7.

ALTERNATING CURRENT

Single Correct Answer Type

1. A resistor 30 Ω, inductor of reactance 10 Ω and capacitor of reactance 10 Ω are connected in series to an

AC voltage source 𝑒 = 300√2 sin(𝜔𝑡). The current in the circuit is
a) 10√2 A b) 10 A c) 30√11 A d) 30/√11 A
2. The natural frequency (𝜔0 ) of oscillations in L - C circuit is given by
1 1 1 1
a) b) √𝐿𝐶 c) d) √𝐿𝐶
2𝜋 √𝐿𝐶 2𝜋 √𝐿𝐶
3. An ac source of angular frequency 𝜔 is fed across a resistor 𝑟 and a capctior 𝐶 in series. The current
registered is 𝐼. If the frequency of source is changed to 𝜔/3 (maintaining the same voltage), the current in
the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency 𝜔
3 2 1 4
a) √ b) √ c) √ d) √
5 5 5 5
4. When a DC voltage of 200 V is applied to a coil of self-inductance (
2√3
)H, a current of 1 A flows through it.
𝜋
But by replacing DC source with AC source of 200 V, the current in the coil is reduced to 0.5 A. Then the
frequency of AC supply is
a) 100 Hz b) 75 Hz c) 60 Hz d) 50 Hz
5. The power factor of good choke coil is
a) Nearly zero b) Exactly zero c) Nearly one d) Exactly one
6. A resistor of 𝑅 = 6Ω, an inductor of L = 1 H and a capacitor of 𝐶 = 17.36 μF are connected in series with
an AC source. Find the Q - factor.
a) 3.72 b) 40 c) 2.37 d) 80
7. Power dissipated in an 𝐿𝐶𝑅 series circuit connected to an a.c. source of 𝑒𝑚𝑓 𝐸 is
2
1 2 2 1
b) 𝐸 √𝑅 + (𝐿𝜔 − 𝐶𝜔)
2
a) 𝐸 2 𝑅/ [𝑅 2 + (𝐿𝜔 − ) ]
𝐶𝜔
𝑅
1 2 𝐸2𝑅
𝐸 2 [𝑅 2 + (𝐿𝜔 − ) ]
c) 𝐶𝜔 d) 2
𝑅 √𝑅 2 + (𝐿𝜔 − 1 )
𝐶𝜔
8. A virtual current of 4𝐴 and 50 𝐻𝑧 flows in an ac circuit containing a coil. The power consumed in the coil is
240 𝑊. If the virtual voltage across the coil is 100 𝑉 its inductance will be
1 1 1 1
a) 𝐻 b) 𝐻 c) 𝐻 d) 𝐻
3𝜋 5𝜋 7𝜋 9𝜋
9. A lamp consumes only 50% of peak power in an a.c. circuit. What is the phase difference between the
applied voltage and the circuit current
𝜋 𝜋 𝜋 𝜋
a) b) c) d)
6 3 4 2
10. A vertical ring of radius r and resistance R falls vertically. It is in contact with two vertical rails which are
joined at the top, figure. The rails are without friction and resistance. There is a horizontal uniform
magnetic field of magnitude B perpendicular to the plane of the ring and the rails. When the speed of the
ring is 𝑣, the current is the section 𝑃𝑄 is

a) Zero 2 𝑅𝑟𝑣 4 𝑅𝑟𝑣 8 𝐵𝑟𝑣

b) c) d)
𝑅 𝑅 𝑅
11. Voltage V and current i in AC circuit are given by
Page|1
 𝑉 = 50 sin(50𝑡)volt
𝜋
𝑖 = 50 sin (50𝑡 + )mA
3
The power dissipated in circuit is
a) 5.0 W b) 2.5 W c) 1.25 W d) zero
12. In an 𝐿𝐶𝑅 series resonant circuit which one of the following cannot be the expression for the Q-factor
𝜔𝐿 1 𝐿1 𝑅
a) b) c) √ d)
𝑅 𝜔𝐶𝑅 𝐶𝑅 𝐿𝐶
13. Which one of the following curves represents the variation of impedance (𝑍) with frequency 𝑓 in series
𝐿𝐶𝑅 circuit
a) Z b) Z c) Z d) Z

f f f f

14. The frequency for which a 5 𝜇𝐹 capacitor has a reactance of 1

𝑜ℎ𝑚
is given by
1000
100 1000 1
a) 𝑀𝐻𝑧 b) 𝐻𝑧 c) 𝐻𝑧 d) 1000 𝐻𝑧
𝜋 𝜋 1000
15. The peak value of an alternating current is 5 A and its frequency is 60 Hz. Find its rms value and time
taken to reach the peak value of current starting from zero.
a) 3.536A; 4.167 ms b) 3.536 A;15 ms c) 6.07 A; 10 ms d) 2.536 A; 4.167 ms
16. The resistance of an R-L circuit is 10 Ω. An emf 𝐸0 applied across the circuit at 𝜔 = 20 rad s −1 . If the
𝑖0
current in the circuit is , what is the value of L?
√2
a) 0.5 H b) 2.25 H c) 3.9 H d) 1.0 H
17. In a circuit, the current lags behind the voltage by a phase difference of 𝜋/2, the circuit will contain which
of the following?
a) Only R b) Only C c) R and C d) Only L
18. In the circuit shown below, the key K is closed at t = 0. The current through the battery is

𝑉𝑅1 𝑅2 𝑉
a) at 𝑡 = 0 and at 𝑡 = ∞
√𝑅12 + 𝑅22 𝑅2
𝑉 𝑉(𝑅1 +𝑅2 )
b) at 𝑡 = 0 and at 𝑡 = ∞
𝑅2 𝑅1 𝑅2
𝑉 𝑉𝑅1 𝑅2
c) 𝑅 at 𝑡 = 0 and at 𝑡 = ∞
2 √𝑅12 + 𝑅22
𝑉(𝑅1 +𝑅2 ) 𝑉
d) at 𝑡 = 0 and at 𝑡 = ∞
𝑅1 𝑅2 𝑅2
19. In a circuit, the value of the alternating current is measured by hot wire ammeter as 10 𝑎𝑚𝑝𝑒𝑟𝑒. Its peak
value will be
a) 10 𝐴 b) 20 𝐴 c) 14.14 𝐴 d) 7.07 𝐴
20. In an electrical circuit 𝑅, 𝐿, 𝐶 and an a.c. voltage source are all connected in series. When 𝐿 is removed from
the circuit, the phase difference between the voltage and the current in the circuit is 𝜋/3. If instead, 𝐶 is
removed from the circuit, the phase difference is again 𝜋/3. The power factor of the circuit is
a) 1/2 b) 1/√2 c) 1 d) √3/2
Page|2
21. The power factor of an AC circuit having resistance R and inductance L (connected in series) and an
angular velocity 𝜔 is
a) 𝑅/𝜔𝐿 b) 𝑅/(𝑅 2 + 𝜔2 𝐿2 )1/2 c) 𝜔 𝐿/𝑅 d) 𝑅/(𝑅 2 − 𝜔2 𝐿2 )1/2
22. A uniformly wound solenoidal coil of self inductance 1.8 × 10−4 H and resistance 6 Ω is broken up into
two identical coils. These identical coils are then connected in parallel across a 12 V battery of negligible
resistance. The time constant of the current in the circuit and the steady state current through battery is
a) 3 × 10−5 s, 8 A b) 1.5 × 10−5 s, 8 A c) 0.75 × 10−4 s, 4 A d) 6 × 10−5 s, 2 A
23. An alternating voltage is connected in series with a resistance 𝑅 and an inductance 𝐿. If the potential drop
across the resistance is 200 𝑉 and across the inductance is 150 𝑉, then the applied voltage is
a) 350 𝑉 b) 250 𝑉 c) 500 𝑉 d) 300 𝑉
24. The number of turns in a secondary coil is twice the number of turns in primary. A leclanche cell of 1.5 V is
connected across the primary. The voltage across secondary is
a) 1.5 V b) 3.0 V c) 240 V d) Zero
25. When the rate of change of current is unity, induced emf is equal to
a) Thickness of coil b) Number of turns in coil c) Coefficient of self- d) Total flux linked with
induction coil
26. A coil of wire of certain radius has 100 turns and a self inductance of 15 mH. The self inductance of a
second similar coil of 500 turns will be
a) 75 mH b) 375 mH c) 15 mH d) None of these
27. The coefficient of induction of a choke coil is 0.1𝐻 and resistance is 12Ω. If it is connected to an alternating
current source of frequency 60 𝐻𝑧, then power factor will be
a) 0.32 b) 0.30 c) 0.28 d) 0.24
28. A square loop of side a placed in the same plane as a long straight wire carrying a current 𝑖. The centre of
the loop is at a distance r from the wire, where 𝑟 >> 𝑎, figure. The loop is moved away from the wire with
a constant velocity 𝑣. The induced emf in the loop is

𝜇0 𝑖 𝑎 𝑣 𝜇 𝑖 𝑎3 𝑣 𝜇0 𝑖 𝑣 𝜇0 𝑖 𝑎2 𝑣
a) b) 0 c) d)
2𝜋𝑟 2 𝜋 𝑟3 2𝜋 2 𝜋 𝑟2
29. Voltage and current in an ac circuit are given by
𝜋 𝜋
𝑉 = 5 sin (100𝜋𝑡 − 6 ) and 𝐼 = 4 sin (100 𝜋𝑡 + 6 )
a) Voltage leads the current by 30° b) Current leads the voltage by 30°
c) Current leads the voltage by 60° d) Voltage leads the current by 60°
30. A coil is wound on a core of rectangular cross-section. If all the linear dimensions of core are increased by
a factor 2 and number of turns per unit length of coil remains same, the self-inductance increases by a
factor of
a) 16 b) 8 c) 4 d) 2
31. The phase angle between 𝑒.m.f. and current in 𝐿𝐶𝑅 series as circuit is
𝜋 π π
a) 0 to 2 b) c) d) π
4 2
32. The primary winding of a transformer has 200 turns and its secondary winding has 50 turns. If the current
in the secondary winding is 40 A, the current in the primary is
a) 10 A b) 80 A c) 160 A d) 800 A
33. The initial phase angle for 𝑖 = 10 sin 𝜔𝑡 + 8 cos 𝜔𝑡 is

Page|3
 4 5 4
a) tan−1 ( ) b) tan−1 ( ) c) sin−1 ( ) d) 90°
5 4 5
34. An inductor is connected to an AC source. When compared to voltage , the current in the lead wires
a) Is ahead in phase by 𝜋 b) Lags in phase by 𝜋
𝜋 𝜋
c) Is ahead in phase by 2
d) Lags in phase by 2
35. An ac supply gives 30 𝑉 𝑟. 𝑚. 𝑠. which passes through a 10 Ω resistance. The power dissipated in it is
a) 90√2 𝑊 b) 90 𝑊 c) 45√2 𝑊 d) 45 𝑊
36. In a series 𝐿𝐶𝑅 circuit, operated with an ac of angular frequency 𝜔, the total impedance is
1/2
1 2
a) [𝑅 2 2 ]1/2
+ (𝐿𝜔 − 𝐶𝜔) b) [𝑅 + (𝐿𝜔 −
2
) ]
𝐶𝜔
−1/2 1/2
1 2 1 2
c) [𝑅 + (𝐿𝜔 −
2
) ] d) [(𝑅𝜔)2 + (𝐿𝜔 − ) ]
𝐶𝜔 𝐶𝜔
37. An 𝐿𝐶𝑅 series circuit is at resonance. Then
a) The phase difference between current and voltage is 90°
b) The phase difference between current and voltage is 45°
c) Its impedance is purely resistive
d) Its impedance is zero
38. The voltage of domestic ac is 220 𝑣𝑜𝑙𝑡. What does the represent
a) Mean voltage b) Peak voltage
c) Root mean voltage d) Root mean square voltage
39. In an ideal transformer, the voltage is stepped down from 11 kV to 220 V. If the primary current be 100 A,
the current in the secondary should be
a) 5 kA b) 1 kA c) 0.5 kA d) 0.1 kA
40. If an 8 Ω resistance and 6 Ω reactance are present in an ac series circuit then the impedance of the circuit
will be
a) 20 𝑜ℎ𝑚 b) 5 𝑜ℎ𝑚 c) 10 𝑜ℎ𝑚 d) 14√2 𝑜ℎ𝑚
41. An alternating current of frequency ‘𝑓’ is flowing in a circuit containing a resistance 𝑅 and a choke 𝐿 in
series. The impedance of this circuit is
a) 𝑅 + 2𝜋𝑓𝐿 b) √𝑅 2 + 4𝜋 2 𝑓 2 𝐿2 c) √𝑅 2 + 𝐿2 d) √𝑅 2 + 2𝜋𝑓𝐿
42. The process by which ac is converted into dc is known as
a) Purification b) Amplification c) Rectification d) Current amplification
43. The frequency of an alternating voltage is 50 𝑐𝑦𝑐𝑙𝑒𝑠/𝑠𝑒𝑐 and its amplitude is 120𝑉. Then the 𝑟. 𝑚. 𝑠. value
of voltage is
a) 101.3𝑉 b) 84.8𝑉 c) 70.7𝑉 d) 56.5𝑉
44. An inductor (L = 100 mH), a resistor (𝑅 = 100 Ω) and a battery (E = 100 V) are initially connected in
series as shown in figure. After a long time the battery is disconnected after short circuiting the points A
and B.
The current in the circuit 1 ms after the short circuit is

a) 1/𝑒 𝐴 b) 𝑒 𝐴 c) 0.1 A d) 1 A
45. 𝑅
has the dimensions to
𝐿

Page|4
 a) Time b) Mass c) Length d) Frequency
46. The instantaneous values of current and emf in an ac circuit are 𝐼 = 1/√2 sin 314 𝑡 𝑎𝑚𝑝 and 𝐸 =
√2 sin(314 𝑡 − 𝜋/6) 𝑉 respectively. The phase difference between 𝐸 and 𝐼 will be
a) −𝜋/6 𝑟𝑎𝑑 b) −𝜋/3 𝑟𝑎𝑑 c) 𝜋/6 𝑟𝑎𝑑 d) 𝜋/3 𝑟𝑎𝑑
47. The variation of the instantaneous current (𝐼)and the instantaneous 𝑒mf (𝐸) in a circuit is as shown in fig.
Which of the following statements is correct
E I

/2 3/2
O  2 t

a) The voltage lags behind the current by 𝜋/2 b) The voltage leads the current by 𝜋/2
c) The voltage and the current are in phase d) The voltage leads the current by 𝜋
48. In a L – R circuit, the value of L is (0.4)H and the value of R is 30 Ω. If in the circuit, an alternating emf of
𝜋
200 V at 50 cycle/s is connected, the impedance of the circuit and current will be
a) 11.4 Ω, 17.5 A b) 30.7 Ω, 6.5 A c) 40.4 Ω, 5 A d) 50 Ω, 4 A
49. In an 𝐴. 𝐶. circuit the current
a) Always leads the voltage b) Always lags behind the voltage
c) Is always in phase with the voltage d) May lead or lag behind or be in phase with the
voltage
50. A 100 V, AC source of frequency 500 Hz is connected to an L-C-R circuit with L=8.1 mH, 𝐶 = 12.5 𝜇F, 𝑅 =
10 Ω all connected in series as shown in figure. What is the quality factor of circuit?

a) 2.02 b) 2.5434 c) 20.54 d) 200.54

51. A constant voltage at different frequencies is applied across a capacitance 𝐶 as shown in the figure. Which
of the following graphs correctly depicts the variation of current with frequency
Signal
Generator
V C

a) I b) I c) I d) I

   

52. If the value of potential in an ac circuit is 10𝑉, then the peak value of potential is
10 20
a) b) 10√2 c) 20√2 d)
√2 √2
53. In the circuit shown in figure switch S is closed at time 𝑡 = 0. The charge which passes through the battery
in one time constant is

𝐸𝐿 𝑒𝐿 𝑒𝑅 2 𝐸 𝐿
a) b) c) d) 𝐸 ( )
𝑒 𝑅2 𝐸𝑅 𝐿 𝑅

Page|5
54. A transformer is used to light 140 W, 24 V lamp from 240 V AC mains. The current in the mains is 0.7 A.
The efficiency of transformer is nearest to
a) 90% b) 80% c) 70% d) 60%
55. In an 𝐿 − 𝑅 circuit to a battery, the rate at which energy is stored in the inductor is plotted against time
during the growth of current in the circuit. Which of the following, figure best represents the resulting
curve?

a)

b)

c)

d)

56. An ac source is rated at 220𝑉, 50 𝐻𝑧. The time taken for voltage to change from its peak value to zero is
a) 50 𝑠𝑒𝑐 b) 0.02 𝑠𝑒𝑐 c) 5 𝑠𝑒𝑐 d) 5 × 10−3 𝑠𝑒𝑐
57. The maximum voltage in DC circuit is 282V. The effective voltage in AC circuit will be
a) 200 V b) 300 V c) 400 V d) 564 V
58. The capacity of a pure capacitor is 1 𝑓𝑎𝑟𝑎𝑑. In dc circuits, its effective resistance will be
a) Zero b) Infinite c) 1 𝑜ℎ𝑚 d) 1/2 𝑜ℎ𝑚
59. An inductive circuit contains a resistance of 10 𝑜ℎ𝑚 and an inductance of 2.0 ℎ𝑒𝑛𝑟𝑦. If an ac voltage of
120 𝑣𝑜𝑙𝑡 and frequency of 60 𝐻𝑧 is applied to this circuit, the current in the circuit would be nearly
a) 0.32 𝑎𝑚𝑝 b) 0.16 𝑎𝑚𝑝 c) 0.48 𝑎𝑚𝑝 d) 0.80 𝑎𝑚𝑝
60. The time taken by an alternating current of 50 Hz in reaching from zero to its maximum value will be
a) 0.5 s b) 0.005 s c) 0.05 s d) 5 s
61. If coefficient of self induction of a coil is 1 H, an emf of 1 V is induced, if
a) Current flowing is 1 A b) Current variation rate is 1 As −1
c) Current of 1 A flows for one sec d) None of the above
62. A parallel plate capacitor C with plates of unit area and separation d is filled with a liquid of dielectric
𝑑
constant K = 2. The level of liquid is 3 initially. Suppose the liquid level decreases at a constant speed 𝑣, the
time constant as a function of time t is.

Page|6
 C
|

d R
𝑑/3
|

6𝜀0 𝑅 (15𝑑 + 9𝑣𝑡)𝜀0 𝑅 6𝜀0 𝑅 (15𝑑 − 9𝑣𝑡)𝜀0 𝑅

a) b) 2 c) d) 2
5𝑑 + 3𝑣𝑡 2𝑑 − 3𝑑𝑣𝑡 − 9𝑣 𝑡 2 2 5𝑑 − 3𝑣𝑡 2𝑑 + 3𝑑𝑣𝑡 − 9𝑣 2 𝑡 2
63. If the coils of a transformer are made up of thick wire, then
a) Eddy currents loss will be more b) Magnetic flux leakage is reduced
c) Joule’s heating loss is increased d) Joule’s heating loss is reduced
64. The peak value of 220 𝑣𝑜𝑙𝑡𝑠 of ac mains is
a) 155.6 𝑣𝑜𝑙𝑡𝑠 b) 220.0 𝑣𝑜𝑙𝑡𝑠 c) 311.0 𝑣𝑜𝑙𝑡𝑠 d) 440 𝑣𝑜𝑙𝑡𝑠
65. An alternating 𝑒mf is applied across a parallel combination of a resistance 𝑅, capacitance 𝐶 and an
inductance 𝐿. If 𝐼𝑅 , 𝐼𝐿 , 𝐼𝐶 are the current through 𝑅, 𝐿 and 𝐶 respectively, then the diagram which correctly
represents the phase relationship among 𝐼𝑅 , 𝐼𝐿 , 𝐼𝐶 and source 𝑒mf 𝐸, is given by
a) IL b) IR c) IC d) IR
E E E E
IR IL IR IC
IC IC IL IL

66. The time constant of the given circuit is

C

R
3R
R

3𝑅𝐶 6𝑅𝐶 5𝑅𝐶 d) None of these

a) b) c)
5 5 6
67. A solenoid has 2000 turns wound over a length of 0.30 m. The area of its cross section is 1.2 × 10−3 m2.
Around its central section, a coil of 300 turns is wound. If an initial current of 2 A in the solenoid is
reversed in 0.25 s, then the emf induced in the coil is equal to
a) 6 × 10−4 V b) 4.8 × 10−2 V c) 6 × 10−2 V d) 48 kV
68. The potential difference 𝑉 and the current 𝑖 flowing through an instrument in an ac circuit of frequency 𝑓
are given by 𝑉 = 5 cos 𝜔𝑡 𝑣𝑜𝑙𝑡𝑠 and 𝐼 = 2 sin 𝜔𝑡 𝑎𝑚𝑝𝑒𝑟𝑒𝑠 (where 𝜔 = 2𝜋𝑓). The power dissipated in the
instrument is
a) Zero b) 10 𝑊 c) 5 𝑊 d) 2.5 𝑊
69. An 𝑒.m.f. 𝐸 = 4 cos(1000𝑡) 𝑣𝑜𝑙𝑡 is applied to an LR-circuit of inductance 3 𝑚𝐻 and resistance 4 𝑜ℎ𝑚𝑠. The
amplitude of current in the circuit is
4 4
a) 𝐴 b) 1.0 𝐴 c) 𝐴 d) 0.8 𝐴
√7 7
70. A coil of inductive reactance 31Ω has a resistance of 8Ω. It is placed in series with a condenser of
capacitative reactance 25Ω. The combination is connected to an a.c. source of 110 𝑣𝑜𝑙𝑡. The power factor
of the circuit is
a) 0.80 b) 0.33 c) 0.56 d) 0.64
71. The expression for magnetic induction inside a solenoid of length 𝐿, carrying a current 𝑖 and having 𝑁
number of turns is
Page|7
 𝜇0 𝑁 𝜇0 𝑁2
a) 𝑖 b) 𝜇0 𝑁𝐿𝑖 c) 4 𝜋 𝑁𝐿𝑖 d) 𝜇0 𝑖
4𝜋 𝐿 𝐿
72. In an 𝐿𝑅-circuit, the inductive reactance is equal to the resistance 𝑅 of the circuit. An 𝑒.m.f. 𝐸 = 𝐸0 cos(𝜔𝑡)
is applied to the circuit. The power consumed in the circuit is
𝐸2 𝐸2 𝐸2 𝐸2
a) 0 b) 0 c) 0 d) 0
𝑅 2𝑅 4𝑅 8𝑅
73. In an AC circuit, the current lags behind the voltage by 𝜋/3. The components of the circuit are
a) R and L b) L and C c) R and C d) Only R
74. The instantaneous value of current in an A.C. circuit is 𝐼 = 2 sin(100 𝜋 𝑡 + 𝜋/3) 𝐴. The current will be
maximum for the first time at
1 1 1 1
a) 𝑡 = 𝑠 b) 𝑡 = 𝑠 c) 𝑡 = 𝑠 d) 𝑡 = 𝑠
100 200 400 600
75. A resistor 𝑅, an inductor 𝐿 and a capacitor 𝐶 are connected in series to an oscillator of frequency 𝑛, if the
resonant frequency is 𝑛𝑟 , then the current lags behind voltage, when
a) 𝑛 = 0 b) 𝑛 < 𝑛𝑟 c) 𝑛 = 𝑛𝑟 d) 𝑛 > 𝑛𝑟
76. During a current change from 2 A to 4 A in 0.05 s, 8 V of emf is developed in a coil. The coefficient of self-
induction is
a) 0.1 H b) 0.2 H c) 0.4 H d) 0.8 H
77. In an 𝐿 − 𝑅 circuit shown in above figure switch S is closed at time 𝑡 = 0. If 𝑒 denotes the induced emf
across inductor and 𝑖, the current in the circuit at any time 𝑡, then which of the following graphs, figure
shows the variation of 𝑒 with 𝑖?

a) b) c) d)

