Title 18. Current Eleciticity Easy 2 Ans.pdf ✅

Resistance of the part AC RAC = 0.1  40 = 4 and RCB = 0.1  60 = 6 In balanced condition =  = 4 6 4 6 X X Equivalent resistance Req = 5 so current drawn from battery i 1A 5 5 = = . 25. (a) (R + G)ig = V  g i V (R + G) = =    = − 6.25k 30 16 10 3 6  Value of R is nearly equal to 6k This is connected in series in a voltmeter. 26. (d) R1 = 80  200 =16000  =16 k Current flowing through V1 = Current flowing through V2 = A 3 3 5 10 16 10 80 − =   . So, potential differences across V2 is 5 10 32 10 160 volt 3 3 2 =    = − V Hence, line voltage V = V1 + V2 = 80 + 160 = 240 V . 27. (d) V = xl  iR = xl 50 10 0.1 10 2 10 10 2 2 3   =            = − − − i i 10 10 A 10 mA 3  =  = − . 28. (c) S G i i g = 1 + 8 . 12 1 2 5  = +  S =  S (In parallel). 29. (d) G S S i ig + =  G S S + = 100 5  19 G S = 30. (a) = − =  − =  3 5 R G(n 1) 50 10 (3 1) 10 . 31. (c) 1 3 58 29 58 29 1 2 1 2 2 1 = − + = − + = l l l l E E 32. (a) − =   = − = − 1 999 10 10 10 3 G i V R g . 33. (d) For conversion of galvanometer (of resistances) into voltmeter, a resistance R is connected in series.  R G V ig + = 1 and R G V ig + = 2 2  R G V R G V + = + 2 1 2  ( ) 2 2( ) 1 2 R G R G G R G R G V V + + − = + + = ( ) 2 R G G + = −  ( ) 2 1 2 1 R G V G V V + = −  V2  2V1 34. (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. 35. (c) . .10%. 10 1 36 4 4 i e G S S i ig = + = + = 36. (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. 37. (c) Potential gradient L R R R r e x h . ( + + ) =   =  + +  = − − 57 1 3 (3 0) 2 10 10 2 3 h h R R . 38. (c) S G i i g = 1 +  = +  =   − 0.02 999 20 20 1 10 1 3 S S . 39. (a) Resistance of voltmeter should be high. 40. (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. 41. (c) Potential gradient L R R R r e x h . ( + + ) = 0.2 10 20 (20 10 0) 3  = + + = 42. (d) 1 2 (6 2) (6 2) 1 2 1 2 2 1 = − + = − + = l l l l E E 43. (c) Manganin or constantan are used for making the potentiometer wire. 44. (a) G ig R V1 R1 = 16k R2 = 32k V2 V V1 = 80V
45. (a) S G i i g = 1 + S S G V i G g 40 1 800 10 100 10 40 1 . 3 3 = +     = +  − −  S =10 . 46. (a) G S S i i g + =  3 3 100 10 100 10 10 − −   +  = S S 90 S = 1000  = = 11.11  90 1000 S . 47. (c) Before connecting the voltmeter, potential difference across 100 resistance Vi V V 11 10 (100 10) 100  = + = Finally after connecting voltmeter across 100  Equivalent resistance =  +  90 (100 900 ) 100 900 Final potential difference Vf V V 10 9 (90 10) 90  = + = % error = 100 − i i f V V V 100 1.0. 11 10 10 9 11 10  = − = V V V 48. (b) Potential gradient = R r L e R ( ). . + = (3 3) 5 10 3 +   . = 1V/ m = 10 mV / cm. 49. (c) S S G i i g 100 1 10 1 1 5 = +  = + −   =  −3 5 10 10 100 S . 50. (d) 10 1 40 4 36 4 4 = = + = + = G S S i ig 51. (a) R R R V i + = + +  =  = 2 6 6 3 6 3 6 2  R = 1  . 52. (b) G S S i i g + =  + S = 50 5 10 0.01  = = 0.05 999 50 S . 53. (d) R l l S . 