EASY QUESTIONS
1. If resistance of each wire in the network shown is r, the equivalent resistance between A & C is equal to
(a) r (b) 2 r (c) 3 2r (d) 2 3r C A E D B
Sol.: For point A & C, loop BCD shorted
Hence RAC = r r r r 3 2 3 2 (c) E
2. In the circuit shown each capacitor has capacitance C. The emf of the battery is and the Sw is closed. The total heat generated in the wire once the switch Sw is
opened is (a) C2 (b) 6 2 C (c) 12 2 C
(d) No heat will be dissipated
Sw
C C C
Sol.: As the charge distribution remains same on opening the switch, no charge will flow in the
circuit. So heat dissipated is zero. (d)
E
3. In the circuit shown in figure, equivalent resistance between A and B is (a) 8 (b) 15 (c) 2 3 (d) 2 B 4 2 1 A 2 2 4
Sol.: Equivalent circuit diagram of the circuit is
B 2 A 2 4 1 2 4 A B 6 4 3 2
So 2 3 eq R (c) E
4. The resistance of hexagon circuit between A and B represented in figure is (a) r (b) 0.5 r (c) 2r (d) 3r r r r r r r r r r r A B
Sol.: From figure (i) it is evident that the potential difference between points a, b and c is zero.
The equivalent circuit is as shown in figure (ii). r r r r r r rgf de 2 2 2 2 2 2 2 2 2 r r r r r rAB (b) r r r r r r r r r r A B a b c d g f e g f r r r a r b d e r r r r b c b A B E
5. In the given circuit, each resistor has resistance R. The equivalent resistance between A and B is
(a) 4 R (b) 4R (c) 4 3R (d) 3 4R A B Sol.: (a) E
6. A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be (Assuming potential difference is same in both cases)
(a) one fourth (b) halved (c) doubled (d) four times
Sol.:
R H 1
R becomes half so heat generate will be doubled.
(c)
E
7. In the circuit shown the potential difference between points C and B will be
(a) (8/9) volt (b) (4/3) volt (c) (2/3) volt (d) 4 volt 5 B 5 5 5 5 5 C A D + – 2V
Sol.: 15 2 e R V
I A (I is current in each branch)
V 3 4 B C V V (b) E
8. The current through 2 resistor is (a) zero (b) 1 amp (c) 2 amp (d) 4 amp 2 5 10 10V 20V Sol.: (a) E
9. The equivalent resistance between points A and B in the circuit shown is
(a) 4 (b) 6 (c) 10 (d) 8 4 A B 8 8 4 6 4 8 E
10. There are n similar resistors each of resistance R. The equivalent resistance comes out to be x
when connected in parallel. If they are connected in series, the resistance comes out to be
(a) x /n2 (b) n2x (c) x/n (d) nx Sol.: In parallel x 1 = R n
and series Reff = nR = n2x (b)
E
11. In the balanced wheatstone bridge circuit as shown in the figure, when the key is pressed, what will be the change in the reading of the galvanometer? (a) no change (b) increased (c) decreased (d) zero G R R R R
Sol.: Under balanced condition S R Q P
Here resistances are in same proportion
Hence, there will not be any deflection in galvanometer on pressing the key. It remain same.
E
12. In the circuit shown in figure, the reading of voltmeter will be (a) 0.8 V (b) 1.33 V (c) 1.6 V (d) 2.00 V 20 V 80 80 2V, r = 0 Sol.: (b) E
13. In the circuit shown in figure
(a) current in wire AF is 1A (b) current in wire CD is 1A (c) current in wire BE is 2A (d) none of the above
4 4 4 2V A B C D F 2V E 2V Sol.: By KVL in loop 1 24i8i20 i = 0 (d) 4 4 4 2V E 2V 2V i i 2i 1 E
14. A battery of internal resistance 4 is connected to the
network of resistance as shown. In order to give the maximum power to the network, the value of R should be (a) 9 4 (b) 9 8 (c) 2 (d) 18 4 R R R 6R R R 4R E
Sol.: Given circuit is balance wheat stone bridge hence no current will flow through 6 resistance.
So equivalent resistance will be 2R. For maximum power 2R = 4 R = 2
(c)
E
15. A cell of emf E is connected across a resistance R. The potential difference between the terminals of the cell is found to be V. The internal resistance of the cell must be
(a) R V V E ) ( 2 (b) E R V E ) ( 2 (c) V R V E ) ( (d) (E – V) R
Sol.: As V = E – I.r & I = r R E r = V R V E ) ( (c) E
16. The resistance across AB is (a) 8 5 R (b) 8 7 R (c) 1 R (d) 2 R R R R R R D C A B
Sol.: The circuit can be rearranged
Now 2R and R are parallel
R R R RAB 1 3 2 1 1 = R R 1 5 3 RAB = 8 5 R (a) 2 R R A R B R C E
17. The equivalent resistance of the network shown in the figure between the base terminals is
(a) 3 (b) 3 2 1 (c) 3 2 2 (d) 2 1 1 1 1 1 Sol.: Req = 3 8 1 3 2 1 1 2 1 2 1 1 (c) E
18. n identical cells, each of emf and internal resistance r, are joined in series to form a closed circuit as shown. The potential difference across any one cell is
(a) zero (b) (c) n (d) n n 1
r
r
r Sol.: r nr n I , V = Ir = 0 (a)E
19. In the given circuit it is observed that the current I is independent of the value of the resistance R6.
Then the resistance value must satisfy (a) R1R2R5 = R3R4R6 (b) ) ( 1 ) ( 1 1 1 4 3 2 1 6 5 R R R R R R (c) R1R4 = R2R3 (d) R1R3 = R2R4 = R5R6 R2 R1 R5 R3 R4 R6 I
Sol.: This is condition for balance wheatstone bridge 3 2 4 1 4 3 2 1 R R R R R R R R (c) E
20. The resistances 500 and 1000 are connected in series with a battery of 1.5 volt. The voltage across the 1000 resistance is measured by a voltmeter having a resistance of 1000 . The reading in the voltmeter would be
(a) 1.5 volt (b) 1.0 volt
(c) 0.75 volt (d) 0.5 volt Sol.: (c) 500 V 1000 1.5 V 1000 E
21. A set of n identical resistors, each of resistance R ohm when connected in series has an
effective resistance of x ohm. When the resistors are connected in parallel, the effective resistance is y ohm. What is the relation between R, x and y?
