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1, , ELECTRICITY, , CHAPTER, , CONTENTS, , , , , , , , , , , , Electric Current, Flow of Current in Metal, Electric Symbols, Ohm’s Law, Resistance of conductor, Resistivity, Combination of Resistances, Electric Energy, Electrical Power, , ELECTRIC CURRENT (CHARGE IN, MOTION), , Definition : The quantity of electric charge, flowing through cross section of a given, conductor in one second is called current., Thus, if Q is the charge which flows through a, conductor in time t, then the current (I) is given, by, Current (I) =, , Ch arge (Q), Time ( t ), , The electric current (or current) is a scalar, quantity., , The direction of flow of the positive charge taken, as conventional direction of the electric current., When we consider the flow of electric current in, an ordinary conductor, such as a copper wire, the, direction of current is taken as opposite to the, direction of the flow of electrons., ,  FLOW OF CURRENT IN A METAL, Metals show a very different kind of bonding, called metallic bonding. According to this, bonding, the outermost electrons are not bound to, any particular atom, and move freely inside the, metal randomly as shown in fig. So, these, electrons are free electrons. These free electrons, move freely in all the directions. Different, electrons move in different directions and with, different speeds. So there is no net movement of, the electrons in any particular direction. As a, result, there is no net flow of current in any, particular direction., Metallic wire, , Electrons move randomly,, so no flow of current, , , Fig. Flow of electrons inside a metal wire, when no potential is applied across its ends, , , ,  Unit of current, The SI unit of charge (Q) is coulomb (C), and, that of time (t) is second (s). So,, SI unit of current, =, , Electrons move from low, potential to high potential, , I coulomb, = 1 C s–1 = 1 ampere, 1sec ond, , The unit coulomb per second (Cs–1) is called, ampere (A), ,  Direction of Electric Current :, , +Battery–, High potential, , Low potential, , Fig. Flow of electrons inside a metal wire, when the two ends of a wire are connected to, the two terminals of a battery

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 Unit of resistance :, ,  ELECTRIC SYMBOLS, Many different kinds of equipments or, components are used in setting up electrical, circuits. To draw the diagrams of electrical, circuits on paper these equipments/components, are shown by their symbols. Here are some, symbols used in the electric circuit diagrams., , 4., 5., 6., , Components, Electric cell, Battery, Plug key, (switch open), Plug key, (switch closed), A wire joint, Wires crossing, joining, , Symbols, +, , –, , +, , –, , 8., 9., , A resistor of resistance R, or, , or, , rheostat, +A–, , Ammeter, Voltmeter, , +V–, , Fuse, , , OHM'S LAW,  Definition : According to the Ohm’s law at, constant temperature, the current flowing through, a conductor is directly proportional to the, potential difference across the conductor., Thus, if I is the current flowing through a, conductor and V is the potential difference (or, voltage) across the conductor, then according to, Ohm’s law., , or, I =, , (when T is constant), , V, R, , ...(i), , where R is a constant called the resistance of the, conductor., Equation (i) may be written as,, V=I×R, , , , Now, if,, , V = 1 volt and I = 1 ampere, , Then,, , R=, , 1 volt, 1ampere, , Current flowing through a conductor is, directly proportional to the potential, difference across the conductor., , or, , resistance, , I V, , V, I, , Thus, 1 ohm is defined as the resistance of a, conductor which allows a current of 1 ampere to, flow through it when a potential difference of 1, volt is maintained across it., , , , without, , Electric bulb, , 10., 11., 12., , R=, ,  Results of Ohm’s law, , 7., , Variable, , From Ohm’s law,, , Potential difference (volts), , S.N, 1., 2., 3., , The SI unit of resistance (R) is ohm. Ohm is, denoted by the Greek letter omega ()., , ...(ii), , , , IV, , Slope = Resistance (R), Current (amperes), , When the potential difference in a circuit is, kept constant, the current in inversely, proportional to the resistance of the, conductor., I  1/R, The ratio of potential difference to the current, is constant. The value of the constant is equal, to the resistance of the conductor (or resistor)., V/I = R, , RESISTANCE OF CONDUCTOR, The movement of electron gives rise to the, flow of current through metals. The moving, electrons collide with each other as well as, with the positive ions present in the metallic, conductor. These collisions tend to slow, down the speed of the electrons and hence, oppose the flow of electric current., The property of a conductor by virtue of, which it opposes the flow of electric current, through it is called its resistance.

