oppose each, other are balanced over 6m and 2m respectively on the, potentiometer wire. The e.m.f.’s of the cells are in the ratio of, (a) 1 : 2, (b) 1 : 1, (c) 3 : 1, (d) 2 : 1, The material of wire of potentiometer is, , (a) 10 , , 4, , (b) 9.9 ohm, (d) 4.4 ohm, , The potential difference across the 100 resistance in the following, circuit is measured by a voltmeter of 900  resistance. The, percentage error made in reading the potential difference is, , 121., , 2, , (a) 10 4 V/m, , (b) 10 6 V/m, , (c) 10 2 V/m, , (d) 10 8 V/m, , Two resistances of 400  and 800  are connected in series with 6, volt battery of negligible internal resistance. A voltmeter of, resistance 10,000  is used to measure the potential difference, across 400 . The error in the measurement of potential difference, (Med.) 2002], in volts[AMU, approximately, is, (a) 0.01, (b) 0.02, (c) 0.03, (d) 0.05, , [

Page 51 :
Current Electricity 1081, 122., , A galvanometer, having a resistance of 50  gives a full scale deflection for, a current of 0.05 A. The length in meter of a resistance wire of area of, cross-section 2.97× 10 cm that can be used to convert the galvanometer, into an ammeter which can read a maximum of 5 A current is (Specific, resistance of the wire = 5 × 10 7 m), (a) 9, (b) 6, (c) 3, (d) 1.5, An ammeter reads upto 1 ampere. Its internal resistance is 0.81, ohm. To increase the range to 10 A the value of the required shunt, is, [AIEEE 2003], –2, , 123., , 2, , (a) 0.09 , 124., , 125., , 126., , 30 E, 100, , (b), , 30 E, 100. 5, , For the post office box arrangement to determine the value of, , C, , (b) C and D, A, , (d) B and C, 1, , 131., , 1, , B1 has 25 divisions.CA, A galvanometer of 50 ohm resistance, 1 current of, 4  10 ampere gives a deflection of one division. To convert this, galvanometer into a voltmeter having a range of 25 volts, it should, be connected with a resistance of, –4, , [CBSE PMT 2004], , 132., , 30 E, (100  0 .5), , (d), , 30(E  0 .5 i), , where i is the current in the potentiometer, 100, , (a) 2500  as a shunt, , (b) 2450  as a shunt, , (c) 2550  in series, , (d) 2450  in series, , In a metre bridge experiment null point is obtained at 20 cm from, one end of the wire when resistance X is balanced against another, resistance Y. If X < Y, then where will be the new position of the null, point from the same end, if one decides to balance a resistance of, 4X against Y, , Resistance of 100 cm long potentiometer wire is 10, it is connected, to a battery (2 volt) and a resistance R in series. A source of 10 mV, gives null point at 40 cm length, then external resistance R is, , (a) 50[MP, cm PMT 2003], , (b) 80 cm, , (a) 490 , , (c) 40 cm, , (d) 70 cm, , [AIEEE 2004], , (b) 790 , , (c) 590 , (d) 990 , The e.m.f. of a standard cell balances across 150 cm length of a wire, of potentiometer. When a resistance of 2  is connected as a shunt, , (c), , 0 .1 , , 133., , In the circuit given, the correct relation to a balanced Wheatstone, bridge is, [Orissa PMT 2004], (a), , P R, , Q S, , (b), , P S, , Q R, , (c), , P S, , R Q, , (b) 1 , , 2, , (d) 0 .5 , , What is the reading of voltmeter in the following figure, 10 V, , (b) 2 V, , 1000 , , (c) 5 V, , V, A, , 500  A, , (b) 90 , , (c) 1000 , , (d) 10 , , G, S, , Q, , 134., , A galvanometer coil of resistance 50 , show full deflection of, 100 A . The shunt resistance to be added to the galvanometer, to, work as an ammeter of range 10 mA is, [Pb PET 2000], , 500 , , The current flowing in a coil of resistance 90  is to be reduced by, 90%. What value of resistance should be connected in parallel with, it, [MP PMT 2004], (a) 9 , , R, , P, , (d) None of these, , [MP PMT 2004], , (a) 3 V, , (d) 4 V, , D, , (c) A and D, , (c), , (a), , 129., , (d) 4050, , B, , [MP PET 1993], , 128., , (c) 2010, , (a) B and C, , with the cell, the balance point is obtained at 100 cm . The internal, resistance of the cell is, , 127., , (b) 4960, , unknown, [EAMCET, 2003]resistance the unknown resistance should be connected, between, [IIT-JEE (Screening) 2004], , (b) 0.03 , , (c) 0.3 , (d) 0.9 , The length of a wire of a potentiometer is 100 cm, and the emf of its, standard cell is E volt. It is employed to measure the e.m.f of a, battery whose internal resistance is 0.5 . If the balance point is, obtained at l = 30 cm from the positive end, the e.m.f. of the battery, is, [AIEEE 2003], (a), , 130., , (a) 5040, , 135., , (a) 5  in parallel, , (b) 0.5  in series, , (c) 5  in series, , (d) 0.5  in parallel, , In given figure, the potentiometer wire AB has a resistance of 5 , and length 10 m. The balancing length AM for the emf of 0.4 V is, R=45, , (a) 0.4 m, , The maximum current that can be measured by a galvanometer of, resistance 40  is 10 mA. It is converted into a voltmeter that can, read upto 50 V. The resistance to be connected in series with the, galvanometer is ... (in ohm), [KCET 2004], , 5V, , (b) 4 m, (c) 0.8 m, (d) 8 m, , M, , A, 0.4V, , B

Page 52 :
1082 Current Electricity, 136., , A potentiometer consists of a wire of length 4 m and resistance 10, . It is connected to cell of emf 2 V. The potential difference per, unit length of the wire will be, [Pb. PET 2002], , 137., , 138., , 139., , (a) 0.5 V/m, (b) 10 V/m, (c) 2 V/m, (d) 5 V/m, A voltmeter essentially consists of, [UPSEAT 2004], (a) A high resistance, in series with a galvanometer, (b) A low resistance, in series with a galvanometer, (c) A high resistance in parallel with a galvanometer, (d) A low resistance in parallel with a galvanometer, In a potentiometer experiment the balancing with a cell is at length, 240 cm. On shunting the cell with a resistance of 2 , the balancing, length becomes 120 cm. The internal resistance of the cell is, (a) 4 , , (b) 2 , , (c) 1 , , (d) 0.5 , , 2., , 140., , 141., , 142., , 143., , (c) 2.4 V, , (d) 1.41 V, , (b) 1000.1 mA, , (c) 10.01 mA, , (d) 1.01 mA, , (b) 6, , (c) 9, , (d) 12, , (b) 120 , , (c) 124 , , (d) 114, , (d) 270 , , Two uniform wires A and B are of the same metal and have, equal masses. The radius of wire A is twice that of wire B . The, total resistance of A and B when connected in parallel is, , 3., , (b) 7.9 , , (c) 5.9 , , (d) 6.9 , , 4  when the resistance of wire B is 4.25 , , Twelve [DCE, wires2002;, of equal, AIEEElength, 2005] and same cross-section are connected, in the form of a cube. If the resistance of each of the wires is R ,, then the effective resistance between the two diagonal ends would, be, [J & K CET 2004], (a) 2 R, , 5, R, 6, , (d) 8 R, 4., , You are given several identical resistances each of value R  10 , [IIT-JEE, 2005]of carrying maximum current of 1 ampere. It is, and (Screening), each capable, , required to make a suitable combination of these resistances to, produce a resistance of 5  which can carry a current of 4, amperes. The minimum number of resistances of the type R that, will be required for this job, [CBSE PMT 1990], , (a) 4, , [KCET 2005], , (b) 10, , (c) 8, 5., , (d) 20, , The resistance of a wire is 10 6  per metre. It is bend in the, form of a circle of diameter 2 m . A wire of the same material is, [BCECE 2005], , connected across its diameter. The total resistance across its, diameter AB will be, , Potentiometer wire of length 1 m is connected in series with 490 , resistance and 2V battery. If 0.2 mV/cm is the potential gradient,, then resistance of the potentiometer wire is, (a) 4.9 , , 4  when the resistance of wire A is 4.25 , , (d) 4  when the resistance of wire B is 4.25 , , (c), , If resistance of voltmeter is 10000 and resistance of ammeter is, 2 then find R when voltmeter reads 12V and ammeter reads 0.1 A, (a) 118 , , 90 , , (c), , Two resistances are connected in two gaps of a metre bridge. The, balance point is 20 cm from the zero end. A resistance of 15 ohms is, connected in series with the smaller of the two. The null point shifts, to 40 cm. The value of the smaller resistance in ohms is, (a) 3, , (c), , (b) 12 R, , A moving coil galvanometer of resistance 100 is used as an, ammeter using a resistance 0.1 The maximum deflection current, in the galvanometer is 100A. Find the minimum current in the, circuit so that the ammeter shows maximum deflection, (a) 100.1 mA, , (b) 45 , , (b) 5  when the resistance of wire A is 4.25 , , [DCE 2002], , (b) 1.8 V, , 7 .5 , , (a), , With a potentiometer null point were obtained at 140 cm and 180, cm with cells of emf 1.1 V and one unknown X volts. Unknown emf is, (a) 1.1 V, , (a), , A, , B, , [DCE 2005], , 6., , (a), , 4,   10 6 , 3, , (b), , 2,   10 6 , 3, , (c), , 0.88  10 6 , , (d) 14  10 6 , , In the figure shown, the capacity of the condenser C is 2 F . The, current in 2  resistor is, , [IIT 1982], 2, , 1., , In an electrical cable there is a single wire of radius 9 mm of, copper. Its resistance is 5  . The cable is replaced by 6 different, insulated copper wires, the radius of each wire is 3 mm . Now the, , 2F, , 3, 4, , total resistance of the cable will be, [CPMT 1988], , + –, 6V, , 2.8

Page 53 :
Current Electricity 1083, (a) 9 A, (c), 7., , (b) 0.9 A, , 1, A, 9, , 12., , 1, A, 0 .9, , (d), , P, , When the key K is pressed at time t  0 , which of the following, statements about the current I in the resistor AB of the given circuit, is true, [CBSE PMT 1995], A, , If in the circuit shown below, the internal resistance of the battery is, 1.5  and V and V are the potentials at P and Q respectively, what, is the potential difference between the points P and Q, (a), , Q, , Zero, , (b) 4 volts (V > V ), P, , B, , +, , Q, , (c) 4 volts (V > V ), Q, , K, , 2V, , 1000, , 1.5 , –, , 20 V, , 3, , P, , 2, , P, , (d) 2.5 volts (V > V ), Q, , 1000, , 1F, , 13., , C, , P, , Two wires of resistance R and R have, coefficient, of, 2  temperature, 3, Q, resistance  1 and  2 , respectively. These are joined in series. The, 1, , 2, , effective temperature coefficient of resistance is, (a) I = 2 mA at all t, (b) I oscillates between 1 mA and 2mA, , (a), , 1   2, , (c), ,  1 R1   2 R 2, , (c) I = 1 mA at all t, (d) At t = 0 , I = 2 mA and with time it goes to 1 mA, 8., , A torch bulb rated as 4.5 W, 1.5 V is connected as shown in the, figure. The e.m.f. of the cell needed to make the bulb glow at full, intensity is, [MP PMT 1999], , 14., , to an external resistance R, it is observed that the potential, difference across the first cell becomes zero. The value of R will be, [MP PET 1985; KCET 2005; Kerala PMT 2005], , 1, , (c) 2.67 V, , 9., , In the circuit shown in the figure, the current through, 2, , 9V, , [IIT 1998], , 2, , 8, , 2, , 2, , 15., , 4, , 8, , 2, , (a) The 3 resistor is 0.50A (b) The 3 resistor is 0.25 A, , 16., , (c) The 4 resistor is 0.50A (d) The 4 resistor is 0.25 A, 10., , (b) r1  r2, , (c), , r1  r2, 2, , (d), , (a) 3, , (b) 4, , (c) 6, , (d) 5, , r1  r2, 2, , When connected across the terminals of a cell, a voltmeter measures, 5V and a connected ammeter measures 10 A of current. A resistance, of 2 ohms is connected across the terminals of the cell. The current, flowing through this resistance will be, (a) 2.5 A, , (b) 2.0 A, , (c) 5.0 A, , (d) 7.5 A, , In the circuit shown here, E = E = E = 2 V and R = R = 4 ohms. The, current flowing between points A and B through battery E is, 1, , 2, , 3, , 1, , (b) 2 amp from A to B, A, , (c) 2 amp from B to A, , 17., , E2, , B, , E3, , R2, , In the circuit shown below E = 4.0 V, R = 2 , E = 6.0 V, R = 4 , and R = 2 . The current I is [MP PET 2003], 1, , 3, , 1, , r, , r, , r, r, , A, , (c) 1.25 A, , r, , I1, , E1 = 4 V, R3 = 2 , , C, , I2, , R2 = 4 , , (d) 1.0 A, , r, B, , 2, , R1 = 2 , , (b) 1.8 A, , (c) r / 3, , 2, , 1, , [Similar to CBSE PMT 1999; RPET 1999], , (a) 1.6 A, , r, , (a) (4/3) r, , R1, , (a) Zero, , (d) None of the above, , In the circuit shown, the value of each resistance is r, then, equivalent resistance of circuit between points A and B will be, , (b) 3r / 2, , 2, , E1, , There are three resistance coils of equal resistance. The maximum, number of resistances you can obtain by connecting them in any, manner you choose, being free to use any number of the coils in any, way is, , (d) 8r / 7, , r1  r2, , 2, , [ISM Dhanbad 1994], , 11., , (a), E(r=2.67), , 3, , R12  R 22, , Two cells of equal e.m.f. and of internal resistances r1 and, , (b) 1.5 V, , (d) 13.5 V, , R1 R 2 1 2, , (d), , R1  R 2, , r2 (r1  r2 ) are connected in series. On connecting this combination, , 4.5 W, 1.5 V, , (a) 4.5 V, ,  1 2, , (b), , 2, , E2 = 6 V, , [

Page 54 :
1084 Current Electricity, 18., , A microammeter has a resistance of 100  and full scale range of, , R5, , 50 A . It can be used as a voltmeter or as a higher range ammeter, , 19., , I, , provided a resistance is added to it. Pick the correct range and, resistance combination, , R1, , [SCRA 1996; AMU (Med.) 2001; Roorkee 2000], , R2, , (a) 50 V range with 10 k  resistance in series, , (a), , R1 R 2 R5  R 3 R 4 R 6, , (b) 10 V range with 200 k  resistance in series, , (b), , (c) 10 mA range with 1  resistance in parallel, , 1, 1, 1, 1, , , , R5 R 6, R1  R2 R3  R4, , (c), , R1 R 4  R 2 R 3, , (d) 10 mA range with 0 .1  resistance in parallel, , (d), , R1 R 3  R 2 R 4  R5 R6, , The potential difference across 8 ohm resistance is 48 volt as shown, in the figure. The value of potential difference across X and Y points, will be, [MP PET 1996], , 24., , (a) 160 volt, , (c) V / 3, 20, , (b) 128 volt, , Two resistances R1, , 30, , 24, , 8, , 48V, , 25., , 1, , Yand R are made of different materials. The, 2, , temperature coefficient of the material of R1 is  and of the, material of R 2 is   . The resistance of the series combination of, , (c), 21., , , , , (b), , 2  2, , , (d), , 23., , , , 26., , , , , (a), (b), , 5, 15, A and, A, 26, 26, , (c), , 4, 12, A and, A, 25, 25, , (d), , 3, 9, A and, A, 25, 25, , 2V, , 2R, , (b) 6, , (c) 8, , (d) 9, , What is the equivalent resistance between the points A and B of the, network, [AMU (Engg.) 2001], 3, , 2, , A, , 2, , 2, , 4, 10, , 1, 1, , (c) 6 , , [CBSE PMT 1992], , (d), 27., , 1.8, , 5, , 57, , 5, , 2.2, , B, , The effective resistance between points P and Q of the electrical, circuit shown in the figure is, [IIT-JEE (Screening) 2002], , [Roorkee 1999], , 2R, , 2 Rr /(R  r), , (b) 8 R (R  r) /(3 R  r), , P, , (c), , 2r  4 R, , 2R, 2R, , r, , r, Q, , P, , 2R, , (d) 5 R / 2  2r, , Q, , 1, , 57, , 7, , (b) 8 , , (a), , 3V, , [IIT-JEE (Screening) 2001], , (a) 4, , (a), , (d) 0.1 m/s, , A wire of resistance 10  is bent to form a circle. P and Q are, points on the circumference of the circle dividing it into a quadrant, and are connected to a Battery of 3 V and internal resistance 1  as, shown in the figure. The currents in the two parts of the circle are, 6, 18, A and, A, 23, 23, , C, , [MP PMT 1997], , An ionization chamber with parallel conducting plates as anode and, cathode has 5  10 7 electrons and the same number of singlycharged positive ions per cm 3 . The electrons are moving at 0.4 m/s., The current density from anode to cathode is 4 A / m 2 . The, velocity of positive ions moving towards cathode is, (a) 0.4 m/s, (b) 16 m/s, (c) Zero, , 22., , V, , A wire of length L and 3 identical cells of negligible internal, resistances are connected in series. Due to current, the temperature, of the wire is raised by  T in a time t. A number N of similar, cells is now connected in series with a wire of the same material and, cross–section but of length 2 L. The temperature of the wire is, raised by the same amount  T in the same time t. the value of N, is, , R1 and R 2 will not change with temperature, if R1 / R2 equals, , (a), , R, , 60, , (d) 2V / 3, , (c) 80 volt, , 20., , V, , (b) V / 2, , 3, , (d) 62 volt, , R4, , In the given circuit, with steady current, the potential drop across, the capacitor must be, [IIT-JEE (Screening) 2001], (a) V, , X, , R3, , R6, , 28., , 2R, , 2R, , In the circuit element given here, if the potential at point B, V = 0,, then the potentials of A and D are given as, B, , [AMU (Med.) 2002], , In the given circuit, it is observed that the current I is independent, of the value of the resistance R . Then the resistance values must, satisfy, [IIT-JEE (Screening) 2001], 6, , 2.5 , , 1.5 , , 1 amp, , A, , B, , 2V, , C, , D

Page 55 :
Current Electricity 1085, (a), , VA  1.5 V, VD  2 V (b), , (a) 4/9, , VA  1.5 V, VD  2 V, , VA  1.5 V, VD  0.5 V (d) VA  1.5 V, VD  0.5 V, The equivalent resistance between the points P and Q in the, 3, network given here is equal to (given r   ), 2, , 29., , [AMU (Med.) 2002], , (a), , 1, , 2, , (b) 1 , (c), , R, , (d) 18, 36., , 31., , Q, , (b) 0.6 A, 6 V, , r, r, , (c) 6/11 A, 60/11 V, , r, , The current in a conductor varies with time t as I  2t  3 t, where I is in ampere and t in seconds. Electric charge flowing, through a section of the conductor during t = 2 sec to t = 3 sec is, (a) 10 C, (b) 24 C, (c) 33 C, (d) 44 C, A group of N cells whose emf varies directly with the internal, resistance as per the equation E = 1.5 r are connected as shown in, the figure below. The current I in the circuit is, 2, , N, , (b) 5.1 amp, , N, , 38., , (a) x, , R, (2   ), 2, , W, , G, , A, , Potential difference across the terminals of the battery shown in, figure is (r = internal resistance of battery), r =1, , 10 V, , (d) Zero, 40., , 4, , As the switch S is closed in the circuit shown in figure, current, passed through it is, [MP PMT 1994], 20 V, , (a) 4.5 A, , 2, , 3, , B, , 5V, , 4, , B, , A, , (b) 6.0 A, , 2, , (c) 3.0 A, S, , (d) Zero, , 10, , Z, , O, , (c) 6 V, , 10, , (b) 4 , , , Y, , (b) 10 V, , (a) 3 , , 35., , X, , (a) 8 V, , The resistance of a wire of iron is 10 ohms and temp. coefficient of, C, B, resistivity is 5  10 3 / C . AAt 20xC it carries, 30 milliamperes, of, current. Keeping constant potential difference between its ends, the, temperature of the wire is raised to 120C . The current in, milliamperes that flows in the wire is, (a) 20, (b) 15, (c) 10, (d) 40, Seven resistances are connected as shown in the figure. The, equivalent resistance between A and B is, [MP PET 2000], , (d) 5 , , R, (2   ), 4 2, , 4, (2   ), [IIT-JEE (Screening), 2003], R, , R2, , (d) 2x, , (c) 4.5 , , A wire of resistor R is bent into a circular ring of radius r., Equivalent resistance between two points X and Y on its, circumference, when angle XOY is , can be given by, , (d), , (b) x/4, , 34., , (d) None of these, , –5, , (c) R (2 – ), , 39., , R1, , (c) 4  10 , , (b), , 3, , In the shown arrangement of the experiment of 4the meter bridge if, AC corresponding to null deflection of galvanometer is x, what, would be its value if the radius of the wire AB is doubled, , (c) 4x, , (b) 2  10 , , (a), , r4, , (d) 1.5 amp, , (a) 5.6  10 , –5, , r3, , (c) 0.15 amp, , 4, , Length of a hollow tube is 5m, it’s outer diameter is 10 cm and, thickness of it’s wall is 5 mm. If resistivity of the material of the, 2003], tube is [Orissa, 1.7  10JEE, m then resistance of tube will be, –5, , r2, , rN, , A, , –8, , 2, , r1, , (a) 0.51 amp, , 37., , [KCET 2003], , 1, , 33., , V, , 6, , (d) 11/6 A, 11/60 V, , N, , 32., , 4R, , (a) 6 A, 60 V, , r, , (d) 2 , 30., , R, , In the circuit shown here, the readings of the ammeter and, voltmeter are, [Kerala PMT 2002], 6 V, 1, , P, , 3, , 2, , R, , 6R, , r, r, , R, , E, , (c) 2, , r, , r, , R, , (b) 8/9, , (c), , 41., , In the following circuit a 10 m long potentiometer wire with, resistance 1.2 ohm/m, a resistance R and an accumulator of emf 2 V, are connected in series. When the emf of thermocouple is 2.4 mV, then the deflection in galvanometer is zero. The current supplied by, the accumulator will be, 1, , 5, , 8, , 6, , 6, , A battery of internal resistance 4 is connected to the network of, resistances as shown. In order to give the maximum power to the, network, the value of R (in  ) should be, , [IIT101995], (a) 4 , A, , i, , +, , –, , R1, , –4, , A, , 5m, , B, G, , Hot, Junction, , Cold, Junction