(a) R =
) (x y
xy
(b) R = (y – x) (c) R = xy (d) R = (x + y)
Sol.: For series connection x = nR.
For parallel connection y = n R . Therefore xy = nR × n R = R2. (c)
E
22. In the circuit shown in figure, the current
through
(a) the 3 resistor is 0.50 A (b) the 3 resistor is 0.25 A (c) the 4 resistor is 0.50 A (d)the 4 resistor is 0.25 A 3 A 2 C 2 8 8 4 9V 2 B 2 D 2
Sol.: The equivalent resistance between points A and B to the right of AB is 4 . Therefore, total
resistance = 3 + 4 + 2 = 9 . Current I = 9 V/9 = 1 A. This current is equally divided in the 8 resistor between A and B and the remainder 8 resistor. Hence current in AC = 0.5 A. This current is equally divided between the 8 resistor in CD and the circuit to the right of CD. Therefore, current in the 4 resistor = 0.25 A.
(d)
E
23. In the arrangement of resistances shown in the figure, the
potential difference between the points B and D will be zero when the unknown resistance X is
(a) 4 (b) 2 (c) 3
(d) e.m.f. of the cell is needed to find out X
A C B D 3 1 12 4 X 1 1 Sol.: 2 / 1 4 16 X , X = 2 (b) E
24. The current I drawn from the 5 V source will be
(a) 0.33 A (b) 0.5 A (c) 0.67 A (d) 0.17 A 5V I 10 20 10 5 10 Sol.: 10 5 20 10
. So it is a balance wheat stone bridge.
10 45 15 30 e R , 2 1 10 5 I A (b) E
25. Five cells, each of e.m.f. E and internal resistance r are connected in series. If due to over
sight, one cell is connected wrongly, then the equivalent e.m.f. and internal resistance of the combination, is
Sol.: EMF = (41)E3E. Internal resistance = 5r
(c) E
26. Five equal resistors, each equal to R are connected as shown in the
following figure; then the equivalent resistance between points A and B is:
A B
(a) R (b) 5R (c) R/5 (d) 2R/3
Sol.: It is a case of wheat stone bridge. (a)
E
27. A wire has resistance 12 is bent in the form of a circle. The effective resistance between the two points on any diameter of the circle is
(a) 12 (b) 24 (c) 6 (d) 3 Sol.: Rtotal = 12 RAB upper = 6 = RAB lower Combination 2 1 1 1 1 R R Reff 6 1 6 1 1 eff R Reff 3 (d) A B E
28. When cells are connected in series
(a) the emf increases (b) the potential difference decreases (c) the current capacity increases (d) the current capacity decreases
Sol.: (a) E
29. Which of the following has the maximum resistance?
(a) voltmeter (b) milivoltmeter (c) ammeter (d) miliammeter
Sol.: (a) E
30. A conductor with rectangular cross-section has dimensions (a 2a 4a) as shown in figure. Resistance across AB is x, across CD is y and across EF is z. Then (a) x yz (b) x yz (c) yzx (d) xz y C 4a A B 2a F E D a Sol.: A l R ,
a a a a x 2 2 4 ,
a a a a y 8 2 4 then xzy
a a a a z 2 4 2 (d) E31. A wire l = 8m long of uniform cross-sectional area A = 8 mm2, has a conductance of G = 2.45 –1. The resistivity of material of the wire will be
(a) 2.1 10–7 m (b) 3.1 10–7 m (c) 4.1 10–7 m (d) 5.1 10–7 m Sol.: Gl A l RA 8 45 . 2 10 8 6 4.1 10–7 meter (c) E
32. A galvanometer of resistance 400 can measure a current of 1mA. To convert it into a voltmeter of range 8V the required resistance is
(a) 4600 (b) 5600 (c) 6600 (d) 7600
Sol.: ig
GR
V , 103
400
8R , R = 7600 (d)
E
33. An ammeter reads upto 1A. Its internal resistance is 0.81 . To increase the range to 10A, the value of the required shunt is
(a) 0.03 (b) 0.3 (c) 0.9 (d) 0.09 Sol.:
10 1
81 . 0 1 g g I I G I S 0.09 (d) E34. The resistance of the series combination of two resistances 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) 4 (b) 3 (c) 2 (d) 1
Sol.: For two resistances R1 and R2
2 1 R R S (in series), 2 1 1 1 R R P (in parallel) According to S = nP, 2 1 2 1 2 1 R R R R n R R If n is minimum R1 R2 R then n = 4 (a)
E
35. A wire of resistance 4 is stretched to twice its original length. What is the resistance of the wire now?
(a) 1 (b) 14 (c) 8 (d) 16
Sol.: Volume of wire remains constantA1l1 A2l2, A1l1 A2
2l1 So, A 1 2 A2, 1 1 1 A l R , 2 2 2 A l R , 2 1 R R 1 2 2 1 A A l l , R2 16 (d) E36. The net resistance between points P and Q in the circuit shown in the figure is
(a) R/2 (b) 2R/5 (c) 3R/5 (d) R/3 R R R R Q P Sol.: (b) E
37. The equivalent resistance between points M and N is (a) 2 (b) 3
(c) 2/3 (d) none of the above
1 1 1 1 1 1 M N
Sol.: When a battery is connected between points M and N. NO current is found is PQO. Hence
this section may be removed from the circuit.
3 2 1 2 1 2 eff R Q M N P I I O M N O s (c)E
38. The potentiometer wire AB is 600 cm long. At what distance from A should the jockey J touch the wire to get zero deflection in the galvanometer?