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, , , , , Resistance is denoted by the letter R., , The, , SI unit of resistance is ohm. The ohm is, denoted by the Greek letter () called omega., , , , heating elements of heaters, toasters, electric, iron etc., , RESISTIVITY, R, , Resistance is a scalar quantity., , R , , 1, a, , So,, , R , , , a, , or, , R=×, , ,  Factors on which resistance of conductor, depends, , , Effect of the length on the resistance of a, conductor, The resistance of a conductor is directly, proportional to the length. That is, Resistance of a conductor.  Length of the cond., , , , Effect of the area of cross-section on the, resistance of a conductor, The resistance of a conductor is inversely, proportional to its area of cross-section., That is,, Resistance of a conductor ;, 1, R, , Area of cross  sec tion (a ) of the conductor, , * If the area of cross-section of the conductor, is doubled, its resistance gets halved., , , Effect of temperature on the resistance of a, conductor, The resistance of all pure metals increases, with a rise in temperature. The resistance of, alloys increases very slightly with a rise in, temperature. For metal when temperature, increases resistance increases and for, semiconductors when temperature increases, resistance decreases., , , , Effect of the nature of material on the, resistance of a conductor, Some materials have low resistance, whereas, some others have much higher resistance. In, general, an alloy has higher resistance than, pure metals which from the alloy., * Copper, silver, aluminium etc., have very, low resistance., * Nichrome, constantan etc., have higher, resistance. Nichrome is used for making, , , a, , ...(i), , where  (rho) is called resistivity of the material, of conductor., If,, l = 1 m and a = 1 m2, Then, , R = , , ...(ii), , Thus, if we take 1 metre long piece of a, substance having a cross-sectional area of 1, meter2, then the resistance of that piece of the, substance is called its resistivity., Resistivity of a substance can also be defined as, follows :, The resistance offered by a cube of a substance, having side of 1 metre, when current flows, perpendicular to the opposite faces, is called its, resistivity., ,  Units of resistivity, From equation (i), we can write, =, , Ra, , , So, SI unit of resistivity () =, , ohm  m 2, = ohm.m, m, , Thus, the SI unit of resistivity is ohm. m (or  . m), ,  Classification of Material on Basis of, Resistivity,  Substances showing very low resistivities :, The substances which show very low, resistivities allow the flow of electric current, through them. these type of substances are, called conductors., For example, copper, gold, silver, aluminium, and electrolytic solutions are conductors.,  Substances having moderate resistivity:, The substances which have moderate

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resistivity offer appreciable resistance to the, flow of electric current through them., Therefore, such substances are called, resistors. For example, alloys such as, nichrome, manganin, constantan and carbon, are typical resistors.,  Substances having very high resistivity:, The substances which have very high, resistivities do not allow electricity to flow, through them. The substances which do not, allow electricity to pass through them are, called insulators. For example, rubber,, plastics, dry wood, etc. are insulators., , V1 = IR1, V2 = IR2 and V3 = IR3 ...(ii), If, V is the potential difference across the, combination of resistances then,, V = V1 + V2 + V3, , If, R is the equivalent resistance of the circuit,, then, V = IR, ...(iv), Using Eqs. (i) to (iv) we can write,, IR = V = V1 + V2 + V3, = IR1 + IR2 + IR3, , I, , R1, , R2, , R3, , I, , –, Key + Battery, , When a series combination of resistances is, connected to a battery, the same current (I) flows, through each of them., , , Law of combination of resistances in series, : The law of combination of resistances in, series states that when a number of, resistances are connected in series, their, equivalent resistance is equal to the sum of, the individual resistances. Thus, if R1, R2, R3, ..., etc. are combined in series, then the, equivalent resistance (R) is given by,, R = R1 + R2 + R3 + ..., , , , ....(i), , or,, , IR = I (R1 + R2 + R3), , or,, , R = R1 + R2 + R3, , Therefore, when resistances are combined in, series, the equivalent resistance is higher than, each individual resistance., , COMBINATION OF RESISTANCES,  Series Combination, When two or more resistances are joined end-toend so that the same current flows through each, of them, they are said to be connected in series., , , , Some results about series combination :, , (i) When two or more resistors are connected in, series, the total resistance of the combination, is equal to the sum of all the individual, resistances., (ii) When two or more resistors are connected in, series, the same current flows through each, resistor., (iii)When a number of resistors are connected in, series, the voltage across the combination, (i.e. voltage of the battery in the circuit), is, equal to the sum of the voltage drop, (or potential difference) across each, individual resistor., ,  Parallel Combination, When two or more resistances are connected, between two common points so that the same, potential difference is applied across each of, them, they are said to be connected is parallel., , Derivation of mathematical expression of, resistances in series combination : Let, R1,, R2 and R3 be the resistances connected in, series, I be the current flowing through the, circuit, i.e., passing through each resistance,, and V1, V2 and V3 be the potential difference, across R1, R2 and R3, respectively. Then,, from Ohm’s law,, , ...(iii), , R1, I, , I1, I2, , R2, , I, , When such a combination of resistance is, connected to a battery, all the resistances have, the same potential difference across their ends.