78. Let C be the capacitance of a capacitor discharging through a resister R. Suppose 𝑡1 is the time taken for
the energy stored in the capacitor to reduce to half its initial value and 𝑡2 is the time taken for the charge to
𝑡1
reduce to one-fourth its initial value. Then the ratio 𝑡2
will be
a) 1 1 1 d) 2
b) c)
2 4
79. The phase difference between the current and voltage of 𝐿𝐶𝑅 circuit in series combination at resonance is
a) 0 b) 𝜋/2 c) 𝜋 d) −𝜋
80. The impedance of a circuit consists of 3 𝑜ℎ𝑚 resistance and 4 𝑜ℎ𝑚 reactance. The power factor of the
circuit is
a) 0.4 b) 0.6 c) 0.8 d) 1.0
81. A 220 𝑉, 50 𝐻𝑧 ac source is connected to an inductance of 0.2 𝐻 and a resistance of 20 𝑜ℎ𝑚 in series. What
is the current in the circuit
a) 10 𝐴 b) 5 𝐴 c) 33.3 𝐴 d) 3.33 𝐴
82. A transformer is having 2100 turns in primary and 4200 turns in secondary. An AC source of 120 V, 10 A is
connected to its primary. The secondary voltage and current are
a) 240 V,5 A b) 120 V, 10 A c) 240 V, 10 A d) 120 V, 20 A
83. If instantaneous current is given by 𝑖 = 4 cos(𝜔𝑡 + 𝜙) 𝑎𝑚𝑝𝑒𝑟𝑒𝑠, then the 𝑟. 𝑚. 𝑠 value of current is
a) 4 amperes b) 2√2 amperes c) 4√2 amperes d) Zero amperes
84. When an AC source of emf 𝑒 = 𝐸0 sin(100𝑡) is connected across a circuit, the phase difference between
𝜋
the emf e and the current i in the circuit is observed to be 4 , as shown in the diagram. If the circuit
consists possibly only of R – C or R – L or L – C in series, find the relationship between the two elements

Page|8
 a) 𝑅 = 1 k Ω, 𝐶 = 10 𝜇F b) 𝑅 = 1 k Ω, 𝐶 = 1 𝜇F c) 𝑅 = 1 k Ω, 𝐿 = 10 H d) 𝑅 = 1 k Ω, 𝐿 = 1 H
85. If a current 𝐼 given by 𝐼 sin (𝜔𝑡 − 𝜋) flows in an ac circuit across, which an ac potential of 𝐸 = 𝐸 sin 𝜔𝑡
0 2 0
has been applied, then the power consumption 𝑃 in the circuit will be
𝐸0 𝐼0 𝐸 𝐼
a) 𝑃 = b) 𝑃 = √2𝐸0 𝐼0 c) 𝑃 = 0 0 d) 𝑃 = 0
√2 2
86. A resistance 𝑅, inductance 𝐿 and capacitor 𝐶 are connected in series to an oscillator of frequency 𝑓. If
resonant frequency is 𝑓, then current will lag the voltage when
a) 𝑓 = 0 b) 𝑓 < 𝑓𝑟 c) 𝑓 = 𝑓𝑟 d) 𝑓 > 𝑓𝑟
87. A generator produces a voltage that is given by 𝑉 = 240 sin 120 𝑡, where 𝑡 is in seconds. The frequency
and 𝑟. 𝑚. 𝑠. voltage are
a) 60 𝐻𝑧 and 240 V b) 19 𝐻𝑧 and 120 V c) 19 𝐻𝑧 and 170 V d) 754 𝐻𝑧 and 70 V
88. A 50 V AC is applied across an R-C (series) network. The rms voltage across the resistance is 40 V, then the
potential across the capacitance would be
a) 10 V b) 20 V c) 30 V d) 40 V
89. An alternating voltage 𝑒 = 200 sin 100 𝑡 is applied to a series combination 𝑅 = 30 Ω and an inductor of
400 mH. The power factor of the circuit is
a) 0.01 b) 0.2 c) 0.05 d) 0.6
90. Is it possible

15 A
10 A

5A

a) Yes b) No
c) Cannot be predicted d) Insufficient data to reply
91. The figure shows variation of 𝑅, 𝑋𝐿 and 𝑋𝐶 with frequency 𝑓 in a series 𝐿, 𝐶, 𝑅 circuit. Then for what
frequency 𝑓 in a series 𝐿, 𝐶, 𝑅 circuit. Then for what frequency point, the circuit is inductive
XC XL

f
AB C
a) 𝐴 b) 𝐵 c) 𝐶 d) All points
92. In the inductive circuit given in the figure, the current rises after the switch is closed. At instant when the
current is 15 𝑚𝐴, then potential difference across the inductor will be

Page|9
 a) Zero b) 240𝑉 c) 180𝑉 d) 60𝑉
93. If 𝐿 and 𝑅 represent inductance and resistance respectively, then dimension of 𝐿/𝑅 will be
a) [𝑀𝐿0 𝑇 0 ] b) [𝑀0 𝐿0 𝑇 −1 ] c) [𝑀0 𝐿0 𝑇 −2 ] d) [𝑀0 𝐿𝑇 −2 ]
94. Two identical incandescent light bulbs are connected as shown in figure. When the circuit is an 𝐴𝐶 voltage
source of frequency 𝑓, which of the following observation will be correct

a) Both bulbs will glow alternatively

1
b) Both bulbs will glow with same brightness provided 𝑓 = 2𝜋 √(1/𝐿𝐶)
c) Bulb 𝑏1 will light up initially and goes off, bulb 𝑏2 will be ON constantly
d) Bulb 𝑏1 will blink and bulb 𝑏2 will be ON constantly
95. When a coil carrying a steady current is short circuited, the current in it, decreases 𝜂 time in time 𝑡0 . The
time constant of the circuit is
𝑡0 𝑡0 𝑡0
a) b) c) 𝑡0 In 𝜂 d)
𝐼𝑛 𝜂 𝜂−1 𝜂
96. Following figure shows an ac generator connected to a “block box” through a pair of terminals. The box
contains possible 𝑅, 𝐿, 𝐶 or their combination, whose elements and arrangements are not known to us.
Measurements outside the box reveals t

𝑒 = 75 sin(sin 𝜔𝑡) 𝑣𝑜𝑙𝑡,

𝑖 = 1.5 sin(𝜔𝑡 + 45°) 𝑎𝑚𝑝. The wrong statement is
a) There must be a capacitor in the box b) There must be an inductor in the box
c) There must be a resistance in the box d) The power factor is 0.707
97. An L – C – R circuit of 𝑅 = 100 Ω is connected to an AC source 100 V, 50 Hz. The magnitude of phase
difference between current and voltage is 30°. The power dissipated in the L – C – R circuit is
a) 50 W b) 86.6 W c) 100 W d) 200 W
98. In a circuit L, C and R are connected in series with an alternating voltage source of frequency 𝑓. The
current leads the voltage by 45°. The value of C is
1 1 1 1
a) b) c) d)
2𝜋 𝑓 (2𝜋𝑓𝐿 + 𝑅) 𝜋𝑓 (2𝜋𝑓𝐿 + 𝑅) 2𝜋𝑓 (2𝜋𝑓𝐿 − 𝑅) 𝜋𝑓 (2𝜋𝑓𝐿 − 𝑅)
99. If the total charge stored in the 𝐿𝐶 circuit is 𝑄0 , then for 𝑡 ≥ 0
𝜋 𝑡
a) The charge on the capacitor is 𝑄 = 𝑄0 cos ( 2 + )
√𝐿𝐶
𝜋 𝑡
b) The charge on the capacitor is 𝑄 = 𝑄0 cos ( 2 − )
√𝐿𝐶
𝑑2 𝑄
c) The charge on the capacitor is 𝑄 = −𝐿𝐶 2
𝑑𝑡
1 𝑑2 𝑄
d) The charge on the capacitor is 𝑄 =
√𝐿𝐶 𝑑𝑡 2
100. In L – C – R series circuit the resonance condition in terms of capacitive reactance (𝑋𝐶 ) and inductive
reactance (𝑋𝐿 ) is
a) 𝑋𝐶 + 𝑋𝐿 = 0 b) 𝑋𝐶 = 0 c) 𝑋𝐿 = 0 d) 𝑋𝐶 − 𝑋𝐿 = 0
P a g e | 10
101. What is the average value of the AC voltage over one complete cycle?
a) Zero 2𝑉 𝑉
b) 𝑉max c) max d) max
𝜋 2
102. A current of 10 A in the primary coil of a circuit is reduced to zero. If the coefficient f mutual inductance is
3H and emf induced in secondary coil is 30 kV, time taken for the change of current is
a) 103 s b) 102 s c) 10−3 s d) 10−2 s
103. A square metal wire loop 𝑃𝑄𝑅𝑆 of side 10 cm and resistance 1 Ω is moved with a constant velocity 𝑣𝑐 in a
uniform magnetic field of induction 𝐵 = 2 Wbm2, as shown in figure. The magnetic field lines are
perpendicular to the plane of the loop (directed into the paper). The loop is connected to network 𝐴𝐵𝐶𝐷 of
resistors each of value 3 Ω. The resistance of the lead wires 𝑆𝐵 an𝑑 𝑅𝐷 are negligible. The speed of the
loop so as to have a steady current of mA in the loop is

a) 2 ms −1 b) 2 × 10−2 ms−1 c) 20 ms−1 d) 200 ms−1

104. A conducting rod 𝑃𝑄 𝑜𝑓 𝑙𝑒𝑛𝑔𝑡ℎ 𝐿 = 1.0 m is moving with a uniform speed 𝑣 = 2.0 ms−1 in a uniform
magnetic field = 4.0 T directed into the paper. A capacitor of capacity 𝐶 = 10 𝜇𝐹 is connected as shown in
figure. Then,

a) 𝑞𝐴 = −80𝜇 𝐶 𝑎𝑛𝑑 𝑞𝐵 = +80𝜇 𝐶 b) 𝑞𝐴 = +80𝜇 𝐶 𝑎𝑛𝑑 𝑞𝐵 = −80𝜇 𝐶

d) Charge stored in the capacitor increases
c) 𝑞𝐴 = 0 = 𝑞𝐵
expotentially with time
105. For a large industrial city with much load variations , the DC generator should be
a) Series wound b) Shunt wound c) Mixed wound d) Any
106. The self inductance of the motor of an electric fan is 10𝐻. In order to impart maximum power at 50 𝐻𝑧, it
should be connected to a capacitance of
a) 1 𝜇𝐹 b) 2 𝑚𝐹 c) 4 𝑚𝐹 d) 8 𝑚𝐹
107. In L – C – R circuit, an alternating emf of angular frequency𝜔 is applied then the total impedance will be
1
1/2 −
1 2 2 2
a) [(𝑅𝜔)2 + (𝐿𝜔 − ) ] b) [𝑅 2 + (𝐿𝜔 − 1 ) ]
𝐶𝜔 𝐶𝜔
1/2
2
c) [𝑅 2 2 ]1/2
+ (𝐿𝜔 − 𝐶𝜔) d) [𝑅 2 + (𝐿𝜔 − 1 ) ]
𝐶𝜔
108. 2.5 𝜇𝐹 capacitor and 3000-𝑜ℎ𝑚 resistance are joined in series to an ac source of 200 𝑣𝑜𝑙𝑡 and 50𝑠𝑒𝑐 −1
𝜋
frequency. The power factor of the circuit and the power dissipated in it will respectively be
a) 0.6, 0.06 𝑊 b) 0.06, 0.6 𝑊 c) 0.6, 4.8 𝑊 d) 4.8, 0.6 𝑊
109. For series 𝐿𝐶𝑅 circuit, wrong statement is
a) Applied 𝑒.m.f. and potential difference across resistance are in same phase
b) Applied 𝑒.m.f. and potential difference at inductor coil have phase difference of 𝜋/2
c) Potential difference at capacitor and inductor have phase difference of 𝜋/2
d) Potential difference across resistance and capacitor have phase difference of 𝜋/2
110. An ideal coil of 10 H is connected in series with a resistance of 5 Ω and a battery of 5 V. 2s after the
P a g e | 11
 connection is made, the current flowing (in ampere) in the circuit is
a) (1 − 𝑒) b) e c) 𝑒 −1 d) (1 − 𝑒 −1 )
111. If a current of 3 A flowing in the primary coil is reduced to zero in 0.001 s, the induced emf in between the
two coils is 15000 V, the coefficient of mutual induction is
a) 0.5 H b) 5 H c) 1.5 H d) 10 H
112. The power factor of 𝐿𝐶𝑅 circuit at resonance is
a) 0.707 b) 1 c) Zero d) 0.5
113. At time t = 0, a battery of 10 V is connected across points A and B in the given circuit. If the capacitors have
no charge initially, at what time (in second) does the voltage across them become 4 V?
(Take ln 5 = 1.6, ln 3 = 1.1)
2M Ω 2μF

A B

2M Ω 2μF

a) 2 b) 3 c) 2.5 3
d)
2
114. An air cored coil has a self-inductance of 0.1 H. A soft iron core of relative permeability 100 is 1/10 th. The
value of self-inductance now becomes
a) 1 mH b) 10 mH c) 0.4 H d) 0.8 H
115. The armature of a shunt wound motor can with stand current up to 8A before it overheats and it damaged.
If the armature resistance is 0.5 Ω, minimum back emf that must be motor is connected to a 120 V line is
a) 120 V b) 116 V c) 124 V d) 4 V
116. In the circuit shown below what will be the readings of the voltmeter and ammeter? (Total impedance of
circuit 𝑍 = 100 Ω)

a) 200 V, 1 A b) 800 V, 2 A c) 100 V, 2 A d) 220 V, 2.2 A

117. In the non-resonant circuit, what will be the nature of the circuit for frequencies higher than the resonant
frequency
a) Resistive b) Capacitive c) Inductive d) None of the above
118. In AC circuit a resistance of 𝑅 Ω is connected in series with an inductance L. If the phase difference
between the current and voltage is 45°, the inductive reactance will be
a) R/2 b) R/4 c) R d) None of the above
119. The current in series 𝐿𝐶𝑅 circuit will be maximum when 𝜔 is
a) As large as possible b) Equal o natural frequency of 𝐿𝐶𝑅 system
c) √𝐿𝐶 d) √1/𝐿𝐶
120. Two conducting circular loops of radii 𝑅1 and 𝑅2 are placed in the same plane with their centres
coinciding. If 𝑅1 > 𝑅2 , the mutual inductance M between them will be directly proportional to
𝑅1 𝑅2 𝑅2 𝑅2
a) b) c) 1 d) 2
𝑅2 𝑅1 𝑅2 𝑅1
121. Which of the following quantities remains constant in a step-down transformer ?
a) Current b) Voltage c) Power d) None of these
122. The voltage of an ac source varies with time according to the equation 𝑉 = 100 sin 100𝜋𝑡 cos 100𝜋𝑡 where
𝑡 is in second and 𝑉 is in volts. Then

P a g e | 12
 a) The peak voltage of the source is 100 𝑣𝑜𝑙𝑡𝑠
b) The peak voltage of the source is 50 𝑣𝑜𝑙𝑡𝑠
c) The peak voltage of the source is 100/√2 𝑣𝑜𝑙𝑡𝑠
d) The frequency of the source is 50 𝐻𝑧
123. At high frequency, the capacitor offer
a) More reactance b) Less reactance c) Zero reactance d) Infinite reactance
124. A circuit has a resistance of 12 Ω and an impedance of 15 Ω. The power factor of the circuit will be
a) 0.8 b) 0.4 c) 1.25 d) 0.125
125. An inductance of 1 𝑚𝐻 a condenser of 10 𝜇𝐹 and a resistance of 50 Ω are connected in series. The
reactances of inductor and condensers are same. The reactance of either of them will be
a) 100 Ω b) 30 Ω c) 3.2 Ω d) 10 Ω
126. The current flowing in a step down transformer 220 V to 22 V having impedance 220 Ω is
a) 1 A b) 0.1 A c) 2 mA d) 0.1 mA
127. If 𝐸 = 100 sin(100𝑡) volt and 𝐼 = 100 sin (100𝑡 + ) 𝑚𝐴 are the instantaneous values of voltage and
𝜋
3
current, then the 𝑟. 𝑚. 𝑠. values of voltage and current are respectively
a) 70.7𝑉, 70.7𝑚𝐴 b) 70.7𝑉, 70.7𝐴 c) 141.4𝑉, 141.4𝑚𝐴 d) 141.4𝑉, 141.4𝐴
128. An ideal choke draws a current of 8 A when connected to an AC supply of 100 V, 50 Hz. A pure resistor
draws a current of 10 A when connected to the same source. The ideal choke and the resister are
connected in series and then connected to the AC source of 150 V, 40 Hz. The current in the circuit
becomes
15 b) 8 A c) 18 A d) 10 A
a) A
√2
129. If 𝐴 and 𝐵 are identical bulbs, which bulb glows brighter
100 mH A

10 pF B

a) 𝐴 b) 𝐵 c) Both equally bright d) Cannot say

130. A 280 𝑜ℎ𝑚 electric bulb is connected to 200𝑉 electric line. The peak value of current in the bulb will be
a) About one ampere b) Zero c) About two ampere d) About four ampere
131. If 𝐸0 represents the peak value of the voltage in an ac circuit, the 𝑟. 𝑚. 𝑠 value of the voltage will be
𝐸 𝐸 𝐸0 𝐸0
a) 0 b) 0 c) d)
𝜋 2 √𝜋 √2
132. In L – R circuit, resistance is 8 Ω and inductive reactance is 6 Ω , then impedance is
a) 2 Ω b) 14 Ω c) 4 Ω d) 10 Ω
133. The root mean square value of the alternating current is equal to
a) Twice the peak value b) Half the peak value
1
c) times the peak value d) Equal to the peak value
√2
134. What will be the phase difference between virtual voltage and virtual current, when the current in the
circuit is wattles
a) 90° b) 45° c) 180° d) 60°
135. Power factor is maximum in an 𝐿𝐶𝑅 circuit when
a) 𝑋𝐿 = 𝑋𝐶 b) 𝑅 = 0 c) 𝑋𝐿 = 0 d) 𝑋𝐶 = 0
136. The output current versus time curve of a rectifier is shown in the figure. The average value of output
current in this case is

P a g e | 13
 Current

I0

Time

a) 0 𝐼0 2𝐼
b) c) 0 d) 𝐼0
2 𝜋
137. In a series resonant L – C – R circuit, the voltage across R is 100 V and 𝑅 = 1 𝑘 Ω with 𝐶 = 2𝜇F. The
resonant frequency 𝜔 is 200 rads −1 . At resonance the voltage across L is
a) 2.5 × 10−2 V b) 40 V c) 250 V d) 4 × 10−3 V
138. In the previous question, if the direction of 𝑖 is reversed, (𝑉𝐵 − 𝑉𝐴 ) will be
a) 20 V b) 15 V c) 10 V d) 5 V
139. The instantaneous voltage through a device of impedance 20 Ω is 𝑒 = 80 sin 100 𝜋𝑡. The effective value of
the current is
a) 3 A b) 2.828 A c) 1.732 A d) 4 A
140. In an R-C circuit while charging, the graph of lnI versus time is as shown by the dotted line in the adjoining
diagram where I is the current. When the value of the resistance is doubled, which of the solid curves best
represents the variation of ln I versus time?

a) P b) Q c) R d) S
141. The resistance of a coil for dc is in ohms. In ac, the resistance
a) Will remain same b) Will increase c) Will decrease d) Will be zero
142. The current 𝑖 in the circuit shown here varies with time 𝑡 is

a) b) c) d)

143. A circuit has a resistance of 11Ω, an inductive reactance of 25Ω and a capacitative resistance of 18Ω. It is
connected to an ac source of 260𝑉 and 50𝐻𝑧. The current through the circuit (in amperes) is
a) 11 b) 15 c) 18 d) 20
144. The reading of ammeter in the circuit shown will be
A

XC = 5
V 110 V

XL = 5 R = 55

a) 2𝐴 b) 2.4 𝐴 c) Zero d) 1.7 𝐴

145. A step-up transformer is used on a 120 V line to provide a potential difference of 2400 V. If the primary
coil has 75 turns, the number of turns in the secondary coil is
P a g e | 14
 a) 150 b) 1200 c) 1500 d) 1575
146. A coil of inductance 300 mH and resistance 2 Ω is connected to a source of voltage 2V. The current reaches
half of its steady state value in
a) 0.05 s b) 0.1 s c) 0.15 s d) 0.3 s
147. An alternating 𝑒.m.f. of angular frequency 𝜔 is applied across an inductance. The instantaneous power
developed in the circuit has an angular frequency
a) 𝜔/4 b) 𝜔/2 c) 𝜔 d) 2𝜔
148. A 10 𝑜ℎ𝑚 resistance, 5 𝑚𝐻 coil and 10 𝜇𝐹 capacitor are joined in series. When a suitable frequency
alternating current source is joined to this combination, the circuit resonates. If the resistance is halved,
the resonance frequency
a) Is halved b) Is doubled c) Remains unchanged d) In quadrupled
149. There is a 5Ω resistance in an ac, circuit. Inductance of 0.1𝐻 is connected with it in series. If equation of ac
𝑒.m.f. is 5 sin 50𝑡, then the phase difference between current and 𝑒.m.f. is
𝜋 𝜋 𝜋 d) 0
a) b) c)
2 6 4
150. In the alternating current shown in the figure, the currents through inductor and capacitor are 1.2 𝑎𝑚𝑝
and 1.0 𝑎𝑚𝑝 respectively. The current drawn from the generator is

a) 0.4 𝑎𝑚𝑝 b) 0.2 𝑎𝑚𝑝 c) 1.0 𝑎𝑚𝑝 d) 1.2 𝑎𝑚𝑝

151. In a region of uniform magnetic induction 𝐵 = 10 𝑡𝑒𝑠𝑙𝑎, a circular coil of radius 30 𝑐𝑚 and resistance
−2

𝜋 2 𝑜ℎ𝑚 is rotated about an axis which is perpendicular to the direction of 𝐵 and which forms a diameter of
the coil. If the rotates at 200 𝑟𝑝𝑚 the amplitude of the alternating current induced in the coil is
a) 4𝜋 2 𝑚𝐴 b) 30 𝑚𝐴 c) 6 𝑚𝐴 d) 200 𝑚𝐴
152. In an L – C – R circuit, capacitance is changed from C to 2C. For the resonant frequency to remain
unchanged, the inductance should be changed from L to
a) 4L b) 2L c) 𝐿/2 d) 𝐿/4
153. A bulb and a capacitor are in series with an ac source. On increasing frequency how will glow of the bulb
change
a) The glow decreases b) The glow increases
c) The glow remain the same d) The bulb quenches
154. An alternating voltage is represented as 𝐸 = 20 sin 300𝑡. The average value of voltage over one cycle will
be
a) Zero 20
b) 10 𝑣𝑜𝑙𝑡 c) 20√2 𝑣𝑜𝑙𝑡 d) 𝑣𝑜𝑙𝑡
√2
155. The magnet in figure rotates a shown on a pivot through its center. At the instant shown, what are the
directions of the induced currents.

a) 𝐴 to 𝐵 and 𝐶 to 𝐷 b) 𝐵 to 𝐴 and 𝐶 to 𝐷
c) 𝐴 to 𝐵 and 𝐷 to 𝐶 d) 𝐵 to 𝐴 and 𝐷 to 𝐶
156. A magnet is suspended lengthwise from a spring and while it oscillates, the magnet moves in and out of the
coil C connected to a galvanometer G. Then as the magnet oscillates.
a) G shows no deflection b) G shows deflection on one side
c) Deflection of G to the left and right has constant d) Deflection of G to the left and right has decreasing
amplitude amplitude