(a) 320 cm (b) 120 cm (c) 20 cm (d) 450 cm E r A B J r R=15r E/2 G Sol.: In case of zero deflection in galvanometer
2 E VAJ , 2 E iRAJ , 2 600 15 15 E AJ r r r E AJ = 320 cm (a) E
39. The emf of the battery shown in the figure is (a) 6 V (b) 12 V (c) 18 V (d) 8 V 6 2 1 1 2 2 I=1.5A
Sol.: According to KVL E ir0 (r is effective resistance in circuit) 0 5 . 1 4 E E = 6 volt (a) E
40. In the figure, the steady state current in 2 resistance is (a) 1.5 A (b) 0.9 A (c) 0.6 A (d) zero 2 3 A B 6 V 2.8 4 C = 0.2 F
Sol.: In steady state, current through battery I =
2 . 1 8 . 2 6 = 1.5 A I2 = 1.5 3 2 3 = 0.9 A (b)
E
41. The charge on the capacitor in the figure is (a) 2 C (b) 2/3 C (c) 4/3 C (d) zero 1F 4 1 2V/0.5 Sol.: Inet =
A r 3 4 2 / 3 2 1 2 . VT 3V 4 5 . 0 2 4 2 . C 3 4 3 4 F 1 CV V Q (c) E42. Each of the resistance in the network shown in the figure below is equal to R. The resistance between the terminals A and B is
(a) R (b) 5R (c) 3R (d) 5/3R M R R R R R A B O N
Sol.: Resistance between M and N can be removed (Balanced whetstone bridge) Reff = R
(a)
E
43. Kirchoff’s second law is based on the law of conservation of
(a) momentum (b) charge
(c) energy (d) sum of mass and energy
Sol.: A charge if taken around a closed loop work done is zero (c)
E
44. The current i in the figure below is (a) 1/5 A (b) 1/10 A (c) 1/15 A (d) 1/45 A 30 30 30 i + – 2V Sol.: Reff = 20 , A R E i eff 10 1 (b)
E
45. The time constant of an RC circuit shown in the figure is (a) 3RC (b) 2/3 RC (c) 6RC/5 (d) 2 RC R R 3R C Sol.: R R R R R Reff 5 6 3 2 3 2 , 5 6 .R RC C eff (c) E
46. What is the current through the resistor R in the circuit shown below? The emf of each cell is Em and internal
resistance is r (a) r R Em 2 (b) r R Em 2 (c) r R Em 2 2 (d) R r Em 2 2 r r R + – + – Sol.: 2 2 2 r R r R Reff , Eeff Em, r R E R E I m eff eff 2 2 (d) E
47. Current I3 in the given circuit shown in the figure is (a) A 11 5 (b) A 11 7 (c) A 11 2 (d) none of these R 3V 3 I3 2 2V I2 I1 1V 1
Sol.: Applying Kirchoff’s law I A
11 5 3
(a) E
48. Six resistors each of resistance R are connected as shown in figure. What is the effective resistance between points A and B? R A R R R R B R (a) 3 R (b) R (c) 3R (d) 6R
Sol.: (a) E
49. The current at which a fuse wire melts does not depend on (a) cross-sectional area (b) length (c) resistivity (d) density
Sol.: (b) E
50. In the circuit shown in figure the heat produced in the 5 resistor due to a current flowing in it is 10 calories per second. The heat produced in the 4 resistor is
4 6
5
(a) 1 cal s–1 (b) 2 cal s–1 (c) 3 cal s–1 (d) 4 cal s–1
Sol.: Let I be current through 5
10 5 2 I … (i)
current through 4 will be
2
I
Heat produced in 4 resistance 4 4 2 I 2 (b) E
51. In the circuit shown in the figure, the current through
(a) the 3 resistor is 0.50 A (b) the 3 resistor is 0.25 A (c) the 4 resistor is 0.50 A (d) the 4 resistor is 0.25 A 3 2 2 2 2 2 8 8 4 A C B D 9V Sol.: (a) E
52. Figure shows currents in a part of an electrical circuit. The current i is
(a) 1 A (b) 1.3 A (c) 1.7 A (d) 3.7 A 2A 2A P Q R 1A 1.3A i Sol.: (c)
E
53. The meter bridge circuit shown in figure is balanced when jockey J divides wire AB in two parts AJ and BJ in the ratio of 1: 2. The unknown resistance Q has value
(a) 1 (b) 3 (c) 4 (d) 7 A B J P Q G 1.5 Sol.: JB Q AJ R R R 5 . 1 , 1 2 AJ JB Q R R R 1.5, RQ 3 (b) E
54. n identical cells, each of emf and internal resistance r, are joined in series to form a closed circuit. The potential difference across any one cell is
(a) zero (b) (c) n (d) n n 1
Sol.: Current in circuit i =
r nr n
The equivalent circuit of one cell is shown in the figure p.d. across the cell
= VA - VB = – + ir = – + . 0 r r (a) r B A – + E
55. In the circuit shown, P R, the reading of the galvanometer is same with switch S open or closed. Then (a) I R IG (b) I P IG (c) IQ IG (d) IQ IR P Q S R G V
Sol.: As P R and reading of galvanometer is same so wheat bridge must be balanced and in that case IR = IG
E
56. The current I drawn from the 5 V source will be
(a) 0.33 A (b) 0.5 A (c) 0.67 A (d) 0.17 A 5V I 10 20 10 5 10 Sol.: 10 5 20 10 .
So it is a balance wheat stone bridge. 10 45 15 30 e R , 2 1 10 5 I A (b) E
57. In the steady state in the circuit shown
(a) potential difference across C1 is 4 V (b) potential difference across 10 is 2V (c) potential difference across C2 is 4 V (d) charge on C1 or C2 is 0 C 10F 10 4F 14V C1 C2 Sol.: 14 4 10 q q q = 40 C.
Potential difference across
10 40 1 C = 4 V (a) E
58. Find the current supplied by the battery as shown in the figure.
(a) 1.5 amp (b) 5 amp (c) 1.2 amp (d) 2.4 amp 4 6 4 6 24V
Sol.: Circuit become simple, then i
20 =24 i1.2 amp 24V 4.8
E
59. What is the equivalent resistance between A and B? (Each resistor has resistance R)
(a) 3 4R (b) 3 5R (c) 5 4R (d) 4 3R B A
Sol.: In the figure
5 4 eff R R (c) B A 2R/3 2R/3 R R E
60. The ammeter will read the value of current
(a) 3A (b) A 3 10 (c) 30 A (d) 3 100 A 5 5 5 5 5 5 B A 5 A 10V Sol.: 3 10 AB
R (use wheat stone bridge)
3 / 10 10 I = 3A (a) E
61. Each cell has emf and internal resistance r in the figure. Find the current through resistance R
(a) r 4 (b) r 3 (c) r (d) zero R A B
Sol.: Potential difference between A and B is zero the current through R is zero.