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, , , , Q = I × t = 10 × 10–6A × 1s = 10 × 10–6C, We known,, Charge on an electron = 1.6 × 10–19C, So, No. of electrons striking the TV screen, , Derivation of mathematical expression of, parallel combination :, Let, V be the potential difference across the, two common points A and B. Then, from, Ohm’s law, Current passing through R1,I1 = V/R1 ...(i), Current passing through R2,I2 = V/R2 ...(ii), Current passing through R3,I3 = V/R3 ...(iii), If R is the equivalent resistance, then from, Ohm’s law, the total current flowing through, the circuit is given by,, I = V/R, ...(iv), and, I = I1 + I2 + I3, ...(v), Substituting the values of I,I1,I2 and I3 in Eq. (v),, V, V, V, V, =, +, +, R R1, R2, R3, , per second =, , (b) Charge striking the screen per min, = (6.25 × 1014 × 60) × 1.6 × 10–19C, = 6.0 × 10–3C, Ex.2, , Sol., , 1, 1, 1, 1, =, +, +, R R1, R2, R3, , The equivalent resistance of a parallel, combination of resistance is less than each of, all the individual resistances.,  Important, results, about, parallel, combination :, (i) Total current through the circuit is equal to, the sum of the currents flowing through it., (ii) In a parallel combination of resistors the, voltage (or potential difference) across each, resistor is the same and is equal to the applied, voltage i.e. v1 = v2 = v3 = v :, (iii) Current flowing through each resistor is, inversely proportional to its resistances, thus, higher the resistance of a resistors, lower will, be the current flowing through it., , , , Sol., , A TV set shoots out a beam of electrons. The, beam current is 10A., (a) How many electrons strike the TV screen, in each second ?, (b) How much charge strikes the screen in a, minute?, Beam current, I = 10 µA = 10 ×10–6A, Time, t = 1 s, So,, (a) Charge flowing per second,, , A current of 10A exists in a conductor., Assuming that this current is entirely due to, the flow of electrons (a) find the number of, electrons crossing the area of cross section, per second, (b) if such a current is maintained, for one hour, find the net flow of charge., Current,, I = 10 A, Charge flowing through the circuit, in one second, Q = 10 C (, , Ch arge, = Current), Time, , (a) We know, Charge on an electron, = 1.6 × 10–19C, So, No. of electrons crossing per second, =, , 10 C, 1.6 10 19, , = 6.25 × 1019, , (b) Net flow of charge in one hour, = Current × Time, = 10 A × 1 h, 10 A × (1 × 60 × 60 s) = 36000 C, Ex.3, , Sol., ,  SOLVED EXAMPLES , Ex.1, , 1.6 10 19 C, , = 6.25 × 1014, , ...(vi), , Cancelling common V term, one gets, , 10  10 -6 C, , Ex.4, , Sol., , A current of 5.0 A flows through a circuit for, 15 min. Calculate the amount of electric, charge that flows through the circuit during, this time., Given :, Current, I = 5.0 A, Time, t = 15 min. = 15 × 60 s = 900 s, Then, Charge that flows through the circuit,, Q = Current × Time, = 5.0A × 900 s, = 4500 A.s = 4500 C, A piece of wire is redrawn by pulling it until, its length is doubled. Compare the new, resistance with the original value., Volume of the material of wire remains same., So, when length is doubled, its area of crosssection will get halved. So, if l and a are the

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original length and area of cross-section of, wire,, Original value of the resistance, R = ×, , Ex.5, Sol., , , a, , and,, New value of the resistance,, 2, , R’ = ×, =  × 4 = 4R, a/2, a, Calculate the resistance of 100 m long copper, wire. The diameter of the wire is 1 mm., Using the relationship,, R=×, We have,, , r=, R=, , Sol., , 6, I, , Sol., , Ex.7, , Sol., , , , = × 2, a, r, , 1, mm = 0.5 × 10–3 m, 2, , 1.6  10 6 ohm.cm  100m, 3.141  (0.5 10 3 m) 2, , I, , 20 V, + –, , If four resistances each of values 1 ohm are, connected in series. Calculate equivalent, resistance., , Total circuit resistance = 10 , , In series,, R1 = R2 = R3 = R4 = 1ohm, putting values, we get,, Rs = 1 + 1 + 1 + 1 = 4, , Circuit current,, , Putting values, we get,, , Rs =, , From Ohm’s law,, , Rs =, , V, RS, , I = 2 ampere or (2A), , Putting values, we get,, , V1 = 2 × 6 = 12 volts, , Potential difference across 6 resistance = 12 V, , Suppose a 6-volt battery is connected across a, lamp whose resistance is 20 ohm the current, in the circuit is 0.25 A, calculate the value of, the resistance from the resistor which must be, used., Lamp resistance,, R = 20 ohm, Extra resistance from resistor,, R=?, (to be calculated), For R and R’ in series,, Total circuit resistance,, Rs = R + R’, From relation, (Ohm’s law), , Ex.9, , A resistance of 6 ohms is connected in series, with another resistance of 4 ohms. A potential, , 10 V, , 6V, , 5, , R, V volt, + –, , I, , (i) What is the current through the 5 ohm, resistance ?, (ii) What is the current through R ?, (iii) What is the value of R ?, , 6, Rs =, 0.25, , = 24 ohm, Rs = R + R’, R’ = Rs – R, = 24 – 20 = 4 ohm, Extra resistance from resistor,, R’ = 4 ohm., , Two resistances are connected in series as, shown in the diagram., , I, , V, I, , But, Hence, , Ex.8, , 4, , We use proper symbols for electrical, components.Resistances are shown connected, in series, with 20 V battery across its positive, and negative terminals. Direction of current, flow is also shows from positive terminal of, the battery towards its negative terminal., Potential difference, V = 20 V, Potential difference across 6 , V1 = ? (to be calculated), , R = 2.04 ohm, Ex.6, , difference of 20 volts is applied across the, combination. Calculate the current through, the circuit and potential difference across the, 6 ohm resistance., For better understanding we must drawn a, proper circuit diagram. It is shown in fig., , (iv) What is the value of V ?, Sol., , First resistance,, , R1 = 5 , , (i) Current through 5 ohm resistance, I = ?, (ii) Current through R,, I=?, (iii) Value of second resistance,, R=?, (iv) Potential difference applied by the, battery,