Page 56 :
1086 Current Electricity, (b) 8  10 A, –4, , (c) 4  10 A, , (a) 4, , (b) 1, , (c) 3, , (d) 2, , –3, , 48., , (d) 8  10 A, –3, , 42., , In the following circuit, bulb rated as 1.5 V, 0.45 W. If bulbs glows, with full intensity then what will be the equivalent resistance, between X and Y, , 3a, , 6V, , (a) 0.45 , , 2a, , a, , (b) 1 , , 3, , R, , X, , A, , Y, , (c) 3 , (d) 5 , 43., , B, , Consider the circuits shown in the figure. Both the circuits are, taking same current from battery but current through R in the, 1, second circuit is, th of current through R in the first circuit. If R, 10, is 11 , the value of R, , 49., , 1, , (a) 9.9 , , (c) 8.8 , , R, , (d) 7.7 , , (a) R = R = R, (b) R > R > R, (c) R < R < R, (d) Information is not sufficient, In the following star circuit diagram (figure), the equivalent, resistance between the points A and H will be, , (a), , In order to quadruple the resistance of a uniform wire, a part of its, length was uniformly stretched till the final length of the entire wire, was 1.5 times the original length, the part of the wire was fraction, equal to, , (c) 1 / 10, , 50., , In the circuit shown in figure reading of voltmeter is V when only S, is closed, reading of voltmeter is V when only S is closed and, reading of voltmeter is V when both S and S are closed. Then, 1, , 2, , 2, , (b) V > V > V, 2, , 1, , 1, , 46., , 2, , 51., , 3R, , R, , S1, 6R, , 2, , S2, , 2i, 5, , (b), , 3i, 5, , (c), , 4i, 5, , (d), , i, 5, , XE, , 1, , (c) 2 A, , (c), , 2, , 52., 3, , E, r, , r, F, r, , J, , H, , r, , r, , r, , E, F, , B, , i, , i, C, , A, , D, , Y, , 4, , 2, A, 5, , 5, 10, 5V, , 5V, , A, , C, , 12, , E, , F, , The reading of the ideal voltmeter in the adjoining diagram will be, A, , (a) 4 V, , (d) 3 A, , (b) 8 V, , 50V, 12 cells each having same emf are connected, in series with some, , (c) 12 V, , cells wrongly connected. The arrangement is connected in series, with an ammeter and two cells which are in series. Current is 3 A, when cells and battery aid each other and is 2 A when cells and, battery oppose each other. The number of cells wrongly connected, is, , 5, , 1, A, 5, , (d) 0 A, , 2, , (b) 4 A, , 47., , (b), , Current through wire XY of circuit shown is, (a) 1 A, , r C, , r, , In the circuit of adjoining figure the current through 12  resister, will be, , V, , 3, , 1, , B, , 72°, , D, , (a) 1 A, , 3, , (d) V > V > V, 1, , i, r, , r, , 2, , (c) V > V > V, 3, , C, , B, , (a), , 1, , 2, , 1, , 1, , B, , A, , I of 10 . The, In the adjoining circuit diagramG each resistance is, current in the arm AD will be, , 0.5l, , (d) 1 / 4, , 3, , A, , (d) 0.243 r, , l, , (a) V > V > V, , C, , (c) 0.486 r, , (b), , 3, , B, , (b) 0.973 r, , (a) 1 / 8, , 45., , A, , (a) 1.944 r, , R, , R2, , (b) 1 / 6, , C, , A, i/10, , E, , B, , R1, , i, , i, , (b) 11 , , 44., , Following figure shows cross-sections through three long conductors, of the same length and material, with square cross-section of edge, lengths as shown. Conductor B will fit snugly within conductor A,, and conductor C will fit snugly within conductor B. Relationship, between their end to end resistance is, , 10V, , 20, , V, , (d) 14 V, 4V, , 10, , B, , N, , C

Page 57 :
Current Electricity 1087, 53., , The resistance of the series combination of two resistance is S., When they are joined in parallel the total resistance is P. If S = nP,, then the minimum possible value of n is, , (a) AB, (b) BC, (c) CD, , [AIEEE 2004], , (a) 4, (c) 2, 54., , (d) DE, , (b) 3, 5., , (d) 1, , I-V characteristic of a copper wire of length L and area of crosssection A is shown in figure. The slope of the curve becomes, , A moving coil galvanometer has 150 equal divisions. Its current, sensitivity is 10 divisions per milliampere and voltage sensitivity is 2, divisions per millivolt. In order that each division reads 1 volt, the, resistance in ohms needed to be connected in series with the coil, will be, [AIEEE 2005], (a) 99995, (c) 10, , I, , (b) 9995, , 3, , (d) 10, , (a) More if the experiment is performed at higher temperature, O, , 5, , (b) More if a wire of steel of same dimensionVis used, (c) More if the length of the wire is increased, (d) Less if the length of the wire is increased, 6., , 1., , thermal power produced in the conductor, then which of the, following graph is incorrect, (a) vd, (b) P, , Which of the adjoining graphs represents ohmic resistance, [CPMT 1981; DPMT 2002], , (a), , (b) V, , V, , I, , (c), , E, , (d) P, , (d), V, , 7., I, , Variation of current passing through a conductorI as the voltage, applied across its ends as varied is shown in the adjoining diagram., If the resistance (R) is determined at the points A, B, C and D, we, will find that, [CPMT 1988], (a) R = R, C, , (b) R > R, B, , C, , D, , V, , C, , vd, , i, , The two ends of a uniform conductor are joined to a cell of e.m.f. E, and some internal resistance. Starting from the midpoint P of the, conductor, we move in the direction of current and return to P. The, potential V at every point on the path is plotted against the distance, covered (x). Which of the following graphs best represents the, resulting curve, V, , D, , V, , (a), , (b), , B, , E, , A, , (c) R > R, , <E, , A, B, , (d) None of these, 3., , E, , (c) P, , I, , V, , 2., , E denotes electric field in a uniform conductor, I corresponding, current through it, v d drift velocity of electrons and P denotes, , X, , V, , (c), , i, The voltage V and current I graph for a conductor at two different, temperatures T1 and T2 are shown in the figure. The relation, , X, , (d), , V, , E, <E, , between T1 and T2 is, [MP PET 1996; KCET 2002], , (a), , T1  T2, , (b) T1  T2, (c), , T1, V, , 4., , (a)  and  are both negative, , T1  T2, , From the graph between current I and voltage V shown below,, identify the portion corresponding to negative resistance, , E, C, B, D, A, , [CPMT 1988], , Rt, , (b)  and  are both positive, , I, , I, , X, , The resistance R t of a conductor varies with temperature t as, shown in the figure. If the variation is represented by, Rt  R0 [1  t  t 2 ] , then, , T2, , (d) T1  T2, , X, , 8., , V, , (c)  is positive and  is negative, [CBSE PMT 1997], , (d)  is negative and  are positive, , t

Page 58 :
1088 Current Electricity, 9., , Variation of current and voltage in a conductor has been shown in, the diagram below. The resistance of the conductor is., V, , 6, 5, 4, 3, 2, 1, , (c), , 14., 1, , (a) 4 ohm, (b) 2 ohm, (c) 3 ohm, (d) 1 ohm, Resistance as shown in figure is negative at, I, , T2, , V, , (b) sin, [CPMT 1997], , T1, , cot 2, , (c), , , , (d) tan , 15., , B, , 11., , cos 2, , (a), , C, , A, , The V-i graph for a conductor at temperature T1 and T2 are as, shown in the figure. (T2  T1 ) is proportional to, , 2 3 4 5 6, , i, , 10., , (d), , V B, (a) A, (b), (c) C, (d) None of these, For a cell, the graph between the potential difference (V) across the, terminals of the cell and the current (I) drawn from the cell is, shown in the figure. The e.m.f. and the internal resistance of the cell, are, , i, , A cylindrical conductor has uniform cross-section. Resistivity of its, material increase linearly from left end to right end. If a constant, current is flowing through it and at a section distance x from left, end, magnitude of electric field intensity is E, which of the following, graphs is correct, (a) E, (b) E, , x, , O, , V(Volts), , x, , O, , (c) E, , 2.0, , , , (d), E, , 1.5, 1.0, 0.5, , 12., , 0, , (a), , 2V ,0.5 , , (c), ,  2V ,0.5 , , 16., , The graph which represents the relation between the total resistance, R of a multi range moving coil voltmeter and its full scale deflection, V is, R, , R, , 17., , V, , (a), , sin, , (b) cos , , (c), , tan , , (d) cot , , A battery consists of a variable number 'n' of identical cells having, internal resistances connected in series. The terminals of battery are, short circuited and the current i is measured. Which of the graph, below shows the relation ship between i and n, i, , (a), , (b), , O, , log I, , (b), , log I, , (a) (i), (b) (ii) (iv), (iii), (c) (iii), (d) (iv), When a current I is passed through a wire of constant resistance, it, produces a potential difference V across its ends. The graph drawn, between log I and log V will be, , 18., , O, , n, , n, , (d) i, , (c) i, , V, , V, , O, , n, , O, , n, , In an experiment, a graph was plotted of the potential difference V, between the terminals of a cell against the circuit current i by, varying load rheostat. Internal conductance of the cell is given by, V, x, , log V, , V, , log V, , y, , log I, , log V, log I, , (a), , i, , R, , R, , 13., , x, , O, , The V-i graph for a conductor makes an angle  with V-axis. Here, V denotes the voltage and i denotes current. The resistance of, conductor is given by, , V, , (ii), , (i), , x, , O, , 4, 1 2 3(b) 4 2V, 5 ,0I.(amperes), (d)  2V,0.4 , , i, , log V, , A

Page 59 :
Current Electricity 1089, 4., , (a) xy, (c), 19., , (b), , Assertion, , : In the following circuit emf is 2V and internal, resistance of the cell is 1  and R = 1, then, reading of the voltmeter is 1V., , y, x, , V, E=2V, , x, y, , (d) (x – y), r=1, , A, , V-i graphs for parallel and series combination of two identical, R=1, , resistors are as shown in figure. Which graph represents parallel, combination, B, , V, , A, , 20., , 5., , (a) A, , (b)i B, , (c) A and B both, , (d) Neither A nor B, , Q, , Shunt, , S, , S, , 1, , Assertion, , Reason, Assertion, , Reason, , R, , O, , 1, [AIIMS 1995], : There is no current in the metals in the absence of, electric field., : Motion of free electron are randomly., , P, , Q, , (a) P, (c) R, , Assertion, , [AIIMS 1994], , 6., , 7., , R, , 2, , 3, (b), Range, , Q, (d) S, , 8., 4, , Assertion, Reason, , Ampere, , 9., , Assertion, Reason, , 10., , Assertion, Reason, , Read the assertion and reason carefully to mark the correct option out of, the options given below :, , (a), , (c), (d), (e), , If both assertion and reason are true and the reason is the correct, explanation of the assertion., If both assertion and reason are true but reason is not the correct, explanation of the assertion., If assertion is true but reason is false., If the assertion and reason both are false., If assertion is false but reason is true., , 1., , Assertion, , (b), , Reason, 2., , Assertion, Reason, , : The resistivity of a semiconductor increases with, temperature., : The atoms of a semiconductor vibrate with larger, amplitude at higher temperatures thereby, increasing its resistivity, [AIIMS 2003], : In a simple battery circuit the point of lowest, potential is positive terminal of the battery, : The current flows towards the point of the higher, potential as it flows in such a circuit from the, negative to the positive terminal., , 11., , Assertion, Reason, , 12., , Assertion, , 13., , Reason, Assertion, Reason, , 14., , Assertion, Reason, , 15., , Assertion, , [AIIMS 2002], , 3., , Assertion, Reason, , : The temperature coefficient of resistance is positive, for metals and negative for p-type semiconductor., : The effective charge carriers in metals are, negatively charged whereas in p-type semiconductor, they are positively charged., [AIIMS 1996], , 2,  1 A and R =, 2, , :, , Reason, , The ammeter has range 1 ampere without shunt. the range can be, varied by using different shunt resistances. The graph between, shunt resistance and range will have the nature, , V  E  ir where E = 2 V, i , , Reason, , 16., , : Electric appliances with metallic body have three, connections, whereas an electric bulb has a two pin, connection., : Three pin connections reduce heating of connecting, wires., : The drift velocity of electrons in a metallic wire will, decrease, if the temperature of the wire is, increased., : On increasing temperature, conductivity of metallic, wire decreases., : The electric bulbs glows immediately when switch, is on., : The drift velocity of electrons in a metallic wire is, very high., : Bending a wire does not effect electrical resistance., : Resistance of wire is proportional to resistivity of, material., : In meter bridge experiment, a high resistance is, always connected in series with a galvanometer., : As resistance increases current through the circuit, increases., : Voltameter measures current more accurately than, ammeter., : Relative error will be small if measured from, voltameter., : Electric field outside the conducting wire which, carries a constant current is zero., : Net charge on conducting wire is zero., : The resistance of super-conductor is zero., : The super-conductors are used for the transmission, of electric power., : A potentiometer of longer length is used for, accurate measurement., : The potential gradient for a potentiometer of longer, length with a given source of e.m.f. becomes small., : The e.m.f. of the driver cell in the potentiometer, experiment should be greater than the e.m.f. of the, cell to be determined., , Reason, , : The fall of potential across the potentiometer wire, should not be less than the e.m.f. of the cell to be, determined., , Assertion, , : A person touching a high power line gets stuck, with the line.

Page 60 :
1090 Current Electricity, Reason, 17., , : The current carrying wires attract the man towards, it., , 41, , a, , 42, , c, , 43, , b, , 44, , d, , 45, , c, , 46, , d, , 47, , c, , 48, , b, , 49, , b, , 50, , d, , Assertion, , : The connecting wires are made of copper., , 51, , d, , 52, , c, , 53, , d, , 54, , a, , 55, , c, , Reason, , : The electrical conductivity of copper is high., , 56, , d, , 57, , c, , 58, , c, , 59, , d, , 60, , c, , 61, , d, , 62, , c, , 63, , d, , 64, , c, , 65, , c, , 66, , c, , 67, , b, , 68, , c, , 69, , d, , 70, , b, , 71, , a, , 72, , c, , 73, , a, , 74, , b, , 75, , a, , 76, , c, , 77, , c, , 78, , b, , 79, , c, , 80, , a, , 81, , a, , 82, , b, , 83, , b, , 84, , d, , 85, , d, , 86, , a, , 87, , a, , 88, , a, , 89, , b, , 90, , b, , Electric Conduction, Ohm's Law and Resistance, , 91, , b, , 92, , c, , 93, , b, , 94, , d, , 95, , a, , 1, , a, , 2, , c, , 3, , b, , 4, , b, , 5, , c, , 96, , d, , 97, , b, , 98, , b, , 99, , d, , 100, , a, , 6, , a, , 7, , a, , 8, , a, , 9, , d, , 10, , c, , 101, , c, , 102, , a, , 103, , b, , 104, , d, , 105, , a, , a, , 107, , b, , 108, , d, , 109, , bc, , 110, , b, , 11, , d, , 12, , d, , 13, , a, , 14, , c, , 15, , a, , 106, , 16, , a, , 17, , c, , 18, , b, , 19, , c, , 20, , b, , 111, , d, , 112, , c, , 113, , a, , 114, , a, , 115, , d, , 116, , a, , 117, , d, , 118, , c, , 119, , d, , 120, , c, , 121, , b, , 122, , b, , 123, , b, , 124, , c, , 125, , b, , 126, , a, , 127, , c, , 128, , b, , 129, , c, , 130, , a, , 131, , a, , 132, , a, , 133, , c, , 134, , a, , 135, , b, , 137, , a, , 138, , b, , 139, , c, , 140, , b, , 21, , d, , 22, , b, , 23, , b, , 24, , b, , 25, , d, , 26, , c, , 27, , b, , 28, , b, , 29, , b, , 30, , a, , 31, , c, , 32, , d, , 33, , b, , 34, , d, , 35, , c, , 36, , b, , 37, , b, , 38, , c, , 39, , a, , 40, , d, , 136, , b, , 41, , b, , 42, , b, , 43, , a, , 44, , b, , 45, , c, , 141, , b, , 46, , a, , 47, , b, , 48, , b, , 49, , c, , 50, , a, , 51, , c, , 52, , c, , 53, , b, , 54, , b, , 55, , b, , 56, , a, , 57, , a, , 58, , a, , 59, , c, , 60, , c, , 61, , a, , 62, , b, , 63, , b, , 64, , c, , 65, , c, , 66, , d, , 67, , a, , 68, , b, , 69, , d, , 70, , d, , 71, , a, , 72, , a, , 73, , c, , 74, , b, , 75, , 76, , c, , 77, , c, , 78, , c, , 79, , d, , 81, , a, , 82, , d, , 83, , b, , 84, , 86, , b, , 87, , c, , 88, , a, , 91, , a, , 92, , c, , 93, , 96, , b, , 97, , c, , 98, , 101, , c, , 102, , a, , 103, , 106, , d, , 107, , d, , 108, , Kirchhoff's Law, Cells, 1, , b, , 2, , d, , 3, , c, , 4, , a, , 5, , a, , 6, , b, , 7, , a, , 8, , a, , 9, , b, , 10, , a, , 11, , c, , 12, , d, , 13, , a, , 14, , d, , 15, , b, , b, , 16, , c, , 17, , c, , 18, , c, , 19, , d, , 20, , b, , 80, , b, , 21, , c, , 22, , c, , 23, , b, , 24, , d, , 25, , a, , b, , 85, , c, , 26, , a, , 27, , b, , 28, , b, , 29, , a, , 30, , b, , 89, , a, , 90, , d, , 31, , a, , 32, , c, , 33, , b, , 34, , a, , 35, , a, , b, , 94, , a, , 95, , b, , 36, , b, , 37, , a, , 38, , b, , 39, , b, , 40, , c, , a, , 99, , c, , 100, , d, , 41, , d, , 42, , d, , 43, , d, , 44, , a, , 45, , c, , d, , 104, , b, , 105, , b, , 46, , c, , 47, , b, , 48, , a, , 49, , a, , 50, , d, , d, , 51, , b, , 52, , d, , 53, , b, , 54, , c, , 55, , a, , 56, , b, , 57, , c, , 58, , a, , 59, , d, , 60, , b, , 61, , c, , 62, , c, , 63, , c, , 64, , b, , 65, , a, , 66, , c, , 67, , a, , 68, , d, , 69, , b, , 70, , a, , 71, , a, , 72, , d, , 73, , c, , 74, , b, , 75, , b, , 76, , b, , 77, , c, , 78, , c, , 79, , d, , 80, , d, , 81, , a, , 82, , d, , 83, , c, , 84, , c, , 85, , a, , a, , 109, , d, , 110, , 111, , d, , 112, , d, , 113, , a, , 114, , a, , 115, , c, , 116, , a, , 117, , a, , 118, , b, , 119, , c, , 120, , a, , 121, , d, , 122, , a, , 123, , a, , 124, , d, , 125, , c, , 126, , b, , 127, , c, , 128, , a, , 129, , a, , 130, , c, , 131, , c, , 132, , b, , 133, , c, , Different Measuring Instruments, , Grouping of Resistances, 1, , c, , 2, , d, , 3, , a, , 4, , c, , 5, , b, , 1, , a, , 2, , c, , 3, , d, , 4, , d, , 5, , c, , 6, , c, , 7, , c, , 8, , b, , 9, , a, , 10, , b, , 6, , c, , 7, , a, , 8, , d, , 9, , c, , 10, , c, , d, , 12, , c, , 13, , d, , 14, , a, , 15, , d, , 11, , d, , 12, , d, , 13, , b, , 14, , d, , 15, , b, , 11, , 16, , d, , 17, , c, , 18, , c, , 19, , b, , 20, , d, , 16, , c, , 17, , a, , 18, , b, , 19, , c, , 20, , a, , 21, , a, , 22, , a, , 23, , b, , 24, , b, , 25, , c, , 21, , b, , 22, , b, , 23, , a, , 24, , a, , 25, , a, , 26, , b, , 27, , d, , 28, , d, , 29, , d, , 30, , c, , 26, , a, , 27, , a, , 28, , a, , 29, , b, , 30, , b, , 31, , b, , 32, , d, , 33, , a, , 34, , b, , 35, , c, , 31, , b, , 32, , b, , 33, , b, , 34, , b, , 35, , c, , 36, , d, , 37, , d, , 38, , b, , 39, , c, , 40, , b, , 36, , c, , 37, , b, , 38, , b, , 39, , d, , 40, , b