(d)
E
62. If emf in a thermocouple is T T2 then the neutral temperature of the thermocouple is
Sol.: For neutral temperature 0 dT d 2T 0 Then , 2 T (c) E
63. The charge flowing through a resistance R varies with time t as 2
bt at
Q . The total heat produced in R from t = 0 to the time when value of Q becomes again zero is
(a) b R a 6 3 (b) b R a 3 3 (c) b R a 2 3 (d) b R a3 Sol.: H dH
a bt
dt b a 2 2 / 0
(a) E64. In the steady state in the circuit shown
(a) potential difference across C1 is 4 V (b) potential difference across 10 is 2V (c) potential difference across C2 is 4 V (d) charge on C1 or C2 is 0 C 10F 10 4F 14V C1 C2 Sol.: 14 4 10 q q q = 40 C
Potential difference across
10 40 1 C = 4 V (a) E
65. The charge flowing through a resistance R varies with time t as Qatbt2. The total heat produced in R from t = 0 to the time when value of Q becomes again zero is
60 10 15 5 1A 1A i (a) b R a 6 3 (b) b R a 3 3 (c) b R a 2 3 (d) b R a3 Sol.:
b R a dt R bt a dt R I H b a b a 3 2 3 2 / 0 2 / 0
(b)E
66. The current–voltage (I-V) graphs for a given metallic wire at two different temperatures T1 and T2 are shown in the figure. It follows from the graphs that
(a) T1 > T2 (b) T1 < T2 (c) T1 = T2 V T1 T2 I
(d) T1 is greater or less than T2 depending on whether the resistance R of the wire is greater or less than the ratio V/I.
Sol.: (a)
E
67. The potential difference between points A and B in the following circuit diagram will be
(a) 8V (b) 6V (c) 4V (d) 2V 5 5 5 5 5 5 2V A B Sol.: (c) E
68. The current in the arm CD in the circuit shown in the figure will be
(a) i 1 i2 (b) i 2 i3 (c) i 1 i3 (d) i1i2i3 i1 i2 A i3 B C D Sol.: (b) E
69. In the given circuit resistance of voltmeter is 400 and its reading is 20V. Find the value of emf of battery (a) 130/3 volt (b) 65 volt
(c) 40 volt (d) 33.6 volt E
V
200 300
Sol.; Current in voltmeter, amp. 20 1 400 20 I Current in amp. 15 1 300 Current in 60 7 300 35 15 1 20 1 200 V 3 130 20 60 7 200 E (a)
E
D70. In the given circuit, find the equivalent resistance
between point A and B.
(a) 18 (b) 12 (c) 20 (d) 27 3 6 9 4 5 A B 6 12 18
Sol.: Equivalent circuit is
12 eq R (b) 3 6 9 A B 6 12 18 E
71. In the given circuit diagram, current through the battery is R V 2 3 , if and only if (a) R1 = R2 = R (b) R1 > R2 (c) R1 < R2 (d) always. R R R R R R R1 R2 V Sol.: Current through the battery is independent on R1 and R2.
(d)
E
78. In the given circuit diagram. Find the value of current in resistance R. (a) 2 A (b) A 2 3 (c) 1 A (d) 4 A 1 2 6 6V 12V R = 9
Sol.: Potential difference across R = 18 V
So 9 18 I 2A (a) E
79. In the given circuit, the equivalent resistance between point A and B is
(a) 3 10 (b) 3 5 (c) 57 24 (d) 24 57 3 2 5 6 10 A B 10 10
Sol.: Equivalent ciruit is balanced Wheat- stone bridge as shown RAB = 3 10 (a) 2 2 10/3 A B 10 10 E
80. The current in branch CD of given circuit is,
(a) zero (b) 1 A (c) 2 A (d) 3 A A B C D E F 8V 4 3 4V 2 12V
Sol.: The equivalent emf of 12V and 8V battery =
3 / 1 2 / 1 3 8 2 12 = 2 3 16 36 = 4 V 5 6 3 2 3 2 eq r
The equivalent circuit is
4V 6/5 4V 4 C D I = 0 (a) E
81. Two sources of emf 6V and internal resistance 3 and 2 are connected to an external resistance R as shown. If potential difference across source A is zero, then value of R is A B R 6V,3 6V,2 (a) 1 (b) 2 (c) 3 (d) 4 Sol.: R I 5 12 and 0 5 12 3 6 R R 5 36 6 R = 1 (a) E
82. The equivalent resistance between points A and B is (a) 2R (b) R 4 3 (c) R 3 4 (d) R 5 3 R R R R A B
Sol.: Circuit can be rearranged as follows
5 3 2 3 2 3 R R R R R Req A R R B R R (d) E
83. In the circuit shown, current through 3 resistance is (a) 1 amp (b) 2 amp
(c) 3 amp (d) 4 amp
6 6V
3
Sol.: Current through battery
3 6 3 6 6 i 3 amp Current through 3 is i
6 3 6 3 2 amp (b) E84. The circuit as shown in figure. The ratio of
current i1/ i2 is (a) 2 (b) 8 (c) 0.5 (d) 4 8 2 4 3 2 8 8 1 V 16 V 8 2 i 1 i
Sol.: The simplified circuit can be drawn as
4 8 V 16 V 8 A 4 A 2 2A A I A I 2 1 4 2 1 (b)
E
85. In the circuit shown in the figure, reading of voltmeter is V1 when only S1 is closed, reading of voltmeter is V2 when only S2 is closed and reading of voltmeter is V3 when both S1 and S2 are closed. Then:
(a) V3V2 V1 (b) V2 V1 V3 (c) V3V1 V2 (d) V1 V2 V3 S1 S2 3R 6R R V E Sol.: In series P.D. R
When only S1 is closed, V E 0.75E
4 3
1
When only S2 is closed, V E 0.86E
7 6
2
And when S1 and S2 are closed, combined resistance of 6R and 3R is 2R. V E 0.67E 3 2 3 V2 V1 V3 (b) E
86. The resistance of a wire is 10. Its length is increased by 10% by stretching. The new resistance will now be nearly
(a) 12 (b) 1.2 (c) 13 (d) 11
Sol.: Since R I2
If length is increased by 10% resistance is increases by almost 20% Hence new resistance R = 10 + 20% of 10 = 10 + 10
100 20