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V=?, , (c) From relation, , V, (i) From Ohm’s law, R =, I, , We have,, , I=, , 1, 1, 1, 1, 6  3 1, 10, = +, +, =, =, Rp, 5 10 30, 30, 30, , V V1, =, R R1, , Rp = 3 ohm., , 10, I = = 2 ampere, 5, , Ex.11 Resistors R1 = 10 ohms, R2 = 40 ohms,, R3 = 30 ohms, R4 = 20 ohms, R5 = 60 ohms, , Current through 5 resistance = 2 ampere (2A)., (ii), Since R is in series with 5  same, current will flow through it,, Current through R = 2 A., (iii) From Ohm’s law,, , R=, , V, I, , and a 12 volt battery is connected as shown., Calculate :, (a) the total resistance and (b) the total current, flowing in the circuit., Sol., , The situation is shown in (figure)., , V2, I, , R1, , 6, R2 = = 3 ohms, 2, , R2, , R2 =, , R3, R4, R5, , Resistance R has value = 3 ohms., (iv) From relation, V = V1 + V2, , 12 V, + –, , V = 10 + 6 = 16 volts, V = 16 volts, , For R1 and R2 in parallel, , Ex.10 Resistors R1, R2 and R3 having values 5,, 10, and 30respectively are connected in, parallel across a battery of 12 volt. Calculate, (a) the current through each resistor (b) the, total current in the circuit and (c) the total, circuit resistance., Sol., Here,, R1 = 5R2 = 10, R3 = 30 , V = 12 V, (a) I1 = ?, I2 = ?, I3 = ?, (b) I = I1 + I2 + I3 = ?, (c) Rp = ?, (a) From relation, (Ohm’s law),, , Putting values, we get,, , , , , , (b) Total current,, , 1, 1, 1, 1, =, +, +, R1, R2, R3, Rp, , R=, , V, I, , V, I=, R, V 12, I1 =, = =2.4 A, R1 5, , 1, 1, 1, 1, 4 1, 5 1, 1, =, +, =, +, =, =, =, R P1, R1, R2, 10 40, 40, 40 8, , or, , R P1 = 8 ohm, , For R3, R4 and R5 is parallel, 1, 1, 1, 1, 1, 1, 1, =, +, +, =, +, +, R4, R5, R P2 R 3, 30 20 60, , =, or, , 2  3 1, 6, 1, =, =, 60, 60 10, , R P2 = 10 ohm., , (a) For R P1 and R P2 in series., Total resistance,, , R = R P1 + R P2, , Putting values, we get,, Total resistance,, , R = 8 + 10 = 18, R = 18 ohms. Ans., , (b) From relation, (Ohm’s law) R =, , I2 =, , V 12, = = 1.2 A, R 2 10, , We have,, , I3 =, , V 12, =, = 0.4 A, R 3 30, , Putting values, we get,, , I = I1 23, I = 2.4 + 1.2 + 0.4 = 4 A, , Total current,, , I=, I=, , V, I, V, R, , 12 2, =, = 0.67, 18 3, , I = 0.67 A., , Ans