Page 61 :
Current Electricity 1091, 41, , a, , 42, , b, , 43, , c, , 44, , d, , 45, , a, , 46, , b, , 47, , c, , 48, , a, , 49, , b, , 50, , a, , 51, , b, , 52, , c, , 53, , b, , 54, , b, , 55, , a, , 56, , b, , 57, , d, , 58, , c, , 59, , c, , 60, , d, , 61, , a, , 62, , a, , 63, , d, , 64, , a, , 65, , d, , 66, , b, , 67, , a, , 68, , b, , 69, , c, , 70, , c, , 71, , d, , 72, , c, , 73, , a, , 74, , c, , 75, , a, , 76, , d, , 77, , b, , 78, , a, , 79, , b, , 80, , a, , 81, , c, , 82, , c, , 83, , a, , 84, , a, , 85, , b, , 86, , d, , 87, , c, , 88, , d, , 89, , c, , 90, , a, , 91, , d, , 92, , d, , 93, , d, , 94, , c, , 95, , d, , 96, , a, , 97, , c, , 98, , a, , 99, , d, , 100, , d, , 101, , c, , 102, , d, , 103, , c, , 104, , c, , 105, , a, , 106, , c, , 107, , c, , 108, , d, , 109, , c, , 110, , a, , 111, , a, , 112, , a, , 113, , c, , 114, , b, , 115, , c, , 116, , d, , 117, , a, , 118, , b, , 119, , d, , 120, , c, , 121, , d, , 122, , c, , 123, , a, , 124, , a, , 125, , b, , 126, , b, , 127, , d, , 128, , d, , 129, , b, , 130, , c, , 131, , d, , 132, , a, , 133, , c, , 134, , d, , 135, , d, , 136, , a, , 137, , a, , 138, , b, , 139, , d, , 140, , a, , 141, , c, , 142, , a, , 143, , a, , Critical Thinking Questions, 1, , a, , 2, , a, , 3, , c, , 4, , c, , 5, , c, , 6, , b, , 7, , d, , 8, , d, , 9, , d, , 10, , b, , 11, , d, , 12, , d, , 13, , c, , 14, , b, , 15, , b, , 16, , b, , 17, , b, , 18, , b, , 19, , a, , 20, , d, , 21, , d, , 22, , a, , 23, , c, , 24, , c, , 25, , b, , 26, , b, , 27, , a, , 28, , d, , 29, , b, , 30, , b, , 31, , d, , 32, , a, , 33, , a, , 34, , b, , 35, , c, , 36, , c, , 37, , a, , 38, , a, , 39, , d, , 40, , a, , 41, , a, , 42, , b, , 43, , a, , 44, , a, , 45, , b, , 46, , c, , 47, , b, , 48, , a, , 49, , b, , 50, , a, , 51, , d, , 52, , b, , 53, , a, , 54, , b, , Graphical Questions, 1, , a, , 2, , d, , 3, , a, , 4, , c, , 5, , d, , 6, , c, , 7, , b, , 8, , b, , 9, , d, , 10, , a, , 11, , b, , 12, , d, , 13, , a, , 14, , c, , 15, , b, , 16, , d, , 17, , d, , 18, , b, , 19, , a, , 20, , b

Page 62 :
1096 Current Electricity,  If length becomes 3 times so Resistance becomes 9 times, , i.e. R'  9  20  180, , Assertion and Reason, 1, , d, , 2, , d, , 3, , b, , 4, , a, , 5, , a, , 6, , c, , 7, , b, , 8, , c, , 9, , a, , 10, , c, , 11, , a, , 12, , a, , 13, , b, , 14, , a, , 15, , a, , 16, , d, , 17, , a, , 12., , 14., , (d) Resistivity is the property of the material. It does not depend, upon size and shape., (a) Because with rise in temperature resistance of conductor, increase, so graph between V and i becomes non linear., (c) Because V-i graph of diode is non-linear., , 15., , (a), , 13., , e El, e V,   or v d  .  (Since V  El), m l, m l, , vd , , vd  E, , 16., , (a) Resistance of conductor depends upon relation as R , , 17., , (c), , 18., , (b) Volume  Al  3  A , , Electric Conduction, Ohm's Law and Resistance, 1., , (a) Number of electrons flowing per second, n i,   4.8 / 1.6  10 19  3  1019, t e, , 2., , (c), , J,  vd  J (current density), vd , ne, J1 , , 1, , ., , With rise in temperature rms speed of free electron inside the, conductor increase, so relaxation time decrease and hence, resistance increases, i, , q 4,   2 ampere, t, 2, , 9, 3,   l l 2, l,  l2  , 3, , , A, 3 /l, 3, , , Now R  , , 2i, i, i, and J 2 ,   J1 ;  (vd )1  (vd )2  v, A, 2A A, , 3, l, , ne 62.5  10 18  1.6  10 19, ,  10 ampere, t, 1, , 3., , (b) Order of drift velocity  10 4 m / sec  10 2 cm / sec, , 19., , (c), , 4., , (b) Density of Cu  9  10 3 kg / m 3 (mass of 1 m of Cu), , 20., , (b) In twisted wire, two halves each of resistance 2 are in, 2, parallel, so equivalent resistance will be,  1 ., 2, , 21., , (d) In stretching of wire R , , 22., , (b), , 3, ,  6.0  10 atoms has a mass = 63  10 kg, 23, , –3, ,  Number of electrons per m are, 3, , , , 6 .0  10 23, 63  10  3, ,  9  10 3  8 .5  10 28, , Now drift velocity  v d , , , i, neA, , 1.1, 8.5  10, , 28, , 5., 6., , (c) Because 1 H.P. = 746 J/s = 746 watt, (a), , R 0, R 2 l, Rl , , , %  2  0 .1  0 .2%, R, R, l, , R, , l, ,  50  10, , 8, , , , 50  10 2, ,  10, , 6, , , , (a), , 8., , (a) Resistivity of some material is its intrinsic property and is, constant at particular temperature. Resistivity does not depend, upon shape., (d), , A, , (50  10  2 ) 2, ,  1 (1  t1 ), 1 (1  0 .00125  27), ,  ,  2 (1  t 2 ), 2, (1  0 .00125  t),  t  854C  T  1127 K, , 10., , (c), , l, 2l, l, R1 ,  R2 , i.e. R 2 , A, 2A, A,  R1  R 2, , 11., , L, ,  0 .7 , , A, , (d) In case of stretching of wire R  l 2, , r4, ,  1, 22, (1  10  3 ) 2, 7, , R, , 1, 1, 1, R 2  2, A, r, d, , [d = diameter of wire], , 23., , (b), , 24., , (b) i = q  1.6  10 19  6.6  10 15  10.56  10 4 A  1mA, , 25., , (d), , 26., 27., , (c), (b) In semiconductors charge carries are free electrons and holes, , 28., , (b) Net current inet  i()  i(), , 2, , 7., , 9., , R, , 1, ,   2.2  10 6 ohm-m., ,  1 .6  10 19    (0.5  10  3 )2, ,  0.1  10 3 m / sec, , i, , R, , , , , , l, r2, , 2, , , , n()q(), t, , n(), t, , , , R1, l, r2, 1 5 2,  1  22       l 2  20m, R2, l 2 r1, 1 l2  1 , , +2e, , n()q(), ,  2e , , +, , t, , n(), t, , –, , –e, , e, , inet, , = 3.2  10  2  1.6  10 + 3.6  10  1.6  10, = 1.6 A (towards right), 18, , –19, , 18, , –19

Page 63 :
Current Electricity 1097, 29., , (b) In the absence of external electric field mean velocity of free, , 47., , (b) Because as temperature increases, the resistivity increases and, , 1, hence the relaxation time decreases for conductors   ., , , , 48., , (b) In VI graph, we will not get a straight line in case of liquids., , 49., , (c), , 32., , (a) With rise in temperature specific resistance increases, (c) For metallic conductors, temperature co-efficient of resistance, is positive., (d), , 33., , (b) Length l = 1 cm  10 2 m, , 50., , (a) Since R  l 2  If length is increased by 10%, resistance is, increases by almost 20%, , 3 KT,  Vrms  T, m, , electron (V ) is given by Vrms , rms, , 30., 31., , R, , l, A, , Hence new resistance R'  10  20% of 10, , 1 cm, ,  10 , 100 cm, , 51., , (c), , 1 cm, , Resistance R = 3  10 , –7, , 34., , 10 2, 10  2, , –2, , 2, , –4, , 2, , 53., , (b) Current density J , , vd , , i, 20,  1.25  10 3 m / s,  29, 6, nAe 10  10  1.6  10 19, , 36., , (b) Specific resistance k , , 37., , (b), , hence i1  i2 . So, , (c), , J1 r22, ,  9 :1, J 2 r12, , V,  R and R  temperature, i, , 54., , (b) As, , 55., 56., 57., , (b) R  l 2  If l doubled then R becomes 4 times., (a) Temperature coefficient of a semiconductor is negative., (a) The reciprocal of resistance is called conductance, , 58., , (a), ,  R = R = R., , 59., , (c) Ohm’s Law is not obeyed by semiconductors., , i, 1.344, vd ,  6, nAe 10  1.6  10 19  8.4  10 22, , 60., , (c) Drift velocity vd , , R, , , , E, j, , R, l d, l, l,  2  1  1   2, R2 l2  d1, A d, , 2, , 38., , J, i r2, i, i,  2  1  1  22, A r, J 2 i2 r1, , But the wires are in series, so they have the same current,, , –7, , –3, , (c), , l, 64  10 6  198,  r  0.024 cm, 7 , 22, A,  r2, 7, , (c), , –7, , l = 100 cm = 1 m, A = 1 cm = 10 m , so resistance R = 3  10, 1,   4 = 3  10 , 10, 35., , R, , 52., , = 3  10 , , (d) In the above question for calculating equivalent resistance, between two opposite square faces., 2, , R150, [1   (150)], . Putting R150  133 and, , R 500, [1   (500)], ,   0.0045 / C, we get R 500  258 , , Area of cross-section A = 1 cm  100 cm, = 100 cm = 10 m, 2, , 20,  10  12 ., 100, , 2, , , L  2d ,  , ,  1, , 4L  d , , 2, , 1, , 1.344,  0.01cm / s  0.1mm / s, 10  1.6  8.4, , Resistance , , Potential difference, Current, , V, ,  lne, , ; vd, , diameter., 61., , (a) Using R T2  R T1 [1  (T2  T1 )], , 39., , 1, (a) Internal resistance , Temperatur e, ,  R100  R 50 [1  (100  50)], , 40., , (d) Charge = Current × Time =5 × 60 = 300 C, ,  7  5 [1  (  50)]   , , 41., , (b) By R  l / A, , 42., 43., , (b), (a), , 44., , (b), , 45., , (c) For semiconductors, resistance decreases on increasing the, temperature., , 46., , (a), , R, , l, a, , R, , for first wire and R’=, , l, n, l, , ., A ne 2 A, , 62., , l R,  for second wire., 4a 4, , does not depend upon, , (7  5),  0 .008 / o C, 250, , 63., , (b) This is because of secondary ionisation which is possible in the, gas filled in it., (b), , 64., , (c), , R1 (1  t1 ), 50, (1  3.92  10 3  20), , , , R2 (1  t2 ), 76.8, (1  3 .92  10  3 t),  t  167C, i,  i  vd A  i  vd r 2, neA, , 65., , (c) From v d , , 66., , (d) Resistivity depends only on the material of the conductor.

Page 64 :
1098 Current Electricity, 67., , (a) A particular temperature, the resistance of a superconductor is, 1 1, zero  G    , R 0, , 68., , (b) Net current i  i  i , , –, , e, , (n ) (q  ) (n ) (q  ), , t, t, , –, , 77., , (c) Human body, though has a large resistance of the order, of, K (say 10k  ), is very sensitive to minute currents even as, low as a few mA. Electrons, excites and disorders the nervous, system of the body and hence one fails to control the activity, of the body., , 78., , (c), ,  4.2 = R 0 (1  0.004  100)  1.4 R 0  R 0  3 ., , +, , Ne+, , R t  R 0 (1   t), , 2, , 79., , i, , (d), , (n ), (n ),  i   e   e, t, t, , , ,  2.9  10  1.6  10, 18, , 19, ,  1.2  10  1.6  10, 18, , 19, ,  i  0.66 A, 69., , (d) If E be electric field, then current density j = E, Also we know that current density j , , i, A, , (d), , 71., , (a), , R, , l, and mass m = volume (V)  density (d) = (A l) d, A, , Since wires have same material so  and d is same for both., Also they have same mass  Al = constant  l , 2, , , , r , R1 l1 A2  A2 ,   2,  , , r , R2 l2 A1  A1 ,  1, , , , 34  r ,     R2  544 , R 2  2r , , (a), , R, , (a), , R1  r2, , R 2  r1, , 82., , (d), , R1 (1  t1 ), 5, (1    50), 1, ,  ,  , per o C, R2 (1  t2 ), 6 (1    100), 200, , (Loss of weight)1, A,  1  2; so A has more loss of, (Loss of wight)2, A2, weight., (c) Q = it = 20 × 10 × 30 = 6× 10 C, , 74., , 75., , 76., , –6, , (b) i , , it 1 .6  10 3  1, ne,  10 16 .,  n , t, e, 1 .6  10 19, , (c) Drift velocity vd , 2, , , , i, 1, 1,  vd  or vd  2, neA, A, d, , vP  dQ , d /2, 1, 1,   , ,    v P  vQ ., v Q  d P , 4, 4,  d , 2, , Q,  Q  1.6  10 19  5  1015  0.8 mA ., T, , 83., , (b) i , , 84., , (b), , 85., , (c), , 86., , (b) Resistivity, , riron, rCopper, , , , iron, 1  10 7, ,  2 .4 ., copper, 1 .7  10  8, , i  e  1.6  10 19  6.8  1015  1.1  10 3 amp., , of, , the, , material, , of, , the, , RA 3  10 3  (0 .3  10 2 )2, = 27  10 9    m, , , l, 1, , Resistance of disc R , = 27  10 9  , 87., , (Thickness), (Area of cross section), , (10 3 ),  2.7  10 7 .,   (1  10  2 )2, , (c) By using R t  R 0 (1  t), 3  R 0  R 0 (1  4  10 3 t)  t  500 o C ., , 88., , (a), , i  6  10 15  1.6  10 19  0.96mA., , 89., , (a), , i, , 90., , (d), , R, , V 100  0.5, ,  10  0.25  ., i, 10  0.2, , 91., , (a), , R, , 2, 50  10 2, V, l,    1  10 6 m ., ,  , 4, i, A, (1  10  3 )2, , 92., , (c), , 93., , (b) i , , –4, , (b) Ge is semiconductor and Na is a metal. The conductivity of, semiconductor increases and that of the metals decreases with, the rise in temperature., , 4, , 1, , ,  5  R0 1 ,  50   R0  4 ., 200, , , , So,, , 73., , 4, , , R  nr , R,  ,     R2  4 ., , R2  r , n, , , Again by Rt  R0 (1  t), , R, A, A, R, l,  1  2 ( , L constant)  1  2  2, A, R2, A1, A2, R1, , Now, when a body dipped in water, loss of weight,  V L g  AL L g, , 25 9 1, 1, : :,  25 : 3 :  125 : 15 : 1 ., 1 3 5, 5, , 81., , 4, , 72., , 2, , (b), , 1, A, , 4, , 2, ,  l   l   l , l2,  R1 : R2 : R3   1  :  2  :  3 , m,  m1   m 2   m 3 , , 80., , Hence j is different for different area of cross-sections. When j, is different, then E is also different. Thus E is not constant. The, j, drift velocity v d is given by v d , = different for different j, ne, values. Hence only current i will be constant., 70., , R, , ne, n  1 .6  10 19,  16  10 3 ,  n  1017, t, 1, , V Q, Vt 20  2  60,  Q , ,  240 C ., R, t, R, 10, , rod

Page 65 :
Current Electricity 1099, 2, , 94., , (a), , R  l2 , , 95., , (b), , 96., , (b) Vd , , 2, , R 1  l1 , R  l , R,   ,     4  R2  ., R 2  l 2 , R2  l/2 , 4, , 115., , (c) Same mass, same material i.e. volume is same or Al = constant, 2, , Also, R  , , i, 40, , neA 10 29  10 6  1.6  10 19, , (c), , Vd , , i, 5 .4, , nAe 8.4  10 28  10 6  1 .6  10 19, , 2, , (a), , 99., , (c), , R 1  l1 , 10  5 , 1,   , ,  R 2  160 .,  , R 2  l 2 , R 2  20 , 16, 1, , 100., 101., 102., , (d), (c), (a), , R  91  10 2  9.1k ., , 103., , (d), , R, , 117., , (a) As steady current is flowing through the conductor, hence the, number of electrons entering from one end and outgoing from, the other end of any segment is equal. Hence charge will be, zero., , (b) n , , l2 l2, l2, l2,  R1 : R 2 : R 3  1 : 2 : 3, m, m1 m 2 m 3, R1 : R 2 : R 3 , , 1  10, , 3, , 1 .6  10 19, , 9 4 1, : :  27 : 6 : 1 ., 1 2 3, , 1, A, 1, , C , R l, l, , dQ, t, 5,  dQ  idt  Q   t 2 i dt  0 (1.2t  3) dt, 1, dt, , i, , 5, ,  1 .2 t 2, , ,  3 t   30 C, 2, ,  0, , i, i, v i,  vd  2   1, v ' i2, ne r 2, r, , 121. (d), , r , v,   2   v '  ., r, 2,  1, , (d), , 108., , 8, i, (a) v d , , nAe 8  10 28  (2  10 3 ) 2  1 .6  10 19, , 4, , 4, , R  3r/4 , 81, 256,  , , ,  R2 , R, , , , R2  r , 256, 81, , , Significant figures, , 109., 110., , = 0.156  10 3 m/sec ., (d) Specific resistance doesn’t depend upon length and area., (d) Heating effect of current., , 111., , (d) l , , , , Black, , (a) If suppose initial length l1  100 then l2  100  100  200, R1  l1 ,  100 ,   ,   R 2  4 R1, R2  l2 ,  200 , 2, , R, R  R1, 4R  R,  100  2,  100  1 1  100  300%., R, R1, R1, , (a) Ammeter is always connected in series and Voltmeter is always, connected in parallel., , Brown, , 0, , 10, , 1, ,  R = 10  10 = 100 , 123. (a), 124. (d), 125. (c), (b) R , , l, r2, , 2, , 2, , , , R2, l, r, 1, 2 1,  2  12 =    , R1, l1 r2, 2, 1 2, , R1, , specific resistance doesn't depend upon length,, 2, and radius., ,  R2 , , 127., , (c) By using vd , , 4 .2  3 .14  (0.2  10 3 )2,  1 .1m, 48  10  8, , (d) For conductors, resistance  Temperature and for semi1, conductor, resistance , Temperatur e, , 2, , Multiplier, , 1, , 126., , , , R'  n 2 R  R'  16 R, , 122. (a), , 2, , 107., , R r 2, , 4, , 1, , (d), , R 1  r2, , R 2  r1, , R 2  r1 , R, 2,     2     R2  16 R, R1  r2 , R 1, , (a) In stretching,, , Brown, , 106., , 114., , (c), ,  6 .25  10 15 ., , (b) vd , , 113., , 119., , Conductance C , , 4, , 105., , 112., , 118. (b), , 120., , , , 104., , I  ne qe  np q p  1mA towards right, , (a), , 2, , ; where  = Relaxation time., , When lamp is switched on, temperature of filament increase,, hence  decrease so R increases, R, , 24  d ,   16  R2  1.5  ., , R2  d /2 , , 116., , = 0.4  10 3 m/sec  0.4 mm /sec ., 98., , 4, , 4, , , , = 2.5  10 3 m/sec ., 97., ,  A  d , R, l, A, l,  1  1  2   2   2 , R, l, A, A, 2, 2, 1,  A1   d1 , , i, 100, , neA 10 28  1 .6  10 19    (0 .02)2, 4, ,  2  10 4 m / sec, , 128. (a), , 129. (a), , l, l, . For highest resistance 2, r2, r, which is correct for option (a), R, , should be maximum,, , Red, brown, orange, silver red and brown represents the first, two significant figures., Multiplier, , Tolerance, , Red, , Significant figures, Brown, , Orange, , Silver, , 2, , 1, , 10, ,  10%, ,  R = 21  10 3  10%, , 3