P a g e | 15
157. Current growth in two 𝐿 − 𝑅 circuits (ii) and (iii) is as shown in figure (i). Let 𝐿1 , 𝐿2 , 𝑅1 and 𝑅2 be the
corresponding values in two circuits. Then

a) 𝐿1 > 𝐿2 b) 𝐿1 < 𝐿2 c) 𝑅1 > 𝑅2 d) 𝑅1 = 𝑅2

158. An electric heater rated 220 V and 550 W is connected to A.C. mains. The current drawn by it is
a) 0.8 𝐴 b) 2.5 𝐴 c) 0.4 𝐴 d) 1.25 𝐴
159. A resistor and a capacitor are connected in series with an AC source. If the potential drop across the
capacitor is 5 V and that across resistor is 12 V, then applied voltage is
a) 13 V b) 17 V c) 5 V d) 12 V
160. An inductor of 2 H and a resistance of 10 Ω are connected in series with a battery of 5 V. the initial rate of
change of current is
a) 0.5 As −1 b) 2.0 As −1 c) 2.5 As −1 d) 0.25 As −1
161. A conducting wire frame is placed in a magnetic field, which is directed into the paper, figure. The
magnetic field is increasing at a constant rate. The directions of induced current in wires 𝐴𝐵 and 𝐶𝐷 are

a) 𝐴 to B and 𝐶 to 𝐷 b) 𝐵 to 𝐴 and 𝐶 to 𝐷
c) 𝐴 to 𝐵 and 𝐷 to 𝐶 d) 𝐵 to 𝐴 and 𝐷 to 𝐶
162. A pure inductive coil of 30 mH is connected to an AC source of 220 V, 50 Hz. The rms current in the coil is
a) 50.35 A b) 23.4 A c) 30.5 A d) 12.3 A
163. In an ac circuit, 𝑉 and 𝐼 are given by
𝜋
𝑉 = 100 sin (100 𝑡) 𝑣𝑜𝑙𝑡𝑠, 𝐼 = 100 sin (100𝑡 + 3 ) 𝑚𝐴. The power dissipated in circuit is
a) 104 𝑤𝑎𝑡𝑡 b) 10 𝑤𝑎𝑡𝑡 c) 2.5 𝑤𝑎𝑡𝑡 d) 5 𝑤𝑎𝑡𝑡
164. Two connectric and coplanar circular coils have radii 𝑎 and 𝑏 as shown in figure. Resistance of the inner
coil is R. Current in the other coil is increased from 𝑂 to 𝑖, then the total charge circulating the inner coil is

𝜇0 𝑖𝑎𝑏 𝜇 𝑖 𝑎 𝜋𝑏 2 𝜇 𝑖𝑏 𝜇 𝑖 𝑎2
a) b) 0 c) 0 d) 0
2𝑅 2𝑎𝑏 2𝜋𝑅 2 𝑅𝑏
165. A circuit area is 0.01 m2 is kept inside a magnetic field which is normal to its plane. The magnetic field
changes from 2 T to 1 T in 1 millisecond. If the resistance of the circuit is 2Ω. The amount of heat evolved is
a) 0.05 J b) 50 J c) 0.50 J d) 500 J
166. In an AC circuit the emf(e) and the current (i) at any instant are given respectively by
𝑒 = 𝐸0 sin 𝜔𝑡
𝑖 = 𝐼0 sin(𝜔𝑡 − ϕ)
The average power in the circuit over one cycle of AC is
P a g e | 16
 𝐸0 𝐼0 𝐸 𝐼 𝐸 𝐼
a) b) 0 0 sin 𝜙 c) 0 0 cos 𝜙 d) 𝐸0 𝐼0
2 2 2
167. A copper rod of mass m slides under gravity on two smooth parallel rails 𝑙 distance apart and set at an
angle θ to the horizontal. At the bottom, the rails are joined by a resistance R, figure. There is a uniform
magnetic field B perpendicular to the plane of the rails. The terminal velocity of the rod is

𝑚g𝑅 tanθ 𝑚g𝑅 cotθ 𝑚g𝑅 sinθ 𝑚g𝑅 cosθ

a) 2 2
b) 2 2
c) 2 2
d)
B 𝑙 𝐵 𝑙 𝐵 𝑙 𝐵2 𝑙 2
168. Reactance of a capacitor of capacitance 𝐶𝜇𝐹 for ac frequency 400
𝐻𝑧 is 25Ω. The value 𝐶 is
𝜋
a) 50𝜇𝐹 b) 25𝜇𝐹 c) 100𝜇𝐹 d) 75𝜇𝐹
169. A choke coil has
a) High inductance and low resistance b) Low inductance and high resistance
c) High inductance and high resistance d) Low inductance and low resistance
170. Two coils 𝐴 and 𝐵 have coefficient of mutual inductance M = 2H. The magnetic flux passing through coil 𝐴
changes by 4 Wb in 10 s due to change in current in B. Then
a) Change in current in B in this time interval is 0.5 A b) Change in current in B in this time interval is 8 A
c) The change in current in B in this time interval is 2 d) A change in current of 1 A in coil A will produce a
A change in flux passing through B by 4 Wb
171. In an ac circuit the reactance of a coil is √3 times its resistance, the phase difference between the, voltage
across the coil to the current through the coil will be
a) 𝜋/3 b) 𝜋/2 c) 𝜋/4 d) 𝜋/6
172. The phase difference between the voltage and the current in an ac circuit is 𝜋/4. If the frequency is 50 𝐻𝑧
then this phase difference will be equivalent to a time of
a) 0.02 𝑠 b) 0.25 𝑠 c) 2.5 𝑚𝑠 d) 25 𝑚𝑠
173. In 𝐴𝐶 series circuit, the resistance, inductive reactance and capacitive reactance are 3Ω, 10Ω and 14Ω
respectively. The impedance of the circuit is
a) 5Ω b) 4Ω c) 7Ω d) 10Ω
174. The voltage across a pure inductor is represented by the following diagram. Which of the following
diagrams will represent the current
V

a) i b) i c) i d) i

t t t t

175. Q-factor can be increased by having a coil of

a) Large inductance, small ohmic resistance
b) Large inductance, large ohmic resistance
c) Small inductance, large ohmic resistance
d) Small inductance, small ohmic resistance
176. The current which does not contribute to the power consumed in an AC circuit is called
a) non-ideal current b) wattles current

P a g e | 17
 c) convectional current d) inductance current
177. In the circuit shown in figure, a conducting wire 𝐻𝐸 is moved with a constant speed 𝑣 towards legt. Th
⃗ perpendicular to the plane of circuit inwards. The
complete circuit is placed in a uniform magnetic field 𝐵
current in 𝐻𝐾𝐷𝐸 is

a) Anti-clock-wise b) Clock-wise c) Alternating d) Zero

178. The current passing through a choke coil of 5 H is decreasing at the rate of 2 As −1 . The emf developed
across the coil is
a) −10 V b) + 10 V c) 2.5 V d) −2.5 V
179. A light bulb is rated 100 𝑊 for a 220 𝑉 supply. The resistance of the bulb and the peak voltage of the
source respectively are
a) 242 Ω and 311 𝑉 b) 484 Ω and 311 𝑉 c) 484 Ω and 440 𝑉 d) 242 Ω and 440 𝑉
180. If number of turns in primary and secondary coils is increased to two times each, the mutual inductance
a) Becomes 4 times b) Becomes 2 times
c) Becomes 1 /4 times d) Remains unchanged
181. An 𝐿𝐶𝑅 circuit contains 𝑅 = 50 Ω, 𝐿 = 1 𝑚𝐻 and 𝐶 = 0.1 𝜇𝐹. The impedance of the circuit will be
minimum for a frequency of
105 −1 106 −1
a) 𝑠 b) 𝑠 c) 2𝜋 × 105 𝑠 −1 d) 2𝜋 × 106 𝑠 −1
2𝜋 2𝜋
182. A metal rod of resistance 20 Ω is fixed along a diameter of a conducting ring of radius 0.1 m and lies on 𝑥 −
⃗ = (50 T) 𝑘̂. The ring rotates with an angular velocity 𝜔 =
𝑦 plane. There is a magnetic field ⃗𝑩
20 rads −1 about its axis. An external resistance of 10 Ω is connected across the centre of the ring and rim.
The current through external resistance is
1
a) A
1
b) A
1
c) A d) zero
2 3 4
183. A 12 𝑜ℎ𝑚 resistor and a 0.21 henry inductor are connected in series to an ac source operating at 20 𝑣𝑜𝑙𝑡𝑠,
50 cycle/second. The phase angle between the current and the source voltage is
a) 30° b) 40° c) 80° d) 90°
184. The ratio of peak value and 𝑟. 𝑚. 𝑠. value of an alternating current is
a) 1 1
b) c) √2 d) 1/√2
2
185. In an induction coil, the coefficient of mutual inductance is 4H. If current of 5A in the primary coil is cut off
𝑖 1/ 1500s, the emf at the terminals of the secondary coil will be
a) 15 kV b) 60 kV c) 10 kV d) 30 kV
186. The coil of choke in a circuit
a) Increases the current b) Decreases the current
c) De not change the current d) Has high resistance to dc circuit
187. In the L-C-R circuit shown, the impedance is

P a g e | 18
 L C R

1H 20 μF 300 Ω

a) 500 Ω b) 300 Ω c) 100 Ω d) 200 Ω

188. The frequency of ac mains in India is
a) 30 𝑐/𝑠 or 𝐻𝑧 b) 50 𝑐/𝑠 or 𝐻𝑧 c) 60 𝑐/𝑠 or 𝐻𝑧 d) 120 𝑐/𝑠 or 𝐻𝑧
189. In the circuit shown in the figure, the ac source gives a voltage 𝑉 = 20 cos(2000𝑡). Neglecting source
resistance, the voltmeter and ammeter reading will be
6
A

5mH 4 50F

a) 0𝑉, 0.47𝐴 b) 1.68𝑉, 0.47𝐴 c) 0𝑉, 1.4 𝐴 d) 5.6𝑉, 1.4 𝐴

190. An 𝐿𝐶𝑅 series ac circuit is at resonance with 10 𝑉 each across 𝐿, 𝐶 and 𝑅. If the resistance is halved, the
respective voltage across 𝐿, 𝐶 and 𝑅 are
a) 10 V, 10 V and 5 V b) 10 V, 10 V and 10 V c) 20 V, 20 V and 5 V d) 20 V, 20 V and 10 V
191. The readings of ammeter and voltmeter in the following circuit are respectively

a) 2𝐴, 200𝑉 b) 1.5𝐴, 100𝑉 c) 2.7𝐴, 220𝑉 d) 2.2𝐴, 220𝑉

192. A rectangular loop with a sliding connector of length 𝑙 = 1.0 m is situated in a uniform magnetic field B =
2T. Perpendicular to the plane of loop. Resistance of connector is 𝑟 = 2Ω. Two resistance of 6 Ω and 3 Ω
are connected as shown in figure. The external force required to keep the connector moving with a
constant velocity 𝑣 = 2 ms −1 is
× B
6Ω v 3Ω

a) 2 N b) 1 N c) 4 N d) 6 N
193. What is the 𝑟. 𝑚. 𝑠. value of an alternating current which when passed through a resistor produces heat
which is thrice of that produced by a direct current of 2 amperes in the same resistor
a) 6 𝑎𝑚𝑝 b) 2 𝑎𝑚𝑝 c) 3.46 𝑎𝑚𝑝 d) 0.66 𝑎𝑚𝑝
194. A bulb is connected first with dc and then ac of same voltage it will shine brightly with
a) AC b) DC
c) Brightness will be in ratio 1/1.4 d) Equally with both
195. If an alternating voltage is represented as 𝐸 = 141 sin(628 𝑡), then the rms value of the voltage and the
frequency are respectively
a) 141 V, 628 Hz b) 100 V, 50 Hz c) 100 V, 100 Hz d) 141 V, 100 Hz

P a g e | 19
196. Some magnetic flux is changed from a coil of resistance 110 Ω. As a result, an induced current is developed
in it, which varies with time as shown in figure. The magnitude of change in flux through the coil in weber
is

a) 4 b) 8 c) 2 d) 6
197. Two coils 𝐴 and 𝐵 have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of
10−3 Wb to link with 𝐴 and a flux per turn of 0.8× 10−3 Wb through B. The ratio of mutual inductance of
𝐴 and 𝐵 is
a) 0.625 b) 1.25 c) 1.5 d) 1.625
198. 220 V, 50 Hz AC is applied to a resistor. The instantaneous value of voltage is
a) 220√2 sin 100𝜋𝑡 b) 220 sin 100𝜋𝑡 c) 220√2 sin 50𝜋𝑡 d) 220 sin 50𝜋𝑡
199. Two circuits have mutual inductance of 0.09 H. Average emf induced in the secondary by a change of
current from 0 to 20 A in 0.006 s in primary will be
a) 120 V b) 200 V c) 180 V d) 300 V
200. One 10 𝑉, 60 𝑊 bulb is to be connected to 100 𝑉 line. The required induction coil has self inductance of
value (𝑓 = 50 𝐻𝑧)
a) 0.052 𝐻 b) 2.42 𝐻 c) 16.2 𝑚𝐻 d) 1.62 𝑚𝐻
201. What is self inductance of a coil which produces 5V, when current in it changes from 3 A to 2 A in one
millisecond?
a) 5000 H b) 5 mH c) 50 H d) 5 H
202. The self inductance of a choke coils is 10 𝑚𝐻. When it is connected with a 10𝑉 dc source, then the loss of
power is 20 𝑤𝑎𝑡𝑡. When it is connected with 10 𝑣𝑜𝑙𝑡 ac source loss of power is 10 𝑤𝑎𝑡𝑡. The frequency of
ac source will be
a) 50 𝐻𝑧 b) 60 𝐻𝑧 c) 80 𝐻𝑧 d) 100 𝐻𝑧
203. In the circuit shown in figure neglecting source resistance, the voltmeter and ammeter readings will be
respectively

a) 0 V, 3 A b) 150 V, 3 A c) 150 V, 6 A d) 0 V, 8 A
204. A coil has resistance 30 ohm and inductive reactance 20 Ohm at 50 Hz frequency. If ac source, of 200 volt,
100 Hz, is connected across the coil, the current in the coil will be
20
a) 𝐴 b) 2.0𝐴 c) 4.0𝐴 d) 8.0𝐴
√13
205. Eddy current are produced when
a) A metal is kept in varying magnetic field b) A metal is kept in steady magnetic field
c) A circular coil is placed in a magnetic field d) Through a circular coil, current is passed
206. Average power generated in an inductor connected to an AC source is
1 c) Zero d) None of these
a) 𝐿𝑖 2 b) 𝐿𝑖 2
2
207. The resonant frequency of a circuit is 𝑓. If the capacitance is made 4 times the initial values, then the
resonant frequency will become
P a g e | 20
 a) 𝑓/2 b) 2𝑓 c) 𝑓 d) 𝑓/4
208. Three identical ring move with the same speed on a horizontal surface in a uniform horizontal magnetic
field normal to the planes of the rings. The first (𝐴) slips without rolling, the second (B) rolls without
slipping, and the third rolls with slipping
a) The same emf is induced in all the three rings b) No emf is induced in any of the rings
c) In each ring, all points are at the same potential B developes the maximum induced emf, and 𝐴 the
d)
least.
209. Two coils are at fixed locations. When coil 1 has no current and the current in the coil 2 increases at the
rate 15.0 As −1, the emf in coil 1 is 25.0 mV. When coil 2 has no current of 3.6 A. The flux linkage in coil 2
a) 4 mWb b) 6 mWb c) 10 mWb d) 16 mWb
210. In Colpitt oscillator the feedback network consists of
a) Two inductors and a capacitor b) Two capacitors and an inductor
c) Three pairs of R-C circuit d) Three pairs of R-L circuit
211. A choke is preferred to a resistance for limiting current in AC circuit because
a) Choke is cheap b) There is no wastage of power
c) Choke is compact in size d) Choke is a good absorber of heat
212. The induced emf of a generator when the flux of poles is doubled and speed is doubled
a) Becomes half b) Remains same
c) Becomes double d) Becomes 4 times
213. In an AC circuit, the instantaneous values of emf and current are 𝑒 = 200 sin 314𝑡 volt and 𝐼 =
𝜋
sin (314𝑡 + ) amp. The average power consumed in watt is
3
a) 200 b) 100 c) 50 d) 25
214. An emf is 15 V is applied in a circuit coil containing 5 H inductance and 10 Ω resistance. The ratio of
currents at time 𝑡 = ∞ and 𝑡 = 1 s is
𝑒 1/2 𝑒2
a) b) c) 1 − 𝑒 −1 d) 𝑒 −1
𝑒 1/2 − 1 𝑒2 − 1
215. For a series L – C – R circuit, the phase difference between current and voltage at the condition of
resonance will be
𝜋 𝜋 c) Zero d) Nothing can be said
a) b)
2 4
216. Which of the following components of a L – C – R circuit, with AC supply, dissipates energy?
a) L b) R c) C d) All of these
217. An AC voltage source has an output of Δ𝑉 = (200V) sin 2𝜋𝑓𝑡. This source is connected to a 100 Ω resistor.
RMS current in the resistance is
a) 1.41 A b) 2.41 A c) 3.41 A d) 0.71 A
218. In a pure inductive circuit or In an ac circuit containing inductance only, the current
a) Leads the 𝑒.m.f. by 90° b) Lags behind the 𝑒.m.f. by 90°
Sometimes leads and sometimes lags behind the
c) d) Is in phase with the 𝑒.m.f.
𝑒.m.f.
219. An inductance of (200)mH, a capacitance of (10−3 )F and a resistance of 10 Ω are connected in series with
𝜋 𝜋
an AC source 220 V, 50 Hz. The phase angle of the circuit is
𝜋 𝜋 𝜋 𝜋
a) b) c) d)
6 4 2 3
220. The ratio of turns in primary and secondary coils of a transformer is 1 : 20. The ratio of currents in
primary and secondary coils will be
a) 1 : 20 b) 20 : 1 c) 1 : 400 d) 400 : 1
221. A group of electric lamps having a total power rating of 1000 𝑤𝑎𝑡𝑡 is supplied by an ac voltage 𝐸 =
200 sin(310𝑡 + 60°). Then the 𝑟. 𝑚. 𝑠. value of the circuit current is
a) 10 𝐴 b) 10√2 𝐴 c) 20 𝐴 d) 20√2 𝐴
222. The values of L, C and R for a circuit are 1H, 9F and 3Ω. What is the quality factor for the circuit at

P a g e | 21
 resonance?
a) 1 b) 9 1 1
c) d)
9 3
223. In a series resonant circuit, the AC voltage across resistance R, inductor L and capacitor C are 5 V, 10 V and
10 V respectively. The AC voltage applied to the current will be
a) 10 V b) 25 V c) 5 V d) 20 V
224. The impedance of a R-C circuit is 𝑍1 for frequency f and 𝑍2 for frequency 2f. Then,
𝑍1
𝑍2
is
a) Between 1 and 2 b) 2 1 1
c) Between 2 and 1 d)
2
225. A circuit consists of an inductance of 0.5 mH and a capacitor of 20𝜇F. The frequency of the L – C
oscillations is approximately
a) 400 Hz b) 88 Hz c) 1600 Hz d) 2400 Hz
226. A coil of 200Ω resistance and 0.1 𝐻 inductance is connected to an ac source of frequency 200/2π 𝐻𝑧. Phase
angle between potential and current will be
a) 30° b) 90o c) 45° d) 0°
227. For a coil having L = 2 mH, current flows at the rate of 10 As . The emf induced is
3 −1

a) 2 V b) 1 V c) 4 V d) 3 V
228. In the transmission of a.c. power through transmission lines, when the voltage is stepped up 𝑛 times, the
power loss in transmission
a) Increases 𝑛 times b) Decreases 𝑛 times
c) Increases 𝑛 times
2 d) Decreases 𝑛2 times
229. The instantaneous values of current and voltage in an ac circuit are 𝑖 = 100 sin 314 𝑡 𝑎𝑚𝑝 and 𝑒 =
200 𝑖𝑛 (314 𝑡 + 𝜋/3)𝑉 respectively. If the resistance is 1Ω, then the reactance of the circuit will be
a) −200√3 Ω b) √3 Ω c) −200√3 Ω d) 100√3 Ω
230. What is the approximate peak value of an alternating current producing four times the heat produced per
second by a steady current of 2.0 𝐴 in a resistor
a) 2.8 𝐴 b) 4.0 𝐴 c) 5.6 𝐴 d) 8.0 𝐴
231. The power is transmitted from a power house on high voltage ac because
a) Electric current travels faster at higher 𝑣𝑜𝑙𝑡𝑠
b) It is more economical due to less power wastage
c) It is difficult to generate power at low voltage
d) Changes of stealing transmission lines are minimized
232. Two electric bulbs marked 25𝑊 − 220𝑉 and 100𝑊 − 220𝑉 are connected in series to a 440𝑉 supply.
Which of the bulbs will fuse
a) Both b) 100 𝑊 c) 25 𝑊 d) Neither
233. If 25 A current is drawn by 220 V motor and back emf produced is 80 V, the value of armature resistance is
a) 56 Ω b) 5.6 Ω c) 0.56 Ω d) 0.5 Ω
234. Current in the LCR circuit becomes extremely large when
a) Frequency of 𝐴𝐶 supply is increased
b) Frequency of 𝐴𝐶 supply is decreased
c) Inductive reactance becomes equal to capacitive reactance
d) Inductance becomes equal to capacitance
235. The 𝑖 − 𝑣 curve for anti-resonant circuit is
a) i b) i c) i d) i

   
236. The average power dissipated in a pure inductor of inductance 𝐿 when an ac current is passing through it,
P a g e | 22
 is
(Inductance of the coil 𝐿 and current 𝐼)
1 1 d) Zero
a) 𝐿𝐼 2 b) 𝐿𝐼 2 c) 2 𝐿𝑖 2
2 4
237. An inductor of inductance 𝐿 and resistor of resistance 𝑅 are joined in series and connected by a source of
frequency 𝜔. Power dissipated in the circuit is
(𝑅 2 + 𝜔2 𝐿2 ) 𝑉 2𝑅 𝑉 √𝑅 2 + 𝜔 2 𝐿2
a) b) 2 2 2
c) 2 2 2 d)
𝑉 (𝑅 + 𝜔 𝐿 ) (𝑅 + 𝜔 𝐿 ) 𝑉2
238. The network shown in figure is part of a complete circuit. If a certain instant, the current 𝑖 is 5 A and is
decreasing at a rate 103 As −1 , then (𝑉𝐵 − 𝑉𝐴 ) is

a) 20 V b) 15 V c) 10 V d) 5 V
239. For a series L-C-R circuit at resonance, the statement which is not true is
a) Peak energy stored by a capacitor = peak energy stored by an inductor
b) Average power = apparent power
c) Wattles current is zero
d) Power factor is zero
240. In ac circuit of capacitance the current from potential is
a) Forward b) Backward
c) Both are in the same phase d) None of these
241. In a 𝐿𝐶𝑅 circuit having 𝐿 = 8.0 ℎ𝑒𝑛𝑟𝑦, 𝐶 = 0.5 𝜇𝐹 and 𝑅 = 100 𝑜ℎ𝑚 in series. The resonance frequency in
per second is
a) 700 𝑟𝑎𝑑𝑖𝑎𝑛 b) 600 𝐻𝑧 c) 500 𝑟𝑎𝑑𝑖𝑎𝑛 d) 500 𝐻𝑧
242. Which of the following curves correctly represents the variation of capacitive reactance 𝑋𝐶 with frequency
𝑓
a) b) c) d)
Xc Xc Xc Xc

f f f f
243. An AC voltage source of variable angular frequency 𝜔 and fixed amplitude 𝑉0 is connected in series with a
capacitance C and an electric bulb of resistance R (inductance zero). When 𝜔 is increased
a) The bulb glows dimmer b) The bulb glows brighter
c) Total impedance of the circuit is unchanged d) Total impedance of the circuit increases
244. In order to obtain a time constant of 10 s in a 𝑅 − 𝐶 circuit containing a resistance of 103 Ω , the capacity of
the condenser should be
a) 10 𝜇𝐹 b) 100 𝜇𝐹 c) 1000 𝜇𝐹 d) 10000 𝜇𝐹
245. An ac generator, produces an output voltage 𝐸 = 170 sin 377 𝑡 𝑣𝑜𝑙𝑡𝑠, where 𝑡 is in seconds. The frequency
of ac voltage is
a) 50 𝐻𝑧 b) 110 𝐻𝑧 c) 60 𝐻𝑧 d) 230 𝐻𝑧
246. Radio frequency choke uses core of
a) Air b) Iron c) Air and iron d) None of these
247. The natural frequency of an L – C circuit is 125000 cycle/s. Then the capacitor C is replaced by another
capacitor with a dielectric medium of dielectric constant K. In this case, the frequency decreases by 25 kHz.
The value of K is
a) 3.0 b) 2.1 c) 1.56 d) 1.7
248. A low-loss transformer has 230 V applied to the primary and gives 4.6 V in the secondary. Secondary is
connected to a load, which draws 5 A of current. The current (in ampere) in the primary is
a) 0.1 b) 1.0 c) 10 d) 250