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Ex.12 In the circuit diagram given below. find, 8, 7.2 , 12 , , R1, I, , Ex.13 Three resistances are connected as shown in, diagram through the resistance 5 ohms, a, current of 1 ampere is flowing :, 10 , , I, , 6V, –, +, , A, , (i) For total resistance, 8  and 12 are connected in parallel., Their equivalent resistance comes in series, with 7.2 resistance as shown in fig., 7.2 , , 4.8 , , R1, , Sol., , 15 , , (i) What is the current through the other two, resistors?, (ii) What is the potential difference (p.d.), across AB and across AC ?, (iii) What is the total resistance., (i) For current in parallel resistors, For same potential difference across two, parallel resistors,, V = I1R1 = I2R2, , 6V, + –, , With 7.2 and 4.8 in series, Rs = 7.2 + 4.8 = 12 , Total circuit resistance = 12 ohms., (ii) For total current, Total circuit resistance,, R = 12 ohm, Potential difference applied, V = 6 V, I=?, From Ohm’s law, , C, , B, , 1 amp, , (i) total resistance of the circuit, (ii) total current flowing in the circuit, (iii) potential difference across R1, Sol., , 5, , V, R=, I, V, I=, R, 6, I = = 0.5, 12, , Total circuit current = 0.5 A, Ans., (iii) For potential difference across R1, R=, , V, I, , V = IR, V1 = IR1, V1 = 0.5 × 7.2, = 3.6 V, Potential difference across, V1 = 3.6 V. Ans, , i.e., , =, , I1 R 2, =, I 2 R1, , Current divides itself in inverse ration of, the resistances., Also total current, I = I1 + I2, I1 R 2, 15 3, =, =, =, I 2 R1, 10 2, , Also,, , I1 + I2 = 1 amp., I1 = 0.6A, I2 = 0.4 A. Ans., , Current is 0.6 A through 10 , (ii) For p.d. across AB, From Ohm’s law,, , R=, , V, , V = IR, I, , V = 1 × 5 = 5V, P.D. across AB, = 5 V. Ans, For parallel combination of 10 and 15, P.D. across BC, V = I1R1 = 0.6 × 10 = 6 V, P.D. across AC = P.D. across AB + P.D. across, BC., = 5 + 6 = 11 V, (iii) For total circuit resistance, For 10  and 15 in parallel, Rp =, , 10  15, 150, =, = 6 , 10  15, 25, , Total resistance = 5 + 6 = 11 , Total circuit resistance = 11 Ans, V 11, , , Also R  I  1 11, , 

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Ex.14 In the diagram shown below (Fig.), the cells, and the ammeter both have negligible, resistance. The resistors are identical. With, the switch K open, the ammeter reads 0.6 A., What will be the ammeter reading when the, switch is closed ?, , Sol., , Sol., , 5, 5, , V, + –, , A, R, , 3  and 2 in series become 5 .Equivalent, circuit is shown in fig., , 4V, + –, , A, , (i) For total resistance, R1 = R2 = 5 are in parallel., , R, R, , Rp =, , Let the cell have potential difference V and, each resistor have resistance R, With key open, Potential difference, = V, Circuit resistance of two parallel, resistors,, R P1, , Circuit current,, , 5 5, 25, =, = 2.5 , 5  5 10, , Circuit resistance = 2.5 ohm, (ii) For circuit current, Potential difference, V = 4 V, Circuit resistance, Rp = 2.5 , Circuit current,, , R, R, =, = , n1, 2, , I = ? (to be calculated), , From Ohm’s law,, , I1 = 0.6 A, , With key closed, Potential difference = V, Circuit resistance of three, resistors,, R P2, , Circuit current,, , parallel, , R, R, =, = , n2, 3, , I2 = ?, , V, RP, , I=, , 4, = 1.6 A, 2.5, , 4V, + –, , I1R P1, R P2, , R, 3, I2 = 0.6 × ×, = 0.9, 2, R, , Ex.15 In the circuit diagram., , I=, , Ex.16 For the circuit shown in the following, diagram what is the value of, , V = I1 R P1 = I2 R P2, , Circuit current with closed key = 0.9 A., , V, I, , Circuit current, = 1.6 A, Ammeter reads circuit current 1.6 A, , For same potential difference V, , I2 =, , R=, , 6, , 3, , 12, , 3, , (i) current through 6 resistor, (ii) potential difference (p.d.) across 12 , Sol., , (i) For current through 6 , Current from 4 V battery flows through first, parallel branch having 6 and 3 in series., , 5, , Current in this branch, 3, , 2, , I=, A, , 4V, + –, , Find (i) total resistance, (ii) current shown by the ammeter A., , 4, 4, =, = 0.44 A, 63 9, , (ii) For p.d. across 12 , Current through second parallel branch