12
(a)
E
87. The same mass of copper is drawn into two wires 1 mm and 2 mm thick. Two wires are connected in series and current is passed through them. Heat produced in the wire is in the ratio (a) 2 : 1 (b) 1 : 16 (c) 4 : 1 (d) 16 : 1 Sol.: 2 2 2 2 A Vt i t A l i RT i H (V = volume) 14 r H 4 1 2 2 1 r r H H = 1 16 1 2 4 (d)
E
88. 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
X 6 5V G 40cm 60cm A B C
(a) 6, 5 amp (b) 4, 0.1 amp (c) 4, 1.0 amp (d) 12, 0.5 amp
Sol.: Resistance of the part AC
RAC = 0.1 × 40 = 4 and RCB = 0.1 × 60 = 6 In balanced condition 6 4 6 X X = 4
Equivalent resistance Req = 5 so current drawn from battery
5 5 i = 1A (c) E
89. Find the equivalent resistance across AB
(a) 1 (b) 2 (c) 3 (d) 4 2 2 2 2 2 A B Sol.: 2 2 2 2 2 A B 2 2 A B 2 2 2 2 AB R = 1 (a) E
90. The reading of the ammeter in the figure shown is (a) A 8 1 (b) 4 3 A (c) A 2 1 (d) 2A 2V A 2 2 2 2 Sol.: (b)
E
91. The total current supplied to the circuit by the battery is
(a) 1A (b) 2A (c) 4A (d) 6A 2 6 1.5 3 6V
Sol.: Net resistance =
2 3
Then by Kirchoff law 6 =
2 3
i, i = 4 amp (c)
E
92. The magnitude of i in ampere unit is
(a) 0.1 (b) 0.3 (c) 0.6 (d) 0.4
Sol.: (a) E
93. AB is a wire of uniform resistance. The galvanometer G
shows zero current when the length AC= 20 cm and CB = 80 cm. The resistance R is equal to
(a) 2 (b) 8 (c) 20 (d) 40 80 R A C B G
Sol.: By Balanced wheat stone bridge
80 80 20 R R = 20 (c)
MODERATE QUESTIONS
M94. In the shown arrangement of the experiment of the 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?
B G x C R1 R2 A (a) x (b) x/4 (c) 4x (d) 2x Sol.: (a)
M
95. Two cells with the same emf E and different internal resistances r1 and r2 are connected in series to an external resistance R. The value of R so that the potential difference across the first cell be zero is
(a) r1r2 (b) r 1 r2 (c) r 1 r2 (d)
2 2
1 r
r
Sol.: Current in the circuit is
2 1 2 r r R E I , V1 EIr1 0 2 0 2 1 1 r r R Er E Rr1r2 (c) M
96. A battery of internal resistance 4 is connected to the network of resistances as shown in the figure. In order that maximum power can be delivered to the network, the value of R in ohm should be
(a) 9 4 (b) 2 (c) 3 8 (d) 18 R R R 4R R R 6R Sol.: (b) M
97. In the adjoining circuit, when the key K is pressed at time t = 0, which of the following statements about current I in the resistor AB is true?
(a) I = 2 mA at all t
(b) I oscillates between 1 mA and 2 mA (c) I = 1 mA at all t
(d) At t = 0, I = 2 mA and with time it goes to 1 mA
A K 2V 1000 B 1000 1F Sol.: (d) M
98. A, B and C are voltmeters of resistances R, 1.5R and 3R respectively. When some potential difference is applied between X and Y, the voltmeter readings are VA, VB and
VC respectively. A X B C Y (a) VA = VB = VC (b) VA VB = VC (c) VA = VB VC (d) VB VA = VC
Sol.: VA = iR iR R i VB 1.5 3 2
R iR i VC 3 3 A B C i R i/3 2i/3 3R 1.5R (a) M99. A nucleus with mass number 220 initially at rest emits an -particle. If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the -particle.
(a) 4.4 MeV (b) 5.4 MeV (c) 3.4 MeV (d) 5.6 MeV
Sol.: By conservation of linear momentum mv MV 54 v V …(i) By energy conservation J 10 6 . 1 5 . 5 2 1 2 1 2 2 13 MV mv …(ii)
From (i) and (ii)
K.E. of -particle = 2 2 1 mv 5.4 MeV (b) M
100. A galvanometer of resistance 19.5 gives full scale deflection when a current of 0.5 ampere is passed through it. It is desired to convert it into an ammeter of full scale current 20 ampere. Value of shunt is
(a) 0.5 (b) 1 (c) 1.5 (d) 2
Sol.: 19.5 0.5 = S (20 – 0.5) S = 0.5
(a) M
101. A galvanometer of coil resistance 1 is converted into voltmeter by using a resistance of 5 in series and same galvanometer is converted into ammeter by using a shunt of 1. Now ammeter and voltmeter connected in circuit as shown, find the reading of voltmeter and ammeter.
15 30 V A V 12 15 6 4.5
Sol.: RVoltmeter = 6, Rammeter = 0.5 Req = 10 A I 3 10 30
Reading of voltmeter = 1 3 = 3 volt. (a)
M
102. In the arrangement shown, the magnitude of each
resistance is 2. The equivalent resistance between O and A is given by (a) 15 14 (b) 15 7 (c) 3 4 (d) 6 5 A B C D O
Sol.: From symmetry. B and D are points having same
potential so, redrawing the network as
15 14 OA R (a) B, D O A C M
103. A 4μFcapacitor is given 20μCcharge and is connected with an uncharged capacitor of capacitance 2μFas shown in figure. When switch S is closed.
(a) charged flown through the battery is μC 3 40
(b) charge flown through the battery is μC 3 20
(c) work done by the battery is μJ 3 200
(d) work done by the battery is μJ 3 100 10V 2F 20C C S + + + + + + – – – – – – 4F
Sol.: Using Kirchhoff’s loop law and conservation of charge, final distribution of charge on the capacitors will be as shown in the figure.