P a g e | 23
249. If an ac main supply is given to be 220 𝑉. What would be the average 𝑒.m.f. during a positive half cycle
a) 198𝑉 b) 386𝑉 c) 256𝑉 d) None of these
250. A circuit draws 330 W from a 110V, 60 Hz AC line. The power factor is 0.6 and the current lags the voltage.
The capacitance of a series capacitor that will result in a power factor of unity is equal to
a) 31𝜇F b) 54𝜇F c) 151𝜇F d) 201𝜇F
251. If the capacity of a condenser is 1 F, then its resistance in a DC circuit will be
a) Zero b) infinity 1
c) 1 Ω d) Ω
2
252. What is the charge stored by 1 𝜇F as shown in the figure?
2V 0.5 Ω

1 μF 1Ω

1Ω

a) 2.33 μC b) 3.33 μC c) 1.33 μC d) 4.33 μC

253. A transistor-oscillator using a resonant circuit with an inductor 𝐿 (of negligible resistance) and a capacitor
𝐶 in series produce oscillation of frequency 𝑓. If 𝐿 is doubled and 𝐶 is changed to 4𝐶, the frequency will be
a) 𝑓/2√2 b) 𝑓/2 c) 𝑓/4 d) 8𝑓
254. A coil of inductance 0.2 H and 1.0 W resistance is connected to a 90 V source. At what rate will the current
in the coil grow at the instant the coil is connected to the source?
a) 450 As −1 b) 4.5 As −1 c) 45 As −1 d) 0.45 As −1
255. The voltage of an ac supply varies with time (𝑡) as 𝑉 = 120 sin 100𝜋𝑡 cos 100𝜋𝑡. The maximum voltage
and frequency respectively are
120
a) 12 𝑣𝑜𝑙𝑡𝑠, 100 𝐻𝑧 b) 𝑣𝑜𝑙𝑡𝑠, 100 𝐻𝑧 c) 60 𝑣𝑜𝑙𝑡𝑠, 200 𝐻𝑧 d) 60 𝑣𝑜𝑙𝑡𝑠, 100 𝐻𝑧
√2
256. In an AC circuit the emf(e) and the current (i) at any instant are given respectively by
𝑒 = 𝐸0 sin 𝜔𝑡
𝑖 = 𝐼0 sin(𝜔𝑡 − ϕ)
The average power in the circuit over one cycle of AC is
𝐸 𝐼 𝐸 𝐼 𝐸 𝐼
a) 0 0 b) 0 0 sin 𝜙 c) 0 0 cos 𝜙 d) 𝐸0 𝐼0
2 2 2
257. From figure shown below a series L – C – R circuit connected to a variable frequency 200 V source. 𝐶 =
80 𝜇𝐹 and 𝑅 = 40 Ω. Then the source frequency which drive the circuit at resonance is

a) 25 Hz b)
25
Hz c) 50 Hz d)
50
Hz
𝜋 𝜋
258. In an ac circuit, the 𝑟. 𝑚. 𝑠. value of current, 𝐼𝑟𝑚𝑠 is related to the peak current, 𝐼0 by the relation
1 1
a) 𝐼𝑟𝑚𝑠 = 𝐼0 b) 𝐼𝑟𝑚𝑠 = 𝐼0 c) 𝐼𝑟𝑚𝑠 = √2𝐼0 d) 𝐼𝑟𝑚𝑠 = 𝜋𝐼0
𝜋 √2
259. The time taken by AC of 50 Hz in reaching from zero to the maximum value is
a) 50 × 10−3 s b) 5 × 10−3 s c) 1 × 10−3 s d) 2 × 10−3 s
260. In an AC circuit the voltage applied is 𝐸 = 𝐸 sin 𝜔𝑡. The resulting current in the circuit is 𝐼 = 𝐼 sin (𝜔𝑡 −
0 0

P a g e | 24
 𝜋
2
). The power consumption in the circuit is given by
𝐸0 𝐼0 b) P = zero 𝐸0 𝐼0
a) 𝑃 = c) 𝑃 = d) 𝑃 = √2𝐸0 𝐼0
√2 2
261. The quality factor of 𝐿𝐶𝑅 circuit having resistance (𝑅) and inductance (𝐿) at resonance frequency (𝜔) is
given by
𝜔𝐿 𝑅 𝜔𝐿 1/2 𝜔𝐿 2
a) b) c) ( ) d) ( )
𝑅 𝜔𝐿 𝑅 𝑅
262. In a circuit containing an inductance of zero resistance, the 𝑒.m.f. of the applied ac voltage leads the
current by
a) 90° b) 45° c) 30° d) 0°
263. In an ac circuit, the current is given by 𝑖 = 5 sin (100 𝑡 − 𝜋) and the ac potential is 𝑉 = 200 sin(100) 𝑣𝑜𝑙𝑡.
2
Then the power consumption is
a) 20 𝑤𝑎𝑡𝑡𝑠 b) 40 𝑤𝑎𝑡𝑡𝑠 c) 1000 𝑤𝑎𝑡𝑡𝑠 d) 0 𝑤𝑎𝑡𝑡
264. The graphs given below depict the dependence of two reactive impedances 𝑋1 and 𝑋2 on the frequency of
the alternating e.m.f. applied individually to them. We can then say that
Impedance

Impedance

X1 X2

Frequency Frequency
a) 𝑋1 is an inductor and 𝑋2 is a capacitor b) 𝑋1 is a resistor and 𝑋2 is a capacitor
c) 𝑋1 is a capacitor and 𝑋2 is an inductor d) 𝑋1 is an inductor and 𝑋2 is a resistor
265. An alternating current of rms value 10 A is passed through a 12 Ω resistor. The maximum potential
difference across the resistor is
a) 20𝑉 b) 90𝑉 c) 169.68 𝑉 d) None of these
266. A series R-C circuit is connected to AC Voltage source. Consider two cases: (A) when C is without a
dielectric medium and (B) when C is filled with dielectric of constant 4. The current 𝐼𝑅 through the resistor
and voltage 𝑉𝑐 across the capacitor are compared in the two cases. Which of the following is/are true?
a) 𝐼𝑅𝐴 > 𝐼𝑅𝐵 b) 𝐼𝑅𝐴 < 𝐼𝑅𝐵 c) 𝑉𝐶𝐴 > 𝑉𝐶𝐵 d) 𝑉𝐶𝐴 < 𝑉𝐶𝐵
267. An ac voltage is applied to a resistance 𝑅 and inductor 𝐿 in series. If 𝑅 and the inductive reactance are both
equal to 3Ω, the phase difference between the applied voltage and the current in the circuit is
a) Zero b) 𝜋/6 c) 𝜋/4 d) 𝜋/2
268. The phase difference between the alternating current and emf is 𝜋/2. Which of the following cannot be the
constituent of the circuit?
a) C alone b) R, L c) L, C d) L alone
269. Two parallel wires 𝐴1 𝐿 𝑎𝑛𝑑 𝐵1 𝑀 placed at a distance 𝑤 are connected by a resistor R and placed in a
magnetic field B which is perpendicular to the plane containing the wires (see figure). Another wire 𝐶𝐷
now connects the two wires perpendicularly and made to slide with velocity 𝑣 through distance L. The
power developed is

P a g e | 25
 𝑙𝑣 𝐵2 𝑙 2 𝑣 2 𝐵𝑤𝑣 𝐵2 𝑤 2 𝑣 2
a) 𝐵 b) c) d)
𝑅 𝑅 𝑅 𝑅
270. The diagram shows a capacitor 𝐶 and a resistor 𝑅 connected in series to an ac source. 𝑉1 and 𝑉2 are
voltmeters and 𝐴 is an ammeter
V1

C R V2
A

Consider the following statements

I. Readings in 𝐴 are always in phase
II. Reading in 𝑉1 is ahead in phase with reading in 𝑉2
III. Reading in 𝐴 and 𝑉1 are always in phase. Which of these statements are/is correct
a) I only b) II only c) I and II only d) II and III only
271. The number of turns in the primary coil of a transformer is 200 and the number of turns in secondary coil
is 10. If 240 V AC is applied to the primary, the output from secondary will be
a) 48 V b) 24 V c) 12 V d) 6 V
272. In a L – R circuit of 3 mH inductance and 4 Ω resistance, emf 𝐸 = 4 cos 1000𝑡 V is applied. The amplitude
of emf is
a) 0.8 A 4 c) 1.0 A 4
b) A d) A
7 √7
273. The resonance frequency of the tank circuit of an oscillator when 𝐿 = 10 mH and 𝐶 = 0.04 𝜇F are
𝜋2
connected in parallel is
a) 250 kHz b) 25 kHz c) 2.5 kHz d) 25 MHz
274. The 𝑟. 𝑚. 𝑠. value of potential difference 𝑉 shown in the figure is

a) 𝑉0 /2 b) 𝑉0 /√3 c) 𝑉0 d) 𝑉0 /√2
275. A 20 𝑣𝑜𝑙𝑡𝑠 ac is applied to a circuit consisting of a resistance and a coil with negligible resistance. If the
voltage across the resistance is 12 𝑉, the voltage across the coil is
a) 16 𝑣𝑜𝑙𝑡𝑠 b) 10 𝑣𝑜𝑙𝑡𝑠 c) 8 𝑣𝑜𝑙𝑡𝑠 d) 6 𝑣𝑜𝑙𝑡𝑠
276. Average power in the L-C-R circuit depends upon
a) Current b) phase difference only
c) Emf d) Current, emf and phase difference
277. The reactance of a coil when used in the domestic ac power supply (220 𝑣𝑜𝑙𝑡𝑠, 50 cycles per second) is
50 𝑜ℎ𝑚𝑠. The inductance of the coil is nearly
a) 2.2 ℎ𝑒𝑛𝑟𝑦 b) 0.22 ℎ𝑒𝑛𝑟𝑦 c) 1.6 ℎ𝑒𝑛𝑟𝑦 d) 0.16 ℎ𝑒𝑛𝑟𝑦
278. If 𝐸0 is the peak emf, 𝐼0 is the peak current and ϕ is the phase difference between them, then the average
power dissipation in the circuit is
1 𝐸0 𝐼0 1 1
a) 𝐸0 𝐼0 b) c) 𝐸0 𝐼0 sin ϕ d) 𝐸0 𝐼0 cos ϕ
2 √2 2 2
279. A coil of resistance 𝑅 and inductance 𝐿 is connected to a battery of emf 𝐸 volt. The final current in the coil
is
𝐸 𝐸 𝐸 𝐸𝐿
a) b) c) √( ) d) √( )
𝑅 𝐿 𝑅2 + 𝐿2 𝑅2
+ 𝐿2
280. An irregular closed loop carrying a current has a shape such that the entire loop cannot lie in a single

P a g e | 26
 plane. If this is placed in a uniform magnetic field, the force acting on the loop
a) Must be zero b) Can never be zero
c) May be zero d) Will be zero only for one particular direction of
the magnetic field
281. Mutual inductance of two coils can be increased by
a) Decreasing the number of turns in the coils b) Increasing the number of turns in the coils
c) Winding the coils on wooden cores d) None of the above
282. In a L-C-R series circuit, the potential difference between the terminals of the inductance is 60 V, between
the terminals of the capacitor is 30 V and that across the resistance is 40 V. Then, supply voltage will be
equal to
a) 50 V b) 70 V c) 130 V d) 10 V
283. The value of alternating emf 𝐸 in the given circuit will be

a) 220 V b) 140 V c) 100 V d) 20 V

284. Choke coil works on the principle of
a) Transient current b) Self induction c) Mutual induction d) Wattless current
285. In an AC circuit, V and I are given by 𝑉 = 150 sin(150t) volt and 𝐼 = 150 sin (150t + π) amp.
3
The power dissipated in the circuit is
a) Zero b) 5625 W c) 150 W d) 106 W
286. An 𝐿𝐶𝑅 series circuit with 𝑅 = 100Ω is connected to a 200 𝑉, 50 𝐻𝑧 a.c. source when only the capacitance
is removed, the current leads the voltage by 60°. When only the inductance is removed, the current leads
the voltage by 60°. The current in the circuit is
2
a) 2𝐴 b) 1𝐴 c) √3 𝐴 d) 𝐴
2 √3
287. An electric bulb has a rated power of 50 W at 100 V. If it is used on an AC source 200 V, 50 Hz, a choke has
to be used in series with it. The should have an inductance of
a) 0.1 mH b) 1 mH c) 0.1 H d) 1.1 H
288. In L – C – R circuit if resistance increases, quality factor
a) Increases finitely b) Decreases finitely c) Remains constant d) None of the above
289. In the given circuit diagram the current through the battery and the charge on the capacitor respectively in
steady state are

6V

l
1Ω

2Ω

2Ω

4Ω

0.5 μF

6 12
a) 1 A and 3 μC b) 17 A and 0 μC c) 7
A and 7
μC d) 11 A and 3 μC

P a g e | 27
290. In a series circuit 𝐶 = 2𝜇𝐹, 𝐿 = 1𝑚𝐻 and 𝑅 = 10Ω. When the current in the circuit is maximum, at that
time the ratio of the energies stored in the capacitor and the inductor will be
a) 1 ∶ 1 b) 1 ∶ 2 c) 2 ∶ 1 d) 1 ∶ 5
291. The inductive reactance of an inductor of ℎ𝑒𝑛𝑟𝑦 at 50 𝐻𝑧 frequency is
1
𝜋
50 𝜋
a) 𝑜ℎ𝑚 b) 𝑜ℎ𝑚 c) 100 𝑜ℎ𝑚 d) 50 𝑜ℎ𝑚
𝜋 50
292. In general in an alternating current circuit
a) The average value of current is zero
b) The average value of square of the current is zero
c) Average power dissipation is zero
d) The phase difference between voltage and current is zero
293. In a series combination 𝑅 = 300 Ω, 𝐿 = 0.9H, 𝐶 = 2.0 μF, ω = 1000 rads−1, the impedance of the circuit is
a) 1300 Ω b) 900 Ω c) 500 Ω d) 400 Ω
294. Three identical coils 𝐴, 𝐵 and 𝐶 are placed with their planes parallel to one another. Coils 𝐴 and 𝐶 carry
currents as shown in figure. Coils 𝐵 and 𝐶 are fixed in position and coil A is moved towards 𝐵. Then,
current induced in 𝐵 is in

a) Clock-wise current
b) Anti-clock-wise current
c) No current is induced in 𝐵
d) Current in induced only when both coils move
295. Which of the following plots may represent the reactance of a series 𝐿𝐶 combination

a
Reactance

c
b
Frequency
d

a) 𝑎 b) 𝑏 c) 𝑐 d) 𝑑
296. In an L-C-R series AC circuit the voltage across L, C and R is 10 V each. If the inductor is short circuited, the
voltage across the capacitor would become
a) 10 V 20 10
b) V c) 20√2 V d) V
√2 √2
297. adsf
a) 122 b) 3 c) 4 d) 5
298. When current in a coil changes from 2 A to 4 A in 0.05s, an emf of 8 V is induced in the coil. Self inductance
of the coil is
a) 0.1 H b) 0.2 H c) 0.4 H d) 0.8 H
299. A telephone wire of length 200 km has a capacitance of 0.014 μF per km. If it carries an AC frequency 5
kHz, what should be the value of an inductor required to be connected in series so that the impedance of
the circuit is minimum?
a) 0.35 mH b) 3.5 mH c) 2.5 mH d) zero
300. The resonance point in 𝑋𝐿 − 𝑓 and 𝑋𝐶 − 𝑓 curves is

P a g e | 28
 XL

P R S
Q f

XC

a) 𝑃 b) 𝑄 c) 𝑅 d) 𝑆
301. The natural frequency of a 𝐿 − 𝐶 circuit is equal to
1 1 1 𝐿 1 𝐶
a) √𝐿𝐶 b) c) √ d) √
2𝜋 2𝜋√𝐿𝐶 2𝜋 𝐶 2𝜋 𝐿
302. In a purely resistive ac circuit, the current
a) Lags behind the 𝑒.m.f. in phase
b) Is in phase with the 𝑒.m.f.
c) Leads the 𝑒.m.f. in phase
d) Leads the 𝑒.m.f. in half the cycle and lags behind it in the other half
303. In a current carrying long solenoid, the field produced does not depend upon
a) Number of turns per unit length b) Current flowing
c) Radius of solenoid d) All of the above
304. A pure inductor of 25 𝑚𝐻 is connected to a source of 220 𝑉. Given the frequency of the source as 50 𝐻𝑧,
the 𝑟𝑚𝑠 current in the circuit is
a) 7 𝐴 b) 14 𝐴 c) 28 𝐴 d) 42 𝐴
305. In a choke coil, the reactance 𝑋𝐿 and resistance R are such that
a) 𝑋𝐿 = 𝑅 b) 𝑋𝐿 >> 𝑅 c) 𝑋𝐿 << 𝑅 d) 𝑋𝐿 = ∞
306. What is the value of inductance L for which the current is a maximum in a series L-C-R circuit with 𝐶 =
10𝜇F and 𝜔 = 1000 s −1 ?
a) 100 mH b) 1 mH
c) Cannot be calculated unless R is known d) 10 mH
307. The peak value of an alternating emf E given by 𝐸 = 𝐸0 cos 𝜔𝑡 is 10 V and its frequency is 50 Hz. At a time
1
𝑡 = 100 s, the instantaneous value of the emf is
a) 10 V b) 5√3 V c) 5 V d) 1 V
308. Power delivered by the source of the circuit becomes maximum, when
1 1 2
a) 𝜔𝐿 = 𝜔𝐶 b) 𝜔𝐿 = c) 𝜔𝐿 = − ( ) d) 𝜔𝐿 = √𝜔𝐶
𝜔𝐶 𝜔𝐶
309. An ac source of variable frequency 𝑓 is connected to an 𝐿𝐶𝑅 series circuit. Which of the graphs in figure
represents the variation of current 𝐼 in the circuit with frequency 𝑓
a) I b) I c) I d) I

0 f 0 f 0 f 0 f

310. In the figure shown, three AC voltmeters are connected. At resonance,

a) 𝑉2 = 0 b) 𝑉1 = 0 c) 𝑉3 = 0 d) 𝑉1 = 𝑉2 ≠ 0
311. In an 𝐿𝐶𝑅 circuit 𝑅 = 100 𝑜ℎ𝑚. When capacitance 𝐶 is removed, the current lags behind the voltage by
𝜋/3. When inductance 𝐿 is removed, the current leads the voltage by 𝜋/3. The impedance of the circuit is
P a g e | 29
 a) 50 𝑜ℎ𝑚 b) 100 𝑜ℎ𝑚 c) 200 𝑜ℎ𝑚 d) 400 𝑜ℎ𝑚
312. An alternating current source of frequency 100 𝐻𝑧 is joined to a combination of a resistance, a capacitance
and a coil in series. The potential difference across the coil, the resistance and the capacitor is 46, 8 and
40 𝑣𝑜𝑙𝑡 respectively. The electromotive force of alternating current source in 𝑣𝑜𝑙𝑡 is
a) 94 b) 14 c) 10 d) 76
313. For the series L – C – R circuit shown in the figure, what is the resonance frequency and the amplitude of
the current at the resonating frequency?

a) 2500 rad𝑠 −1 and 5√2 A b) 2500 rad𝑠 −1 and 5 A

5
c) 2500 rad𝑠 −1 and√2 A d) 250rad𝑠 −1 and 5√2 A
314. The 𝑟. 𝑚. 𝑠. voltage of the wave form shown is
Y
+ 10

0 t

– 10

a) 10 𝑉 b) 7 𝑉 c) 6.37 𝑉 d) None of these

315. In an ac circuit, peak value of voltage is 423 𝑣𝑜𝑙𝑡𝑠. Its effective voltage is
a) 400 𝑣𝑜𝑙𝑡𝑠 b) 323 𝑣𝑜𝑙𝑡𝑠 c) 300 𝑣𝑜𝑙𝑡𝑠 d) 340 𝑣𝑜𝑙𝑡𝑠
316. An LC circuit contains a 20 mH inductor and a 50 𝜇F capacitor with an initial charge of 10 mC. The
resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0. At what time is the energy
stored completely magnetic?
a) t = 0 b) t = 1.57 ms c) t = 3.14 ms d) t = 6.28 ms
317. Two inductors of inductance 𝐿 each are connected in series with opposite magnetic fluxes. What is the
resultant inductance? (Ignore mutual inductance)
a) Zero b) 𝐿 c) 2 𝐿 d) 3 𝐿
318. The current i passed in any instrument in an AC circuit is 𝑖 = 2 sin 𝜔𝑡 A and potential difference applied is
given by 𝑉 = 5 cos 𝜔𝑡 V. Power loss in the instrument is
a) 10 W b) 5 W c) Zero W d) 20 W
319. The inductance of the oscillatory circuit of the ratio station is 10 mH and its capacitance is 0.25 𝜇F. Taking
the effect of resistance negligible, wavelength of the broadcasted waves will be (velocity of light =
3.0 × 108 ms−1 , π = 3.14 )
a) 9.42 × 104 m b) 18.8 × 104 m c) 4.5 × 104 m d) None of these
320. Two inductors 𝐿1 𝑎𝑛𝑑 𝐿2 are connected in parallel and a time varying current flows as shown in figure.
The ratio of currents 𝑖1 /𝑖2 at any time t is

𝐿22 𝐿21
a) 𝐿2 /𝐿1 b) 𝐿1 /𝐿2 c) d)
(𝐿1 + 𝐿2 )2 (𝐿1 + 𝐿2 )2
321. In a series L – C – R circuit, resistance 𝑅 = 10 Ω and the impedance 𝑍 = 10 Ω. The phase difference
between the current and the voltage is

P a g e | 30
 a) 0° b) 30° c) 45° d) 60°
322. In an L – C – R series AC circuit, the voltage across each of the components. L, C and R is 50 V. The voltage
across the L – C combination will be
a) 50 V b) 50√2 V c) 100 V d) zero
323. What is the self inductance of an air core solenoid 1 m long, diameter 0.05m, if it has 500 turns? Take 𝜋 2 =
10.
a) 3.15 × 10−4 H b) 4.8 × 10−4 H c) 5 × 10−4 H d) 6.25 × 10−4 H
324. An alternating voltage (in volt) given by 𝑉 = 200√2 sin(100𝑡) is connected to1𝜇𝐹 capacitor through an AC
ammeter. The reading of the ammeter will be
a) 10 mA b) 20 m A c) 40 mA d) 80 mA
325. The power loss in AC circuit will be minimum when
a) Resistance is high, inductance is high b) Resistance is high, inductance is low
c) Resistance is low, inductance is low d) None of the above
326. An inductance 1 H is connected in series with an AC source of 220 V and 50 Hz. The inductive reactance (in
ohm) is
a) 2 𝜋 b) 50 𝜋 c) 100 𝜋 d) 1000 𝜋
327. A voltage of peak value 283 V and varying frequency is applied to a series L – C – R combination in which
𝑅 = 3 Ω, 𝐿 = 25 mH and 𝐶 = 400𝜇F. The frequency (in Hz)of the source at which maximum power is
dissipated in the above, is
a) 51.5 b) 50.7 c) 51.1 d) 50.3
328. A fully charged capacitor C with initial charge 𝑞0 is connected to a coil of self inductance L at t = 0. The
time at which the energy is stored equally between the electric and the magnetic fields is
𝜋
a) √𝐿𝐶 b) 2𝜋√𝐿𝐶 c) √𝐿𝐶 d) 𝜋√𝐿𝐶
4
329. An inductor 𝐿 and a capacitor 𝐶 are connected in the circuit as shown in the figure. The frequency of the
power supply is equal to the resonant frequency of the circuit. Which ammeter will read zero ampere

a) 𝐴1
b) 𝐴2
c) 𝐴3
d) None of these
330. The current ′𝑖′ in an inductance coil varies with time ‘𝑡’ according to following graph
i

t
(0, 0)