Page 66 :
1100 Current Electricity, 2, , 130., , (c) In stretching R  l 2 , , R2, l, R, 2,  22  2   , R1, R1  1 , l1, , 2, , 6., , Change in resistance 3 R1 3, , , Original resistance, R1, 1, , 7., , 2, , 131. (c), , R1  l1 ,    , If l1  100 then l = 110, R 2  l2 , 2, , R1  100 , ,   R 2  1.21 R1, R 2  110 , , % change, , 1,  10  10....... 10 times, R, , R2  R1,  100  21%, R1, , 1, 1,  100 i.e. R , , R, 100, , 132. (b), 133. (c), , 8., Resistance  , , , , 2,  resistor., 3, 2 1 3,  Total resistance      1, 3 3 3, (c) Lowest resistance will be in the case when all the resistors are, connected in parallel., 1, 1, 1, , , ....... 10 times, R 0 .1 0 .1, , 2, , , , R 1,  , n 3, , This is in series with, ,  R 2  4 R1 . Change in resistance  R2  R1  3R1, Now,, , (c) Resistance of 1 ohm group , , l, A, , R1, , l, A, 2 3 5 5,  1  1  2    , R 2  2 l 2 A1 3 4 4 8, , (b) Resistance across XY , , 2, , 3, , Total resistance, , 2, , 2 8,  2  , 3 3, , 2, , 1., , , , (c) The given circuit can be redrawn as follows, A, , B, , 5, , 5, , 2/3V, , 2/3V, , 5, , 9., C, , 5, , (2  2)  2 8 4,   , 222, 6 3, , 2, , D, , 5, , 5, , C, , For identical resistances, potential difference distributes equally, among all. Hence potential difference across each resistance is, 4, 2, V , and potential difference between A and B is V ., 3, 3, 2., , (d) Equivalent resistance of parallel resistors is always less than any, of the member of the resistance system., , 3., , (a) Each part will have a resistance r  R / 10, , 10., , 3, , 3, , 5., , R AB, , V, 2, 1, , , ampere, R 20 10, , (d) As resistance  Length, , 13., , 12,  4, 3, 4 8 8,  R effective ,  , 4 8 3, (b) Given circuit is equivalent to, , Resistance of each arm , , R, 2, ,  Total equivalent resistance = 4 , , 6, , R,  2R, 2, , B, , 6, 9  6 9  6 18, , , ,  3.6 , 96, 15, 5, , 12., , (30  30)30, 60  30, , ,  20 , (30  30)  30, 90, , (b) Resistance of parallel group , , 3, , A, , 1 10, 10, 100, R, , , , ,  rR ,  0.01R, rR, r, R / 10, R, 100, , i , , Q, , 2, R, R  nx, (b) In parallel, x , n, In series, R + R + R .... n times = nR = n (nx) = n x, (d) The circuit reduces to, 2, , 11., , 1 1 1 1,    .......... .10 times, rR r r r, , (c) R equivalent, , 2, , P, , Let equivalent resistance be rR , then, , 4., , 2, , 2, 6 3,   A, 8/3 8 4, , 2V, A, , Y, 2, , (a) Equivalent resistance of the combination, , , 2/3V, , A, , X, , Current through ammeter, , Grouping of Resistances, , 2V, , A, , 3, , A, , 6, , 3, , C , , C, , 3, , 3, 3, , 3, , B, , B

Page 67 :
Current Electricity 1101, , 14., 15., , So the equivalent resistance between points A and B is equal to, 63, R,  2, 63, (d) Potential difference across all resistors in parallel combination, is same., (b) Current through each arm DAC and DBC = 1A, VD  V A  2 and VD  VB  3  V A  VB  1V, , 25., , (c), , 26., , (b), , 4, , 24, , 20, , 16, , (d) R = r , , 17., , (c) If resistances are R 1 and R 2 then, , A, , 6, , On solving equations (i) and (ii) we get R 2  6 / 5 , 27., , 20, , B, , A, , , , 6, , 6, , 9, , …..(ii), , (c), , 4, , B, , 16, , R1 R 2, 6, …..(i), , R1  R 2, 8, , Suppose R 2 is broken then R1  2, 18., ,  6 12  6 , V p  Vq   ,  (0.5)  (2  4 ) (0.5)  3 V,  3 12  6 , 8, , 3r 5 r, , 2, 2, , 16., , effecti ve, , i1 3,  i1  i2  i2  0.5 A  i1, , i2 3, , 6, 18 24  12, , R AB , , 12, ,  8, , (24  12), (d) The network can be redrawn as follows, 3, , 3, , 3, , B, , A, ,  Req  9 , 19., , (b) Because all the lamps have same voltage., , 20., , (d), , 21., , (a) Current supplied by cell i , , 28., , R series  R1  R 2  R 3  ......, , 2, , 2, 1,  A, 235 5, , 3, , 1 4, R,   4  R 1 , R1  R , 16, , 5, , 29., i, , 2V, , 3 1,  0.6 V, 5, (a) According to the problem, we arrange four resistance as, follows, A, , So potential difference across 3 will be V , , 22., , 10, , Equivalent resistance , R1  R2  9 and, , Y, RR 2, ,  R1  R  2 R  R  2  R 2  2 R, R  R2, , 20 C20,  10 , 40, , R1 R2,  2  R1 R2  18, R1  R2, ,  R 2  R  2  0  R  1 or R  2 ohm, , 31., , (b) Cut the series from XY and let the resistance towards right of, XY be R 0 whose value should be such that when connected, across AB does not change the entire resistance. The, combination is reduced to as shown below., E, , R, , A, , 1 .5,  1 amp, 3/2, , R, , X, , R, 2, , 1, , R, , R, , B, F, , i1, i, , R, , B, , R1  6 , R2  3, , (b) i1  i2 , , X, , R2 = 2, , 10, , R1  R2  (R1  R2 )2  4 R1 R2  81  72  3, , 24., , R1 = 1, , A, , B, , 10, , (b), , 30., , 4 4 66, ,  5 ohm So the, 4 4 66, 20, current in the circuit ,  4 ampere Hence the current, 5, flowing through each resistance = 2 ampere., (c) Let the resultant resistance be R. If we add one more branch,, then the resultant resistance would be the same because this is, an infinite sequence., , , (d) Equivalent resistance, , 10, , D, , 23., , (d) Let the resistance of the wire be R, then we know that, resistance is proportional to the length of the wire. So each of, the four wires will have R/4 resistance and they are connected, in parallel. So the effective resistance will be, , R, , Y, , R, , R, , R, E, , A, i2, , R, , R, , R0, , R, , 1.5V, , B, , F, , C, , R, , D, , R, , 3, , =, , R, , R

Page 68 :
1102 Current Electricity, , Potential difference between C and A,, .......(i), VC  VA  1  1  1V, Potential difference between C and B,, ......(ii), VC  VB  1  3  3 V, , The resistance across EF,  R EF  (R 0  2 R), Thus R AB , , (R 0  2 R)R, R R  2R 2,  0,  R0, R0  2R  R, R0  3 R, , On solving (i) and (ii) VA  VB  2 volt, ,  R 02  2 RR 0  2 R 2  0  R 0  R( 3  1), , 32., , (d) The last two resistance are out of circuit. Now 8  is in, parallel with (1  1  4  1  1) .,  R  8  || 8  , , 33., , Shot Trick : (VA  VB ) , 39., , 8,  4   R AB  4  2  2  8 , 2, , (a) The given circuit can be simplified as follows, 2, , 40., , 2, , 18, , 4.5, , 1 1 1 1 3, 1,  R  ohm,    , R 1 1 1 1, 3, Now such three resistance are joined in series, hence total, 1 1 1, R     1ohm, 3 3 3, (b) To obtain minimum resistance, all resistors must be connected, in parallel., , (c), , 7, , 15V, , Hence equivalent resistance of combination , , 15V, 6, , 1, , , , 0.5, , 6, , 18, , 0.5, , 41., , 10, 8, , 8, , Resistance of thick wire R1  10, , 35., ,  Resistance of thin wire R2  30, , 15, Hence current from the battery i ,  1A, 15, (b) The circuit will be as shown, , 42., , 10,  2A, 5, , R  22, , 2R,  2R  R 2  8  4 R  2R, 2R, ,  R2  4 R  8  0  R , , 5, , A, , 43., , (b) P.d. across the circuit  1 .2 , , I – I1, R2, , 44., , (d) Three resistances are in parallel., 1, 1 1 1, 3, ,    , R' R R R R, , R, , The equivalent resistance R ' , , I1, I, , 45., , A, , 38., , I, and R, 2, (d) Effective resistance between the points A and B is, 32 8, R,  , 12 3, , (b), , Req  5  , Current i , , branch = 1A, , i/2, , 10,  2A, 5, 1, , 3, , A, 3, , i, 10V, , i/2, , 1, , B, ,  R 1, , R, , 3, , 1 14, 1 5, , 2, 2, , Since R cannot be negative, hence R , 46., , (d), , Rl, , R2  R, R2  R, , 1 R,  R2  R  1  R  R, 1 R, ,  R 2  R  1  0 or R , , and current in each, 3, , 2 .88,  0 .48 A, 6, , (c) Similar to Q. No. 30. By formula R  R1 , , Only two values satisfying the above relation are, , 37., , 64,  2.88 volt, 64, , Current through 6 ohm resistance , , (d) According to the figure, (I  I1 )R2  I1 R, , + –, , 4  16  32,  22 3, 2, , R cannot be negative, hence R  2  2 3  5.46, , 8, 4, (c) The current in the circuit , , 5 1 3, 4, 4, Now VC  VE   1  VE   V, 3, 3, , 36., , Total resistance in series = 10 + 30 = 40 , (c) Similar to Q. No. 30, , 10V, , i, , r, 10, , (a) For same material and same length, R2, A, 3,  1 ,  R2  3R1, R1, A2 2, , On further solving equivalent resistance R  15 , , 34., , i, 2, (R2  R1 )  (3  1)  2 V, 2, 2, , 1 5, , 2

Page 69 :
Current Electricity 1103, R, . If two, 10, pieces are connected in series, then their resistance, 2R R, , , 10, 5, If 5 such combinations are joined in parallel, then net, R, R, resistance , , 5  5 25, ,  I, , Hence every new piece will have a resistance, , 47., , (c), , 54., , V 4.8  15, , 2A, R, 36, , (a) Equivalent resistance of the circuit R ,  Current through the circuit i , , 3, , 2, , V, 3, ,  2A, R 3/2, , Rmax  nR and Rmin  R / n , , R max,  n2, R min, , 55., , (c), , 56., , (d) According to the principle of Wheatstone’s bridge, the effective, resistance between the given points is 4., , 6, , B, , Req , , 48., , 4, , 6,  3, 2, , 6, , (b) Current in the given circuit i , , 50,  2A, (5  7  10  3), , 4, , (c), , 58., , (c) Current through 6 resistance in parallel with 3 resistance =, 0.4 A, Potential drop across 4   1.2  4  4.8 V, , 4,   1 .33 , 3, , (d) Equivalent external resistance of the given circuit R eq  4 , E, 10, , 2A, R eq  r (4  1), , i, 2, Hence, (VA  VB )   (R2  R1 )  (2  4 )  2 V ., 2, 2, , 51., , (d) Resistance of each part will be, , R, ., n2, (c) Let equivalent resistance between A and B be R, then, equivalent resistance between C and D will also be R., 1, , A, , (d) Two resistances in series are connected parallel with the third., 1, 1 1 3, 8,   , Hence,  Rp  , Rp 4 8 8, 3, , 60., , (c) Resistances at C and B are not in the circuit. Use laws of, resistances in series and parallel excluding the two resistance., , 61., , (d) After simplifying the network, equivalent resistance obtained, between A and B is 8., , 62., , (c) The circuit consists of three resistances (2 R, 2R and R), connected in parallel., , 63., , (d) Resistance across the battery is, , R, ; such n parts are joined in, n, , parallel so Req , , 52., , 59., , 8,   2.6 , 3, , If two are in parallel then series with third, R eq  6 , , Current given by the cell i , , R, , D, , So total current = 0.8 + 0.4 = 1.2 A, , (b) If all are in series then R eq  12 , , If two are in series then parallel with third, R eq, , 50., , 4, , 57., ,  VA  0  24 V  VA  24 V, , If all are in parallel then R eq , , C, 16, , Potential difference between A and B VA  VB  2  12, , 49., , 4, , A, , 1, 1 1 2 1 3, 2,   , ,  R p  2  I   1 A, Rp 3 6, 6, 6, 2, , 64., , (c) The voltmeter is assumed to have infinite resistance. Hence (1 +, 2 + 1) + 4 = 8., , 65., , (c), , 66., , (c) The given circuit can be redrawn as follows, , C, , 1, , R' , , R, 1, ,  0 .1, n 10, 2, , R, , 2, , 2, , , , 1, , B, , 53., , 2, , D, 2, , R, R ,  2  R or R 2  2 R  2  0, R 1, , 2 4 8,  3 1, 2, (d) 6 and 6 are in series, so effective resistance is 12 which is, in parallel with 3, so, ,  R, , 1, , 2, ,  R eq  5  ., R AB  R1 , , R2 R3, 44,  R4 = 2 ,  2  6 ., R2  R3, 44, , 67., , (b), , 68., , (c) Let equivalent resistance between A and B is R', so given circuit, can be reduced as follows, , 1 1 1, 15, 36,  R,  , , R 3 12 36, 15, , R, , R, , R, , A, R, , 2R, , 2R, , 2R, , B, R, A,  R, , B, , 2R, , R, , 

Page 70 :
1104 Current Electricity, , R'  R , ,  R AB  5  ., , 2R  R ,  R  2  RR   2 R 2  0, (2 R  R ), , 75., , (a), , On solving the equation we get R'  2 R ., (d), , 70., , (b) i , , 71., , (a) Equivalent resistance R  4 , , R AB, ,  eq. (l1  l 2 ), , R, 2, 8, 2,  R =,  2 2 ., 3, 3, 3, 3, , 69., , A, , 76., ,  1 l1, , , ,  2 l2, , 36,  6  and main current, 36, , 100 , , i/2, , R AC , , 200  200,  100 ., 200  200, , (c), 1, , 1, , 1, , 30V, , 10k, , 10k, , 78., 10,  15 k , 2, , (b), ,   same, l  same, A2 , By using R  , , 30,  2  10  3 A, 15, Hence, potential difference between A and B, , Current i =, , –, , 1A, , 1A, , Hence, Req , 79., , ,   10  10 3  10 Volt., , , , (a) Equivalent resistance R , +, , 1A, , 9, , 10, , 1A, , 1A, , 1A, , 1A, , 10, , 9 9 9 9 9, , , , A, , 7, , 5, , 80., , 10, , 10, , , , (a) The circuit can be drawn as follows, A, , 10, , A, , B, , 10, , 3, , B, , 3, , i1, B, , C, , i2, , 3, , i, , 10, , A, , 39  3 R  30  10 R 69  13 R, , 13  R, 13  R, , 5, , C, , , 10  (3  R(3) + R) 30  10 R, 3, 10  3  R, 13  R, , 13 R  R 2  69  13 R  R  69  ., , C, 10, , 3, , A, , R, , Q, , P, , R3, , C, , 5, 5, , 10, , B, , , , 2V, , A, , 10, , Q, , R, , 3, , 9,  9A, 1, Current passes through the ammeter = 5A., (b) The figure can be drawn as follows, D, , R1 R 2, 28, 8, ,  ., R1  R 2 (2  8 ) 5, , 3, 1A, , 9 9 9, , R, A, R, l, 1,  1  2  1   R1  2, A, R 2 A1, 8, 4, , (c) The given circuit can be simplified as follows, , 9,  , 9, , 1A, , 1, r, A1 (as r2  1 ), 2, 4, , 3, , Current i , , 74., , 1, , 1, 1,  2 ., 3, 3, , B, , Equivalent resistance R  10 , , 9V, , 1, , B, , R AB  2 , , A, , i/2, , B, , ., , B, , 100 , , A, 10k, , 73., , l1  l 2, , 100 , , A, , 77., ,  1 l1   2 l 2, , C, , 100 , , i, ,  2  10 3, V  , 2, , ,   eq , , A, , (c) The figure can be drawn as follows, D, , The same potential difference, also develops across 3, resistance., (c), , , , A, , E, 10,  10 = 0.5R + 1.5  R = 17.,  0.5 , Rr, R3, , E 3, i    0.5 A, R 6, Now potential difference across the combination of 3 and,  36 , 6, V  0 .5  ,   1Volt, 36, , 72., , 1l1, l, and R2  2 2 . In series Req  R1  R2, A, A, , R1 

Page 71 :
Current Electricity 1105, R, 2, R 1, , As, resistance is not fractional ,  x  3, R  2, 2 x  6, , 3  (3  3), Equivalent resistance R ,  2, 3  (3  3), , Current i , , Hence, the value of largest resistance = 6., , 2,  3  1,  1 A. So, i1  1  ,  A., 2, 36 3, , Potential difference between A and B =, , 1,  3  1volt., 3, , (a), , 82., , (b) The given figure is balance wheat stone bridge., , 83., , (b), , 84., , (d) Suppose resistance of wires are, , (b) Given circuit is a balanced Wheatstone bridge circuit, hence it, can be redrawn as follows, 12, 4, , and, , R AB , , 12  6,  4 ., (12  6), , (d) The given circuit is a balanced wheatstone bridge circuit. Hence, potential difference between A and B is zero., , 95., , (a) In the following circuit potential difference between, , C and A is VC  V A  1  4  4, 1A, , 48,  0 .16 A, (100  100  80  20), 2A, , C, , 160  40, R,  32 ., 160  40, , 87., , 100, , A, , 2, , 2, , R AB , , On solving equations (i) and (ii) we get, VA  VB  12V ., 96., , B, , (a) Given circuit is a balance Wheatstone bridge circuit., , 89., , (b) All of three resistance are in parallel So, R '  R/n , , 90., , (b), , 91., , (b) Two resistance are in ratio 1 : 2 and third resistance is R, So,, , A, , , , 1 l, A, , , , 12,  4, 3, , 8, 4, , 4, , 4, , B, ,  eff.2l, , (d) As resistance  Length, , 4 8 8,  R effective ,  , 4 8 3, , 2, , 2, , 88., , R eq  R1  R 2 , , 4, , 60, , 2, , 22,  1 ., 22, , B, , 16, , C and B is VC  VB  1  16  16 ……(ii), , 2, , B, , 16, , V, ,  Resistance of each arm , , 2 , , A, , C, 1A, , A, , A, , 4, , ……(i), , Series, 100+60 = 160, , 40, , (a), , 6, , 94., , (d) Potential difference across PQ i.e. p.d. across the resistance of, 20, which is V = i × 20, , (a), , b, , a, , 4, ,  V  0.16  20  3.2V ., 86., , , , 6, , then, , R2, , 6 2  R2, ,  R 2  3 ., 5 2  R2, , and i , , b, , 2, , R1, , 12, , 8, , a, , R1 R 2, 6, . If R 2 breaks then R1  2, , 5 R1  R 2, , 85., , 93., , 7, 1 1,  ,  R  3, 12 4 R, , Hence,, , (3  3)  3, 3,  2  i   1 . 5 A ., 2, (3  3)  3, , (c), , 8, 1, 1 1 1 4  2 1,  R eq   .,    , Req 2 4 8, 8, 7, , 81., , R, , 92., ,  2l, A, , 1, 1, 1, 3 R , ,  1 x  , , x 2x R, 2  R 1, ,  eff. , , 97., , 12, (b) i ,  5 A., (1  1)  0 .4, , 98., , (b) By balanced Wheatstone bridge condition, , R, ., 3, ,  X, , 1   2, 2, , ., , 99., 100., , 16, 4, , X, 0 .5, , 8,  2, 4, , , , (25  5), (d) Current through 2  1 .4 ,   1A, (, 10, , 2, ), , (, 25, , 5, ), , , (a) Since the given bridge is balanced, hence there will be no, current through 9  resistance. This resistance has no effect, and must be ignored in the calculations., 9, 5, , 4, , B, , A, 10, , 8, 14