Charge q flown through the battery = charge on 2μF
capacitor and work done by the battery qV
(b) (c) 10V C 3 80 + + + + + + – – – – – – 2F C 3 20 + + + + + + – – – – – – 4F
M
104. In the circuit shown if point O is earthed, the potential of point X is equal to
(a) 10 V (b) 15 V (c) 25 V (d) 12.5 V 2 2 15V 5 10 V 10 V O 5 V 5 V X 5 Sol.: V0 + 10 –5 + 10 = Vx Vx = 15 V (b) M
105. Figure shows a network of a capacitor and resistors. The charge on capacitor in steady state is (a) 4 C (b) 6 C (c) 10 C (d) 16 C 6V 2 4 4 1F 10 V 8 V 4 V 8
Sol.: Let the potential of the junction be
V. Then 0 4 8 4 4 2 6 V V V 0 8 4 2 12 V V V V 4 24 6 V volt 6V 2 4 4 1F 10 V 8 V 4 V 8 i1 i3 i2
Potential drop across capacitor
10
6 V 16 Charge on capacitor = 16 C (d)M
106. A parallel plate capacitor is connected with a resistance R and
a cell of emf as shown in figure. The capacitor is fully charged. Keeping the right plate fixed, the left plate is moved slowly towards further left with a variable velocity v such that the current flowing through the circuit is constant. Then the variation of v with separation x between the plates is
represented by curve v R x v (a) x v (b) x v (c) x v (d) Sol.:
IR
x A IR C q 0 … (i)
q IR A x0 … (ii) On differentiation of equation (ii) and from (i)
IR
A Ix v 0 2 (b) x I R M107. The electric potential variation around a single closed
loop containing an ideal battery and one or more resistors as shown in figure. If current of 1A flows in the circuit, the circuit can not have
(a) two resistors and two batteries (b) one resistor and three batteries (c) maximum net emf of 6 volt (d) three resistor and one battery
4 6 8 10 V
Sol.: The possible circuit of close loop corresponding to graph are
(i) 1 R R2 V 2 4 V (ii) R 2V 4 V V 2
(iii) V 4 V 2 R1 R2 (iv) R 4 V V 2 4 V (d) M
108. An ammeter is obtained by shunting a 30 galvanometer with a 30 resistance. What additional shunt should be connected across it to double the range?
(a) 15 (b) 10 (c) 5 (d) none of these
Sol.: For ammeter, G I I I S g g 1 g I I S G I 2Ig
New range is doubled, i.e. 4Ig
Now shunt required, G I I I S g g g 4 = 10
This can be obtained by shunting the earlier shunt of 30 with an additional shunt of 15.
(a)
M
109. An ideal ammeter and an ideal voltmeter are connected as shown. The ammeter and voltmeter reading for R1 = 5, R2 = 15, R3 = 1.25 and E = 20V are given as (a) 6.25 A, 3.75 V (b) 3.00 A, 5 V (c) 3.75 A, 3.75 V (d) 3.75 A; 6.25 V 20V 15 5 A V 1.25 R1 R2 R3
Sol.: Req of the circuit = 3 2 1 2 1 R R R R R = 4 5 20 75 100 125 15 5 15 5 5 5 20 eq R E I = 4 A 20V 15 5 A V 1.25 R1 R2 R3 Current in A R R IR R 3 2 1 2 1 P.D. across R3 IR3 5V (b)
M
110. A moving coil voltmeter is generally used in the laboratory to measure the potential difference across a conductor of resistance r, and carrying a current I. The voltameter has a resistance R and will measure the potential difference more accurately as
(a) R approaches r
(b) R becomes larger than r (c) R becomes smaller than r (d) R equals to zero
Sol.: A voltmeter should have high resistance.
(b)
M
111. A potentiometer has a driving cell of negligible internal resistance. The balancing length of a Daniel cell is 5 m. If the driving cell has internal resistance, the balancing length of the same Daniel cell would have been
(a) more (b) less
(c) same (d) cannot be said from the data
47. (a)
M
112. A simple potentiometer circuit is shown in the figure. The internal resistance of the 4V battery is negligible. AB is a uniform wire of length 100 cm and resistance 2. What would be the length AC for zero galvanometer deflection? (a) 78.5 cm (b) 84.5 cm (c) 82.5 cm (d) 80.5 cm G 2.4 C B A 1.5 V 4 V 28. 11 10 1 . 1 1 4 . 4 4 2 4 . 2 4 I If AC = x Then 50 x RAC 5 . 1 50 11 10 x x1.551182.5cm (c) M
113. As the switch S is closed in the circuit shown in figure current passed through it is
(a) 4.5 A (b) 6.0 A (c) 3.0 A (d) zero S 2 4 2 B A 20V 5V
Sol.: Let V be the potential of the junction as
shown in the figure, applying junction law. 402V 5V 2V or V = 9Volt 2 3 V i 4.5 Amp S 2 4 2 B A 20V 5V (a) M
114. A potentiometer wire AB is 100cm long and has total resistance of 10. Find the value of unknown resistance R so that null point is obtained at a distance 40 cm from A. (a) 1 (b) 2 (c) 3 (d) 4 10V 40cm A C B R r = 1 E = 5V G Sol.: 1 1 5 5 40 100 10 R R = 4 (d) M
115. A miliammeter of range 10mA and resistance 9 is joined in a circuit as shown. The meter gives full scale deflection for current I when A and B are used as its terminals, i.e. current enters at A and leaves at B (C is left isolated). The value of I is (a) 100 mA (b) 900 mA (c) 1 A (d) 1.1 A 9 ,10mA 0.1 0.9 A B C Sol.: 3
3
10 10 1 1 . 0 9 . 9 10 10 or I = 1A (c) 9 10mA 0.1 0.9 A B CM
116. A resistance R carries a current i. The power lost to the surroundings is ( - 0). Here is a constant, is temperature of the resistance and 0 is the temperature of the atmosphere. If the coefficient of linear expansion is . The strain in the resistance is
(a) i2R (b) iR (c) 2 2 R i
(d) proportional to the length of the resistance wire
Sol.: Under steady state condition power developed = power loss
or
0
2 R i i R 2 0 Now, strain = = R i2 (a) M117. Two identical batteries, each having emf of 1.8V and of equal internal resistances are connected as shown in the figure. Potential difference between A and B will be equal to : (Ignore the resistance of lead wires)
(a) 3.6 V (b) 1.8 V
(c) zero (d) none of these
A B Sol.: r I 2 6 . 3 , VAB 1.8Ir0 (c) M
118. A milliammeter of range 10 mA has a coil of resistance 1. To use it is an ammeter of range 1A, the required shunt must have a resistance of
(a) 101 1 (b) 100 1 (c) 99 1 (d) 9 1 Sol.: ig = 10 mA = 0.01 A r = 1 I = 1A VA – VB = igr = (I – ig)S S = 99 1 01 . 0 1 1 01 . 0 ) ( g g i I r i (c) S B I A ig (I = ig)
M
119. A 100 W bulb B1, and two 60 W bulbs B2 and B3, are connected to a 250 V source, as shown in the figure. Now W1, W2 and W3 are the output powers of the bulbs B1, B2 and B3, respectively. Then.