Which of the following plots shows the variation of voltage in the coil
a) V b) V c) V d) V

(0, 0) t (0, 0) t (0, 0) t (0, 0) t

331. What is the average power dissipation in an ideal capacitor in AC circuit?

P a g e | 31
 1 c) Zero
a) 2𝐶𝑉 2 b) 𝐶𝑉 2 d) 𝐶𝑉 2
2
332. Same current is flowing in two alternating circuits. The first circuit contains only inductance and the other
contains only a capacitor. If the frequency of the emf of AC is increased, the effect on the value of the
current
a) Increases in the first circuit and decreases in the other
b) Increases in both the circuits
c) Decreases in both the circuits
d) Decreases in the first circuit and increases in the other
333. Consider a short magnetic dipole of magnetic length 10 cm. Its geometric length is
a) 12 cm b) 5 c) 3 d) 4
334. In a series resonant R-L-C circuit, the voltage across R is 100 V and the value of = 1000 Ω . The capacitance
of the capacitor is 2 × 10−6 F; angular frequency of AC is 200 rad s−1 . Then the potential difference across
the inductance coil is
a) 100 V b) 40 V c) 250 V d) 400 V
335. Find the time required for a 50 Hz alternating current to become its value from zero to the rms value
a) 10.0 ms b) 2.5 ms c) 15.0 ms d) 5.0 ms
336. Alternating current cannot be measured by DC ammeter because
a) AC cannot pass through DC ammeter
b) AC changes direction
c) Average value of current for complete cycle is zero
d) DC ammeter will get damaged
337. An ac circuit consists of an inductor of inductance 0.5 𝐻 and a capacitor of capacitance 8 𝜇𝐹 in series. The
current in the circuit is maximum when the angular frequency of ac source is
a) 500 𝑟𝑎𝑑/𝑠𝑒𝑐 b) 2 × 105 𝑟𝑎𝑑/𝑠𝑒𝑐 c) 4000 𝑟𝑎𝑑/𝑠𝑒𝑐 d) 5000 𝑟𝑎𝑑/𝑠𝑒𝑐
338. The armature of a DC motor has a resistance of 20 Ω. It draws a current of 1.5 A when run by 220 V DC.
The value of peak emf induced in it will be
a) 150 V b) 170 V c) 190 V d) 180 V
339. The maximum value of AC voltage in a circuit is 707 V. Its rms value is
a) 70.7 V b) 100 V c) 500 V d) 707 V
340. The impedance of a circuit, when a resistance R and an inductor of inductance L are connected in series in
an AC circuit of frequency 𝑓, is
a) √𝑅 + 2𝜋 2 𝑓 2 𝐿2 b) √𝑅 + 4𝜋 2 𝑓 2 𝐿2 c) √𝑅 2 + 4𝜋 2 𝑓 2 𝐿2 d) √𝑅 2 + 2𝜋 2 𝑓 2 𝐿2
341. A 4𝜇F capacitor, a resistance of 2.5 mΩ is in series with 12 V battery. Find the the time after which the
potential difference across the capacitor is 3 times the potential difference across the resistor. [Given ln
(2)=0.693]
a) 13.86 s b) 6.93 s c) 7 s d) 14 s
342. An inductor of inductance 𝐿 = 400 mH and resistors of resistances 𝑅1 = 4Ω and 𝑅2 = 2Ω are connected to
battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S
is closed at t = 0. The potential drop across L as a function of time is

12 −3𝑡
a) 6𝑒 −5𝑡 V b) 𝑒 V c) 6(1 − 𝑒 −𝑡/0.2 ) V d) 12 𝑒 −5𝑡 V
𝑡
343. A capacitor 50𝜇F is connected to a supply of 220 V and angular frequency 50 rad s −1. The value of rms
current in the circuit is
a) 0.45 A b) 0.50 A c) 0.55 A d) 0.60 A
344. In an ideal choke, ratio of its inductance L to its DC resistance R is
a) Infinity b) Zero c) Unity d) hundred
P a g e | 32
345. In a series L – C – R circuit the frequency of a 10 V AC voltage source is adjusted in such a fashion that the
reactance of the inductor measures 15 Ω and that of the capacitor 11 Ω. If 𝑅 = 3 Ω, the potential difference
across the series combination of L and C will be
a) 8 V b) 10 V c) 22 V d) 52 V
346. In the adjoining ac circuit the voltmeter whose reading will be zero at resonance is

a) 𝑉1 b) 𝑉2 c) 𝑉3 d) 𝑉4
347. In an ac circuit with voltage 𝑉 and current 𝐼, the power dissipated is
1
a) 𝑉𝐼 b) 𝑉𝐼
2
1
c) 𝑉𝐼 d) Depends on the phases between 𝑉 and 𝐼
√2
348. Two coils have mutual inductance 0.005 H. The current changes in the first coil according to equation 𝑖 =
𝑖0 sin 𝜔𝑡 where 𝑖0 = 10 A and 𝜔 = 100𝜋 rads−1 . The maximum value of emf in second coil is
a) 2 𝜋 b) 5 𝜋 c) 𝜋 d) 4 𝜋
349. A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic
induction 𝐵 ⃗ , figure. At the position 𝑀𝑁𝑄, the speed of the ring is 𝑣. The potential difference developed
across the ring is

a) Zero
1
b) 2 𝐵 𝑣𝜋 𝑅 2 , and 𝑀 is at a higher potential
c) 𝜋 𝑅 𝐵 𝑣, and 𝑄 is at a higher potential
d) 2 𝑅 𝐵 𝑣, and 𝑄 is at a higher potential
350. What is the self-inductance of a coil which produces 5 V when the current changes from 3 A to 2 A in one
millisecond?
a) 5000 H b) 5 mH c) 50 H d) 5 H
351. The number of turns of primary and secondary coils of a transformer is 5 and 10 respectively and mutual
inductance of the transformer is 25 H. Now, number of turns in primary and secondary are made 10 and 5
respectively. Mutual inductance of transformer will be
a) 25 H b) 12.5 H c) 50 H d) 6.25 H
352. An alternating 𝑒.m.f. is applied to purely capacitive circuit. The phase relation between 𝑒.m.f. and current
flowing in the circuit is or
In a circuit containing capacitance only
a) 𝑒.m.f. is ahead of current by 𝜋/2 b) Current is ahead of 𝑒.m.f. by 𝜋/2
c) Current lags behind 𝑒.m.f. by 𝜋 d) Current is ahead of 𝑒.m.f. by 𝜋
353. A capacitor and an inductance coil are connected in separate AC circuits with a bulb glowing in both the
circuits. The bulb glows more brightly when

P a g e | 33
 a) An iron rod is introduced into the inductance coil
b) The number of turns in the inductance coil is increased
c) Separation between the plates of the capacitor is increased
d) A dielectric is introduced into the gap between the plates of the capacitor
354. Which of the following statement is incorrect?
a) In a L – C – R series AC If the net reactance of c) At resonance, the d) Below resonance,
circuit, as the frequency an L – C – R series AC impedance of an AC voltage leads the
of the source increases, circuit is same as its circuit becomes purely current while above it,
b)
the impedance of the resistance, then the resistive. current leads the
circuit first decreases current lags behind the voltage
and then increases voltage by 45°
355. In an AC series circuit, the instantaneous current is maximum when the instantaneous voltage is
maximum. The circuit element connected to the source will be
a) Pure inductor b) Pure capacitor
c) Pure resistor d) Combination of capacitor and an inductor
356. 𝐿, 𝐶 and 𝑅 represent physical quantities inductance capacitance and resistance respectively. The
combination representing dimension of frequency is
a) LC 𝐿 −1/2 𝐶
b) (𝐿𝐶)−1/2 c) ( ) d)
𝐶 𝐿
357. A 𝐿𝐶𝑅 series 𝐴. 𝐶. circuit is tuned to resonance. The impedence of the circuit is now
21/2
1
a) 𝑅 b) [𝑅 2 + ( − 𝜔𝐿) ]
𝜔𝐶
1/2 1/2
1 2 1 2
c) [𝑅 2 + (𝜔𝐿)2 + ( ) ] d) [𝑅 2 + (𝜔𝐿 − ) ]
𝜔𝐶 𝜔𝐶
358. In pure inductive circuit, the curves between frequency 𝑓 and reciprocal of inductive reactance 1/𝑋𝐿 is

1 1 1
1
a) XL b) XL c) XL d) XL

f f f f

359. In AC circuit in which inductance and capacitance are joined in series. Current is found to be maximum
when the value of inductance is 0.5 H and the value of capacitance is 8 𝜇F. The angular frequency of
applied alternating voltage will be
a) 4000 Hz b) 5000 Hz c) 2 × 105 Hz d) 500 Hz
360. In an AC circuit the instantaneous values of emf and current are
𝑒 = 200 sin 300 𝑡 volt
𝜋
and 𝑖 = 2 sin (300𝑡 + 3 )amp.
The average power consumed in watt is
a) 200 b) 100 c) 50 d) 400
361. In a series L-C-R circuit 𝑅 = 200 Ω and the voltage and the frequency of the main supply is 220V and 50 Hz
respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30°. On
taking out the inductor from the circuit the current leads the voltage by 30°. The power dissipated in the L-
C-R circuit is
a) 305 W b) 210 W c) Zero d) 242 W
362. A resistance of 40 𝑜ℎ𝑚 and an inductance of 95.5 𝑚𝑖𝑙𝑙𝑖ℎ𝑒𝑛𝑟𝑦 are connected in series in a 50 𝑐𝑦𝑐𝑙𝑒𝑠/
𝑠𝑒𝑐𝑜𝑛𝑑 ac circuit. The impedance of this combination is very nearly
a) 30 𝑜ℎ𝑚 b) 40 𝑜ℎ𝑚 c) 50 𝑜ℎ𝑚 d) 60 𝑜ℎ𝑚
363. In the adjoining figure the impedance of the circuit will be

P a g e | 34
 90 V
XL = 30  XC =20

a) 120 𝑜ℎ𝑚 b) 50 𝑜ℎ𝑚 c) 60 𝑜ℎ𝑚 d) 90 𝑜ℎ𝑚

364. Two coil 𝑋 and 𝑌 are placed in a circuit such that a current changes by 2 A in coil 𝑋 and magnetic flux
change of 0.4 Wb occurs in 𝑌. The value of mutual inductance of the coils is
a) 0.8 H b) 0.2 Wb c) 0.2 H d) 5 H
365. A 0.7 ℎ𝑒𝑛𝑟𝑦 inductor is connected across a 120𝑉 − 60 𝐻𝑧 ac source. The current in the inductor will be
very nearly
a) 4.55 𝑎𝑚𝑝 b) 0.355 𝑎𝑚𝑝 c) 0.455 𝑎𝑚𝑝 d) 3.55 𝑎𝑚𝑝
366. What will be the self inductance of a coil, to be connected in a series with a resistance of 𝜋√3Ω such that
the phase difference between the emf and the current at 50 𝐻𝑧 frequency is 30°
a) 0.5 ℎ𝑒𝑛𝑟𝑦 b) 0.03 ℎ𝑒𝑛𝑟𝑦 c) 0.05 ℎ𝑒𝑛𝑟𝑦 d) 0.01 ℎ𝑒𝑛𝑟𝑦
367. A coil of inductance 𝐿 has an inductive reactance of 𝑋𝐿 in an AC circuit in which the effective current is 𝑙.
The coil is made from a super-conducting material and has no resistance. The rate at which power is
dissipated in the coil is
a) 0 b) 𝐼𝑋𝐿 c) 𝐼 2 𝑋𝐿 d) 𝐼𝑋𝐿2
368. Time constant of LC circuit is
1 1 𝐿𝐶
a) b) 2 2
c) d) 2𝜋 √𝐿𝐶
2𝜋 𝐿𝐶 2𝜋 𝐿 𝐶 2𝜋
369. In the circuit shown below, the ac source has voltage 𝑉 = 20 cos(𝜔𝑡) volts with 𝜔 = 2000 𝑟𝑎𝑑/𝑠𝑒𝑐. The
amplitude of the current will be nearest to

a) 2𝐴 b) 3.3𝐴 c) 2/√5 𝐴 d) √5 𝐴
370. The following series 𝐿 − 𝐶 − 𝑅 circuit, when driven by an 𝑒. 𝑚. 𝑓. source of angular frequency 70 kilo-
radians per second, the circuit effectively behaves like

a) Purely resistive circuit b) Series 𝑅 − 𝐿 circuit

c) Series 𝑅 − 𝐶 circuit d) Series 𝐿 − 𝐶 circuit with 𝑅 = 0
371. 𝐿, 𝐶 and 𝑅 denote inductance, capacitance and resistance respectively. Pick out the combination which
does not have the dimensions of frequency
1 𝑅 1 𝐶
a) b) c) d)
𝑅𝐶 𝐿 √𝐿𝐶 𝐿

P a g e | 35
 7.ALTERNATING CURRENT

: ANSWER KEY :
1) b 2) a 3) a 4) d 189) d 190) d 191) d 192) a
5) a 6) b 7) a 8) b 193) c 194) d 195) c 196) c
9) b 10) d 11) c 12) d 197) a 198) a 199) d 200) a
13) c 14) a 15) a 16) a 201) b 202) c 203) d 204) c
17) b 18) b 19) c 20) c 205) a 206) c 207) a 208) a
21) b 22) a 23) b 24) d 209) b 210) b 211) d 212) d
25) c 26) b 27) b 28) d 213) c 214) b 215) c 216) b
29) c 30) b 31) a 32) a 217) a 218) b 219) b 220) b
33) a 34) d 35) b 36) b 221) b 222) c 223) c 224) b
37) c 38) d 39) c 40) c 225) c 226) c 227) a 228) d
41) b 42) c 43) b 44) a 229) b 230) c 231) b 232) c
45) d 46) a 47) b 48) d 233) b 234) c 235) b 236) d
49) d 50) b 51) b 52) b 237) b 238) b 239) d 240) a
53) a 54) b 55) c 56) d 241) c 242) b 243) b 244) d
57) a 58) b 59) b 60) b 245) c 246) a 247) c 248) a
61) b 62) a 63) d 64) c 249) a 250) b 251) b 252) c
65) c 66) b 67) b 68) a 253) a 254) a 255) d 256) c
69) d 70) a 71) d 72) c 257) b 258) b 259) b 260) b
73) a 74) d 75) d 76) b 261) a 262) a 263) d 264) c
77) c 78) c 79) a 80) b 265) c 266) b 267) c 268) c
81) d 82) a 83) b 84) a 269) d 270) b 271) c 272) a
85) d 86) d 87) c 88) c 273) b 274) d 275) a 276) d
89) d 90) a 91) c 92) c 277) d 278) d 279) a 280) a
93) b 94) a 95) a 96) b 281) b 282) a 283) c 284) b
97) b 98) c 99) c 100) d 285) b 286) a 287) d 288) b
101) a 102) c 103) b 104) b 289) d 290) d 291) c 292) a
105) c 106) a 107) d 108) c 293) c 294) b 295) d 296) d
109) c 110) d 111) b 112) b 297) c 298) b 299) a 300) c
113) a 114) c 115) b 116) c 301) b 302) b 303) c 304) c
117) b 118) c 119) d 120) d 305) b 306) a 307) b 308) b
121) c 122) b 123) b 124) a 309) d 310) a 311) b 312) c
125) d 126) b 127) a 128) a 313) b 314) a 315) c 316) b
129) a 130) a 131) d 132) d 317) c 318) a 319) a 320) a
133) c 134) a 135) a 136) c 321) a 322) d 323) d 324) b
137) c 138) b 139) b 140) b 325) c 326) c 327) d 328) a
141) b 142) a 143) d 144) c 329) c 330) b 331) c 332) d
145) c 146) b 147) d 148) c 333) a 334) c 335) d 336) c
149) c 150) b 151) c 152) c 337) a 338) c 339) c 340) c
153) b 154) a 155) a 156) d 341) a 342) d 343) c 344) a
157) b 158) b 159) a 160) c 345) a 346) d 347) d 348) b
161) d 162) b 163) c 164) d 349) d 350) b 351) a 352) b
165) a 166) c 167) c 168) a 353) d 354) d 355) c 356) b
169) a 170) c 171) a 172) c 357) a 358) c 359) d 360) b
173) a 174) d 175) a 176) b 361) d 362) c 363) c 364) c
177) d 178) b 179) b 180) a 365) c 366) d 367) a 368) d
181) a 182) b 183) c 184) c 369) a 370) c 371) d
185) d 186) b 187) a 188) b
P a g e | 36
 7.ALTERNATING CURRENT

: HINTS AND SOLUTIONS :

1 (b) or 𝑓 = 50 𝐻𝑧
𝑒 = 300√2 sin 𝜔𝑡 5 (a)
𝑅
𝑒0 300√2 cos 𝜙 = 𝑍 . In choke coil 𝜙 = 90° so cos 𝜙 ≈ 0
𝐼0 = =
𝑍 √(30)2 + (10 − 10)2 6 (b)
{∵ 𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2 } 1 𝐿 1 1
Q – factor = 𝑅 √𝐶 = 6 √17.36×10−6 = 40
300√2
= 30
= 10√2 A 7 (a)
𝐼
∴ Current 𝐼 = 02 = 10 A 𝐸2𝑅 𝐸2𝑅
√ 𝑃 = 𝐸𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙 = = 2
2 (a) 𝑍2 1
[𝑅 2 + (𝜔𝐿 − 𝜔𝐶 ) ]
Natural frequency is nothing but resonant
8 (b)
frequency. 𝑃 240 𝑉 100
In this case 𝑋𝐿 = 𝑋𝐶 𝑅= 2 = = 15Ω; 𝑍 = = = 25Ω
𝑖𝑟𝑚𝑠 16 𝑖 4
1
⇒ 𝜔0 𝐿 = Now 𝑋𝐿 = √𝑍 2 − 𝑅 2 = √(25)2 − (15)2 = 20Ω
𝜔0 𝐶
1 20 1
⇒ 𝜔02 = ∴ 2𝜋𝑣𝐿 = 20 ⇒ 𝐿 = = 𝐻𝑧
𝐿𝐶 2𝜋 × 50 5𝜋
1 9 (b)
⇒ 𝜔0 = 1
√𝐿𝐶 𝑃 = 𝑉0 𝑖0 cos 𝜙 ⇒ 𝑃 = 𝑃𝑃𝑒𝑎𝑘 . cos 𝜙
1 2
⇒ 2𝜋𝑓 = 1 1 𝜋
√𝐿𝐶 ⇒ (𝑃𝑝𝑒𝑎𝑘 ) = 𝑃𝑝𝑒𝑎𝑘 cos 𝜙 ⇒ cos 𝜙 ⇒ 𝜙 =
1 2 2 3
⇒ 𝑓= 10 (d)
2𝜋√𝐿𝐶 When a ring moves in a magnetic field in a
3 (a)
direction perpendicular to its plane, we replace
At angular frequency 𝜔, the current in 𝑅𝐶 circuit
the ring by a diameter (2r) perpendicular to the
is given by
direction of motion. The emf is induced across
𝑉𝑟𝑚𝑠
𝑖𝑟𝑚𝑠 = … (i) this diameter. Current flow in the ring will be
2
√𝑅2 + ( 1 ) through the two semicircular portions in parallel.
𝜔𝐶
𝑖 𝑉𝑟𝑚𝑠 𝑉𝑟𝑚𝑠
Induced emf = 𝐵 (2 𝑟)𝑣.
Also 𝑟𝑚𝑠
2
= 2
= 9
…(ii) Resistance of each half of ring = R /2
1 √𝑅 2 + 2 2
√𝑅2 +( 𝜔 𝐶) 𝜔 𝐶
3
As the two halves are in parallel, therefore,
From equation (i) and (ii), we get equivalent resistance = R /4
𝐵(2𝑟)𝑣
1 ∴ Current in the section =
5 3 𝑋𝐶 3 𝑅/4
3𝑅 2 = 2 2 ⇒ 𝜔𝐶 = √ ⇒ =√ 8𝐵𝑟𝑣
𝜔 𝐶 𝑅 5 𝑅 5 𝐼=
𝑅
4 (d) 13 (c)
200
Resistance of coil(𝑅) = = 200 Ω 1 2
1
200 𝑍 = √𝑅 2 + (2𝜋𝑓𝐿 − )
Current, 𝐼 = 2𝜋𝑓𝐶
√𝑅2 +𝑋𝐿2
From above equation at 𝑓 = 0, 𝑧 = ∞
200 1
or 0.5 = When 𝑓 = (resonant frequency) ⇒𝑍=𝑅
√𝑅2 +𝑋𝐿2 2𝜋√𝐿𝐶
1
or 𝑅 2 + (2𝜋𝑓𝐿)2 = (400)2 For 𝑓 > 2𝜋√𝐿𝐶
⇒ 𝑍 starts increasing
2 𝑖. 𝑒., for frequency 0 − 𝑓𝑟 , 𝑍 decreases and for 𝑓𝑟 to
2√3
or (2𝜋𝑓 × ) = (400)2 − (200)2
𝜋 ∞, 𝑍 increases
= 120000 This is justified by graph 𝑐
or 4𝑓√3 = 200√3 14 (a)

P a g e | 37
 1 1 1 Power factor cos 𝜙 =
𝑅
=1
𝑋𝐶 = ⇒ = 𝑍
2𝜋𝑣𝐶 1000 2𝜋 × 𝑣 × 5 × 10−6
100 21 (b)
⇒𝑣= 𝑀𝐻𝑧 From the relation, tan 𝜙 =
𝜔𝐿
𝜋 𝑅
15 (a) Power factor cos 𝜙 =
1
𝑖0 5 √1+tan2 𝜙
𝑖𝑟𝑚𝑠 = = = 3.536 A 1
√2 √2 = 2
√1+(𝜔𝐿)
16 (a) 𝑅
𝑅
𝑋𝐿 = 𝜔𝐿. =
√𝑅2 +𝜔2 𝐿2
𝑋𝐿 10
or 𝐿= 𝜔
= 20 = 0.5 H 22 (a)
17 (b) 1 1 1 2 𝐿
= + = 𝑜𝑟 𝐿𝑃 =
When a circuit contains inductance only, then the 𝐿𝑝 𝐿 𝐿 𝐿 2
current lags behind the voltage by the phase Where L is inductance of each part
𝜋 1.8 ×10−4
difference of 2 or 90°. = = 0.9 × 10−4 H
2
While in a purely capacitive circuit, the current 𝐿 0.9×10−4
𝜋 ∴ 𝐿𝑃 = 2 = 2
= 0.45 × 10−4H
leads the voltage by a phase angle of 2 or 90°. 6
Resistance of each part, 𝑟 = 2 = 3Ω
In a purely resistive circuit current is in phase
1 1 1 2
with the applied voltage. Now, = + =
𝑟𝑃 3 3 3
18 (b) ∴ 𝑟𝑃 = 3/2Ω
At t = 0, inductor behaves like an infinite 𝐿𝑃 0.45×10−4
Time constant of circuit = = = 8A
resistance. So at 𝑟𝑃 3/2
𝑉 23 (b)
𝑡 = 0, 𝑖 =
𝑅2 The applied voltage is given by 𝑉 = √𝑉𝑅2 + 𝑉𝐿2
And at 𝑡 = ∞, inductor behaves like a conducting 𝑉 = √(200)2 + (150)2 = 250 𝑣𝑜𝑙𝑡
wire, 24 (d)
𝑉 𝑉(𝑅1 +𝑅2 )
𝑖=𝑅 = 𝑅1 𝑅2 The voltage across secondary in zero, as
eq

19 (c) transformer does not work on DC supply.