Page 72 :
1106 Current Electricity, , R AB , , 101., , 107., , (b) Let current through 5  resistance be i. Then, , 108., , 10,  2 .1  0 .6 A, 35, (d) Let the value of shunt be r. Hence the equivalent resistance of, Sr, branch containing S will be, S r, i  25  (2.1  i)10  i , , 9  18, 6, 27, , (c) Potential difference between B and D is zero, it means, Wheatstone bridge is in balanced condition, B, , In balance condition,, , X, , 6, , 21, , 3, , 8, 15, , 3, , 8X, , 109., , (8  X ), , (b, c), , 2R, , C, R, , R, A, , 6, , 103., , 104., , 4, , (10  10)  (10  10),  10 , (10  10)  (10  10), , C, , A, , B, , , , 105., , 110., 111., , 1, R, (d) Equivalent resistance between P and Q, , (b) i , , 1, 1, 1, 1, 48, ,  ,  RPQ , , R PQ (6  2) 3 (4  12), 25, , Current between P and Q; i = 1.5A, So, potential difference between P and Q, 48,  2.88 V ., 25, (c) Given circuit is a balanced Wheatstone bridge i.e. potential, difference between B and D is zero. Hence, no current flows, between B and D., (a) The given circuit is a balanced Wheatstone bridge, hence it can, be redrawn as follows, VPQ  1 .5 , , 2, , A, , B, , 112., , , 2, , D, , 2, , 113., , B, , Clearly, the circuit is aD balanced Wheatstone Dbridge. So, effective resistance between A and B is 2  ., , 7, , (a) By the concept of balanced Wheatstore bridge, the given circuit, can be redrawn as follows, 5, , 10, , 15, , 2R, , R, , Between A and C circuit becomes equivalent to balanced, Wheatstone bridge so R AC  R ., , C, 2, , A, , 3, , 4, , A, , 30, , B, 6, , A, , B, , 20, , 114., , 30, , 30  60,  R AB ,  20 , (30  60), (a) The given circuit is a balanced Wheatstone bridge type, hence, it can be simplified as follows, , A' D, A' 4, , ,   A'  5 , B C, 5, 4, , 115., , A' (5 ) is obtained by connecting a 10  resistance in, parallel with A., (d) Given circuit is a balanced Wheatstone bridge circuit. So there, will be no change in equivalent resistance. Hence no further, current will be drawn., , 116., , (a) No current flow through vertical resistances, , 5, 2, , 3, , A, , B, 4, , 6, , , , R AB, , 14, , Req , , A D, 10 4,  ,  (Unbalanced), B C, 5, 4, , 60, , 106., , 8, , 7  14, 14, , ., (7  14 ), 3, (a) For a balance Wheatstone bridge., , , 10, , D, , 1, 1, 1, 1, R, ,  ,  R BD , R BD 2 R R 2 R, 2, , (b) This is a balanced Wheatstone bridge circuit. So potential at B, and D will be same and no current flows through 4R, resistance., (d) The equivalent circuits are as shown below, C, , R, , B, , A, , P R, 21 D 18, So, ,  X  8, , , 8X, Q S, 6, 3, (8  X ), (a) This is a balanced Wheatstone bridge. Therefore no current will, flow from the diagonal resistance 10 , ,  Equivalent resistance , , D, , R, 4, , 6, , 102., , R, , 6, , 4, , 15, , 18, , B, , C, , P Sr /(S  r), . This gives r  8 , , Q, R, , 10, , , 3, , 10, , 3, , 3, , 9, , 3, , B, , A, , 3, , 3, , 3, , , , B, , A, , 9

Page 73 :
Current Electricity 1107, , R PR, R AB , , 9, ., 2, , R 3R, , R R, , 2  3 R,  ||   R   3, R, 3, R, 3 2, 11, , , 3, 2, , Hence it is clear that R PQ is maximum., , 117., , (d) The given circuit is a balanced Wheatstone bridge., , 118., , (c) The given circuit can be redrawn as follows, A, , 127., , (c) Given circuit can be redrawn as follows, 2, , R, , 6, , 6V, , R, , R, , 1.5, , , , 3, , 6V, , 3, , R, 1.5, , R, , , , B, , 6V, , 1.5, , 6, 4A, 1 .5, ,  i, , 3 3, , Equivalent resistance between A to B is R., 119., 120., , (d) Equivalent resistance of the given circuit is 3  ., (c), , R, , R, , R, , 129., , (c), , R, 2R/3, , (b) For balanced Wheatstone bridge, , P R, , Q S, , R 2  10  R1 ., , B, , For, , second, , balancing, , R R,  R ||    , 3 2, , (a) Equivalent resistance of the given network R eq  75 , i R1(50), , 5, R, 5, 6, , R, 5, R  R 11, 6, , R1(50), R4, , 60 R3, , (50), , (30), , i1  i2 , , , , 133., , (c), , i, , 3, 60, 3, 60, 2, , , , , A, 150 (30  60) 150 90 150, , 2, 2,  30  V  0.4 V, 150, 5, , 10, 10, ,  5A, 1.5  1 || 1 1 .5  0 .5, , Parallel, , R 4R, , R , R, 3  4 R,  ||  R    2, R 4R, 2 , 3, 11, , 2, 3, , 3, 75, , 3, 3, , 75  2 150, , V4  i4  R 4 , , (a), , 3, 2, 6, , 2, , , Req  4 , , 134., , 50, , (50), , R5(30), , Current through R4 , , 132., , i2, , i1, R2, ,  Total current through battery i , , condition, , Resistance between Q and R, , Resistance between P and R, , i2, , i1, R2, , 3V, ,  R1  20 , , R, , 1, 1, 1, 1, 20, 19, 1 1 1, , , ,  Req , ,    , Req R1 R2 R3, 20, 2 4 5, 19, , 131., , R/2, R 6, Req ,    1.5 , 2, 4 4, (a) Resistance between P and Q, , R QR, , R=4, , (a), , (b) Given R  6  . When resistor is cut into two equal parts and, connected in parallel, then, , R PQ, , , , 130., , 12, x 6,  x  6, , (1 / 2) (1 / 2), (b) For maximum energy equivalent resistance of combination, should be minimum., 10  R1, 50, , (c) For first balancing condition, R2, 50, , R1, R1 40, 2, , , , 10  R1, 3, R2 60, , 126., , A, , 2R/3, , , , 125., , A, , B, , 122., , , , 2, , R, , (b), , 124., , 2, , , 3 2, 1,     , , 4 3, 3, , , B, , 2R, ., 3, , 121., , 123., , r, i1, R, l,  2  2   1, i2, R1, l1  r2, , R=4, , R, , R, R, , 2R/3, , B  A, , R, , Hence R eq , , 2R/3, , R, , R, , A, , (b), , R, , R, , R, , 128., , (a) The equivalent resistance between C and D is, , 2

Page 74 :
1108 Current Electricity, 141., , 1, 1 1 1 2, 3, or R '   1 .5 ,    , R' 6 6 3 3, 2, , Now the equivalent resistance between A and B as R'  1.5 , and 2.5  are connected in series, so, R"  1. 5  2.5  4 , , Kirchoff's Law, Cells, , Now by ohm's law, potential difference between A and B is, given by VA  VB  iR  2  4.0  8 volt, 135., , (b) Since voltmeter records 5V, it means the equivalent. Resistance, of voltmeter and 100  must be 50, because in series grouping, if resistances are equal, they share equal potential difference. It, conclude that resistance of voltmeter must be 100 ., , 1., ,  E1 ,  12 , , X  E  ,  X  2  X = 100 , 500, , X, , ,  500  X , , (b) The given circuit can be redrawn as follows, A, , R, , F, , C, R, , R, , 2., , (d) Since E1 (10 V )  E2 (4 V ), So current in the circuit will be clockwise., , R, , D, , (b) For no current through galvanometer, we have, , R, , B, , 10V, , Equivalent resistance between A and B is R and, V, R, (b) The given network is a balanced Wheatstone bridge. It's, 18, equivalent resistance will be R , , 5, , So current from the battery i , 137., , (a)  R AB , , R/2, , R/2, , A, , 138., 139., , (b) i , , B, , i, R, 1, 5 (R  2),  1  2 ,  R = 8, , R, i2, R1, 4, R, , , , Applying Kirchoff's voltage law,  1  i  10  4  2  i  3i  0  i  1 A(a to b via e), , V 10  4, ,  1 .0 ampere, R, 6, , 3., , (c) For maximum power, external resistance = internal resistance., , 4., 5., , (a) 0.9 (2 + r) = 0.3 (7 + r)  6 + 3r = 7 + r  r = 0.5 , (a) Since both the resistors are same, therefore potential difference, E,  V V  E  V , 2, , 6., , (b) Let the current in the circuit  i , Across the cell, E  V  ir  r , , 1, 1, 1, , , 50 50 75, , 7., , R1, R4, , i, E, , R2, , 50  75, 50  75 75, R, , ,   18.75 , 75  75  50, 200, 4, , 8., 9., 10., , (a) For maximum energy, we have, External resistance of the circuit, , 2A, , (b) When resistances 4  and 12 are connected in series,  4  12  16, , 1A, , A, , B, , 1.3A, , C, , 2A, , t, , When these resistance are connected in parallel., , At junction C, i  iBC  1.3  3  1.3  1.7 amp, 11., 12., , r, 2, , (a) Kirchhoff's first law is based on the law of conservation of, charge., (b) Kirchhoff's second law is based on the law of conservation of, energy., (a) According to Kirchhoff's first law, , At junction B, iAB  iBC  1  3 A, , ,  R1  R  100  18.75  118.75, , 1, 1 1, 4  12 4  12,  ,  RP , ,  3, RP 4 12, 4  12, 16, , EV EV EV, , , R, i, V/R  V , , At junction A, iAB  2  2  4 A, , This resistance is in series with R1, , Rresultant, , V, R, , = Equivalent internal resistance of the circuit i.e. R , , R3, , 75  75  50, , 50  75, , 140., , 4V, , (c) In given circuit three resistance R2 , R4 and R3 are parallel., 1, 1, 1, 1, , , , R R 2 R4 R3, , b, , 3, ,  Current , , V, V, 5V, , , R 18 / 5 18, , R/2 R, , 2, 4, , 2, , i, , current i , , 136., , E1 e E2, , 1, , a, , E, , (c) In charging V > E., (d) In open circuit of a cell V = E

Page 75 :
Current Electricity 1109, 13., 14., 15., , (a) Zero (Circuit open means no current and hence no potential, difference across resistance)., (d) Zero (No potential difference across voltmeter)., (b) Let the e.m.f. of cell be E and internal resistance be r. Then, E, E, and 0 .25 , 0.5 , (r  5), (r  2), On dividing, 2 , , (c) In short circuiting R = 0, so V = 0, , 17., , (c) Short circuit current iSC , , 19., , (d) (4  r)i  2.2, , 21., , 1, 2, , 34., 35., , E, 1 .5,  15 ,  r  0.06, R r, 0 .04  r, , (c), , 36., 37., , E  2.2 volt, V  1.8 volt, R  5 R, , 23., , , , E, (b) In parallel, equivalent resistance is low  i , r, , R, , n, , , 24., , (d) Internal resistance  distance , , 25., , (a) Total e.m.f. = nE, Total resistance R + nr  i , , 26., , (a) Current through R is maximum when total internal resistance, of the circuit is equal to external resistance., (b) Cells are joined in parallel when internal resistance is higher, then a external resistance. (R <<r), i, , 28., , 29., , E 2,   0.5 , i, 4, , (c) By Kirchhoff's current law., (b) For power to be maximum, External resistance = Equivalent internal resistance of the, circuit, E, 1.5, (a) i  ,  30 A, r, 0.05, 12, (a) i ,  2A, (4  2), (b) V2  V1  E  ir  5  2  0.5  4volt, , nE, R  nr, , (a) If m = Number of rows, and n = Number of cells in a row, Then m  n = 100, , .....(i), , Also condition of maximum current is R , , , , , , , , , 1,  concentrat ion, Area, , 1n,  n = 25 m, m, On solving (i) and (ii) m = 2, ,  25 , , (b) According to Kirchhoff's law iCD  i2  i3, , 39., ,  E , (b) Since i  , , we get,  R r, E, 2r, E, 0 .25 , 5 r, 0 .5 , , ......(i), .....(ii), , Dividing (i) by (ii), we get 2 , , r, R, n, , nr, m, , .....(ii), , 38., , E, , 2E, (b) In series , i1 , 2  2r, E, 2E, In parallel, i2 , , r, 4, r, 2, 2, 2E, 2E, , Since i1  i2 ,  r  2, 4  r 2  2r, (a) Applying Kirchhoff law, 20, A, (2  2)  (0.1  0.3  0.2)i  i , 3, , (b) Here two cells are in series., Therefore total emf = 2E., Total resistance = R + 2r, 2E, 2  1.45, 2.9 29,  i, , , ,  1.611 amp, R  2r 1.5  2  0.15 1.8 18, (a) E  V  ir, ,  V2  4  V1  4  10  14 volt, , E, ,  2 .2, , r    1 R  ,  1   5  1 .1, V, ,  1 .8, , , 27., , 20, 0, 3, , Energy loss inside the source  i2r  (2)2  2  8W, , (c) The voltage across cell terminal will be given by, 2, E,  3.9  1.95 V, , R , (3 .9  0.1), R r, , 22., , 31., , 32., 33., , Putting the value of i in (i), we get r = 0.4 ohm., (b) Let the internal resistance of cell be r, then, i, , 30., , After short-circuiting, V  0;  r , , ......(i), , and 4 i  2  i , , 20., , E, 1 .5,  3,  r  0.5 , r, r, , 50, 50, 5,  r, i,  10 ,  1.1, R r, 4 .5, 4 .5, , (c), , Potential difference across B  2  0 .3 , , 5 r,  r  1, 2r, , 16., , 18., , Hence potential difference across A, 20 4,  2  0 .1 ,  V (less than 2V), 3, 3, , 5 r,  r  1, 2r, , E,  E  1. 5 V, 2 1, r, (c) Because Eeq  E and req , 2, (d) In parallel combination Eeq  E  6 V, ,  0 .5 , , 40., 41., 42., , (d) Suppose current through different paths of the circuit is as, follows., 28, , 1, , 54, 6V, , 2, , i3, 8V, , 12 V

Page 76 :
1110 Current Electricity, Applying Kirchoff’s voltage law for the loop ADEFA., 40i2  40i3  80  40  0, , After applying KVL for loop (1) and loop (2), 1,  i1   A, 2, , We get 28i1  6  8, , 1, A, 3, , 54 i2  6  12  i2  , , and, , 5, Hence i3  i1  i2   A, 6, , 44., 45., , 5 X  2  10, (d) VAB  4 ,  X  20, X  10, (a) After short circuiting, R becomes meaningless., (c) V  E  IR  15  10  0.05  14.5 V, , 46., , (c) In series i , , 47., , (b), , 48., , (a), , 43., , 49., , , ,  40i2  40(i1  i2 )  120, , , , i1  2i2  3, , On solving equation (ii) and (iii) i1  0.4 A ., 57., 58., , (c) V  E  ir = 12  60  5  10 2 = 9V., (a) Applying Kirchoff's voltage law in the loop, i, A, , nE, n  1 .5,  0 .6 ,  n = 10, nr  R, n  0 .5  20, , W, W, 1000,  Vi  V , ,  1.38 V, t, it 2  6  60, (a) Applying Kirchoff's voltage law in the given loop., 4V i 8V 2, , 1, , P, , 2V, , (d) V  E  ir  1.5  2  0.15  1.20Volt ., , 60., , (b) i , , 61., , (c), , E, 4, 1 ,  r  2, R r, 2r, Short circuit when terminals of battery connected directly then, E 4, current flows which is iSC    2 A ., r 2, , 9, , i, , 22, 4, , A, 1  1.9  0.9 3.8, , For cell A E  V  ir  V  2 , , 1, 2i  8  4  1  i  9i  0  i  A, 3, 1,  9  3V, 3, , 62., , (c) By using i ,  0.5 , , 50., , (d) Because cell is in open circuit., , 51., , (b) In parallel combination Eeq  E  12V, , 52., , (d), , 53., , (b) i , , 20, , 59., , 4V, , Potential difference across PQ , , B, , i, , 10i  5  20i  2  0  i  0.1 A, , Q, , i, , 5V, , 10, , 2, , P, , …….(iii), , 4,  1 .9  0 ., 3.8, , E, R r, , E,  E  5.5  0. 5r, 11  r, , ..…(i), , E,  E  4. 5  0.9r, 5r, On solving these equation, we have r  2.5 , , and 0 .9 , E, 6, ,  12 amp., r, 0 .5, , 63., , (c), , 54., , (c) Strength = 5 × 18 = 90AH., , 64., , (b) W  qV  6  10 6  9  54  10 6 J ., , 55., , (a), , 65., , (a), , 66., , (c), , 67., 68., , (a), (d) Applying Kirchhoff law in the first mesh, 10, 10  5  i  i , 2A, 5, (b) Applying Kirchhoff law in the first mesh, .....(i), 10  5i1  i, , E, 5, ,  1A, R  r 4.5  0.5, , V  E  ir  5  1  0.5  4.5 Volt, , 56., , (b) The circuit can be simplified as follows, C, , B, i1, A, , 30, , i3, , i3, , i2, F, , D, 40, , 40V, , 40, , 80V, , E, , Applying KCL at junction A, i3  i1  i2, , 69., , V2, ; for P to be maximum Req should be less. Hence, R eq, option (a) is correct., P, , Pmax , , E2, (2)2, ,  2W, 4 r 4  0.5, , i – i1, , .….(i), , i, i1, , Applying Kirchoff’s voltage law for the loop ABCDA, A, , 30i1  40i3  40  0, , ,  30i1  40(i1  i2 )  40  0, , , , 7i1  4 i2  4, , .….(ii), , 4, 5, , 10V, , r = 1, , i, , ..…(ii)

Page 77 :
Current Electricity, Applying in the second mesh, 5i1  4 i  4 i1, , ......(ii), , Solving equation (i) and (ii), we get i1 , 70., , 80., 81., 40, A, 29, , i  i1  i2  1 .... (i), , nE, nr, R, m, ,  i1  3i, , 2, , 72., , (d) Watt hour efficiency , , , 73., , 74., 75., , 76., , (b) i , , 2, , 1, ,  20 , , , 3,  i  1 , , 20, ,  60 ,  3, , , (c) From Kirchoff's junction Law, , 9  10, E, ,  40, , r    1 R  ,  1  9 ,  3, 30, V, ,  30, , , 10, , (Branch, current), , 14  5  15,  0.875  87.5%, 15  8  10, , (b) In the given case cell is in open circuit (i = 0) so voltage across, the cell is equal to its e.m.f., (b) The internal resistance of battery is given by, , 1A, , 2, ,  Resistance of opposite branch , main current , , Total resistance, , , , Discharging energy, Charging energy, ,  4 2i5 3  0 i  2 A, , B, ,  i = i = (3 i) = 6i, On solving equation (i), (ii) and (iii) we get i = 0.1 A, Short Trick : Branch current =, , 5000  0 .15, 750,  1.5 A, , i, 5000  0 .25, 512.5, 500 , 100, , (a), , 5, , I2, , ...(ii), , 1, , E  0.15 V and r  0.25 , , 71., , I1, , 1A, , For loop (2), – (15 + 5) i + 10 i = 0, , Here m  100, n  5000, R  500 , , , , 1, , 15, , A, ,  60 i  (15  5)i1  0, , 60, , I, , For Loop (i), , (a) Given problem is the case of mixed grouping of cells, So total current produced i , , (d), (a) Applying Kirchoff's law in following figure., At junction A :, , 1111, , 60, , I, , 1A, , 5, , 15, 10, , 83., 84., , = 0.1 A, 20/3 , (d) Maximum current will be drawn from the circuit if resultant, resistance of all internal resistances is equal to the value of, external resistance if the arrangement s mixed. In series,, R  nr and in parallel, the external resistance is negligible., (c) On applying Kirchoff's current law i = 13 A., (c) Total cells = m  n = 24, .... (i), mr, For maximum current in the circuit R , n, m,  3   (0 .5)  m = 6n, ..... (ii), n, On solving equation (i) and (ii), we get m = 12, n =2, , 85., ,  E , (a) Power dissipated  i2 R  ,  R,  R r, , 82., , E2R, E,  P  i2 R  P , rR, (r  R)2, , Power is maximum when r = R  Pmax  E 2 / 4 r, , 2, , 77., , 78., , (c) Since the current coming out from the positive terminal is, equal to the current entering the negative terminal, therefore,, current in the respective loop will remain confined in the loop, itself., current through 2 resistor = 0, (c) Reading of voltmeter,  E eq , , 79., , 2, , 2, ,  E ,  E ,  R1  , ,  , ,  R  r  R2,  R1  r ,  2, , ,  R1 (R22  r2  2 R2r)  R2 (R12  r 2  2 R1r), , E1 r2  E 2 r1 18  1  12  2, ,  14 V, r1  r2, 12, ,  R22 R1  R1r 2  2 R2r  R12 R2  R2r 2  2 R1 R2r,  (R1  R2 )r 2  (R1  R2 )r 2  (R1  R2 )R1 R2, , 2E, (d) i , R  R1  R 2, ,  r  R1 R2, , From cell (2) E  V  iR2  0  iR2, , Different Measuring Instruments, , R, , 1., i, , E, R1, (1), ,  E, , E, R2, (2), , 2E,  R 2  R  R 2  R1, R  R1  R 2, , 2., 3., , (a) In meter bridge experiment, it is assumed that the resistance of, the L shaped plate is negligible, but actually it is not so. The, error created due to this is called, end error. To remove this, the resistance box and the unknown resistance must be, interchanged and then the mean reading must be taken., (c) To convert a galvanometer into an ammeter a low value, resistance is to be connected in parallel to it called shunt., (d) Balance point has some fixed position on potentiometer wire. It, is not affect by the addition of resistance between balance, point and cell.