100       − = Initially, l cm l l 10 25 100 30    =      − = Finally, l cm l l 30 75 100 10    =      − = So, shift = 50cm. 54. (c) Potential gradient (x) A i = 10 V/m 10 0.1 10 2 6 7 − − − =  = 55. (d) Before connecting voltmeter potential difference across 400 resistance is Vi 6 2V (400 800 ) 400  = + = After connecting voltmeter equivalent resistance between A and B =  +  = 384 .6 (400 10,000 ) 400 10,000 Hence, potential difference measured by voltmeter Vf 6 1.95V (384 .6 800 ) 384 .6  = + = Error in measurement = − = 2 −1.95 Vi Vf = 0.05V. 56. (c) S S G i i g 50 1 0.05 5 = 1 +  = +  A l S  = =  99 50  l 3m 5 10 2.97 10 10 99 50 7 2 4 =    =  − − − . 57. (a) S G i i g = 1 +  S 0.81 1 1 10 = +  S = 0.09 . 58. (a) From the principle of potentiometer V  l  L l E V = ; where V = emf of battery, E = emf of standard cell, L = Length of potentiometer wire 100 30 E L El V = = . 59. (b) l L R R R r e E h  + + = . ( ) 0.4 1 10 (10 0) 2 10 10 3   + +   = − R  R = 790 60. (b) Using         = −1 2 1 l l r R  =       = −1 1 100 150 2 61. (d) Resistance between A and B 3 1000 (1500 ) 1000 500 =  = So, equivalent resistance of the circuit Req 3 2500 3 1000 = 500 + = V 10 100 Vi V 100 900 Vf 10 6V 10,000 A 400 B 800 V V 500 500 A B C 1000 10V
 Current drawn from the cell (2500 / 3) 10 i = A 250 3 = Reading of voltmeter i.e. potential difference across AB 4 V 3 1000 250 3 =  = 62. (d) 10 i ig =  Required shunt =  − = − = 10 (10 1) 90 (n 1) G S 63. (b) − =   = − 40 4960 10 10 50 g 3 i 64. (c) Post office box is based on the principle of Wheatstone's bridge 65. (d) Full deflection current 4 25 4 10 − ig =   A 4 100 10 − =  Using G I V R g = − 50 100 10 25 4 −  = − = 2450  in series. 66. (a) In balancing condition, 1 1 2 1 2 1 100 l l l l R R − = =  4 1 80 20 = = Y X .....(i) and l l Y X − = 100 4 .....(ii)  l l − = 4 100 4  l = 50 cm 67. (c) 68. (d) G i i i S g g          − = 50 (10 10 100 10 ) 100 10 3 6 6   −   = − − −  0.5  (in parallel) 69. (d) l L R R R r e E h  + + = . ( )   l + + = 10 5 (5 45 0) 5 0.4  l = 8 m 70. (a) Potential difference per unit length V m L V 0.5 / 4 2 = = = 71. (a) 72. (b)         = − 1 2 1 l l r R       = − 1 120 240 2 = 2 73. (d) l V E = ; E is constant (volt. gradient).  2 2 1 1 l V l V =  140 180 1.1 V =  V 1.41 V 140 180 1.1 =  = 74. (a) IG G = (I − IG )S  I = G I S G       1 +  I = 100.1 mA 75. (c) Let S be larger and R be smaller resistance connected in two gaps of meter bridge.  R R R l l S 4 20 100 100 20 = −  =      − = .....(i) When 15 resistance is added to resistance R, then ( 15) 4 6 ( 15) 40 100 40  + = +      − S = R R .... (ii) From equations (i) and (ii) R = 9 76. (a) According to following figure Reading of voltmeter = Potential difference between A and B = i (R + 2) 12 = 0.1 (R + 2)  R = 118 . 77. (a) Potential gradient L R R R r e x h . ( + + ) =  ( 490 0) 1 2 10 0.2 10 2 3 R R  + + =  − −  R = 4.9 . 1. (a) Initially : Resistance of given cable 3 2 (9 10 ) −   =   l R ... (i) Finally : Resistance of each insulated copper wire is 3 2 (3 10 ) ' −   =   l R . Hence equivalent resistance of cable           = =  −3 2 6 (3 10 ) 1 6 '   R l Req ....(ii) On solving equation (i) and (ii) we get Req = 7.5  2. (a) 4         = A B B A r r R R  16 1 2 1 4  =      = B A R R  RB = 16 RA V A 2 A B i=0.1A l  9 mm l 