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I=, P.D. across 12 , , V=, , 4, × 12 = 3.2 V., 15, ,  ELECTRIC ENERGY, When a potential difference is applied across, a wire, current starts flowing in it. The free, electrons collide with the positive ions of the, metal and lose energy. Thus energy taken, from the battery is dissipated. The battery, constantly provide energy to continue the, motion of electron and hence electric current, in the circuit. This energy is given to ions of, the metal during collision and thus, temperature of wire rises. Thus, energy taken, from the battery gets transferred in to heat., This energy is called electrical energy. This, effect is also called 'Heating Effect of, Current'., If, R = Resistance of wire, I = Current in wire, V = Potential difference across wire., Flow of charge in 't' time = It., Energy dissipated W = Vq = VIt, , , , , , , V = IR,, , , , , ,  W = VIt = I2Rt =, , 1 watt = 1 joule/sec, 1 HP = 746 watt, unit of electrical energy = watt second,, kilowatt hour, , 4, 4, =, A, 12  3 15, , 1 kilowatt hour (kwh) = 3.6 × 106 Joule, ,  Points to Be Remember, , , Current =, , , , , , , V2, t = Vq, R, , This energy is equal to work done by battery, or heat produced in the wire., , , , , , P, , W, V2,  I 2 R  IV , t, R, , units of power = joule/sec, watt, horse power, , Charge, or, Time, , I=, , Q, t, , The SI unit of current is ampere (A) : 1A = 1 C/s, The current flowing through a circuit is, measured by a device called ammeter., Ammeter is connected in series with the, conductor. The direction of the current is, taken as the direction of the flow of positive, charge., Ohm’s law : At any constant temperature, the, current (I) flowing through a conductor is, directly proportional to the potential (V), applied across it., Mathematically,, I = V/R or V = IR, Resistance : Resistance is the property of a, conductor by virtue of which it opposes the, flow of electricity through it. Resistance is, measured in ohms. Resistance is a scalar, quantity., Resistivity : The resistance offered by a cube, of a substance having side of 1 meter, when, current flows perpendicular to the opposite, faces, is called its resistivity (). The SI unit, ,  ELECTRICAL POWER, The rate of loss of energy in an electrical, circuit is called electrical power. It, is denoted by'P', , Current : The rate of flow of charge (Q), through a conductor is called current ., Current (I) is given by,, , of resistivity is ohm.m., Equivalent resistance : A single resistance, which, can replace a combination of, resistances so that current through the circuit, remains the same is called equivalent, resistance.

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, , Law of combination of resistances in series :, When a number of resistance are connected in, series, their equivalent resistance is equal to, the sum of the individual resistances., If R1, R2, R3, etc. are combined in series, then, the equivalent resistance (R) is given by,, R = R1 + R2 + R3 + ....., The equivalent resistance of a number of, resistances connected in series is higher than, each individual resistance., , , , Law of combination of resistances in, parallel : When a number of resistances are, connected in parallel, the reciprocal of the, equivalent resistance is equal to the sum of, the reciprocals of the individual resistances., If R1, R2, R3, etc. are combined in parallel,, then the equivalent resistance (R) is given by., , 1, 1, 1, 1, =, +, +, R R1, R2, R3, , The equivalent resistance of a number of, resistances connected in parallel is less than, each of all the individual resistances.

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EXERCISE - 1, Q.20, , AVery Short Answer Type Questions, Q.1, , Define current., , Q.2, , Define one ampere., , Q.3, , What is the conventional direction of electric, current? How does it differ from the direction, of flow of electrons?, , Q.4, , What do you mean by elementary charge?, , Q.5, , State Ohm’s law., , Q.6, , Define one ohm., , Q.7, , Write the formula for resistance of a wire of, length l and cross-section A., , Q.8, , Define specific resistance., , Q.9, , Write the unit of specific resistance., , Q.10, , Distinguish, resistivity., , between, , A Piece of wire is redrawn by pulling it until, its length is doubled. Compare the new, resistance with the original value., , CLong Answer Type Questions, , resistance, , Q.21, , Define charge. What do you understand by, positive and negative charge ? Write down, the expression for force between two charges., , Q.22, , State Ohm’s law. How it can be, experimentally?, , Q.23, , Describe the conditions for constituting an, electric current. Explain the mechanism of, flow of electrons in a conductor., , Q.24, , Derive the expression for the equivalent, resistance when two resistors are joined in, series., , Q.25, , Derive the expression for the equivalent, resistance when two resistors are joined in, parallel., , and, , Q.11, , Two resistors R1 and R2 are joined in series., Find the equivalent resistance., , Q.12, , Two resistors R1 and R2 are joined in parallel., Find the equivalent resistance., , BShort Answer Type Questions, Q.13, , Define current. Is it a scalar quantity or a, vector quantity ? What is meant by the, conventional direction of current ?, , Q.14, , Define resistance., , Q.15, , On what factors does the resistance of a, conductor depend?, , Q.16, , Define resistivity. Write the formula for, resistivity., , Q.17, , What is the formula for the combination of, resistances when they are combined in :, (i) series and, (ii) parallel ?, , Q.18, , Why is the series arrangement not used for, domestic circuits?, , Q.19, , How does the resistance of a wire vary with, its cross-sectional area ?, , verified, , DNumerical Problems, Q.26, , If the charge on an electron be 1.6 × 10–19 C,, how many electrons should pass through a, conductor in 1 second to constitute 1ampere, current ?, , Q.27, , How many electrons pass through a lamp in, one minute if the current be 200 mA?, (Charge on an electron, e = 1.6 × 10–10 C)., , Q.28, , A conductor carries a current of 0.2A. Find, the amount of charge that will pass through, the cross-section of the conductor in 30 s., How many electrons will flow in this timeinterval?, (Charge on an electron, e = 1.6 × 10–19 C.), , Q.29, , The potential difference between the two, points of a wire carrying 2 amperes current is, 0.1 volt. Calculate the resistance between, these points.