Hot wire ammeter reads 𝑟𝑚𝑠 value of current. 25 (c)
𝐿𝑑𝑖 𝑑𝑖
Hence its peak value = 𝑖𝑟𝑚𝑠 × √2 = 14.14 𝑎𝑚𝑝 From 𝑒 = 𝑑𝑡
, when 𝑑𝑡 = 1, 𝑒 = 𝐿
20 (c) 26 (b)
𝐿2 𝑁22
𝑋𝐶 𝜋 =
𝐿1 𝑁12
= tan
𝑅 3 𝑁22 500 2
∴ 𝐿2 = 𝐿1 = 1.5 ( ) = 375 mH
𝑁12 100
27 (b)
𝑅 𝑅
cos 𝜙 = =
𝑍 √𝑅 2 + 𝜔 2 𝐿2
12
𝜋 = ⇒ cos 𝜙
𝑋𝐶 = 𝑅 tan … (i) √(12)2 + 4 × 𝜋 2 × (60)2 × (0.1)2
3
𝑋𝐿 𝜋 = 0.30
= tan 28 (d)
𝑅 3
Magnetic field intensity at a distance r from the
straight wire carrying current is
𝜇 𝑖
0
𝐵 = 2𝜋𝑟
As area of loop, 𝐴 = 𝑎2
𝜋 And magnetic flux, 𝜙 = 𝐵𝐴
𝑋𝐿 = 𝑅 tan … (ii) 𝜇0 𝑖𝑎2
3 ∴ 𝜙=
Net impedance 𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2 = 𝑅 2𝜋𝑟
The induced emf in the loop is

P a g e | 38
 𝑑𝜙 𝑑 𝜇0 𝑖 𝑎 2 𝑣
𝑒 = | 𝑑𝑡 | = |𝑑𝑡 | ∴ 𝑍 = √𝑅 2 + 4𝜋 2 𝑓 2 𝐿2
2𝜋𝑟
2
𝜇0 𝑖𝑎 𝑑𝑟 𝜇 𝑖𝑎 𝑣2 43 (b)
𝑒= 2 𝜋 𝑟 2 𝑑𝑡
= 20 𝜋 𝑟2 𝑉0 120
𝑑𝑟 𝑉𝑟𝑚𝑠 = = = 84.8 𝑉
Where 𝑣 = 𝑑𝑡 is velocity . √2 1.414
29 (c) 44 (a)
𝜋 −𝜋 𝜋 This is a combined example of growth and decay
Phase difference ∆𝜙 = 𝜙2 − 𝜙1 = 6 − ( 6 ) = 3 of current in an L – R circuit.
30 (b)
𝜇0 𝑁 2 𝐴
As 𝐿 = 𝑙
2×2×4
∴ 𝐴 → 2
times = 8 times
32 (a)
𝑛 50
𝑖𝑃 = 𝑛 𝑠 𝑖𝑠 = 200 × 40 = 10 A
𝑃
33 (a)
Current 𝑖 = 𝑖0 sin(𝜔𝑡 + 𝜙)
𝑖𝑝 = 𝑖0 sin 𝜔𝑡 cos 𝜙 + 𝑖0 cos 𝜔𝑡 sin 𝜙 The current through circuit just before shorting
Thus, 𝑖0 cos 𝜙 = 10 the battery,
𝐸
𝑖0 sin 𝜙 = 8 𝐼0 = 𝑅 = 1 A
4
Hence, tan 𝜙 = (as inductor would be shorted in steady state)
5
34 (d) After this decay of current starts in the circuit
In a purely inductive circuit, current is according to the equation 𝐼 = 𝐼0 𝑒 −𝑡/𝜏
𝜋 Where 𝜏 = 𝐿/𝑅.
𝑖 = 𝑖0 sin (𝜔𝑡 − )
2

Which shows that the current lags behind the emf

𝜋
by a phase angle of 2 or 90° or the
𝜋
emf leads the current by a phase angle of or 90°. 0−3 )/(1 00×1 0−3 /1 00)
2 𝐼 = 1 × 𝑒 −(1×1 = (1/𝑒) A
35 (b) 45 (d)
2 (30)2
𝑉𝑟𝑚𝑠 𝑅 𝑒/𝑖 1
𝑃= = = 90 𝑊 = 𝑒𝑑𝑡/𝑑𝑖 = 𝑑𝑡 = frequency.
𝑅 10 𝐿
37 (c) 46 (a)
In series LCR, the impedance of the circuit is given Phase difference relative to the current
by 𝜋 𝜋
𝜙 = (314𝑡 − ) − (314 𝑡) = −
6 6
𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2
47 (b)
At resonance, 𝑋𝐿 = 𝑋𝐶
At 𝑡 = 0, phase of the voltage is zero, while phase
∴𝑍=𝑅 𝜋 𝜋
of the current is − , 𝑖. 𝑒., voltage leads by
At resonance, the phase difference between the 2 2

current and voltage is 0°. Current is maximum at 48 (d)

resonance 𝑍 2 = 𝑅 2 + (2𝜋𝑓𝐿)2
0.4 2
39 (c) = (30)2 + (2𝜋 × 50 × )
𝜋
𝐸𝑃 𝑖𝑃 1100×100
𝑒= 𝐸𝑠
= 220
= 500 A=0.5 kA = (900 + 1600) = 2500
40 (c) or 𝑍 = 50 Ω
𝑉 200
Impedance 𝑍 = √𝑅 2 + 𝑋 2 = √(8)2 + (6)2 = 10Ω Also, 𝐼= = =4A
𝑍 50
41 (b) 50 (b)
𝑍 = √𝑅 2 + 𝑋𝐿2 , 𝑋𝐿 = 𝜔𝐿 and 𝜔 = 2𝜋𝑓 We know that Q - factor of series resonant circuit
is given as
P a g e | 39
 𝑄=
𝜔𝑟 𝐿 59 (b)
𝑅
Here, 𝐿 = 8.1 mH, 𝐶 = 12.5 𝜇F, 𝑅 = 10Ω, 𝑓 = 𝑍 = √𝑅 2 + 𝑋𝐿2 = √102 + (2𝜋 × 60 × 2)2 = 753.7
500 Hz
120
𝜔𝑟 𝐿 2𝜋𝑓𝐿 ∴𝑖= = 0.159 𝐴
∴ 𝑄= = 753.7
𝑅 𝑅 60 (b)
2×𝜋×500×8.1×10−3 8.1𝜋
= 10
= 10
= 2.5434 An alternating current is one whose magnitude
51 (b) changes continuously with time between zero and
1
For capacitive circuits 𝑋𝐶 = 𝜔𝐶 a maximum value and whose direction reverses
𝑉 periodically. The relation between frequency (𝑓)
∴𝑖= = 𝑉𝜔𝐶 ⇒ 𝑖 ∝ 𝜔 and time (𝑇) is.
𝑋𝐶
52 (b)
𝑉0 = √2 𝑉𝑟𝑚𝑠 = 10√2
53 (a)
In L-R circuit, the growing current at time 𝑡 is
𝑡
𝐸 𝐿
given y 𝑖 = 𝑖0 [1 − 𝑒 −𝜏 ] where 𝑖0 = 𝑅 𝑎𝑛𝑑 𝜏 = 𝑅
∴ Charge passed through the battery in one time
constant is
𝜏 𝜏
𝑞 = ∫0 𝑖𝑑𝑡 = ∫0 𝑖0 (1 − 𝑒 −𝑡/𝑟 )𝑑𝑡
𝑡
𝑖 𝑒 −𝑡
0
𝑞 = 𝑖0 𝜏 − [ −2/𝜏 ] = 𝑖0 𝜏 + 𝑖0 𝜏[𝑒 −1 − 1]
0
𝑖0 𝜏
= 𝑖0 𝜏 − 𝑖0 𝜏 + 𝑒
1 1
𝑇= = = 0.02 𝑠
𝑖0 𝜏 (𝐸/𝑅)(𝐿/𝑅) 𝑒𝑙 𝑓 50
𝑞 = = = 2
𝑒 𝑒 𝑒𝑅 As is clear from the figure time taken to reach the
54 (b) maximum value is
𝑃𝑖 = 240 × 0.7 = 168 W, 𝑃0 = 140 W 𝑇 0.02
4
= 4
= 0.005 s
𝑃0 140
𝜂 = × 100 = × 100 ≈ 80% 61 (b)
𝑃𝑖 168 𝑒 1
55 (c) From 𝑒 = 𝐿𝑑𝐼/𝑑𝑡 ⇒ 𝑑𝐼 = 𝐿 = 1 = 1 As −1
Energy stored in a inductor 𝐿 carrying 62 (a)
Current 𝑖 is 𝑈 = 2 𝐿 𝑖 2
1 After time t, thickness of liquid will remain
𝑑
Rate at which energy is stored ( − 𝑣𝑡).
3
𝑑𝑈 1 𝑑𝑖 𝑑𝑖 Now, time constant as function of time
= 𝑑𝑡
= 2 𝐿 2𝑖 (𝑑𝑡) = 𝐿𝑖 (𝑑𝑡)
𝑑𝑈 𝜏𝑐 = 𝐶𝑅
At 𝑡 = 0, 𝑖 = 0, ∴ 𝑑𝑡
=0
𝑑𝑖 𝜀0 (1). 𝑅
At 𝑡 = ∞, 𝑖 = 𝑖0 (constant) , ∴ =0 = 𝑑 𝑑/3−𝑣𝑡
(Applying 𝐶
𝑑𝑡 (𝑑 − + 𝑣𝑡) +
56 (d) 3 2
1
Required time 𝑡 = 𝑇/4 = 4×50 = 5 × 10−3 𝑠𝑒𝑐 𝜀0 𝐴
= 𝑡)
57 (a) 𝑑−𝑡+𝑘
Maximum voltage is AC circuit 6𝜀0 𝑅
𝑉0 = 282 𝑉 =
5𝑑 + 3𝑣𝑡
𝑉0 282 63 (d)
𝑉= =
√2 √2 When wire is thick, its resistance reduces.
282 28200 Therefore, Joules’ heating loss is reduced.
𝑉= =
1.41 141 64 (c)
𝑉 = 200 𝑉
Peak value = 220√2 = 311 𝑉
58 (b)
65 (c)
1 1 𝜋
𝑋𝐶 = = =∞ 𝐼𝐿 lags behind 𝐼𝑅 by a phase of 2 , while 𝐼𝐶 leads by
2𝜋𝑣 𝐶 0
P a g e | 40
 𝜋 𝑋𝐿
a phase of ∴ tan 𝜙 =
2
66 (b) 𝑅
Y
Time constant of R – C circuit is 𝜏 = 𝑅𝐶 L K

|
Here effective resistance of the circuit
2𝑅×3𝑅 6𝑅
= 2𝑅+3𝑅 = 5 VL
6𝑅 6𝑅𝐶
∴ 𝜏= ×𝐶 = )
5 5

|
X
O A
67 (b)
𝑀𝑑𝑖 𝜇0 𝑁1 𝑁2 𝐴 𝑑𝑖 VR |
𝑒= = ( )
𝑑𝑡 𝑙 𝑑𝑡 Y'
4𝜋×10−7 ×2000×300×1.2 ×10−3 (4)
= 0.3×0.25
= 4.8 × 10 −2
V 74 (d)
68 (a) Current will be max at first time when
𝜋 100𝜋𝑡 + 𝜋/3 = 𝜋/2 ⇒ 100 𝜋𝑡 = 𝜋/6 ⇒ 𝑡
𝑉 = 5 cos 𝜔𝑡 = 5 sin (𝜔𝑡 + 2 ) and 𝑖 = 2 sin 𝜔𝑡
= 1/600 𝑠
Power = 𝑉𝑟.𝑚.𝑠. × 𝑖𝑟.𝑚.𝑠. × cos 𝜙 = 0
𝜋 𝜋 75 (d)
[Since 𝜙 = 2 , therefore cos 𝜙 = cos 2 = 0]
The current will lag behind the voltage when
69 (d) reactance of inductance is more than the
𝑉 4 reactance of condenser.
𝑖= = = 0.8 𝐴
𝑍 √42 + (1000 × 3 × 10−3 )2 1 1
Thus, 𝜔𝐿 > 𝜔𝐶 or 𝜔 >
√𝐿𝐶
70 (a) 1
𝑋𝐿 = 31Ω, 𝑋𝐶 = 25Ω, 𝑅 = 8Ω or 𝑛 > 2𝜋 or 𝑛 > 𝑛𝑟 where 𝑛𝑟 = resonant
√𝐿𝐶
Impedance of series 𝐿𝐶𝑅 is frequency
𝑍 = √(𝑅)2 + (𝑋𝐿 − 𝑋𝐶 )2 76 (b)
𝑑𝐼 𝑒𝑑𝑡 8(0.05)
= √(8)2 + (31 − 25)2 = √64 + 36 = 10Ω 𝑒=𝐿 𝐿= = = 0.2 H
𝑑𝑡 𝑑𝐼 (4−2)
𝑅 8
Power factor, cos 𝜙 = 𝑍 = 10 = 0.8 77 (c)
71 (d) In 𝐿 − 𝑅 circuit, current at any time 𝑡 is given by
𝐸 𝑅 𝐸 𝐸 𝑅
𝜙 𝜇0 𝑁1 𝐴𝑖 𝜇0 𝑁 2 𝑖 𝑖 = (1 − 𝑒 − 𝐿 𝑡 ) = − 𝑒 − 𝐿 𝑡
𝐵= = = 𝑅 𝑅 𝑅
𝐴 𝐿𝐴 𝐿 𝑑𝑖 𝐸 −𝑅 𝑡 𝑅 𝐸 𝑅
72 (c) = 𝑒 𝐿 ( ) = 𝑒 −𝐿 𝑡
𝐸0 𝑖0 𝑅 𝑑𝑡 𝑅 𝐿 𝐿
𝑅
𝑃 = 𝐸𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙 = × × Induced emf = 𝐿 𝑑𝑡 = 𝐸𝑒 − 𝐿 𝑡
𝑑𝑖
√2 √2 𝑍
𝐸0 𝑅 𝐸0𝐸02 𝑅 𝑅
⇒ × × ⇒𝑃= From Eq. (i), 𝑖𝑅 = 𝐸 − 𝐸𝑒 − 𝐿 𝑡
√2 𝑍√2 𝑍 2𝑍 2 Using Eq. (ii), 𝑖𝑅 = 𝐸 − 𝑒 𝑜𝑟 𝑒 = 𝐸 − 𝑖𝑅
𝐸02
Given 𝑋𝐿 = 𝑅 so, 𝑍 = √2𝑅 ⇒ 𝑃 = 4𝑅 Therefore, graph between 𝑒 and 𝑖 is a straight line
73 (a) with negative slope and positive intercept. The
Since, current lags behind the voltage in phase by choice (c) is correct.
a constant angle, then circuit must contain R and 78 (c)
1 𝑞2 1 2 𝑞2
L. 𝑈= = (𝑞0 𝑒 −𝑡/𝜏 ) = 0 𝑒 −2𝑡/𝜏 (where 𝜏 =
2 𝐶 2𝐶 2𝐶
𝐶𝑅)
𝑈 = 𝑈 𝑒−2𝑡/𝜏
𝑖
1
𝑈 = 𝑈 𝑒−2𝑡1/𝜏
2 𝑖 𝑖
1
= 𝑒 −2𝑡1 /𝜏
We find that in R – L circuit, voltage leads the 2
𝜏
current by a phase angle 𝜙, where ⇒ 𝑡1 = ln 2
𝐴𝐾 𝑂𝐿 2
tan 𝜙 = 𝑂𝐴 = 𝑂𝐴 Now 𝑞 = 𝑞 𝑒−𝑡/𝜏
0
𝑉 𝐼 𝑋
= 𝐿= 0 𝐿 1
𝑞 =𝑞 −𝑡 /𝜏
𝑉𝑅 𝐼0 𝑅 2 0 0𝑒 2

P a g e | 41
 𝑡2 = 𝜏 ln 4 = 2𝜏 ln 2 𝑅
cos 𝜙 =
𝑡1 1 √𝑅 2 + 𝜔 2 𝐿2
∴ = 30
𝑡2 4 =
√(30)2 +(100)2 ×(400×10−3 )2
79 (a) 30 30
At resonance 𝐿𝐶𝑅 series circuit behaves as pure = = 50 = 0.6
√900+1600
resistive circuit. For resistive circuit 𝜙 = 0° 90 (a)
80 (b) Yes, in 𝐴𝐶 if branch 𝐴𝐵 has 𝑅, 𝐵𝐶 has a capacitor
𝑍 = √𝑅 2 + 𝑋 2 = √42 + 32 = 5 𝐶, and 𝐵𝐷 has a pure inductance 𝐿
C
𝑅 3
∴ cos 𝜙 = = = 0.6
𝑍 5 15 A
81 (d) 10 A B
A
220 220
𝑖= = = 3.33 𝐴
√(20)2 + (2 × 𝜋 × 50 × 0.2)2 66 5A
82 (a) D
𝑛 4200
𝐸𝑠 = 𝑛 𝑠 𝐸𝑃 = 2100 × 120 = 240V 91 (c)
𝑃
𝑛𝑠 2100 At 𝐴 ∶ 𝑋𝐶 > 𝑋𝐿
𝑖𝑠 = 𝑖
𝑛𝑃 𝑃
= 4200
× 10 = 5 A
At 𝐵 ∶ 𝑋𝐶 = 𝑋𝐿
83 (b)
At 𝐶 ∶ 𝑋𝐶 < 𝑋𝐿
𝑖0 4
𝑖𝑟.𝑚.𝑠. = = = 2√2 𝑎𝑚𝑝𝑒𝑟𝑒 92 (c)
√2 √2 Here: Current in the circuit
84 (a)
𝜋 (𝑖) = 15 𝑚𝐴 = 15 × 10−3 𝐴
As the current i leads the emf e by 4 , it is an R – C
Resistance 𝑅 = 4000 𝑉𝑜𝑙𝑡
circuit Applied voltage in the circuit = 240 𝑉
𝑋𝐶
tan 𝜙 = 𝑅
At any instant, the 𝑒𝑚𝑓 of the battery is equal to
𝜋
1 the sum of potential drop on the resistor and the
𝜔𝐶
or tan = 𝑒𝑚𝑓 developed in the induction coil
4 𝑅
∴ 𝜔𝐶𝑅 = 1 𝑑𝑖
Hence, 𝐸 = 𝑖𝑅 + 𝐿 𝑑𝑡
As 𝜔 = 100 rads−1 𝑑𝑖
The product of C – R should be 100 s −1.
1 240 = 15 × 10−3 × 4000 + 𝐿
𝑑𝑡
𝑑𝑖
85 (d) Hence, 𝐿 𝑑𝑡 = 240 − 60 = 180 𝑉
Phase angle 𝜙 = 90°, so power 𝑃 = 𝑉𝑖 cos 𝜙 = 0 93 (b)
86 (d) 𝐿/𝑅 represents time constant of R-L circuit.
1
Current lags the voltage if 𝜔𝐿 > 𝜔𝐶 Therefore, its dimensions are [𝑀0 𝐿0 𝑇 1 ].
1 94 (a)
𝑓> ⇒ 𝑓 > 𝑓𝑟
2𝜋√𝐿𝐶 This is a parallel circuit, For oscillation, the energy
87 (c) in 𝐿 and 𝐶 will be alternately maximum
𝜔 120 × 7 95 (a)
𝑣= = = 19 𝐻𝑧
2𝜋 2 × 22 The current in a coil is given by
240 𝑖 = 𝑖0 𝑒 −𝑡/𝜏
𝑉𝑟.𝑚.𝑠 = = 120√2 = 170 𝑉
√2 Now, 𝑖 =
𝑖0
𝑖𝑛 𝑡 = 𝑡0
88 (c) 𝜂
𝑖0
For an R – C circuit ∴ 𝜂
= 𝑖0 𝑒 −𝑡0 /𝜏
−𝑡0 /𝜏
Applied voltage, 𝑉 = √𝑉𝑅2 + 𝑉𝐶2 𝑒 = 𝜂 −1
Taking log of both sides,
𝑡0
∴ 50 = √(40)2 + 𝑉𝐶2 − log 𝑒 𝑒 = −1 log 𝑒 𝜂
𝜏
𝑡0
⇒ 𝑉𝐶 = 30 V = log 𝑒 𝜂
𝜏
89 (d) 𝜏 = 𝑡0 / log 𝑒 𝜂 = 𝑡0 /In𝜂
Power factor 96 (b)

P a g e | 42
 Since voltage is lagging behind the current, so 𝑉 = 𝐵𝑙𝑣 = 4(1) (2) = 8 volt
there must be no inductor in the box This is the potential to which the capacitor is
97 (b) charged.
Average power dissipated in an AC circuit As 𝑞 = 𝐶𝑉
𝑃𝑎𝑣 = 𝑉rms 𝐼rms cos 𝜙 …(i) ∴ 𝑞 = (10 × 10−6 )8 = 10−5 𝐶 = 80𝜇 𝐶
Where the term cos 𝜙 is known as power factor. As magnetic force on electron in the conducting
Given, 𝑉rms = 100 V, 𝑅 = 100 Ω, ϕ = 30° rod 𝑃𝑄 is towards 𝑄, therefore, 𝐴 is positively
𝑉rms 100
∴ 𝐼rms = = =1A charged and 𝐵 is negatively charged
𝑅 100
Putting the values in Eq. (i), we get 𝑖𝑒, 𝑞𝐴 = +80𝜇 𝐶 and 𝑞𝐵 = −80 𝜇 𝐶
𝑃𝑎𝑣 = 100 × 1 × cos 30° 105 (c)
√3
The DC generator must be mixed wound to with
= 100 2 stand the load variation.
= 50√3 = 86.6 W 106 (a)
98 (c) For imparting max power
𝜔𝐿 − 𝜔𝐶
1 1
𝑋𝐿 = 𝑋𝐶 ⇒ 𝜔𝐿 =
tan 𝜙 = 𝜔𝐶
𝑅 1 1 1
𝜙 being the angle by which the current leads the 𝐶= 2 = =
𝜔 𝐿 (2𝜋𝑓) × 𝐿 (100𝜋)2 × 10
2
voltage.
= 1 × 10−6 = 1𝜇𝐹
Given, 𝜙 = 45°
1 108 (c)
𝜔𝐿 −
𝜔𝐶
∴ tan 45° = 1 2
𝑅 𝑍 = √𝑅 2 + ( )
1 2𝜋𝑣𝐶
𝜔𝐿 − 𝜔𝐶
⇒ 1=
𝑅 1
1 = √(3000)2 + 2
⇒ 𝑅 = 𝜔𝐿 − (2𝜋 × 50 ×
2.5
× 10−6 )
𝜔𝐶 𝜋
1
⇒ 𝐶=
𝜔(𝜔𝐿 − 𝑅) ⇒ 𝑍 = √(3000)2 + (4000)2 = 5 × 103 Ω
1 𝑅 3000
= So power factor cos 𝜙 = 𝑍 = 5×103 = 0.6 and
2𝜋𝑓(2𝜋𝑓𝐿−𝑅)
100 (d) power
2
In an L – C – R circuit in resonance condition 𝑉𝑟𝑚𝑠 cos 𝜙
𝑃 = 𝑉𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙 = ⇒𝑃
𝑋𝐿 = 𝑋𝐶 𝑍
or 𝑋𝐶 − 𝑋𝐿 = 0 (200)2 × 0.6
= = 4.8𝑊
102 (c) 5 × 103
𝑑𝑖 110 (d)
As 𝑒 = 𝑀 𝑑𝑡
Rise of current in L – R circuit is given by
10
∴ 30 × 103 = 3 × ,
𝑑𝑡
30 −3
𝑑= 3 = 10 s
30×10
103 (b)
Wheatstone bridge is balanced. Current through
𝐴𝐶 is zero. Effective resistance R of bridge is
1 1 1 1
𝑅
= 6 + 6 = 3 , 𝑅 = 3Ω
Total resistance = 1 + 3 = 4Ω
Induced emf
𝑒 = 𝑖𝑅 = 𝐵𝑙𝑣 𝐼 = 𝐼0 (1 − 𝑒 −𝑡/𝜏 )
𝐸 5
𝑖𝑅 1 × 10−3 × 4 Where, 𝐼0 = 𝑅 = 5 = 1 A
∴ 𝑣= = 𝐿 10
𝐵𝑙 2 × 0.1 Now, 𝜏=𝑅= =2s
5
= 2 × 10−2 ms−1
After 2s, ie, at t = 2 s
104 (b)
Rise of current 𝐼 = (1 − 𝑒 −1 )A
Motional emf across 𝑃𝑄
P a g e | 43
111 (b) flux passing through secondary coil is
As 𝑒 = 𝑀𝑑𝐼/𝑑𝑡, 𝜇0 𝑖1
𝑒𝑑𝑡 15000×0.001
𝜙 = 𝐵𝐴 = (𝜋𝑅22 )
∴ 𝑀= =− = 5H 2𝑅1
𝑑𝐼 3 𝜙 𝜇 𝜋𝑅2
112 (b) Now, 𝑀 = 𝑖 2 = 02𝑅 2
1 1
At resonance, 𝐿𝐶𝑅 circuit behaves as purely 𝑅22