(a) W1 > W2 = W3 (b) W1 > W2 > W3 (c) W1 < W2 = W3 (d) W1 < W2 < W3 B1 B2 B3 250 V
Sol.: Resistances of bulbs B2 and B3 are equal but that of B1 is smaller than their resistance, hence resistance of path of bulb B3 is less than that of the series combination of bulbs B2 and B3, therefore power consumed by B3 will be maximum possible.
Since B1 and B2 are in series, therefore current through them will be the same. Since resistance of B2 is greater, therefore W2 > W1.
(d) M
120. In the circuits shown in the figure, the heat produced in
the 5- resistor due to the current flowing through it, is 10 cal s1. The heat generated in the 4- resistor is (a) 1 cal s–1 (b) 2 cal s1
(c) 3 cal s–1 (d) 4 cal s1
4 6
5
Sol.: Heat produce in 5 resistance is 5i 12 t i1 = 2i/3
where i is the total current again heat produce through 4 is 4i 22 t i2 = i/3 given that 5i12 t 10 therefore 10 9 20 2 t i 2 9 2 t i
Heat produce in 4 is equal to
2 9 9 4 9 4 2 t i = 2 (b)
M
121. For what value of R, power developed across 6 resistor is equal to the power developed across 24 resistor (a) 12 (b) 6 (c) 24 (d) 8 6 R 24 Sol.: We have, I1 = 24 R RI I2 = 24 24 R I P1 = 24 ) 24 ( 2 2 2 R I R P2 = I 2.6 P2 = P1 R = 24 (c) I I2 R 24 I1 6 M
122. If a cell produces the same amount of heat in two resistors R1 and R2 in the same time separately, the internal resistance of the cell is
(a) (R1 + R2)/2 (b) R1 R2 (c) R1 R2 /2 (d) (R1 – R2)/2 Sol.: Let r is the internal resistance of cell
Case I: current in loop I1 = r R 1 H1 = I12 R1 = 2 1 1 2 ) (R r R
Case II: Current in loop I2 = r R 2 H2 = I22 R2 2 2 2 2 ) (R r R i.e., 2 2 2 2 2 1 1 2 ) ( ) ( R r R r R R r R r R R R 2 1 2 1 r = R1R2 (b)
M
123. Eight identical resistances r, each are connected along edges
of a pyramid having square base ABCD as shown in figure. The equivalent resistance between A and D is
(a) 15 2r (b) 15 r (c) 15 4r (d) 15 8r A D C B O
Sol.: The circuit can be represented as C1 O C2 So we can arrange the circuit in following way r r r r q 1 2 1 3 2 2 1 Re 1 = r r r 1 2 1 8 3 Req = 15 8 r (d) B C D A C1 C2 O r 2r 2r r r r M
124. What is the potential difference between points C and D in the circuit shown in figure in steady state? (a) 3.6 V (b) 7.2 V (c) 10.8 V (d)12 V 12 V 3 1 6 C = 1 F 1 C = 2 F 2 B A D C I Sol.: 1.2A 1 6 3 12 V I , VAB = 12 – 1.2 × 1 = 10.8 V, VAD = 6 × 1.2 = 7.2 V F F C C C C Ceff 6 2 1 2 1 10 3 2 3 2 2 1 2 1 , Q = VAB × Ceff = 10.8 × 10 6 3 2 = 7.2 × 10-6 C V C Q VAC 3.6 10 2 10 2 . 7 6 6 2 , VCD = VAD – VAC = 7.2 – 3.6 = 3.6 V (a)
M
125. A battery of emf E and internal resistance r is connected to a resistor of resistance r1 and Q joules of heat is produced in a certain time t. When the same battery is connected to another resistor of resistance r2, the same quantity of heat is produced in the same time t, the value of r is (a) 2 2 1 r r (b) 1 2 2 r r (c) 2 1 (r1 + r2) (d) r1r2 Sol.: I1 = r r E 1 , Q1 = I2 r1 t = 2 1 r r E × r1 t Q2 = r r E 2 × r2t, 2 2 2 2 1 1 ) ( ) ( r r r r r r , r = r1r2 (d) M
126. In the network shown in the figure, each resistance is 1 ohm.
The effective resistance between A and B is
(a) (4/3) (b) (3/2) (c) 7 (d) (8/7) A B Sol.: 7 8 2 3 8 2 3 8 e R (d) A B 1 1 1 1 1 1 1 M
127. A ammeter is to be constructed which can read current upto 2.0 A. If the coil has a resistance of 25 and takes 1 mA for full-scale deflection, what should be the resistance of the shunt used?
(a) 2.25 10–2 (b) 1 10–2 (c) 1.25 10–2 (d) 1.25 10–4 Sol.: I 2A So, 1 10–3 25 = 2 R R= 1.25 10–2 A 25 R 2A I 1mA (c)
M
128. Figure shows a network of eight resistors numbered 1 to 8, each equal to 2, connected to a 3V battery of negligible internal resistance. The current I in the circuit is (a) 0.25 A (b) 0.5 A (c) 0.75 A (d) 1.0 A 3V A 1 B 4 C 6 D 2 3 5 8 7 E F
Sol.: No current will flow through 3 and 5.