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Q.30, , A resistance of 12 ohm is connected in, parallel with another resistor X. The resultant, resistance of the combination is 4.8 ohms., What is the resistance X?, , Q.31, , Three resistances 12 ohms each are connected, in parallel. Three such combinations are, connected in series. What is the total, resistance?, , Q.32, , How will you connect three resistors of 3 ,, 4 and 7 respectively so as to obtain a, resultant resistance, of 3.5  ?, , Q.33, , Find the current through the circuit shown in, figure. Also find the potential difference, across the 20-resistor., 20, , Q.36, , 10, , Q.37, , Three resistors of resistances 10 , 20  and, 30  are connected in parallel with a 6-V, cell. Find (a) the current through each, resistor, (b) the current supplied by the cell,, and (c) the equivalent resistance of the circuit., , Q.38, , When two resistors are joined in series, the, equivalent resistance is 90 . When the same, resistors are joined in parallel, the equivalent, resistance is 20. Calculate the resistances of, the two resistors., , Q.39, , Consider the circuit shown in figure., Calculate the current through the 3- resistor., , 15, , Find (a) the equivalent resistance, (b) the, current passing through the cell, and (c) the, current passing through the 30- resistor in, the circuit shown in figure., 15, 30, , 6V, , Q.35, , Find the current supplied by the cell in the, circuit shown in figure., 5, , 10, 10, 5V, , +A –, , 20, , 7V, , Q.34, , Figure shows a part of an electric circuit. The, reading of the ammeter is 3.0 A. Find the, currents through the 10-and 20-resistors., , i 4, , i1 3, i2, 6, 12V, , Q.40, , (a) How will you join three resistors of, resistances 4, 6 and 12 to get an, equivalent resistance of 8 ?, (b) What would be the highest and the lowest, equivalent resistances possible by joining, these resistors ?

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EXERCISE - 2, , , Single Correct Answer Type Questions, , Q.1, , How many electrons in 1 s constitute a current, of 1 A?, (A) 6.25 × 1018, (B) 6.25 × 1012, 11, (C) 6.25 × 10, (D) 6.25, , Q.2, , 1 Coulomb is equal to (A) 1 amp × 1 sec, (B) 1 amp / 1 sec, (C) 1 joule × 1 amp, (D) 1 joule / 1 sec, , Q.3, , Q.4, , Q.5, , When a body is negatively charged by friction,, it means(A) the body has acquired excess of electrons, (B) the body has acquired excess of protons, (C) the body has lost some electrons, (D) the body has lost some neutrons, If a charged body attracts another body, the, charge on the other body(A) must be negative, (B) must be positive, (C) must be zero, (D) may be negative or positive or zero, A suitable unit for expressing the strength of, electric field is (A) V/C, (B) C/m, (C) N/C, (D) C/N, , Figure shows, current in a part of electrical, circuit, then the value of current is-, , 2A, , 1A, , 2A, , iA, , (A) 1.7 A, (C) 13 A, Q.11, , (B) 3.7 A, (D) 1.0 A, , When the temperature of a metallic conductor is, increased its resistance-, , (A) always decrease, (B) always increase, (C) may increase or decrease, (D) remain the same, Q.12, , Specific resistance of wire depends upon-, , (A) its length, (C) its dimensions, Q.13, , (B) its cross-section area, (D) Its material, , The unit of resistivity is-, , (A) ohm, (C) ohm meter–1, , (B) ohm mete, (D) mho metre–1, , A piece of wire of resistance 4 is bent through, 180° at its mid point and the two halves are, twisted together, then resistance is (A) 1 , (B) 2 , (C) 5 , (D) 8 , , Q.16, , (B) e/It, , Three resistance each of 8  are connected to a, triangle. The resistance between any two, terminals will be:, (A) 12 , (B) 2 , , (D) Ite, , (C) 6 , , What constitutes current in a metal wire ?, , (B) Protons, (D) Molecules, , If I is the current through a wire and 'e' is the, charge of electron then the number of electrons, in t seconds will be given by -, , Ie, t, (C) It/e, , Q.10, , Q.15, , Q.7, , (A), , (A) the direction of flow of negative charges, (B) the direction of flow of atoms, (C) the direction of flow of positive charges, (D) the direction of flow of molecules, , A wire of resistance R is cut into n equal parts., These parts are then connected in parallel. The, equivalent resistance of combination will be –, (A) nR, (B) R/n, (C) n/R, (D) R/n2, , One ampere equals (A) 106A, (B) 10–6A, –3, (C) 10 A, (D) 10mA, , Q.8, , Conventionally, the direction of the current is, taken as-, , Q.14, Q.6, , (A) Electrons, (C) Atoms, , Q.9, , (D), , 16, , 3