resistive circuit. For purely resistive circuit power ∴ 𝑀∝

𝑅1
factor = 1 121 (c)
113 (a) Power remains constant in a ideal step down
Voltage across the capacitors will increase from 0 transformer.
to 10 V exponentially. The voltage at time t will be 122 (b)
given by 𝑉 = 50 × 2 sin 100𝜋𝑡 cos 100𝜋𝑡 = 50 sin 200𝜋𝑡
𝑉 = 10(1 − 𝑒 −𝑡/𝜏𝐶 ) ⇒ 𝑉0 = 50 𝑣𝑜𝑙𝑡𝑠 and 𝑣 = 100𝐻𝑧
Here 𝜏𝐶 = 𝐶net 𝑅net = (1 × 106 )(4 × 10−6 ) = 4 s 123 (b)
∴ 𝑉 = 10(1 − 𝑒 −𝑡/4 ) Capacitive reactance is given by
Substituting 𝑉 = 4 volt, we have, 1
𝑋𝐶 = 𝜔 𝐶
−𝑡/4
4=10(1 − 𝑒 )
3
Where C is capacitance and 𝜔 the angular
𝑒 −𝑡/4 = 0.6 = 5 frequency (𝜔 = 2𝜋𝑓).
Taking log both sides we have, 1
𝑡 ∴ 𝑋𝐶 =
2𝜋𝑓𝐶
− = In 3 − In 5
4 ⇒ 𝑋𝐶 ∝ 𝑓
1
or t =4(In 5 − In 3) = 2s.
114 (c) Hence, when frequency 𝑓 increases capacitive
𝜇 𝑁 𝐴2 2
𝜇 𝜇 𝑁 𝐴 reactance decreases.
From 𝐿 = 0 𝑙 = 0 𝑟𝑙 ,
124 (a)
1 𝑅
When 𝜇 = 1000 and 𝑁 becomes 10 Power factor= cos 𝜙 =
𝑍
1 2 12 4
∴ 𝐿 become𝑠 1000 × ( ) = 10 times = = = 0.8
10 15 5
𝑖𝑒, 𝐿 = 10 × 0.1 = 1H 125 (d)
115 (b) Given 𝜔𝐿 =
1
⇒ 𝜔2 =
1
𝐸−𝑉 𝜔𝐶 𝐿𝐶
From 𝑅 = 1 1
𝑖 Or 𝜔 = = = 104
120−𝑉 √10−3 ×10×10−6 √10−8
0.5 = 8 𝑋𝐿 = 𝜔𝐿 = 10 × 10 4 −3
= 10 Ω
𝑉 = 116 𝑉 126 (b)
117 (b) 𝑖𝑠 =
𝐸𝑠 22
= 220 = 0.1 A
𝑍
In non resonant circuits
1 127 (a)
Impedance 𝑍 = 2
, with rise in The instantaneous value of voltage is
√ 12 +(𝜔𝐶− 1 )
𝑅 𝜔𝐿
𝐸 = 100 sin(100𝑡) 𝑉 … (i)
frequency 𝑍 decreases, 𝑖. 𝑒., current increases so Compare it with 𝐸 = 𝐸0 sin(𝜔𝑡) 𝑉
circuit behaves as capacitive circuit We get
118 (c) 𝐸0 = 100𝑉, 𝜔 = 100𝑟𝑎𝑑𝑠 −1
Phase difference in R – L circuit, The rms value of voltage is
𝑋𝐿
𝜙 = tan−1 𝑅
𝐸0 100
𝑋𝐿
𝐸𝑟𝑚𝑠 = = 𝑉 = 70.7𝑉
or tan 45° = √2 √2
𝑅
The instantaneous value of current is
or 𝑋𝐿 = 𝑅 𝜋
119 (d) 𝐼 = 100 sin (100𝑡 + ) 𝑚𝐴
3
At resonant frequency current in series 𝐿𝐶𝑅 Compare it with
circuit is maximum 𝐼 = 𝐼0 sin(𝜔𝑡 + 𝜙)
120 (d) We get
Magnetic field at the centre of primary coil 𝐵 = 𝐼0 = 100𝑚𝐴, 𝜔 = 100rads−1
𝜇0 𝑖1 /2𝑅1 . Considering it to be uniform, magnetic
P a g e | 44
 The rms value of current is 𝑉𝑅
𝐼=
𝑅
𝐼0 100 100
𝐼𝑟𝑚𝑠 = = 𝑚𝐴 = 70.7𝑚𝐴 = = 0.1 A
√2 √2 1000
128 (a) So, voltage across L is given by
Resistance , 𝑅 =
100
= 10 Ω 𝑉𝐿 = 𝐼𝑋𝐿 = 𝐼𝜔𝐿
10 1
But 𝜔𝐿 = 𝜔𝐶
Inductive reactance , 𝑋𝐿 = 2𝜋𝑓𝐿
100 1
= 2𝜋 × 50 × 𝐿 ∴ 𝑉𝐿 =
8 𝜔𝐶
0.1
1 = 200×2×10−6 = 250 V
⇒ 𝐿= H
8𝜋 138 (b)
1
𝑋𝐿′ = 2𝜋𝑓 ′ 𝐿 = 2𝜋 × 40 × 8𝜋 = 10 Ω When the direction of current is reversed, moving
2 from 𝐵 to 𝐴.
Impedance of the circuit is 𝑍 = √𝑅 2 + 𝑋𝐿′
𝑉𝐵 − 𝑉𝐴 = [5 × 10−3 (−103 ) + 15 + 1 × 5]
= √(10)2 + (10)2 = 15 volt
= 10√2 Ω 139 (b)
𝑉 150 15 The instantaneous voltage through the given
Current in the circuit is 𝑖 = 𝑍 = 10 2 = 2 A
√ √ device
129 (a) 𝑒 = 80 sin 100𝜋𝑡
∵ (𝑋𝐶 ) >> (𝑋𝐿 ) Comparing the given instantaneous voltage with
130 (a) standard instantaneous voltage
200 5 5
𝑖𝑟𝑚𝑠 = = 𝐴. So 𝑖0 = 𝑖𝑟𝑚𝑠 × √2 = × √2 ≈ 1𝐴 𝑒 = 𝑒0 sin 𝜔𝑡.
280 7 7
132 (d) We get 𝑒0 = 80 𝑉
In series L – R circuit, impedance is given by Where 𝑒 0 is the peak value of voltage
𝑍 = √𝑅 2 + 𝑋𝐿2 Impedance (𝑍) = 20Ω
𝑒
Where R is the resistance and 𝑋𝐿 the inductive Peak value of current 𝐼0 = 𝑍0
reactance. 80
= 20 = 4A
Given, 𝑅 = 8Ω, 𝑋𝐿 = 6Ω
Effective value of current (root mean square value
∴ 𝑍 = √(8)2 + (6)2 of current).
= √64 + 36 𝐼
𝐼𝑟𝑚𝑠 = 0
√2
= √100 = 10 Ω 4
= 2 = 2√2 = 2.828 A
134 (a) √
If the current is wattles then power is zero. Hence 140 (b)
𝑡
phase difference 𝜙 = 90° 𝐸
Charging current, 𝐼 = 𝑅 𝑒 −𝑅𝐶
135 (a)
Taking log both sides,
In 𝐿𝐶𝑅 circuit; in the condition of resonance 𝑋𝐿 = 𝐸 𝑡
𝑋𝐶 , 𝑖. 𝑒., circuit behaves as resistive circuit. In Log 𝐼 = log (𝑅) − 𝑅𝐶
resistive circuit power factor is maximum When R is doubled, slope of curve increases. Also
136 (c) at t = 0, the current will be less. Graph Q
𝑇/2 𝑇/2
∫0 𝑖 𝑑𝑡 ∫0 𝐼0 sin(𝜔𝑡)𝑑𝑡 represents the best.
𝐼𝑎𝑣 = 𝑇/2 = 141 (b)
∫0 𝑑𝑡 𝑇/2
The coil has inductance 𝐿 besides the resistance 𝑅.
𝜔𝑇
2𝐼0 − cos 𝜔𝑡 𝑇/2 2𝐼0 cos ( 2 ) cos 0° Hence for ac it’s effective resistance √𝑅 2 + 𝑋𝐿2 will
= [ ] = [− + ]
𝑇 𝜔 0 𝑇 𝜔 𝜔 be larger than it’s resistance 𝑅 for dc
142 (a)
2𝐼0 2𝐼0 2𝐼0
= [− cos 𝜋 + cos 0°] = [1 + 1] = In 𝑅-𝐶 circuit current increases exponentially
𝜔𝑇 2𝜋 𝜋
with time, so correct graph will be (a)
137 (c)
1 143 (d)
At resonance, 𝜔𝐿 =
𝜔𝐶 𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2 = √(11)2 + (25 − 18)2
Current flowing through the circuit,
= 13 Ω

P a g e | 45
 260 200
Current 𝑖 = 13
= 20 𝐴 2𝜋 × 60
× 1 × 10−2 × 𝜋 × (0.3)2
⇒ 𝑖0 = = 6 𝑚𝐴
144 (c) 𝜋2
Given 𝑋𝐿 = 𝑋𝐶 = 5Ω, this is the condition of 152 (c)
resonance. So 𝑉𝐿 = 𝑉𝐶 , so net voltage across 𝐿 and In the condition of resonance
𝐶 combination will be zero 𝑋𝐿 = 𝑋𝐶
1
145 (c) or 𝜔𝐿 = 𝜔𝐶 …(i)
𝑛𝑠 𝐸𝑠 2400 Since, resonant frequency remains unchanged,
= = = 20
𝑛𝑃 𝐸𝑃 120 So, √𝐿𝐶 =constant
𝑛𝑠 = 20 𝑛𝑃 = 20 × 75 = 1500.
or LC = constant
146 (b)
∴ 𝐿1 𝐶1 = 𝐿2 𝐶2
The current at any instant is given by
L .R
⇒ 𝐿 × 𝐶 = 𝐿2 × 2 𝐶
𝐿
⇒ 𝐿2 =
2
153 (b)
2V
This is because, when frequency 𝑣 is increased,
1
𝐼 = 𝐼0 (1 − 𝑒 −𝑅𝑡/𝐿 ) the capacitive reactance 𝑋𝐶 = 2𝜋𝑣𝐶 decreases and
𝐼0 hence the current through the bulb increases
= 𝐼0 (1 − 𝑒 −𝑅𝑡/𝐿 )
2 155 (a)
1
= (1 − 𝑒 −𝑅𝑡/𝐿 ) In the rotation of magnet, 𝑁 pole moves closer to
2
1 coil 𝐶𝐷 and 𝑆 pole moves closer to coil 𝐴𝐵. As per
𝑒 −𝑅𝑡/𝐿 = Lenz’s law, 𝑁 pole should develop at the end
2
𝑅𝑡 corresponding to 𝐶. Induced current flows from
𝐿
=1n2
𝐶 𝑡𝑜 𝐷. Again 𝑆 pole should develop at the end
𝐿 3 00 × 1 0−3
∴ 𝑡= 1𝑛2= × 0.6 93 corresponding to 𝐵. Therefore, induced current in
2 2 the coil flows from 𝐴 to 𝐵.
= 1 50 × 0.6 93 × 1 0−3
156 (d)
= 0.10395s = 0.1 s
As a given pole (𝑁 or 𝑆) of suspended magnet
147 (d)
goes into the coil and comes out of its, current is
The instantaneous values of emf and current in
induced in the coil in two opposite directions.
inductive circuit are given by 𝐸 = 𝐸0 sin 𝜔𝑡 and
𝜋
Therefore, galvanometer deflection goes to left
𝑖 = 𝑖0 sin (𝜔𝑡 − 2 ) respectively and right both. As amplitude of oscillation of
𝜋 magnet goes on decreasing, so does the amplitude
So, 𝑃inst = 𝐸𝑖 = 𝐸0 sin 𝜔𝑡 × 𝑖0 sin (𝜔𝑡 − 2 )
𝜋 𝜋 of deflection.
= 𝐸0 𝑖0 sin 𝜔𝑡 (sin 𝜔𝑡 cos − cos 𝜔𝑡 sin ) 157 (b)
2 2
= 𝐸0 𝑖0 sin 𝜔𝑡 cos 𝜔𝑡 As is for Fig. (i), steady state current for t = both
1 the circuits is same. Therefore,
= 𝐸0 𝑖0 sin 2𝜔𝑡 (sin 2𝜔𝑡 = 2 sin 𝜔𝑡 cos 𝜔𝑡) 𝑉 𝑉
2 =𝑅
𝑅1
Hence, angular frequency of instantaneous power 2
Or 𝑅1 = 𝑅2
is 2𝜔
Again, from the same figure, we observe that
148 (c)
1 𝐿1 𝐿2
Resonance frequency = 2𝜋 𝐿𝐶 does not depend on 𝜏1 < 𝜏2 ∴ <
√ 𝑅1 𝑅2
resistance As 𝑅1 = 𝑅2 , therefore, 𝐿1 < 𝐿2 .
149 (c) 158 (b)
𝑅 𝑅 5 𝑃 = 𝑉𝐼
cos 𝜙 = = =
𝑍 √𝑅 2 + 𝜔 2 𝐿2 √25 + (50)2 × (0.1)2 550
𝐼= = 2.5 𝐴
5 1 220
= = ⇒ 𝜙 = 𝜋/4 159 (a)
√25 + 25 √2
151 (c) Let the applied voltage be 𝑉 volt.
𝑉0 𝜔𝑁𝐵𝐴 (2𝜋𝑣)𝑁𝐵(𝜋𝑟 2 )
Amplitude of 𝑎𝑐 = 𝑖0 = 𝑅
= 𝑅
= 𝑅

P a g e | 46
 R C
𝜇 𝑖
0
∴ Change in flux, 𝑑𝜙 = (2𝜋𝑏 ) 𝜋 𝑎2
|
𝑑𝜙
VR V C
As 𝑑𝑞 = 𝑅
∴ Total charge circulating the inner coil is
Here, 𝑉𝑅 = 12 𝑉, 𝑉𝐶 = 5 𝑉 𝜇 𝑖 𝜋𝑎 2 𝜇0 𝑖 𝑎 2
= (2 𝜋0 𝑏) 𝑅
= 2𝑅𝑏
∴ 𝑉 = √𝑉𝑅2 + 𝑉𝐶2 = √(12)2 + (5)2
165 (a)
= √144 + 25 = √169 = 13V Induced emf produced in coil
160 (c) 𝑒=
−𝑑𝜙
=
−𝑑
(𝐵𝐴)
𝑅𝑡 𝑑𝑡 𝑑𝑡
𝑖 = 𝑖0 (1 − 𝑒 − 𝐿 ) 𝑑𝐵 1
∴ |𝑒| = 𝐴
= 0.01 ×
𝑑𝑖 𝑑 𝑑 𝑅𝑡 𝑖0 𝑅 −𝑅𝑡 𝑑𝑡 1 × 10−3
⇒ = 𝑖0 − (𝑖0 𝑒 − 𝐿 ) = 0 + 𝑒 𝐿 |𝑒| = 10 V
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝐿
Initially, t = 0 Current produced in coil,
|𝑒| 10
𝑑𝑖 𝑖0 × 𝑅 𝐸 5 𝑖= = =5A
⇒ = = = = 2.5 A𝑠 −1 𝑅 2
𝑑𝑡 𝐿 𝐿 2 Heat evolved = 𝑖 2 𝑅𝑡
161 (d) = (5)2 × (2) × 1 × 10−3 = 0.05 J
As the magnetic field directed into the paper is 166 (c)
increasing at a constant rate, therefore, induced Power = Rate of work done in one complete cycle.
current should produce a magnetic field directed 𝑊
or 𝑃𝑎𝑣 = 𝑇
out of the paper. Thus current in both the loops
(𝐸 𝐼 cos 𝜙)𝑇/2
must be anti-clock-wise. or 𝑃𝑎𝑣 = 0 0
𝑇
𝐸0 𝐼0 cos 𝜙
or 𝑃𝑎𝑣 = 2
Where cos 𝜙 is called the power factor of an AC
circuit.
167 (c)
As area of loop on right side is more, therefore,
Terminal velocity of the rod is attained when
induced emf o right side of loop will be more
magnetic force on the rod (𝐵𝑖𝑙) balances the
compared to the emf induced on the left-side of
component of weight of the rod(mg sin θ) , figure.
the loop
𝑑𝜙 𝑑𝐵
[∴ 𝑒 = − 𝑑𝑡 = −𝐴 𝑑𝑡 ]
162 (b) in
mg s
Fm
Given, 𝐿 = 30 mH
𝑉𝑟𝑚𝑠 = 220 V R
𝑓 = 50𝐻𝑧
Now, 𝑋𝐿 = 𝜔𝐿 = 2𝜋𝑓𝐿
= 2 × 3.14 × 50 × 30 × 10−3
𝑖𝑒 , 𝐵𝑖𝑙 = mg sin θ
= 9.42 Ω 𝑒
𝐵 (𝑅) 𝑙 = mg sin θ
The rms current in the coil is
𝑉𝑟𝑚𝑠 220𝑉 𝐵𝑙
𝑖𝑟𝑚𝑠 = = =23.4 A (𝑒) = mg sin θ
𝑋𝐿 9.42 Ω 𝑅
𝐵𝑙
163 (c) 𝑅
(𝐵𝑙𝑣𝑟 ) = mg sin θ
𝑃 = 𝑉𝑟.𝑚.𝑠. × 𝑖𝑟.𝑚.𝑠. × cos 𝜙 mg 𝑅 sin θ
𝑟𝑇 = 𝐵2 𝑙2
100 100 × 10−3 𝜋
= × × cos 168 (a)
√2 √2 3
4 −3
1 1 1
10 × 10 1 10 𝑋𝐶 = ⇒𝐶= =
= × = = 2.5 𝑤𝑎𝑡𝑡 2𝜋𝑣𝐶 2𝜋𝑣𝑋𝐶 2 × 𝜋 × 400 × 25
2 2 4 𝜋
164 (d) = 50 𝜇𝐹
Initial flux linked with inner coil when 𝑖 = 0 is 170 (c)
zero. Final flux linked with inner coil when 𝑖 = Here, 𝑀 = 2H, 𝑑𝜙 = 4 Wb, 𝑑𝑡 = 10 s
𝜇 𝑖 As 𝜙 = 𝑀 𝑖
0
𝑖 𝑖𝑠 (2𝜋𝑏 ) 𝜋 𝑎2
𝑑𝜙 = 𝑀 𝑑𝑖
P a g e | 47
 𝑑𝜙 4 Therefore, charge stored in the capacitor is
Or 𝑑𝑖 = 𝑀
=2=2A
constant. Hence current in the circuit 𝐻𝐾𝐷𝐸 is
Also, 𝑑𝜙 = 𝑀 (𝑑𝑖) = 2(1)
zero.
= 2 Wb
178 (b)
171 (a)
𝑒 = −𝐿𝑑𝐼/𝑑𝑡 = −5 × (−2) = +10 V
𝑋𝐿 √3 𝑅
tan 𝜙 = = = √3 ⇒ 𝜙 = 60° = 𝜋/3 179 (b)
𝑅 𝑅 (Rated voltage)2
172 (c) Resistance of a bulb = Rated power
𝑇 (1/50) 𝜋 1
Time difference = ×𝜙 = ×4 = 𝑠 = (220)2
2𝜋 2𝜋 400 = = 484 Ω
2.5𝑚-𝑠 100
173 (a) Peak voltage of the source, 𝑉0 = 220√2 𝑉 = 311 𝑉
Here, Resistance, 𝑅 = 3Ω 180 (a)
𝜇0 𝑁1 𝑁2 𝐴
Inductive reactance, 𝑋𝐿 = 10Ω As 𝑀 = , therefore, 𝑀 becomes 4 times
𝑙
Capacitive reactance, 𝑋𝐶 = 14Ω 181 (a)
The impedance of the series 𝐿𝐶𝑅 circuit is Impedance of 𝐿𝐶𝑅 circuit will be minimum at
𝑍 = √𝑅 2 + (𝑋𝐶 + 𝑋𝐿 )2 = √(3)2 + (14 − 10)2 resonant frequency so
𝑍 = 5Ω 1 1
𝑉0 = =
174 (d) 2𝜋√𝐿𝐶 2𝜋√1 × 10 × 0.1 × 10−6
−3
5
In purely inductive circuit voltage leads the 10
= 𝐻𝑧
current by 90° 2𝜋
175 (a) 182 (b)
1 𝐿 Here, resistance of rod = 2Ω. 𝑟 = 0.1 m, 𝐵 =
𝑄 factor is given by 𝑅 √𝐶 50 T, along 𝑧 − axis 𝜔 = 20 rads −1 .
So, for large quality factor the inductance should Potential difference between centre of the ring
be large and resistance and capacitance must be and the rim is
1 1
small 𝑉 = 2 𝐵𝜔𝑟 2 = 2 × 50 × 20 × (0.1)2 = 5 V
176 (b) The equivalent circuit of the arrangement is
true power
As, power factor = shown in figure
apparent power
= cos 𝜙 A
𝑅
= 10 Ω
√𝑅2 +(𝑋𝐿 −𝑋𝐶 )2 10 Ω
𝑅
∴ power factor= cos 𝜙 = 𝑍 O

In a non-inductive circuit, 𝑋𝐿 = 𝑋𝐶 10 Ω
𝑅 𝑅
∴ Power factor = cos 𝜙 = = =1 B
√𝑅 2 𝑅 5V
∴ 𝜙 = 0° 10 Ω
5V
This is the maximum value of power factor. In a 5Ω
pure inductor or an ideal capacitor
5V 10 Ω
𝜙 = 90°
∴ Power factor= cos 𝜙 = cos 90° = 0 10 Ω
10 Ω
Average power consumed in a pure inductor or
ideal capacitor Current through external resistance,
𝐸 5 1
𝑃 = 𝐸𝑣 . 𝐼𝑣 cos 90° = zero 𝑖= = = A
𝑅+𝑟 10+5 3
Therefore, current through pure L or pure C, 183 (c)
which consumes no power for its maintenance in 𝜔𝐿 2𝜋 × 50 × 0.21
the circuit is called ideal current or wattles tan 𝜙 = = = 5.5 ⇒ 𝜙 = 80°
𝑅 12
current. 185 (d)
177 (d) 5
𝑒 = 𝐿𝑑𝑖/𝑑𝑡 = 4 × 1/1500 = 30000V=30kV
Potential difference across the capacitor = emf
induced across 𝐻𝐸 = 𝐵𝑙𝑣 which is constant. 187 (a)