So, Req = 6 6 6 6 3 , 3 3 eq R V i 1A (d) M
129. Two heater wires of equal length are first connected in series and then in parallel. The ratio of heat produced in two cases will be
(a) 1 : 2 (b) 1 : 4 (c) 2 : 1 (d) 4 : 1
Sol.: Applied potential difference is same R series = 2R , Rparallel = R/2 Power R V2 , P series R 2 1 Pparallel
/2
1 R , 4 1 2 2 1 R R P P parallel series (b) M130. The effective wattage of 60 W and 40W lamps connected in series is equal to
(a) 24 W (b) 20 W (c) 100 W (d) 80 W Sol.: 2 1 2 1 P P P P Peff Peff 24 W (a) M
131. A heater boils a certain quantity of water in time t1. Another heater boils the same quantity of water in time t2. If both heaters are connected in series, the combination will boil the same quantity of water in time
(a)
1 2
2 1 t t (b)
t 1 t2
(c)
1 2
2 1 t t t t (d) t1t2Sol.: Q = quantity of energy required Q
t
Pseries 2 1 2 1 P P P P Pseries t0 = Q , t Q P P P P 2 0 1 2 1 Solving t0 = t1 + t2 (b) M
132. The filament of an electric heater should have (a) high resistivity and high melting point (b) low resistivity and high melting point (c) high resistivity and low melting point (d) low resistivity and low melting point
Sol.: (a)
M
133. The voltage across a bulb is decreased by 2%. Assuming that the resistance of the filament remains unchanged, the power of the bulb will
(a) decrease by 2% (b) increase by 2% (c) decrease by 4% (d) increase by 4%
Sol.: R V P 2 2 V P [R = constant] (c) M
134. A student has connected a voltmeter, an ammeter and a resistor R as shown. If voltmeter reads 20V and ammeter reads 4A, then R is
(a) = 5 (b) > 5 (c) < 5
(d) > or < 5 depending upon its material.
V – + A 4A + – R Sol.: 20 4
i
R 5 4 20 i R V A R (4-i) i (b)M
135. In the figure, the potentiometer wire of length l = 100 cm and resistance 9 is joined to a cell of emf E1 = 10V and internal resistance r1 = 1. Another cell of emf E2 = 5V and internal resistance r2 = 2 is connected as shown. The galvanometer G will show no deflection when the length AC is (a) 50 cm (b) 55.55 cm (c) 52.67 cm (d) 54.33 cm E1 = 10 V r1 = 1 G A B E2 = 5 V r1 = 2 C Sol.: 1 × 5 100 9 x x = 55.55 cm (b) M
136. An electrical cable of copper has just one wire of radius 9 mm. Its resistance is 5. This single copper wire of the cable is replaced by 6 different well insulated copper wires each of radius 3 mm. The total resistance of the cable will now be equal to
(a) 7.5 (b) 45 (c) 90 (d) 270 Sol.: As R 12 r 9 3 9 5 2 2 R R = 45 Req. = 6 R = 7.5 (a) M
137. A uniform wire of resistance R is shaped into a regular n-sided polygon (n is even). The equivalent resistance between any two corners can have
(a) the maximum value
2
R
(b) the maximum value n R
(c) the minimum value 2 1 n n
R (d) the minimum value n R
Sol.: Resistance between opposite corner is
2
R and
2
R
which is parallely connected.
Maximum value
4
R
For adjacent corner two resistance n R and R n n 1
are parallel connected
So minimum resistance is
21
n n R (c)M
138. In the circuit shown, current through 25V cell is
(a) 7.2 A (b) 10 A (c) 12 A (d) 14.2 A
10V 5V 20V 30V 25V
5 10 5 11
Sol.: Applying KVL in loop ABCDA, ABFEA, ABHGA and ABJIA we get 30 – i1 × 11 = – 25 ... (i)
20 + i2 × 5 = 25. .... (ii) 5 – i3 × 10 = – 25 ... (iii) 10 + i4 × 5 = 25 ... (iv)
i1 = 5A, i2 = 1A, i3 = 3A and i4 = 3A. (c) I G E C A 10V 5V 20V 30V 25V 5 10 5 11 J H F D B i4 i3 i2 i1 i1+i2+i3+i4 M
139. Seven identical lamps of resistances 2200 ohm each are connected to 220 volt line as shown in the figure. What will be the reading in the ammeter?
~
A
(a) (1/10) ampere (b) (3/10) ampere (c) (4/10) ampere (d) (7/10) ampere
Sol.: Current through each resistor will be same current passing through ammeter
10 4 2200 220 4 (c) M
140. In the part of a circuit shown in the figure, the
potential difference between points G and H (VG – VH) will be (a) 0 V (b) 15 V (c) 7 V (d) 3 V G 2A 1A 3A 5V 2 4 3V H 1 Sol.: VG2432221VH V 7 H G V V (c)
M
141. In the circuit shown in the figure, the ratio of VB as to
VC is (a) –2/5 (b) –5/2 (c) 1 (d) 1/3 1 C 2 A B D 5V 10V 2V Sol.: VA VC 5, VA VB 2, 5 2 C B V V (a) M
142. A 3 resistor as shown in the figure, is dipped into a calorimeter containing H2O. The thermal capacity of H2O + calorimeter is 2000 J/K. If the circuit is active for 15 minutes find the rise in temperature of H2O is (a) 2.40 C (b) 2.90C (c) 3.40 C (d) 1.90C 1 6 6V 3 Sol.: 1 2 6 1 6 AB R I 2A 2 3 6 3 6 AB R
Current through 3 resistors I'
I6
/9 4/3 A (mC) T = I2Rt 60 15 3 3 4 2000 2 T T 2.40C (a) M143. In the figure AB is 300 cm long wire having resistance 10 per meter. Rheostat is set at 20. The balance point will be attained at
(a) 1.0 m (b) 1.25 m (c) 1.5 m (d) cannot be determined 2V 0.5 1.5 A G K 6V 20 B Sol.: 50 30 6 AB V 3.6 V
Terminal voltage of cell
2 5 . 1 2 = 1.5 V