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Q.17, , In how many parts (equal) a wire of 100  be, cut so that a resistance of 1 , , Q.18, , Q.23, , connecting them in parallel ?, , consumed in operating ten, 50 watt bulbs for 10, hours per day in a month of 30 days ?, , (A) 10, (C) 100, , (A) 15, (C) 1500, , is obtained by, , (B) 5, (D) 50, , Q.24, The filament of an electric bulb is made of, tungsten because(B) it is cheaper, (C) its melting point is high, , Q.25, , (D) its filament is easily made, , Q.20., , (B) 2 , , (C), , 1, , 4, , (D) 4 , , (A) half, (C) four times, Q.26, , between points A and C will be -, , R, 2R, , R, B, 3, R, 2, 2, (C) R, 3, , (A), , Q.21, , Q.27, , D, , C, , Q.28, , Q.29, , A fuse wire is always connected to the-, , Heating effect of a current carrying conductor is, , (A) Loss of kinetic energy of moving atoms, (B) Loss of kinetic energy of moving, electrons, (C) Attraction between electrons and atoms, (D) Repulsion between electrons and atoms, , Rate of heat generated by electric current in a, , (A) IR, , (B) IR2, , (C) I2R, , (D), , IR, , Two heater wires of equal length are first, connected in series and then in parallel with a, battery. The ratio of heat produced in the two, , Q.30, , The correct relation between heat produce (H), , and electric current I flowing is(A) H  I, , cases is-, , (A) 2 : 1, (C) 4 : 1, , (B) earth wire, (D) none of these, , due to-, , resistive circuit is expressed in-, , Q.22, , An electric iron is based upon the principle of-, , (A) neutral wire, (C) live wire, , (B) 6R, (D) 3R, , (B) joule, (D) Maxwell, , (A) magnetic effect of current, (B) heating effect of current, (C) chemical effect of current, (D) none of these, , R, , R, , (B) one-fourth, (D) double, , Laws of heating are given by-, , (A) faraday, (C) Ohm, , In the given circuit, the effective resistance, , A, , The resistance of a conductor is reduced to half, in the conductor will become., , its length, then the resistance will become-, , 1, , 2, , (B) 55, (D) none of these, , its initial value. ln doing so the heating effects, , If a wire of resistance 1  is stretched to double, , (A), , (B) 150, (D) 15000, , An electric iron draws a current of 4A when, connected to a 220 V mains. Its resistance must, be -, , (A) 40, (C) 100, , (A) its resistance is negligible, , Q.19, , How much electrical energy in kilowatt hour is, , (B) 1 : 2, (D) 1 : 4, , (C) H  I 2, , 1, I, 1, (D) H  2, I, , (B) H 

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Q.31, , The wire having a red plastic covering is a :, (A) live wire, (C) earth wire, , Q.37, , (B) neutral wire, (D) none of these, , In electric fittings in a house:, (A) the live wire goes through the switch, (B) the neutral wire goes through the switch, (C) the earth wire goes through the switch, , Q.32, , Q.33, , (A) earth wire, , (B) neutral wire, , (C) live wire, , (D) None of these, , The wire having a black plastic covering is a, (A) live wire, (C) earth wire, , Q.34, , Q.35, , Q.36, , (D) no wire goes through the switch, , A switch, is always connected to the, , (B) neutral wire, (D) none of these, , The wire having a green plastic covering is a, (A) live wire, , (B) neutral wire, , (C) earth wire, , (D) none of these, , In three pin socket (shoe) the bigger hole is, connected to(A) any wire, , (B) live wire, , (C) neutral wire, , (D) earth wire, , Coming of live wire and neutral wire in direct, contact causes:, (A) short-circuiting, (C) no damage, , (B) over loading, (D) unknown effect, , Q.38, , High power electrical appliances are earthed, to (A) avoid shock, (B) avoid wastage, (C) Make the appliance look beautiful, (D) reduce the bill

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ANSWER KEY, EXERCISE - 1, D. NUMERICAL PROBLEMS :, 26. 6.25 × 1018, 27. 7.5 × 1019, 28. 6C, 3.75 × 1019, 29. 0.05, 30. 8, 31. 12 , 32. 3 & 4 in series corrected, in parallel with 7., 33. 0.2A, 4V, 34. (A) 10 (B) 0.6A (C) 0.2A, 35. 0.5A, 36. 2A, 1A, 37. (A) 0.6 A, 0.3 A, 0.2 A (B) 1.1A (C)  5.5 , , , , 39. 1.33 A, 40. (b) 22 2 , , EXERCISE - 2, Ques, Ans, Ques, Ans, Ques, Ans, , 1, A, 16, D, 31, A, , 2, A, 17, A, 32, C, , 3, A, 18, C, 33, B, , 4, D, 19, D, 34, C, , 5, C, 20, C, 35, D, , 6, A, 21, C, 36, A, , 7, A, 22, D, 37, A, , 8, C, 23, B, 38, A, , 9, C, 24, B, , 10, A, 25, D, , 11, B, 26, B, , 12, D, 27, B, , 13, B, 28, C, , 14, D, 29, B, , 15, A, 30, C