Page 78 :
1112 Current Electricity, 4., 5., 6., , 7., , (d) Resistance of voltmeter should be greater than the external, circuit resistance. An ideal voltmeter has infinite resistance., iG, 100  0 .01, 1, (c) S  g , ,  0.1, i  ig (10  0.01) 10, (c) Equivalent resistance of the circuit Req  100, , G i  ig 10  1 9, S , , , , , S, ig, 1, 1, (i  ig ), , 8., , (d), , 9., , (c) Ammeter is used to measure the current through the circuit., iG, 1  0.018 0.018, (c) S  g, , ,  0.002, (i  ig ), 10  1, 9, , 10., 11., 12., 13., 14., , (d) Potentiometer works on null deflection method. In balance, condition no current flows in secondary circuit., ig G, 10  99, ,  11, (c) Shunt resistances S , (i  ig ) (100  10), (d) By using R , , 15., , (d) R , , 16., , (c), , 18., 19., , V, 100, G  R ,  5  19,995, ig, 5  10  3, , V2  V1,  V2  V1  3 volt, 30 / 100, , V, 5, 5000, G , 2 ,  2  48 , 3, ig, 100, 100 / 10, , , , V1 l1, 6, 300, ,  V =1 V., , , V2 l2, V2, 50, , S, , 2, , ig G, , , , 10  0.01, 10, , ohm, 10  0 .01 999, , 24., , (a), , 25., , (a) Ratio will be equal to the ratio of no deflection lengths i.e., E1 l1 2,  , E2 l2 3, , 26., , (a) Potential gradient , , 27., , (a) Wheatstone bridge is balanced, therefore, , i  ig, , iS, 50  12,  10 ,  12  G  60  G  48 , S G, 12  G, (a) To convert a galvanometer into a voltmeter, a high value, resistance is to be connected in series with it., (b), P R, (c), (For balancing bridge), , Q S, 4  11 44, , 9, 9, , 1 1 1, ,  , S r 6, , 28., , 29., 30., 31., , (a) When the length of potentiometer wire is increased, the, potential gradient decreases and the length of previous balance, point is increased., (b), (b), (b) The actual circuit is same., , 32., , (b)  ig  10% of i , , 33., , (b), , 34., , (b) Suppose resistance R is connected in series with voltmeter as, shown., By Ohm's law, R, ig, ig, G, ig .R  (n  1)V, , , , 9 1 1,  , 44 6 r, ,  r, (a), , V, , 35., 36., , ig  G, , (b) S , , 38., , (b)  ig  (100  90)% of i , , i  ig, , , , 10  10 3  50 50, ,  in parallel., 99, 1  10  3  10, , 37., , R = 4, , , 25 ,  R  ,  2  0.5 , , 100, , , , , r, D, , (n – 1)V, , V, nV, ), G, (c) Ammeter is always connected in series with circuit., (c) If resistance of ammeter is r then, 20  (R  r)4  R  r  5  R  5 , ,  R  (n  1)G (where ig , , C, 6, , 132,  26.4 , 5, , l l, r   1 2,  l2, , Q = 11, , A, , G, 90, i,  S, ,  10 , (n  1) (10  1), 10, , E1 l1  l2 (8  2) 5, , , , E2 l1  l2 (8  2) 3, , B, P = 9, , Potential difference, Length, , P R, 10, or 1 ,  S  10 ohm, , Q S, S, , ig , ,  S , , 20., , 1,  Length ), P.G., (b) In balance condition, potentiometer doesn't take the current, from secondary circuit., (a) Here same current is passing throughout the length of the, wire, hence V  R  l, , (a) Potential gradient = Change in voltage per unit length,  10 , , 17., , 23., , 2, 5, V, V,   0 .5,  0 .005, (15  5  0) 1, m, cm, , ig G, , (b) The sensitivity of potentiometer can be increased by decreasing, the potential gradient i.e. by increasing the length of, potentiometer wire., (Sensitivity , , 22., , 2 .4, current through the circuit i , A, 100, P.D. across combination of voltmeter and 100  resistance, 2 .4, ,  50  1 .2 V, 100, Since the voltmeter and 100  resistance are in parallel, so the, voltmeter reads the same value i.e. 1.2V., e, R, ., (a) Potential gradient , (R  Rh  r) L, , , 21., ,  Required shunt S , S', , R, , i, 10, , G, 900, ,  100 , (n  1) (10  1), , V, 100, G ,  25  9975 , ig, 10  10  3, , 39., , (d), , 40., , (b) Potential gradient x , , V iR, , L, L

Page 79 :
Current Electricity 1113,  x, S, , 2, 15, 3, , , volt / cm, (15  5) 10 2000, , 55., , G, 25, 25, 25, ,  5,  5  2.5  10 4 , i, 5, 10, , 1, 10, 1, 1, ig, 50  10  6, , 41., , (a), , 42., , (b) In balanced Wheatstone bridge, the arms of galvanometer and, cell can be interchanged without affecting the balance of the, bridge., (c) Error in measurement = Actual value – Measured value, , 43., , (a) Potential gradient , , , 56., 57., 58., , V iR iL i, , , , L, L, AL A, , 0 .2  40  10 8,  10  2 V / m, 8  10  6, , G, G, G, i,  S, , , (n  1) (50  1) 49, 50, (d) The resistance of an ideal voltmeter is considered as infinite., (c), , (b) ig  2% of i , , 20×103, , R, , 998 , , V, , V, , 5V, , i, , i, , Actual value = 2V, 2, 1, i, , A, 998  2 500, , 110, Here i , 20  10 3  R, , + –, 2V, , 2, , Since E  V  ir  V  E  ir  2 ,  Measured value , , 110, , ,  V  iR  5  ,   20  10 3, 3,  20  10  R , , 1, 998, 2 , V, 500, 500, , 45., 46., , 998,  4  10  3 volt, 500, (d) The emf of the standard cell must be greater than that of, experimental cells, otherwise balance point is not obtained., (a), (b) In general, ammeter always reads less than the actual value, because of its resistance., R, AC 20,  R  20 , , , 80 BC 80, , 47., , (c) By Wheatstone bridge,, , 48., , (a), , 49., , l l , l 2, (b) r   1 2   R '   1, 5, l,  2 ,  2 , , 59., , 60., , 61., , 62., , ... (i), 63., , On solving (i) and (ii) r = 10 , (a), (b) In the part c b d,, Vc  Vd, 2, , In the part c a d, Vc  Vd,  Va  Vb  Va, 2, (c) In balance condition, no current will flow through the branch, containing S., , 64., , 53., , (b) Resistance in parallel S , , Gig, i  ig, , , , 54., , (b), , Exl, , 66., , (b) ig S  (i  ig )G  ig (S  G)  iG, , , 67., , ig, i, , , , 68., , e, R, , (R  Rh  r) L, , 2.5, 20, V, ,  5  10 5, (20  80  0) 10, mm, , (b) Given ig  2mA, i  20mA, G  180, ig, i, , 69., , G, 8, ,  0 .8, S G 28, , (a) Potential gradient x ,  x, , e, R, V iR,  l,  l  E , (R  Rh  r) L, l, L, , 10, 5,  3  3 V, E, (5  4  1) 5, , X 20, 1,  X    0 .25  ., , 1, 80, 4, (a) Reading of galvanometer remains same whether switch S is, open or closed, hence no current will flow through the switch, i.e. R and G will be in series and same current will flow, through them. IR  IG ., , (d), , (d) Pressing the key does not disturb current in all resistances as, the bridge is balanced. Therefore, deflection in the, galvanometer in whatever direction it was, will stay., , 50  100  10 6, (10  100  10 6 ), ,  S  5  10 4 , , 2, V,  0 .5, 4, m, , 65., , Vc  Va  Va  Vd , , 52., , (a) Since potential difference for full length of wire = 2 V,  P.D. per unit length of wire , , ... (ii), , Vc  Vb  Vb  Vd  Vb , , V, 10, G , 7  3 , ig, 1, , R, , E  l (balancing length), , l 3, and r   1,   10,  3 , , 50., 51., , 10 5,  420 K, 5, (c) Due to the negligible temperature co-efficient of resistance of, constantan wire, there is no change in it's resistance value with, change in temperature., (d) The resistance of voltmeter is too high, so that it draws, negligible current from the circuit, hence potential drop in the, external circuit is also negligible., (a) By connecting a series resistance, ,  10 5  5 R  22  10 5  R  21 , , 998, V, 500, ,  Error  2 , , 44., , 110V, , , , S, 180,  180  S  10 S  S ,  20 , GS, 9, , (c) Resistance of shunted ammeter , , GS, GS

Page 80 :
1114 Current Electricity, Also, , ig .G, i, G, GS, , 1, , ig, S, GS, i, , GS, 0 .05  120, = 0.6 , , GS, 10, (l  l ),  60  50 , r  1 2  R'  ,   6  1.2 , l2,  50 , , , 70., , (c), , 71., , i, G, (d) By using, 1, ig, S, 1000, 1000,  S,  111 , S, 9, V, e, R, , (c) Potential gradient x  , L (R  Rh  r) L, , , , 72., , i, , 100  10  3, , 1, ,  2 .2  10  3 , 73., 74., , 75., 76., , 77., 78., 79., , (a), , V, 18, G ,  12  5988 , ig, 3  10  3, , R, , V, 6, G ,  25  975 (In series)., ig, 6  10  3, , (b), , 86., , (d), , 87., , (c), , 88., , (d) ig  i, , 89., , X, , A, , (a), (c), , 82., , (c), , 83., , (a) When ammeter is connected in parallel to the circuit, net, resistance of the circuit decreases. Hence more current is, drawn from the battery, which damages the ammeter., l l ,  55  50 , (a) r   1 2   R '  r  ,   10  1 ., l,  50 ,  2 , , 2 R  20  R  10 ., i, G, 4, R, R, 1   1 , S  ., ig, S, 1, S, 3, , 25, ., 999, , 6, , 4, , 6, , C, , B, , Resistance of the part AC 5V, R AC  0.1  40  4  and R CB  0.1  60  6 , X 4,   X  4, 6 6, Equivalent resistance R eq  5  so current drawn from, , In balanced condition, , battery i , 90., , 5,  1A ., 5, , (a) (R  G) ig  V  (R  G) , , V, ig, , ig, , 3, ,  6 .25 k , 30  16  10  6, , R, G, ,  Value of R is nearly equal to 6k , This is connected in series in a voltmeter., 91., , (d), , V2, , V1, R1 = 16k, , R2 = 32k, , V1 = 80V, V, , R1  80  200  16000   16 k, , l l , (b) r   1 2  R  0 .5  .,  l1 , (a), e, 2, . R  10  3 ,  10, (b) V  i.R. , (R  Rh  r), (10  R  r), , 80., 81., , S, 25  S, , G, , ig, i GS, S, 2.5, 1, , , , , , ig, S, i G  S 27.5 11, ,  L , i, , V iR, A  i, (a) Potential gradient x  ,  , , L, L, L, A, 10, (d) Here n , 5, 2,  R  (n  1)G  (5  1)2000  8000 , ,  0 .01  10, , (c), , 2.2,  1  R  990 , (10  Rh ), , 80, (c) Total resistance of the circuit ,  20  60 , 2, 2, 1,  Main current i , , A, 60 30, Combination of voltmeter and 80 resistance is connected in, series with 20, so current through 20 and this combination, 1, will be same , A., 30, Since the resistance of voltmeter is also 80, so this current is, equally distributed in 80 resistance and voltmeter, 1, (i.e., A through each), 60, 1, P.D. across 80 resistance ,  80  1 .33 V, 60, , S, GS, ,  1000S  25  S  S , , Current flowing through V1 = Current flowing through V2 =, 80,  5  10  3 A ., 16  10 3, So, potential differences across V2 is, , V2  5  10 3  32  10 3  160 volt, ,  R  19,989 ., , 84., , R, , 85., , Hence, line voltage V  V1  V2  80  160  240V ., 92., , (d) V  xl  iR  xl,  2  10 3,  i  10  , 2,  10, , ,   50  10  2  0 .1, , , ,  i  10  10 3 A  10 mA ., , 93., , (d), , E, , e, R, 2, 10, l ,   0.4  0.16 V ., (10  40  0) 1, (R  Rh  r) L

Page 81 :
Current Electricity 1115, i, G, 5, 12, 1 , , 1,  S  8 . (In parallel)., ig, S, 2, S, , 94., , (c), , 95., , (d), , 96., , (a), , R  G(n  1)  50  10 3 (3  1)  10 5  ., , 97., , (c), , E1 l1  l2 58  29 3, , , , E2 l1  l2 58  29 1, , ig, , R, , (a), , S, 5, S, G, ,  S, , GS, 100 G  S, 19, , , , i, , 112., , (a), , 99., , (d) For conversion of galvanometer (of resistances) into voltmeter,, a resistance R is connected in series.,  ig , , 113., , 100., , Vi , , V1, V2, and ig , RG, 2R  G, , , , 102., , (d) After connecting a resistance R in parallel with voltmeter its, effective resistance decreases. Hence less voltage appears across, it i.e. V will decreases. Since overall resistance decreases so, more current will flow i.e. A will increase., , 103., , i, , (c) Potential gradient x , , , , i, , 10 3, 10, , 1 , , 2, , , , e, R, ., (R  R h  r) L, , 2, 3,   R h  57  ., (3  R h  0) 1, , G, , S, , 1, 20, 20, 1, S ,  0.02 ., S, 999, 10  3, , 104., , (c), , 105., 106., , (a) Resistance of voltmeter should be high., (c) If ammeter is used in place of voltmeter (i.e. in parallel) it may, damage due to large current in circuit. Hence to control this, large amount of current a high resistance must be connected in, series., , 107., , ig, , (c) Potential gradient x , , , 108., 109., , Vf , , (d), , 114., , 110., 111., , (a), , 900, , Vi  Vf, Vi, ,  100, , (b) Potential gradient =, , 100, , 10, , Vf, V, , e .R, 10  3, =, ., (R  r).L (3  3)  5, ,  1V / m  10 mV / cm., , 100, i, G, 1, 100,  S  5  10 3  .,  1   5  1 , ig, S, S, 10, 10, , 115., , (c), , 116., , (d), , 117., , (a), , 118., , (b) ig  i, , 119., ,  100  l , (d) S  , .R,  l , , ig, i, i, , , , S, 4, 4, 1, , , , G  S 36  4 40 10, , V, 6, 6,  R 1 ., 2 , , 63, R, 2, , R, R, 63, 50, S, 0 .01, 5, ,  S, ,  0 .05  ., GS, 10, 50  S, 999, ,  100  l , Initially, 30  ,   10  l  25 cm,  l, ,  100  l , Finally, 10  ,   30  l  75 cm,  l, , So, shift = 50cm., , 3, 20, ,  0 .2, (20  10  0) 10, , (c) Manganin or constantan are used for making the potentiometer, wire., (a), , V, , 10, 9, V V, 11, 10, ,  100  1 .0 ., 10, V, 11, , e, R, ., (R  R h  r) L, , E1 l1  l2 (6  2) 2, , , , E2 l1  l2 (6  2) 1, , Vi, , 90, 9, V , V, (90  10), 10, , % error =, , S, 4, 1, , , i.e. 10%., G  S 36  4 10, , (c), , 100, , Final potential difference, , G, VG,  V2  2 V1  1,  V2  2V1, (R  G), (R  G), , 101., , 10, , 100, 10, V , V, (100  10), 11, , Finally after connecting, voltmeter across 100, Equivalent resistance, 100  900,  90 , (100  900), , (d) If the voltmeter is ideal then given circuit is an open circuit, so, reading of voltmeter is equal to the e.m.f. of cell i.e., 6V., ig, , 1000,  11.11  ., 90, , (c) Before connecting the voltmeter, potential difference across, 100 resistance, , V, 2 R  G 2(R  G)  G, V1, V2, ,  2 , , , V1, RG, (R  G), R  G 2R  G,  2, , S, S,  10  10 3 ,  100  10 3, GS, 100  S, , 90 S  1000  S , , V, 10, G ,  1  999 ., ig, 10  10  3, , 98., , ig  i, , 120., 121., , 0 .1  10 7, i, ,  10  2 V/m, A, 10  6, (d) Before connecting voltmeter potential difference across 400, resistance is, , (c) Potential gradient (x) , , 10,000, , i, G, i.G, G, 100  10 3  40, 40, 1 , 1 , 1, ig, S, Vg, S, S, 800  10  3, , V, ,  S  10 ., A, , 400, , B, , 6V, , 800

Page 82 :
1116 Current Electricity, Vi , , 400,  6  2V, (400  800), , After connecting voltmeter equivalent resistance between A and, 400  10,000, B ,  384.6 , (400  10,000), Hence, potential difference measured by voltmeter, 384.6, Vf ,  6  1 .95 V, (384.6  800), Error in measurement = Vi  Vf  2  1.95 = 0.05V., 122., , (c), , 124., , (a), , i, ig, , 1 , , (a) From the principle of potentiometer V  l, V, l,  ; where V = emf of battery, E = emf of standard, E L, cell, L = Length of potentiometer wire, V, , (b), , E, , El 30 E, ., , L, 100, , 127., , and, , 4X, l, , Y, 100  l, , .....(ii), , (d), , E, , (a), , 138., , l, ,  240, , (b) r  R  1  1   2 ,  1  2 , l, 120, , ,  2, , , 139., , (d), , E, , V1, V, 1 .1, V, 180  1 .1,  2 ,  V, ,  1 .41 V, l1, l2, 140 180, 140, , G, , IG  G  I  IG  S  I =  1   I G  I = 100.1 mA, S, , , (a), , 141., , (c) Let S be larger and R be smaller resistance connected in two, gaps of meter bridge., 100  20,  100  l , R  4R,  S , R , 20,  l , , .....(i), , When 15  resistance is added to resistance R, then, 6,  100  40 , S , (R  15)  (R  15), 4,  40 , , 500 C, , 142., , 2, , Parallel, , 128., , G, 90, i, (d) ig ,  Required shunt S , ,  10 , (n  1) (10  1), 10, , 129., , (b) ig , , 130., , 50,  40  4960 , 10  10  3, (c) Post office box is based on the principle of Wheatstone's bridge, , 131., , (d) Full deflection current ig  25  4  10 4, ,  100  10 4 A, , V, 25, G ,  50  2450  in series., Ig, 100  10  4, , .... (ii), , From equations (i) and (ii) R  9 , (a) According to following figure, A, , 3, 1000, potential difference across AB , , 4V, 250, 3, , Using R , , V, ; E is constant (volt. gradient)., l, , 140., , V, 500 B, , V 2,   0 .5 V / m, L 4, , 137., , 1000, , A, , e, R, 5, 5, .  l  0 .4 , , l, (R  Rh  r) L, (5  45  0) 10, , (a) Potential difference per unit length , , 10V, , 1000 2500,  500 , , 3, 3, , 4, l,  l  50 cm, , 4 100  l, , 136., , 1000  500 1000, , (1500), 3, , So, equivalent resistance of the circuit, ,  Current drawn from the cell, 10, 3, i, , A, (2500 / 3), 250, Reading of voltmeter i.e., , .....(i), , , , l, ,  150, ,  1   1, (b) Using r  R  1  1   2 , 100, l, , , 2, , , , R eq, , X, 20 1, , , Y, 80 4, ,  i , 100  10 6, (d) S   g   G ,  50  0 .5 ,  i  ig , (10  10  3  100  10 6 ), , , (in parallel), , 2, 10, ,  0.4  R = 790, (10  R  0) 1, , (d) Resistance between A and B , , , , (c), , e, R, . l, (R  R h  r) L, ,  10  10  3 , , 126., , 135., , R1, l, l1,  1 , R2, l 2 100  l1, , l=8m, , G, 10, 0 .81, , 1 ,  S  0.09 ., S, 1, S, , , , 125., , 133., 134., , 50   l, 50 2.97  10 2  10 4,  l, , ,  3m ., 99, A, 99, 5  10 7, , (a) In balancing condition,, , , , i, G, 5, 50, 1 , 1, ig, S, 0 .05, S, ,  S, 123., , 132., , B, , A, i=0.1A, , V, , Reading of voltmeter = Potential difference between A and B = i, (R + 2) 12 = 0.1 (R + 2)  R = 118 ., 143., , (a) Potential gradient x , , , , e, R, ., (R  Rh  r) L, , 0 .2  10 3, 2, R, , ,  R = 4.9 ., (R  490  0) 1, 10  2

