Page 1 : wc ote oe, , Important Formulae :, , 1. [= go pee, q, 3. Ohm’s Law, V = IR 4.. R= a, , Notations Used :, Tis current in A, V is voltage in V, Wis work done in J, q or Q is charge, in C, R is resistance in Q, p, is resistivity in Q, 1 is length in m and a is area of cross-section in m2,, , , , Scanned with CamScanner

Page 2 : Numerical Application A current of 10 A flows through a circuit for 900 s, Calculat,, e the, , amount of electric charge that flows., Using, Charge = Current * Time, , Sol., ie., Qelret, wemel Q =10 A x 900 s = 9000 A.s = 9000 C i, : Ng,, Numerical Application Calculate the amount of charge that flows through a conductor ws, a current of 5 A flows through it for 2 min. (C.B.S.E, 2015, , Sol. Given, t = 2 min = 2 x 60s = 120s, Charge, q =I* ft =5x 120 =600C, , Thus, the amount of charge flowing through the conductor is 600 C. haw, , An amount of 100 J of work is done in transferring a charge of, , EEL Numerical Application, he other. Find the e.m.f. of the source., , 5 C from one terminal of a source to t!, , Work done, Sol. Using, e.m.f. = Charge, Ww _ 100J aA., ie, Ve= a =SC =20JC™ =20V Ans., , Numerical Application What is the resistance of a lamp, if it uses 10 A when connected, , to a 220 volt supply ?, , , , Sol. Using, V =I R, we get, Vi 220V, R=7 > 0A = 220 Ans., , Numerical Application The values of current in a given resistor for the corresponding, values of potential difference across the resistor are listed below :, , I (ampere) 1-0 2:0 | 3-0] 4-0, V (volt) 3 6 9 12, , , , , , , , , , , , , , , , , , , , Calculate the constant of Ohm’s law graphically., , Sol. The graph between V and | is plotted as shown in figure. The plot is a straight line passing through, , the origin (0, 0)., , , , Scanned with CamScanner

Page 3 : AV (9-3), , h,, the geP =_ew =, from Slope AI ~ @=1), , , , ie, Constant of Ohm’s Law = 2A" 3o0hm ‘awe:, , , , ce numerical Application 1] Calculate the least count of the given ammeter and voltmeter., 6 (C.B.S.E. 2015, 14), , , , Units 10A, , Least count of ammeter “}j>. of divisions 5 ~ 2A., , ____Units___100V, Least count of voltmeter “}jo. of divisions. 10 ~ 10V, , te the resistance of a resistor on the basis of readings, , Eg Numerical Application Calcula, (C.B.S.E. 2014), , , , , , , , shown., , 1, , 0 3 : p00, , 00 30 ¢, 40 46 S00 30, 0 50 0 40, A Vv, 10A, Sol Least count of ammeter = =) =2A, , Reading of ammeter or current, I = Number of div. x least count, = (6 +4) x 2A = 18A., , 100V, Least count of voltmeter = —79— =10V, Reading of voltmeter or potential difference, v = Number of div. * least count, , = (10 + 8) x 10V = 180 V., Resistance, R =?, We know, V = IxR, , _V_180V, , -¥ 18 Lia., or R=77 Gq 7, , , , f the pointers of a milliammeter and voltmeter, these in his experiment the reading of, , Numer, , . i eae, , Not i, cat Application [NA13] The rest positions 0, t value of current and voltage in, , + 1 use, , Point are as shown in Fig, A. When a student uses, , this ga’, 2% in positions shown in Fig, B. Calculate the correc, *Periment,, , Re, , Scanned with CamScanner

Page 4 : (CEsp ;, , N, mA 30 440mA, [", Oo A, lO oO, , Fig. A Eig. B, 10, , i = least =, - 2 mA and least count of voltmeter = =~ 02y, , Sol. Least count of milliammeter ;, , From figure (A),, Zero error of milliameter = -2 x 2 mA = - 4mA and, Zero error of voltmeter = 3 x 02 V=06V, Corrected reading of milliammeter = 38 + 4 = 42 mA, Corrected reading of voltmeter = 36-06=3V, , , , Numerical Application Electric current I and potential difference V are shown in the, , following figures across a resistor, (i) What are the reading of voltmeter and ammeter, , , , (ii) Calculate resistance (C.B.S.E, 2016), 3, 200 2 ttt. 4, Mga ee i,, 100 i, 1g “sy, = = ¢ =, oz be 2400 gE . £6, Ans. (i) -- Reading of voltmeter = 141 V and reading of ammeter = 200 mA, V_11V 11V 1, i) j R=—= =—_§_— = —9=552 Ans., (ii) Resistance, T 200mA 200x10°A 02, , Numerical Application The potential difference between two terminals of an dle, iron is 220 V and the current flowing through its element is 5A. Calculate the resistance am, wattage of the electric iron. (CBSE: 2016), , Sol. Resistance, R= . = a, Al a, ‘SO, power P = Vx I=220x5=1100W, , . : . °, ea — Application Find the resistance of 1 metre of copper wire that has @ a, ; sectional area of about 2 x 10-2 cm? and resistivity of 1-7 x 10° Q m?, ‘ol. Area of cross-section of wire, a = 2 x 10-2 cm?, = 2x 10% m?, , / 1 17x10 ohmm x 1m, Using, R = p—, we get R= — 77,2, , a 2x10° m Ans, , = 85 x 10-3 ohm, , ath, , Scanned with CamScanner

Page 5 : 189, , , , merical Application (Critical Q) How is the resistan., , Sa at ‘ttn 4, gs Nuh is doubled, (i) its radius is doubled ? Fee oe, enf, , (C.B.S.E. 2016), , , , , , sol. pl, () Using, a Ar We get, R plx2, A, ie R' =2R, ie, resistance will be doubled, if length of the wire is doubled., ft, (i) STA, pl, x Ray, Le. mr, pal goth h®, n(2ry mrt 4 4, , Resistance will decrease by four times. Ans., , zg Numerical Application (NA18] A copper wire of resistivity 1-63 x 10-8 Q-m has cross-section, "area of 10:3 x 10-* cm?. Calculate the length of the wire required to make a 20 Q coil., (C.B.S.E. 2016), , Sol. Here, A = 10:3 x 10+ cm? = 10:3 x 10 x 10-4 m2, Using, RE! 1 RA _20x103%104 «104 m?, sing, S"a We get p 163x105 Q-m, , _ 20x103x10%, , _ Texto =12638m Ans., , 5 Numerical Application A metal wire has diameter of 0-25 mm and electrical resistivity, of 08 x 10°8 O-m., , ( What will be the length of this wire to make a resistance of 5 Q ?, (#) How much will the resistance change, if the diameter of the wire is doubled? (C.B.S.E. 2016), , a 2, 5Qx nx] «10° |, , Ans. () Using, R=) we get, RAL, A p 08x10°, Ax: -6, ac = 306 m, , 08x10", (i) Using, R=?! pe Pe fl, ) ing, R=, we get, R= Toy = TX pF 0, , 4, 1 ol, , and palle p 5 ie, R *Tpe w+-(2), , Scanned with CamScanner