Page 83 :
Current Electricity 1117, 7., , Critical Thinking Questions, 1., , (a) Initially : Resistance of given cable, , R, , l, , ..., ,   (9  10 3 ) 2, , (i), Finally : Resistance of each insulated copper wire is, , l, , R' , , 8., , . Hence equivalent resistance of, ,   (3  10 3 ) 2, R' 1 , l,    , cable R eq , 6, 6    (3  10  3 ) 2, , ,  ….(ii), , , 9., , l, , l, , , , , , (d) At time t = 0 i.e. when capacitor is charging, current, 2, i,  2mA, 1000, When capacitor is full charged, no current will pass through it,, 2, hence current through the circuit i ,  1mA, 2000, P 4 .5, (d) Current in the bulb  ,  3A, V 1.5, 1 .5, Current in 1  resistance ,  1 .5 A, 1, Hence total current from the cell i  3  1.5  4.5 A, By using E  V  ir  E  1.5  4.5  (2.67)  13.5 V, (d) Equivalent resistance of the circuit R  9 , V 9,  Main current i    1 A, R 9, , 9 mm, , 3, , On solving equation (i) and (ii) we get R = 7.5 , , 1A, , eq, , 4, , 2., , (a), , R A  rB , R, 1, 1,    A   ,  RB  16 R A, R B  rA , RB  2 , 16, When R and R are connected in parallel then equivalent, R A RB, 16, , RA, resistance Req , (R A  RB ) 17, A, , 0.25 A, , 8, , 9V, 2, , B, , 8, 2, , 4, , 2, , After proper distribution, the current through 4 resistance is, 0.25 A., 10., 11., , (c) The given circuit can be simplified as follows, , 2, , 0.5 A, , 4, , If R A  4.25 then Req  4  i.e. option (a) is correct., 3., , 2, , (b) Maximum number of resistance  2n 1  2 3 1  4, (d) The given circuit can be simplified as follows., r, , A, ,  R AD , 4., , r, , D, , C, , B, , r, , 5R, 6, , 8r, 3, , A, ,  i'   R '   4   5 ,  i'2 R'  n  i2 R  n              8,  i   R   1   10 , , (c) Resistance across AB , R1  2  10 , , and R    1  10 6 , On solving,, 6., , 12., , R, , 6, , A, , R1, , B, , R, R'  0.88  10 6 , (b) No current flows through the capacitor branch in steady state., Total current supplied by the battery, 6, 3, i ,  ., 2 .8  1 .2 2, 3 3, Current through 2  resistor    0 .9 A, 2 5, , r, , A, , , , r, , B, , , 2r/3, , r, , B, , 2r, , A, , 8r, 7, , , 1, 1 1, 1,   , R ' R R R1, , B, , B, , 2r, , r, r, , r, , A, , 2, , r, , r, , r, , (c) Suppose n resistors are used for the required job. Suppose, equivalent resistance of the combination is R' and according to, energy conservation it's current rating is i'., , 2, , , , r, , r, , r, , Energy consumed by the combination = n  (Energy consumed, by each resistance), , 5., , r, , (d), , A, , B, , R eq, , 5,  , 2, , i, , 20, 5,  1 .5, 2, , 20V, 1.5, ,  5A, , i, , 3, , i, , P, , i/2, X, , 2, , 3, , 2, , i/2, , Potential difference between X and P,, 5, V X  VP     3  7 . 5 V, 2, , Q, , ....(i)

Page 84 :
Current Electricity, V X  VQ , , 5,  2  5V, 2, , …(ii), , 18., , On solving (i) and (ii) VP  VQ  2.5 volt; VQ  VP ., Short Trick : (VP  VQ ) , , i, 5, (R2  R1 )  (2  3)   2 .5, 2, 2, , (c), , On solving equation (i) and (ii) i1  1.8 A ., (b) To convert a galvanometer into an ammeter, a shunt, I, S  g G is connected in parallel with it. To convert a, I  Ig, galvanometer into a voltmeter, a resistance R , ,  VQ  VP, 13., , R t1  R1 (1   1 t) and Rt 2  R2 (1   2 t), , 19., , 14., ,  Req, So  eff, , R   R 2 2,  1 1, R1  R 2, , 3, , X, i, 10, 20, , 24, , r1, , r2, , 48 V, , E, , E, , B, Y, , 24  8, 6, 32, Current between A and B = Current between X and Y, 48, i,  8A, 6, , Resistance between 1, A and B , , 5,  0.5 , 10, , Resistance between X and Y  (3  10  6  1)  20 , , 5,  2A, (2  0 .5), , Current through the resistance i , , 20., , (b) The given circuit can be redrawn, ,  Potential difference between X and Y = 8  20 = 160 V, (d) R1  R2  R1 (1  t)  R2 (1   t), R1 , , R2 , ,  R1  R2  R1  R2  R1t  R2 t , , E1 R1, B, , 8, , i, , (b) Emf E = 5V , Internal resistance r , , A, , i, , V, ,  R  r1  r2, , 16., , 60, , 6, ,  r1  r2  R  2r1, , 15., , 30, , A, , (b) Let the voltage across any one cell is V, then, , , 2E, , V  E  ir  E  r1 , , r, , r, , R,  1 2, , But V = 0, 2 Er1, 0,  E, r1  r2  R, , V,  G is, Ig, , connected in series with it., (a) The given circuit can be redrawn as follows, , Also Req.  Rt1  Rt 2  Req  R1  R2 (R11  R2 2 )t, ,   R   R2 2  , ,  (R1  R2 )1   1 1, .t , , R, , R, , 1, 2,  ,  , , , 1119, , A, , Eeq, , Req, , B, , 21., , (d) Current density of drifting electrons j = nev, n  5  107 cm 3  5  107  106 m 3 ., ,  i, , R2, , E3, , v  0.4 ms 1 , e  1.6  10 19 C  j  3.2  10 6 Am 2, , Current density of ions = (4 – 3.2)  10 = 0 .8  10  6, –6, , E2, , E eq , , E2, , E1 R 2  E 2 R 1 2  4  2  4, ,  2 V and, R1  R 2, 44, , –1, , 22., , 4, 22, Req   2 . Current i ,  2 A from A to B, 2, 2, through E ., (b) Applying Kirchhoff’s law for the loops (1) and (2) as shown in, figure, i1, R1 = 2, For loop (1), 2, , 17., , i1, , i2, , E1 = 4V, (i1 – i2), , 1, , R3 = 2, 2, , R2 = 4, , 2i1  2(i1  i2 )  4  0  2i1  i2  2, , …(i), , For loop (2), 2(i1  i2 )  4 i2  6  0  i1  3i2  3, , …(ii), , i, , P, i1, N, , 3V, 1, , R2 , , 3,  10  7 .5 , 4, , Req , , R1 R2, 2.5  7.5, 15, , =, ., R1  R2 (2.5  7.5) 8, 3, 24, , A, 15,  1 23, 8, ,  R2  24  7.5  18, , , A, So, i = i  , , ,  R1  R2  23  2.5  7.5  23, 1, , M, , i2, , 10, R1 ,  2 .5  and, 4, Resistance of part PMQ;, , Main Current i =, , i2, , E2 = 6V, , This gives v for ions = 0.1 ms ., (a) In the following figure, Resistance of part PNQ;, , A, m2, , Q

Page 85 :
1120 Current Electricity, In a circuit, any circuit element placed between points at the, same potential can be removed, without affecting the rest of, the circuit. Here, by symmetry, points A, B and C are at same, potential, for any potential difference between P and Q., The circuit can therefore be reduced as shown below, , 24 18, 6, , , A., 23 23 23, (c) As I is independent of R 6 , no current flows through R 6 this, , and i2  i  i1 , , 23., , requires that the junction of R 1 and R 2 is at the same potential, , 4R, , as the junction of R 3 and R 4 . This must satisfy the condition, R1, R,  3 , as in the Wheatstone bridge., R2 R4, , 24., , (c) Moving anticlockwise from A, , 2r, , P, R, , V, , 4R, ,  iR  V  2V  2iR  0, A, B, , VA  VB  iR  V  V  iR, ,  Potential drop across C =, 25., , 28., , C, , 2V, , V, 3, , 2R, ,  V A  0  1.5 V  V A  1.5 V, Potential difference between B and C, VB  VC  1  2.5  2.5 V, , i, , (b) Let R and m be the resistance and mass of the first wire, then, the second wire has resistance 2R and mass 2m. Let E = emf of, each cell, S = specific heat capacity of the material of the wire., 3E, For the first wire, current i1 , and i12 Rt  mS T, R, NE, 2R, , For the second wire, i2 , , 2 Rr, ., Rr, (d) Potential difference between A and B, V A  VB  1  1 . 5, , Effective resistance R eq , , V, , V, 3R, , or 3iR  V or i , , and i22 (2 R)t  2mS T ., ,  0  VC  2.5 V  VC  2.5 V, , 29., , Potential difference between C and D, VC  VD  2V   2.5  VD  2  VD  0.5 V., (b) The given circuit can be simplifies as follows, r, , Thus, i1  i2 or N  6 ., 26., , 2, , 3, , A, , r, , , , 1, 5, , 2.2, ,  3, 4, , r, , r, , r, , r, , r, , r, , P, , B, , 2r, , 1, , 2, , Q, , , , 2, , r, , 1, , 1.8, , Q, , r, , 10, , A, , r, , 2, , 2, , 4, , r, , r, r, , P, , (b), , Q, , i, , 2r, , P, , Q, , 2r, 1.8, 2.2, , Short circuited, , B, , , , 2, , 4, , 3, , A, 4, , 30., ,    t , , = t2, , 5, , , , 27., , 2, ,  A,  8 ., , (a), , 2R, , B, 2R, , C, , B, , 32., 33., , r, , 3 3, 2, , t 2, , 3,  3, , Idt   2 tdt  3 t 2 dt ,  2, , 2, , , , , , = (9 – 4) + (27 – 8) = 5 + 19 = 24C., , E  E 2  E 3  .....  E n, (d) i  1, (r1  r2  r3  .......  rn ), 1 .5(r1  r2  r3  ......  rn ),  1.5 A ., (r1  r2  r3  .....  rn ), (a) Balancing length is independent of the cross sectional area of, the wire., , (a), , R1 (1  t1 ), 10, (1  5  10 3  20), , , ,  R2  15 , R2 (1  t2 ), R2 (1  5  10  3  120), , Also, Q, , 2R, , 3, 2, , , , t 3, , , , 2R, , A, , r, P, , B, 5, , 1, , 2R, , (b) dQ = Idt  Q , , 1, , 2, , 31., , R AB, , 2r 2 3,    1 ., 3, 3 2, , R' , , 5, , 34., , i1 R 2, 30 15, , , ,  i2  20 mA, i2, R1, i2, 10, , (b) The given circuit can be simplified as follows, , 2R, , 10, , A, 5, , 3, , 10, 8, , B, 6, , 6

Page 86 :
Current Electricity, , 1121, , R, R,  r(2   ) , (2   ), 2r, 2, R R, , (2   ), R XWY R XZY, R, , (2   ), Req ,  2 2, R  R(2   ), R XWY  R XZY, 4, 2, , 2, 2, (d) Battery is short circuited so potential difference is zero., (a) Let V be the potential of the junction as shown in figure., Applying junction law, we have, , and R XZY , , 5, , A, , , 3, , B, 8, , 5, , 3, , 39., 40., , Now it is a balance Wheatstone bridge., 8, , So,, , 20 V, , A, 8, ,  R AB , 35., , 2, , i2, , i1, , i3, 0V, , (c) The equivalent network is, R, , 20  V 5  V V  0, or, , , 2, 4, 2, or 40 – 2V + 5 – V = 2V or 5V = 45  V = 9V, , 2R, , R, , R, R, 6R, ,  i3 , , R, , , , 4R, , 2R, , 4R, , E, , E, , 4, 4, Clearly, the network, of resistances is a balanced, Wheatstone, bridge. So R AB is given by, , V,  4 .5 A, 2, , E  x l  i l  i , , E, E, 2 .4  10 3, , ,  4  10 4 A ., l l, 1.2  5, , 41., , (a), , 42., , (b) When bulb glows with full intensity, then voltage across it will, be 1.5 V and voltage across 3  resistance will be 4.5 V., , 1, 1, 1, 2 1, 1, , , , ,  R AB  2 R, R AB 3 R 6 R, 6R, 2R, , For maximum power transfer 2R  4   R , 36., , B, , 2, , 8  8 64, ,  4, 8  8 16, , R, , 5V, , 4, , 6V, 4.5 V, 3, , 4,  2, 2, , 1.5 V, , R, , X, , Y, , (c) The given circuit can be redrawn as follows, 6V, 1, , 4 .5 B,  1 .5 A, 3, Same current will flow between X and Y, So VXY  iRXY  1.5  1.5 R XY  R XY  1, , Current through 3  resistance i , V, , A, , B, , i, , A, 6, , 43., , 4, , 6, 6, , A, 6  4  1 11, 6, 60, P.D. between A and B, V ,  10 , V., 11, 11, , Current i , , 37., , Potential difference across R 2,  R2 , , l, ; here A   (r22  r12 ), A, Outer radius r = 5cm, Inner radius r = 5 – 0.5 = 4.5 cm, , (a) By using R   ., 2, , 1, , 5 mm, , r2, , r1, , 44., , 10 cm, , i, 9i, , 10 10, = Potential difference across R, , (a) In figure (b) current through R2  i , , 9, i, R 11, i.e. R2  , i  R, , 10, 10, 9, 9, , 11 11, , R2  R, 1  11 , Req ,  9, (R2  R) 11  11 10, 9, 1, 11, Total circuit resistance ,  R1  R  11  R1  9.9 , 10, (a) Let l be the original length of wire and x be its length stretched, uniformly such that final length is 1.5 l, l, , So R  1 .7  10, , 8, , 5, ,  {(5  10  2 )2  (4.5  10  2 )2 }, , 38., , (a) Here R XWY , , x, , (l – x), ,  5.6  10 5 , , 0.5l, , 1.5l, , R, R,  (r ) , 2r, 2, , l, ,   , r, , , , Then 4 R  , , x, (l  x ), (0.5 l  x ), A, where A' , , (0 .5 l  x ), A, A'

Page 87 :
1122 Current Electricity,  4, , 45., , l, lx, (0 .5 l  x )2, , , A, A, xA, , 1 l 2 x 2 lx, x 1, or 4 l  l  x , or, , , , 4 x, x, x, l, 8, (b) In series : Potential difference  R, 3, When only S is closed V1  E  0 .75 E, 4, 6, When only S is closed V2  E  0 .86 E, 7, and when both S and S are closed combined resistance of 6R, and 3R is 2R, , conductors are equal i.e. R = R = R, (b) Resistance of CD arm = 2r cos 72 = 0.62r, Resistance of CBFC branch, A, , 49., , 46., , 1, 2 .62, , R 1 .24 r, , 1, , I1, , I2, , R , , Equivalent R'  2 R  r  2 , , Ixy, , 1, 3, , 50., , B, , (a), , (i – i1 – i2), ,  i1  0  ixy  3i2  0 i.e. i1  3i2 ......(i), , B, , Also  2(i1  ixy )  4(i2  ixy )  0, , i.e. 2i1  4 i2  6ixy, , r, r, , I, , (i – i1 – i2), , F, , i1, , 10(i  i1 )  10i2  20i1  0, ,  20 (i  i1  i2 )  10 (i1  i2 )  10i2  0, , Resultant e.m.f. of battery  (12  n)E  nE  (12  2n)E, , From equation (i) and (ii) i1 , , ( resistance remains same irrespective of connections of, cells), With additional cells, (a) Total e.m.f. of cells when additional cells help battery = (12, – 2n) E + 2E, Total resistance = 12r + 2r = 14r, ,  3i1  i2  i .......(i), , and in mesh BEFCB, , , Total resistance of cells = 12r, , i, , Applying Kirchoff's law in mesh ABCDA, , Total e.m.f. of such cells  (12  n)E, Total e.m.f. of cells opposing = nE, , (12  2n)E  2 E, 3, 14 r, , r, , (i1 + i2), , D, , .... (iii), , (b) Let n be the number of wrongly connected cells., Number of cells helping one another  (12  n), , , , J, , H, , C, , A, , Solving (i), (ii) and (iii), ixy  2 A, , 51., , 52., , ......(i), 53., , (b) Similarly when additional cells oppose the battery, , 3i1  4 i2  2i, , ......(ii), 2i, i, 2i, , i2   iAD , 5, 5, 5, , (d) Let the current in 12  resistance is i, Applying loop theorem in closed mesh AEFCA, 12i = – E + E = 0  i = 0, E, 10  4, 1, (b) Current flowing in the circuit i  ,  A, R 20  10 5, 1, P.D. across AC   20  4 V, 5, P.D. across AN = 4 + 4 = 8V, (a) If two resistances are R 1 and R 2 then, S  R1  R 2 and P , , (12  2n)E  2 E,  2 ......(ii), 14 r, , R1 R 2, (R 1  R 2 ), ,  RR , From given condition S = nP i.e. (R1  R2 )  n  1 2 ,  R1  R2 , , Solving (i) and (ii), n = 1, 48., , E, , i, , Also VAB  1  i1  2(i1  ixy )  0  50  i1  2(i1  ixy ), , 47., , F, r, , 1.24 r, r, 2.62, , E, r, , r, , i2, , (i – i1), , ....(ii), ,  3i1  2ixy, , r, ,  Equivalent resistance between A and H, 1, 1, 1, R ' 1.946, , ,  Req , , r  0.973r, Req R ' R ', 2, 2, , 4, , A, , 72°, , D, , r C, ,  2 .48, ,  r,  1   1 .946r, 2, ., 62, , , Because the star circuit is symmetrical about the line AH, , 2, , Y, , i, r, , r, , G, , 2, , X, , A, B, , 1 .24 r, 2 .62, , 2, , 2, V3    E  0 .67 E  V2  V1  V3, 3, , (c), , C, , 1, 1, 1, 1  2 .62 , , ,  , , R 2r 0 .62r r  2  0 .62 , , 2, , , , B, , o, , 1, , 1, , l, ; resistances of all the, A, , Hence according to formula R  , ,  (R1  R2 )2  n R1 R2  (R1  R 2 )2  4 R1 R 2  nR1 R 2, , (a) All the conductors have equal lengths. Area of cross-section of, , A is {( 3 a)2  ( 2 a)2 }  a2, , So n  4 , , Similarly area of cross-section of B = Area of cross-section of C, =a, , (R1  R2 )2, . Hence minimum value of n is 4., R1 R2, , 2, , 54., , (b) Voltage sensitivity , , Current sensitivity, Resistance of galvanomet er G

Page 88 :
Current Electricity,  G, , 10, 5., 2, , 150,  15 mA ., 10, V = Voltage to be measured = 150  1 = 150 V., V, 150, Hence R , G ,  5  9995  ., ig, 15  10 3, , Here ig  Full scale deflection current , , Graphical Questions, 1., 2., , 3., , (a) For ohmic resistance V  i  V  Ri (here R is constant), (d) From the curve it is clear that slopes at points A, B, C, D have, following order A > B > C > D., and also resistance at any point equals to slope of the V-i, curve., So order of resistance at three points will be, R A  RB  RC  RD, (a) Slope of the V-i curve at any point equal to resistance at that, point. From the curve slope for T > slope for T,  RT1  RT2 . Also at higher temperature resistance will be, higher so T > T, (c) For portion CD slope of the curve is negative i.e. resistance be, negative.,  l , (d) Slope of V-i curve  R    . But in given curve axis of i,  A, 1, , 1, , 4., 5., , 9., , (d), , 10., , (a), , 11., , (b), , 12., , (d), ,  RV , 13., , 2, , The graph between loge I and loge V will be a straight line, which cut loge V axis and it's gradient will be positive., 14., , (c) As we know, for conductors resistance  Temperature., From figure R  T  tan  T  tan = kT, … (i), and R  T  tan (90 – )  T  cot = kT, ….(ii), From equation (i) and (ii) k (T2  T1 )  (cot   tan  ), 1, , 8., , (b) Since the value of R continuously increases, both  and  must, be positive., Actually the components of the given equation are as follows, Rt, , t, , 2, , 2, , Rt, , 1, , 16., , (d) At an instant approach the student will choose tan will be the, right answer. But it is to be seen here the curve makes the, angle  with the V-axis. So it makes an angle (90 – ) with the, i-axis., So resistance = slope = tan (90 – ) = cot., , 17., , (d) Short circuited current i , , t, , nE E, , nr, r, , i.e. i doesn't depend, , upon n., 18., , 19., , It  is positive,  is negative, the component, t will be shown in, the following graph., , 2, , (b) Let resistivity at a distance 'x' from left end be   (0  ax)., Then electric field intensity at a distance 'x' from left end will, i i(0  ax), be equal to E , where i is the current, , A, A, flowing through the conductor. It means E   or E varies, linearly with distance 'x'. But at x = 0, E has non-zero value., Hence (b) is correct., , (b) Here internal resistance is given by the slope of graph i.e., But conductance , , 0, , 1, , 2, , 15., , t, R, , 1, , 2, ,  cos  sin  (cos 2   sin2  ), (T2  T1 )  , ,  2 cot 2, , sin cos ,  sin cos  ,  (T – T )  cot 2, , 2, , 7., , 1, , o, , 2, ,  EA ,  R  P  E 2 (Symmetric parabola), P  i2 R  , , , , , Also P  i2 (parabola), Hence all graphs a, b, d are correct and c is incorrect., (b) When we move in the direction of the current in a uniform, conductor, the potential difference decreases linearly. When we, pass through the cell, from it's negative to it's positive terminal,, the potential increases by an amount equal to it's potential, difference. This is less than it's emf, as there is some potential, drop across it's internal resistance when the cell is driving, current., , V,  RV  V, ig, , (a) According to ohm's law V  iR,  loge V  loge i  loge R  loge i  loge V  loge R, , 2, , 1  A,  , R  l , i.e. with the increase in length of the wire. Slope of the curve, will decrease., iR i.  neAv d , (c) E ,  vd  E (Straight line), , , L, A, A, , In this case, the value of R will not increase continuously., Hence the correct option is (c)., 1, Slope of V-i curve = resistance. Hence R   1, 1, At point A the slope of the graph will be negative. Hence, resistance is negative., E.m.f. is the value of voltage, when no current is drawn from, 2, the circuit so E = 2V. Also r = slope =  0 .4 , 5, For conversion of a galvanometer into a voltmeter, V, V,  ig ; where R = R + G = Total resistance,  ig , RV, RG, V, , and V are interchanged. So slope of given curve , , 6., , 1123, , 20., , (a), , 1, y, , Resistance, x, , RParallel  RSeries . From graph it is clear that slope of the line, , A is lower than the slope of the line B. Also slope = resistance,, so line A represents the graph for parallel combination., (b) To make range n times, the galvanometer resistance should be, G /n, where G is initial resistance., , Assertion and Reason, R, , 0, , t, , t, , 2, , x, ., y

Page 89 :
1124 Current Electricity, 1., , 2., , 3., , 4., , 5., , 6., , 7., , (d) Resistivity of a semiconductor decreases with the temperature., The atoms of a semiconductor vibrate with larger amplitudes, at higher temperatures thereby increasing it's conductivity not, resistivity., (d) It is quite clear that in a battery circuit, the point of lowest, potential is the negative terminal of the battery and the current, flows from higher potential to lower potential., (b) The temperature co-efficient of resistance for metal is positive, and that for semiconductor is negative., In metals free electrons (negative charge) are charge carriers, while in P-type semiconductors, holes (positive charge) are, majority charge carriers., (a) Here, E  2V , 1 , , 2,  1 A and r  1, 2, , Therefore, V  E  ir  2  1  1  1V, (a) It is clear that electrons move in all directions haphazardly in, metals. When an electric field is applied, each free electron, acquire a drift velocity. There is a net flow of charge, which, constitute current. In the absence of electric field this is, impossible and hence, there is no current., (c) The metallic body of the electrical appliances is connected to, the third pin which is connected to the earth. This is a safety, precaution and avoids eventual electric shock. By doing this the, extra charge flowing through the metallic body is passed to, earth and avoid shocks. There is nothing such as reducing of, the heating of connecting wires by three pin connections., (b) On increasing temperature of wire the kinetic energy of free, electrons increase and so they collide more rapidly with each, other and hence their drift velocity decreases. Also when, temperature increases, resistivity increase and resistivity is, inversely proportional to conductivity of material., , 8., , (c) In a conductor there are large number of free electrons. When, we close the circuit, the electric field is established instantly, with the speed of electromagnetic wave which cause electron, drift at every portion of the circuit. Due to which the current is, set up in the entire circuit instantly. The current which is set, up does not wait for the electrons flow from one end of the, conductor to the another end. It is due to this reason, the, electric bulb glows immediately when switch is on., , 9., , (a) Resistance wire R  , , 10., , (c), , 11., , (a), , 12., , (a), , 13., , (b), , 14., , (a) Sensitivity , , l, . where  is resistivity of material, A, which does not depend on the geometry of wire. Since when, wire is banded, resistivity, length and area of cross-section do, not change, therefore resistance of wire also remain same., The resistance of the galvanometer is fixed. In meter bridge, experiments, to protect the galvanometer from a high current,, high resistance is connected to the galvanometer in order to, protect it from damage., Voltameter measures current indirectly in terms of mass of, ions deposited and electrochemical equivalent of the substance, m, , I , . Since value of m and Z are measured to 3rd, Zt, , , decimal place and 5th decimal place respectively. The relative, error in the measurement of current by voltmeter will be very, small as compared to that when measured by ammeter directly., When current flows through a conductor it always remains, uncharged, hence no electric field is produced outside it., Here assertion and reason both are correct but the reason is, not the correct explanation of assertion., 1,  (Length of wire), Potential gradient, , 15., , 16., , 17., , (a) If either the e.m.f. of the driver cell or potential difference, across the whole potentiometer wire is lesser than the e.m.f. of, the experimental cell, then balance point will not obtained., (d) Because there is no special attractive force that keeps a person, stuck with a high power line. The actual reason is that a, current of the order of 0.05 A or even less is enough to bring, disorder in our nervous system. As a result of it, the affected, person may lose temporarily his ability to exercise his nervous, control to get himself free from the high power line., (a) Due to high electrical conductivity of copper, it conducts the, current without offering much resistance. The copper being, diamagnetic material does not get magnetised due to current, through it and hence does not disturb the current in the, circuit.

Page 90 :
Current Electricity 1125, , Current Electricity, 1., , Figure shows a simple potentiometer circuit for measuring a small, e.m.f. produced by a thermocouple. The meter wire PQ has a, resistance 5  and the driver cell has an e.m.f. of 2 V. If a balance, point is obtained 0.600 m along PQ when measuring an e.m.f. of, 6.00 mV, what is the value of resistance R, (a) 995 , , 9., , (b) 1995 , , 2, , (c) 2995 , , 3., , 4., , 0.600m, , P, , 10., , Q, , Thermocouple, , G, A car has a fresh battery of e.m.f. 12 V and internal, resistance of, 6.00 mV, 0.05 . If the starter motor draws a current of 90 A, the terminal, voltage when the starter is on will be, , (a) 12 V, (b) 10.5 V, (c) 8.5 V, (d) 7.5 V, If the balance point is obtained at the 35 cm in a metre bridge the, resistances in the left and right gaps are in the ratio of, (a) 7 : 13, (b) 13 : 7, (c) 9 : 11, (d) 11 : 9, , (d) 6 A and 7 A, 11., , 10, , (b) 10 , , 15, 8, , 12., , 8, , 4, , In the circuit shown in figure, switch S is initially closed and S is, open. Find V – V, 1, , a, , b, , 2, , 1, , b, , (a) 4 V, , 20, , C, , 3, , B, , 3, , a, 24V, , S1, , The figure here shows a portion of a circuit. What are the, magnitude and direction of the current i in the lower right-hand, wire, 1A, (a) 7 A, 2A, , 2A, , 13., , 3A, , –, , 4A, , i, A carbon resistor has colour strips as violet, yellow brown, and, golden. The resistance is, (a) 641 , (b) 741 , (c) 704 , (d) 407 , A voltmeter of resistance 1000  is connected across a resistance of, 500  in the given circuit. What will be the reading of voltmeter, , B, , C, (a) A-A, (b) x B-B, 2x, (c) C-C, (d) Same for all three pairs, A battery is connected to a uniform resistance wire AB and B is, earthed. Which one of the graphs below shows how the current, density J varies along AB, , 2A, , (c) 6 A, , 8., , 10, , 4x, , (d) 16 V, , (d) 2 A, , 10, , 5, , S2, , (c) 12 V, , (b) 8 A, , 0.5A, , (a) Resistance R = 46 , (b) Current through 20  resistance is 0.1 A, (c) Potential difference across the middle resistance is 2 V, (d) All option are correct, In figure shows a rectangular block with dimensions x, 2x and 4x., Electrical contacts can be made to the block between opposite pairs, of faces (for example, between the faces labelled A-A, B-B and C-C)., Between which two faces would the maximum electrical resistance, be obtained (A-A : Top and bottom faces, B-B : Left and right faces,, C-C : Front and rear faces), , 10F, , (b) 8 V, , 7., , 20V, , In the circuit as shown in figure the, , 25V, , 6, , E=24V, , 2, , R, , Find the equivalent resistance across the terminals of source of e.m.f., 24 V for the circuit shown in figure, (a) 15 , , A, , (c) 5 A and 6 A, , (d) 4 , , 6., , 3, , (b) 4 A and 5 A, , th, , (c) 5 , , 5., , 2, , 2, , (d) None of these, 2., , 8, , 5, , R, , 2V, , (a) 1 V, (b) 2 V, (c) 6 V, (d) 4 V, A beam contains 2  10 doubly charged positive ions per cubic, centimeter, all of which are moving with a speed of 10 m/s. The, current density is, (a) 6.4 A/m, (b) 3.2 A/m, (c) 1.6 A/m, (d) None of these, In the circuit shown, the reading of ammeter when switch S is open, and when switch S is closed respectively are, (a) 3 A and 4 A, 2, S, , +, , A, , (a), , B, , J, , 0, , (b), , J, , Zero at all, points, , A, , B, , J, , 0, , A, , B, , A, , B, , J, , 10 V, , V, , 0, , 0, , A, 500, , 500, , B

Page 91 :
1126 Current Electricity, (c), , 19., , (d), , GR, , (a), , (b), , , , R, G, , GS, Rl, (d), l, S, A potential divider is used to give outputs of 4 V and 8 V from a 12, V source. Which combination of resistances, (R , R , R ) gives the, correct voltages ? R : R : R, , (c), 15., , 1, , 1, , 2, , 2, , 20., , +12V, , +8V, , 21., , R2, , (c) 2 : 2 : 1, , +4V, , (d) 1 : 1 : 2, , R1, , 0 Volt, , (a) R, , R, , 3R, 4, , (b), , R, , A, , R, 2, , (c), , 17., , R, , R, , R, , B, , R, , 22., , R, , 7C/sec, (i), , B, , (a) 1, , (b) 2, , (c) 3, , (d) 4, , A wire has resistance of 24  is bent in the following shape. The, effective resistance between A and B is, 60°, , 16, , 3, , 60°, , A, , B, , 5 cm, 10 cm, , In the circuit shown in figure, find the current through the branch, , BD, R, , A, , (a) 5 A, (b) 0 A, , 6, , 3, , B, , C, , 15 V, 3, , (c) 3 A, , 30 V, , (d) 4 A, , –, 3C/sec, , Two conductors are made of the same material and have the same, length. Conductor A is a solid wire of diameter 1.0 mm. Conductor B is a, hollow tube of outside diameter 2.0 mm and inside diameter 1.0 mm., The resistance ratio R /R will be, , (d) None of these, , Following figure shows four situations in which positive and negative, charges moves horizontally through a region and gives the rate at, which each charge moves. Rank the situations according to the, effective current through the region greatest first, +, , (d) 0.04 A, , (c), , R, R, , 0.03 A, , (b) 10 , , R, , (d) 2R, , (c), , R, R, , R, , (b) 0.02 A, , (a) 24 , , Find equivalent resistance between A and B, , 16., , 0.01 A, , A, , R3, , (b) 1 : 1 : 1, , (a), , 3, , 3, , (a) 2 : 1 : 2, , 18., , into an ammeter reading upto 0.06 A when a shunt of resistance, r is connected across it. What is the maximum current which can, be sent through this galvanometer if no shunt is used, , A cylindrical metal wire of length l and cross sections area S, has, resistance R, conductance G, conductivity  and resistivity . Which, one of the following expressions for  is valid, , 14., , A moving coil galvanometer is converted into an ammeter reading, upto 0.03 A by connecting a shunt of resistance 4 r across it and, , +, , D, , 2C/sec, 4C/sec, , (ii), , 23., , 6C/sec, , –, , –, +, , (a) i = ii = iii = iv, , (b) i > ii5C/sec, > iii > iv, , (c) i = ii = iii > iv, , (d) i = ii(iii)= iii < iv, , +, 1C/sec, (iv), , A battery of 24 cells, each of emf 1.5 V and internal resistance 2 is, to be connected in order to send the maximum current through a 12,  resistor. The correct arrangement of cells will be, (a) 2 rows of 12 cells connected in parallel, , A and B are two square plates of same metal and same thickness but, length of B is twice that of A. Ratio of resistances of A and B is, , (b) 3 rows of 8 cells connected in parallel, , (a) 4 : 1, , (d) All of these, , (b) 1 : 4, , (c) 4 rows of 6 cells connected in parallel, , B, , (c) 1 : 1, (d) 1 : 2, , A, , (SET - 19), 1., , (a) The voltage per unit light of the metre wire PQ is,  6 .00 mV , ,  i.e. 10 mV/ m . Hence potential difference,  0 .600 m , across the metre wire is 10 mV /m  1m  10 mV . The, , current drawn from the driver cell is i , The resistance R , , 10 mV,  2 mA ., 5, , (2 V  10mV ) 1990 mV, ,  995  ., 2 mA, 2 mA, , [

Page 92 :
Current Electricity 1127, 2., , (d) V  E  i.r  12  90  0.05  12  4.5  7.5 V ., , 3., , P R, R, (a) Using Wheatstone principle,  , Q S 100  l, , 4., , 35, 35, 7, , , , 100  35 65 13, (c) Given circuit can be reduced to a simple circuit as shown in, figures below, 6, , 6, , 15, , (a) J  nqv  n(ze)v , , 10., , (b) When switch S is open total current through ammeter., i, , 6, , , , 8, , 8, , 4, , 11., , (d), , 20,  4A ., (3  2), 20,  5A ., 3  (2 || 2), , R, , R, Parallel, , 25 V, 4, , 2, , When switch is closed i , , Parallel, 4, , 2  10 8  2  1.6  10 19  10 5, =6.4A/m, (10  2 )3, , 9., , Parallel, 10, , 2, 3, ,  500  4 V ., 3 250, , Now voltmeter reading  iv  R V , , 0.5 A, , 0.5 A, 10, , Series, , 20, , 10, , R', , , , 4, , R', , , , i.e. Req  5  ., 5., , 1, 1, 1, 1, 20, , , ,  R' , 4, R ' 10 10 20, 5, , 10, , 10, , Now using ohm’s law i , , (b) Switch S 2 is open so capacitor is not in circuit., 1, , 5, , b, ,  R4 , , O, 3, , Current through 20  resistor , , a, , 24 V, , = Potential difference across 20   20  0.1  2V, , 24, 4 A, 33, , 12., , Let potential of point ‘O’ shown in fig. is VO, VO  Va  3  4  12V, , R AA  , , ....(i), , Now current through 5  resistor , , RBB  ., , So V0  Vb  4  1  4 V, .....(ii), (b) By using Kirchoff's junction law as shown below., 13., 2A, 3A, 5A, 3A, , 7., , 6A, , 2A, , 14., , 4A, , (b) Using standard colour codes, Violet = 7, yellow = 4, brown = 1 and gold = 5 % (tolerance), , i, , It is clear maximum resistance will be for contact C-C., (d) Wire AB is uniform so current through wire AB at every across, section will be same. Hence current density J  i / A  at every, point of the wire will be same., 1, (a) Conductivity  , .....(i), , 1, R, ,  GR  1, , So R  74  10 1  5%  740  5%, 8., , 4x, 2  16 , , , x  2x, x, 8x, , and conductance G , , i=8A, , So its value will be nearest to 741  ., (d) Total current through the circuit, , 2x, , 4, , , x  4 x 2x 8 x, , and RCC  , , From equation (i) and (ii) Vb  Va  12  4  8 V., , 2A, , x, , , 2x  4 x 8 x, , Similar for contacts B-B and C-C are respectively, , 24,  4A, 5 1, , 1A, , (c) Let  is the resistivity of the material, Resistance for contact A-A, , then using ohm’s law, , 6., , 25,  50  R  50  4  46 , 0.5, , 0.5  5 2.5, ,  0 .1 A, 20  5, 25, Potential difference across middle resistor, , 3, , Current through 3  resistor , , 25, 25,  0 .5 , R4, R  R', , From equation (i) and (ii)  , , 15., , 10, 3, , A, 1000, 250,  500, 3, , GR, , , (b) Resistors are connected in series. So current through each, resistor will be same, i, , 16., , .....(ii), , 12  8 8  4 4  0, 4, 4, 4, , , , , , R3, R2, R1, R3 R2 R1, , So, R1 : R2 : R3 :: 1 : 1 : 1 ., (c) Given circuit can be redrawn as follows

Page 93 :
1128 Current Electricity, Neglect, Series, , R, , R, , Parallel, , A, , B, , , , A, , R/3, , R, , B, , R, , Parallel, , R/3, , R, , 2R, 3, , R, B, , A, , , , Series, R+R=2R, , B, R/3, , 17., , 2R, 3, , (c) For figure (i) i1  7 A, For figure (ii) i2  4  3  7 A, For figure (iii) i3  5  2  7 A, For figure (iv) i4  6  1  5 A, , 18., , (c), , 19., , (b), , RA , , ig, i, , , , R, l, ,   2l , i.e. A  1 : 1,  and RB , , RB, lt, t, 2l  t, t, S,  ig G  (i  ig )S, GS, ,  ig G  (0.03  ig )4 r, , .....(i), , and ig G  (0.06  ig )r, , .....(ii), , ***, , From (i) and (ii), 0.12  4 ig  0.06  ig  ig  0.02 A ., , 20., , (c) For conductor A, R A , For conductor B, R B , , l, ,, r12, , l,  (r22  r12 ), l, , 1 mm, , r1, , , , 1 mm, , 2 mm, , l, , , r2, , A, 2, , 21., , 2, , B, , d , R A r22  r12  r2 , 2, ,     1   2   1     1  3, , RB, r, d, r12, 1,  1,  1, (b) Given resistance of each part will be, , 6, , 2, , 4, , 12, , 6, , , A, , 6, , 6, , B, , A, , , , eq, , 22., , 6, , 6, , R = 10  , , B, 4as shown 6, (a) The current in the circuit are Aassumed, in the fig., 6, , A, , i1 B, , 15 V, , 3, , 3, , i2, i1, , D, , i1 – i2, , C, 30 V, , 2, , 1, , 2, , 2, , 1, , 2, , 2, , 23., , R, 2, , 2, , 1, , R/3, , R, ,  R eq , , Applying KVL along the loop ABDA, we get, – 6i – 3 i + 15 = 0 or 2i + i = 5, …..(i), Applying KVL along the loop BCDB, we get, – 3(i – i ) – 30 + 3i = 0 or – i + 2i = 10 …..(ii), Solving equation (i) and (ii) for i , we get i = 5 A., (a) Suppose m rows are connected in parallel and each row, contains n identical cells (each cell having E = 15 V and r = 2), For maximum current in the external resistance R, the, nr, necessary condition is R , m, n2,  12 ,  n = 6m, ..... (i), m, Total cells = 24 = n  m, ..... (ii), On solving equations (i) and (ii) n = 12 and m = 2, i.e. 2 rows of 12 cells are connected in parallel., 1, , R, , Neglect, , A, , R+R=2R, , B, , 2