P a g e | 48
 50 to the right side.
𝑋𝐿 = 2𝜋𝑓𝐿 = 2𝜋 ( ) × 1 = 100Ω
𝜋 193 (c)
1 Heat produced by ac = 3 × Heat produced by dc
𝑋𝐶 =
2𝜋𝑓𝐶 2
∴ 𝑖𝑟𝑚𝑠 𝑅𝑡 = 3 × 𝑖 2 𝑅𝑡 ⇒ 𝑖𝑟𝑚𝑠
2
= 3 × 22
1
= ⇒ 𝑖𝑟𝑚𝑠 = 2√3 = 3.46 𝐴
50
2𝜋 ( 𝜋 ) × 20 × 10−6 194 (d)
1
= 500 Ω Brightness ∝ 𝑃consumed ∝ 𝑅. For bulb, 𝑅𝑎𝑐 = 𝑅𝑑𝑐 ,
Impedence Z = √(𝑅)2 + (𝑋𝑐 − 𝑋𝐿 )2 so brightness will be equal in both the cases
= √(300)2 + (400)2 195 (c)
𝐸 = 141 sin 628𝑡
= 500 Ω 𝐸0
∴ 𝐸𝑟𝑚𝑠 =
√2
141
189 (d) = 1.41 = 100 V
𝜔
𝑍 = √(𝑅)2 + (𝑋𝐿 − 𝑋𝐶 )2 ; and 𝑣 = 2𝜋
𝑅 = 10Ω, 𝑋𝐿 = 𝜔𝐿 = 2000 × 5 × 10−3 = 10Ω 628
= 2×3.14 = 100 Hz
1 1
𝑋𝐶 = = = 10Ω, 𝑖. 𝑒. , 𝑍 196 (c)
𝜔𝐶 2000 × 50 × 10−6
= 10Ω Here, 𝑅 = 10 Ω. As is known,
𝑉0 20 𝑑𝜙
Maximum current 𝑖0 = = 10 = 2𝐴 |𝑑𝑞| = = |𝑖 𝑑𝑡| = area under 𝑖 − 𝑡 graphs.
𝑍 𝑅
Hence 𝑖𝑟𝑚𝑠 =
2
= 1.4 𝐴 and 𝑉𝑟𝑚𝑠 = 4 × 1.41 = 𝑑𝜙 (4)(0.1)
√2 ∴ = = 0.2
𝑅 2
5.64 𝑉 𝑑𝜙 = 0.2 𝑅 = 0.2 × 10 = 2 Wb
191 (d) 197 (a)
𝑉 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2 From 𝜙 = 𝑀𝑖
𝑀1 𝜙 10−3 ×200 10
∵ 𝑉𝑅 = 𝑉 ∵ 𝑉𝐿 = 𝑉𝐶 𝑀2
= 𝜙1 = 0.8×10−3 ×400 = 16 = 0.625
2
∴ Reading of voltmeter = 220𝑉 198 (a)
𝐸
Reading of ammeter 𝐼𝑟𝑚𝑠 = 𝑟𝑚𝑠 Here, 𝑉𝑟𝑚𝑠 = 220V, v = 50 Hz
𝑍
220 Peak value of voltage 𝑉0 = √2 𝑉rms = 220√2 V
= = 2.2𝐴 ∴The instantaneous value of voltage is
100
192 (a) 𝑉 = 𝑉0 sin 2𝜋𝑣𝑡 = 220 √2 sin 2𝜋 × 50𝑡
Motion emf induced in the connector = 220√2 sin 100𝜋𝑡
𝑒 = 𝐵𝑙𝑣 = 2(1)(2) = 4 V 199 (d)
This acts as a cell of emf 4 V and internal 𝑀𝑑𝐼 20
𝑒= = 0.09 × = 300V
resistance 2 Ω. 6Ω and 3 Ω resistors are in 𝑑𝑡 0.006
200 (a)
parallel. 𝑃 60
1 1 1 1+2 3 1 Current through the bulb 𝑖 = 𝑉 = 10 = 6𝐴
∴ 𝑅𝑃
=6+3= 6
=6=2
60W, 10V
𝑅𝑃 = 2 Ω L
i

10 V VL
4V 4V i
6Ω 3Ω 2Ω
2Ω 2Ω 100V, 50Hz

𝑉 = √𝑉𝑅2 + 𝑉𝐿2
∴ Current through the connector (𝑖)
=𝑅
𝐸 4
= 2+2 = 1 A. (100)2 = (10)2 + 𝑉𝐿2
𝑃 +𝑟
⇒ 𝑉𝐿 = 99.5 𝑉𝑜𝑙𝑡
Magnetic force on the connector
Also 𝑉𝐿 = 𝑖𝑋𝐿 = 𝑖 × (2𝜋𝑣𝐿)
= 𝐵𝑖𝑙 = (1)(1) = 2 N
⇒ 99.5 = 6 × 2 × 3.14 × 50 × 𝐿
Therefore, to keep the connector moving with a
⇒ 𝐿 = 0.052 𝐻
constant velocity, a force of 2 N has to be applied
P a g e | 49
201 (b) = 6 m Wb
𝑑𝐼 (2 − 3) 210 (b)
𝐿 = ? 𝑒 = 5 𝑉, = = −103 As−1
𝑑𝑡 10−3 In Colpitt oscillator two capacitors are placed
𝑑𝐼
As 𝑒 = −𝐿 𝑑𝑡 across a common inductor and the centre of the
5 two capacitors is tapped.
∴ 5 = −𝐿(−103 ), 𝐿 = 103H = 5mH
211 (d)
202 (c) In an AC circuit, the coil of high inductance and
𝑉2 (10)2
With dc : 𝑃 = ⇒ 𝑅= = 5Ω; negligible resistance used to control current, is
𝑅 20
2 𝑅
𝑉𝑟𝑚𝑠 (10)2 ×5 called the choke coil. The power factor of such a
With ac : 𝑃 = 𝑍2
⇒ 𝑍2 = = 50 Ω2
10 coil is given by
Also 𝑍 2 = 𝑅 2 + 4𝜋 2 𝑣 2 𝐿2 𝑅
Cos 𝜙 =
⇒ 50 = (5)2 + 4(3.14)2 𝑣 2 (10 × 10−3 )2 ⇒ 𝑣 √𝑅2 +𝜔2 𝐿2
𝑅
= 80 𝐻𝑧 ≈ 𝜔𝐿 (as 𝑅 << 𝜔𝐿)
203 (d) As 𝑅 << 𝜔𝐿, cos 𝜙 is very small. Thus, the power
The voltage 𝑉𝐿 and 𝑉𝐶 are equal and opposite so, absorbed by the coil is very small. The only loss of
voltmeter reading will be zero. energy is due to hysteresis in the iron core, which
Also, 𝑅 = 30 Ω, 𝑋𝐿 = 𝑋𝐶 = 25 Ω is much less than the loss of energy in the
𝑉
So, 𝑖= 2 resistance that can also reduce the current if
√𝑅 +(𝑋𝐿 −𝑋𝐶 )2
𝑉 240 placed instead of the choke coil.
= 𝑅 = 30 = 8A 212 (d)
204 (c) 𝑒0 = 𝑁𝐴𝐵𝜔. When 𝐵 and 𝜔 are doubled, 𝑒0
If 𝜔 = 50 × 2𝜋 then 𝜔𝐿 = 20Ω becomes 4 times.
If 𝜔′ = 100 × 2𝜋 then 𝜔′ 𝐿 = 40Ω 213 (c)
200 200 200 200 1
𝐼= = = 𝑉rms = , 𝐼rms =
𝑍 2
√𝑅 + (𝜔 𝐿)′ 2 2
√(30) + (40) 2
√2 √2
𝐼 =4𝐴 200 1 𝜋
∴ 𝑃 = 𝑉rms 𝐼rms cos 𝜙 = cos = 50W
205 (a) √2 √2 3
Eddy currents are produced when a metal is kept 214 (b)
in a varying magnetic field. Here, 𝑖 = 𝑖0 at 𝑡 = ∞. Let 𝑖 be the current at 𝑡 = 1
206 (c) s
𝑅
After the current in the inductor reaches its From 𝑖 = 𝑖0 (1 − 𝑒 −𝐿 𝑡 )
maximum value 𝐼0 , it falls from 𝐼0 to zero. The 10
1
1
energy 𝐿𝐼 2 supplied by the source during build = 𝑖0 (1 − 𝑒 − 5 ×1 ) = 𝑖0 (1 − 𝑒 2 )
2 0
up of current is returned back to the source 𝑖0 𝑒2
∴ = 2
during the fall of current. 𝑖 𝑒 −1
Thus, net power supplied by the source in a 215 (c)
complete cycle is zero. In a series L – C – R circuit, potential difference
207 (a) leads the current by an angle ϕ(let).
𝑋 −𝑋
1 1 ϕ = tan−1 ( 𝐿 𝑅 𝐶 )
𝑓= ⇒𝑓∝
2𝜋√𝐿𝐶 √𝐶 −1
𝜔𝐿 −
1
𝜔𝐶
208 (a) or ϕ = tan ( 𝑅
)
The same emf is induced in all the three rings 1
At resonance, 𝑋𝐿 = 𝑋𝐶 , 𝑖𝑒, 𝜔𝐿 = 𝜔𝐶
because emf is only due to linear motion and does
not depend on spin. Hence, ϕ = tan−1 (0) = 0
209 (b) Therefore, phase difference between current and
𝑑𝑖 voltage at resonance is zero.
As 𝑒 = 𝑀 𝑑𝑡
216 (b)
𝑒 25×10−3 −3
∴ 𝑀= = = 1.67 × 10 H The average power output of the emf source is
𝑑𝑖/𝑑𝑡 15.0
1
As 𝜙 = 𝑀𝑖 𝑃 = |𝑉0 ||𝐼0 | cos 𝜃
2
∴ 𝜙 = 1.67 × 10−3 × 3.6 = 6 × 10−3 Wb Since, 𝑉0 = 𝐼0 𝑅

P a g e | 50
 1 Given, 𝑉𝑅 = 5 V, 𝑉𝐿 = 10 V and 𝑉𝐶 = 10 V
∴ 𝑃 = 𝑅|𝐼0 |2
2 In the L – C – R circuit, the AC voltage applied to
It is clear that only the resistor dissipates energy the circuit will be
in the circuit. The inductor and capacitor both
𝑉 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2
store energy but they eventually return it to the
= √(5)2 + (10 − 10)2 = 5V
circuit without dissipation.
217 (a) VL Y V
𝑉0 200
𝐼0 = = =2A
𝑅 100 V L - VC
𝐼0
𝐼𝑟𝑚𝑠 = = 1.414 A
√2
219 (b)
IV X
The phase angle (𝜃) between I and V is given by VR
𝑋𝐿 −𝑋𝐶
tan 𝜃 = 𝑅
…(i)
Where, 𝑋𝐿 = 2𝜋𝑓𝐿
200 VC
= 2𝜋 × 50 × [ × 10−3 ]
𝜋
= 20 Ω 224 (b)
1 The impedance of R – C circuit for frequency 𝑓1 is
𝑋𝐿 =
2𝜋𝑓𝐶 1
1×𝜋 𝑍1 = √𝑅 2 + 4𝜋2 𝑓2 𝐶 2
=
2𝜋 × 50 × 10−3 The impedance of R – C circuit for frequency 2𝑓 is
=10 Ω 1
and 𝑅 = 10 Ω 𝑍2 = √𝑅 2 + 4𝜋2 (2𝑓2 )𝐶 2
Substituting values of 𝑋𝐿 , 𝑋𝐶 and R in eq.(i), we 1
get or 𝑍2 = √𝑅 2 +
16𝜋2 𝑓2 𝐶 2
20−10 1
tan 𝜃 = =1 𝑍12 𝑅2 +
4𝜋2 𝑓2 𝐶2
10
𝜋 Then, = 1
𝑍22 𝑅2 +
⇒ tan 𝜃 = tan 16𝜋2 𝑓2 𝐶2
4 1+
1
𝜋 𝑍12 4𝜋2 𝑓2 𝐶2 𝑅2
∴ 𝜃= Or 𝑍22
= 1
4 1+
16𝜋2 𝑓2 𝑅2 𝐶2
𝜋 𝑍
The phase angle of the circuit is 4 . Values are greater than 1 then 𝑍1 = lies between 1
2
220 (b)
and 2.
𝑛𝑃 1
= 225 (c)
𝑛𝑠 20 1
𝑖𝑃 𝑛𝑠 𝑓 = 2𝜋
As = √𝐿𝐶
𝑖𝑠 𝑛𝑃 1
𝑖𝑃 ∴ 𝑓=
∴ = 20 ∶ 1 2 × 3.14√0.5 × 10−3 × 20 × 10−6
𝑖𝑠 1
221 (b) = ≈ 1600 Hz
2 × 3.14 × 10−4
1 1 226 (c)
𝑃 = 𝑉0 𝑖0 cos 𝜙 ⇒ 1000 = × 200 × 𝑖0 cos 60°
2 2 200
𝑖0 20 𝑋𝐿 2𝜋𝑣𝐿 2𝜋 × 2𝜋 × 1
⇒ 𝑖0 = 20 𝐴 ⇒ 𝑖𝑟𝑚𝑠 = = = 10√2𝐴 tan 𝜙 = = = =1⇒𝜙
√2 √2 𝑅 𝑅 200
= 45°
222 (c)
𝜔𝐿 1 1 227 (a)
𝑄= = × ×𝐿 𝑑𝐼
𝑒 = 𝐿 𝑑𝑡 = 2 × 10−3 = 2V
𝑅 𝑅 √𝐿𝐶
1 𝐿 228 (d)
= 𝑅 √𝐶 2
𝑉𝑟𝑚𝑠 𝑅
𝑃𝑎𝑣 = 2
1
= ×√ =
1 1 𝑍
3 9 9 229 (b)
223 (c) 𝑉0 = 𝑖0 𝑍 ⇒ 200 = 100 𝑍 ⇒ 𝑍 = 2Ω

P a g e | 51
 Also 𝑍 2 = 𝑅 2 + 𝑋𝐿2 ⇒ (2)2 = (1)2 + 𝑋𝐿2 ⇒ 𝑋𝐿 = As 𝜔 is increased, 𝑋𝐶 will decrease or 𝑍 will
√3Ω decrease. Hence 𝐼𝑟𝑚𝑠 𝑜𝑟 𝑃 will increase.
231 (b) Therefore, bulb glows brighter.
1
Power loss ∝ (Voltage)2 Hence the correct option is (b).
244 (d)
232 (c) 𝑡 10
𝐶 = 𝑅 = 103 = 10−2 𝐹 = 104 𝜇 F
Resistances of both the bulbs are
𝑉 2 2202 245 (c)
𝑅1 = = 2𝜋𝑣 = 377 ⇒ 𝑣 = 60.03 𝐻𝑧
𝑃1 25
𝑉 2
2202 247 (c)
𝑅2 = = 𝑓 = 2𝜋
1
𝑃2 100 √𝐿𝐶
∴ 𝑅2 > 𝑅2 1
or 𝑓∝
√𝐶
Hence 25 𝑊 bulb will fuse
When capacitor C is replaced by another capacitor
233 (b)
𝑉−𝐸 𝑉 − 𝐸 220 − 80
C’ of dielectric constant K, then
𝑖= ,𝑅 = = = 5.6 Ω 𝐶 ′ = 𝐾𝐶
𝑅 𝑖 25
234 (c) 𝑓′ 𝐶
The current in 𝐿𝐶𝑅 circuit becomes maximum at ∴ =√
𝑓 𝐶′
series resonance condition. At this point the total
125000−25000 𝐶
reactance of the circuit is zero. That means the or 125000
= √𝐾𝐶
reactance of inductance becomes equal and 100 1
or = 𝐾
opposite to the reactance by the capacitor 125 √
235 (b) 125 2
or 𝐾=( ) = 1.56
100
For anti-resonant circuit current is minimum at
248 (a)
resonant frequency and at frequencies other than 𝐸𝑆 𝑖𝑃 𝐸 4.6
resonant frequency current rises with frequency 𝐸𝑃
= 𝑖𝑆
⇒ 𝑖𝑃 = 𝐸𝑆 × 𝑖𝑆 = 230 × 5 = 0.1 A
𝑃
237 (b) Frequency is not affected by transformer.
𝑉 𝑅 𝑉 2𝑅 𝑉 2𝑅 249 (a)
𝑃 = 𝑉𝑖 cos 𝜙 = 𝑉 ( ) ( ) = 2 = 2
𝑍 𝑍 𝑍 (𝑅 + 𝜔 2 𝐿2 ) 2 2 2√2
𝑉𝑎𝑣 = 𝑉0 = × (𝑉𝑟𝑚𝑠 × √2) = . 𝑉𝑟𝑚𝑠
238 (b) 𝜋 𝜋 𝜋
Moving from 𝐴 to 𝐵. 2√2
= × 220 = 198 𝑉
𝑉𝐵 − 𝑉𝐴 = [5 × 10−3 (−103 ) + 15 + 1 × 5] 𝜋
= 15 volt 250 (b)
239 (d) Ist case From formula
For a series L – C – R 𝑉2
𝑅= 𝑃
circuit at resonance 110×110 110
Phase difference, 𝜙 = 0° = 330 = 3
Ω
Power factor= cos 𝜙 = 1 Since, current lags the voltage thus, the circuit
241 (c) contains resistance and inductance.
Resonance frequency in 𝑟𝑎𝑑𝑖𝑎𝑛/𝑠𝑒𝑐𝑜𝑛𝑑 is Power factorcos 𝜙 = 0.6
𝑅
1 1 = 0.6
𝜔= = = 500 𝑟𝑎𝑑/𝑠𝑒𝑐 √𝑅2 +𝑋𝐿2
√𝐿𝐶 √8 × 0.5 × 10−6
242 (b) 𝑅 2
1 1 ⇒ 𝑅 2 + 𝑋𝐿2 = ( )
𝑋𝐶 = = 0.6
𝜔𝐶 2𝜋𝑓𝐶 𝑅2
1 ⇒ 𝑋𝐿2 = − 𝑅2
(0.6)2
𝑖. 𝑒. , 𝑋𝐶 ∝
𝑓 𝑅 2 × 0.64
243 (b) ⇒ 𝑋𝐿2 =
0.36
𝑉𝑟𝑚𝑠 0.8 𝑅 4𝑅
𝑍 = √𝑅 2 + 𝑋𝐶2 ∶ 𝐼𝑟𝑚𝑠 = 2
: 𝑃 = 𝐼𝑟𝑚𝑠 𝑅 ∴ 𝑋𝐿 = = …(i)
0.6 3
𝑍
1 IInd case
Where 𝑋𝐶 = 𝜔𝐶 Now cos 𝜙 = 1 (given)
P a g e | 52
 Therefore, circuit is purely resistive, ie, it contains 𝑊
or 𝑃𝑎𝑣 =
𝑇
only resistance. This is the condition of resonance (𝐸 𝐼 cos 𝜙)𝑇/2
or 𝑃𝑎𝑣 = 0 0 𝑇
in which
𝐸 𝐼 cos 𝜙
𝑋𝐿 = 𝑋𝐶 or 𝑃𝑎𝑣 = 0 0 2
4𝑅 4 110 440
∴ 𝑋𝐶 = 3
= 3
× 3 = 9
Ω [From Eq. Where cos 𝜙 is called the power factor of an AC
(i)] circuit.
1
=
440
Ω 257 (b)
2𝜋𝑓𝐶 9
9
The impedance (Z) of an R – L – C series circuit is
𝐶 = 2×3.14×60×440 given by
= 54 𝜇F 1 2
𝑍 = √𝑅 2 + (𝜔𝐿 − 𝜔𝐶 )
251 (b)
Capacitive reactance( 𝑋𝐶 ) is given by As frequency of alternating emf applied to the
1 circuit is increased, 𝑋𝐿 goes on increasing and 𝑋𝐶
𝑋𝐶 = 𝜔𝐶
goes on decreasing.
Where 𝜔 is angular frequency and C the
For a particular value of 𝜔 = (𝜔𝑟 say)
capacitance.
𝑋𝐿 = 𝑋𝐶
Also, 𝜔 = 2𝜋𝑓,where 𝑓 is frequency. 1
In a DC circuit 𝑓 = 0 ∴ 𝜔 = 0 ie, 𝜔𝑟 𝐿 = 𝜔
𝑟𝐶
1 1
𝑋𝐶 = = ∞ or 𝜔𝑟 = 𝐿𝐶
0 √
252 (c) or 2𝜋𝑣𝑟 = 𝐿𝐶
1
√
As continuous flow of DC do not take place 1
through a capacitor, Therefore resistance of the or 𝑣𝑟 = 2𝜋 𝐿𝐶
√
circuit 1
∴ 𝑣=
R = 1+0.5 = 1.5 2 × 3.14 × √5 × 80 × 10−6
1
Current with circuit =
𝐸 2×3.14×√(400×10−6 )
𝐸= 1
𝑅′ =
2 4 2×3.14×2×10−2
= = A 100
1.5 3 =
3.14×4
Potential difference across capacitor =Potential 25 25
difference across 1 Ω resistor = 3.15
= 𝜋 Hz
4
= 3 × 1 = 3V
4 259 (b)
The time taken by AC in reaching from zero to
∴ Charge on capacitor 𝑞 = 𝐶𝑉
4 maximum value is
= 1 × 3 = 1.33 μF 𝑇 1
𝑡 = 4 = 4𝑓
253 (a) 1
1 = 50×4 = 5 × 10−3 s
Frequency of 𝐿𝐶 oscillation = 2𝜋
√𝐿𝐶
260 (b)
1/2
𝑓1 1 𝐿2 𝐶2
⇒ = √𝐿2 𝐶2 = ( ) For given circuit current is lagging the voltage by
𝑓2 √𝐿1 𝐶1 𝐿1 𝐶1
𝜋/2, so circuit is purely inductive and there is no
2𝐿 × 4𝐶 1/2 power consumption in the circuit. The work done
=( ) = (8)1/2
𝐿×𝐶 by battery is stored as magnetic energy in the
𝑓1 𝑓1 𝑓
∴ = 2√2 ⇒ 𝑓2 = or, 𝑓2 = [∵ 𝑓1 = 𝑓] inductor.
𝑓2 2√2 2√2
262 (a)
254 (a)
In a pure inductor (zero resistance), voltage leads
From 𝑒 = 𝐿 𝑑𝐼/𝑑𝑡
𝑑𝐼 𝑒 90 −1
the current by 90° 𝑖. 𝑒. , 𝜋/2
= = = 450 As
𝑑𝑡 𝐿 0.2 263 (d)
255 (d) 𝑃 = 𝑉𝑖 cos 𝜙
𝑉 = 120 sin 100𝜋𝑡 cos 100𝜋𝑡 ⇒ 𝑉 = 60 sin 200𝜋𝑡 𝜋
Phase difference 𝜙 = 2 ⇒ 𝑃 = zero
𝑉max = 60𝑉 and 𝑣 = 100 𝐻𝑧
264 (c)
256 (c) 1
Power = Rate of work done in one complete cycle. We have 𝑋𝐶 = 𝐶×2𝜋𝑓 and 𝑋𝐿 = 𝐿 × 2𝜋𝑓

P a g e | 53
265 (c)
(𝑇/2)𝑉02 + 0 𝑉0
Here 𝐼𝑟𝑚𝑠 = 10 𝐴, 𝑅 = 12 Ω 𝑉𝑟𝑚𝑠 = √ =
𝑇 √2
The maximum current is
𝐼𝑚 = √2 𝐼𝑟𝑚𝑠 = √2(10) = 10√2 𝐴 275 (a)
Maximum potential difference is 𝑉𝑚 = 𝐼𝑚 𝑅 The voltage across a 𝐿 − 𝑅 combination is given
by
= 10√2 × 12 = 169.68 𝑉
𝑉 2 = 𝑉𝑅2 + 𝑉𝐿2
266 (b)
𝑉𝐿 = √𝑉 2 − 𝑉𝑅2 = √400 − 144 = √256 = 16 𝑣𝑜𝑙𝑡
𝑍 = √𝑅 2 + 𝑋𝐶2
276 (d)
1 2
= √(𝑅)2 + (𝜔𝐶 ) The average power of L – C – R circuit
𝑃𝑎𝑣 = 𝑉𝑟𝑚𝑠 . 𝑖𝑟𝑚𝑠 cos 𝜙
In case (b) capacitance (c) will be more.
Hence, the average power depends upon current,
Therefore, impedance Z will be less. Hence
emf and phase difference.
current will be more.
277 (d)
267 (c)
𝑋𝐿 𝑋𝐿 50
tan 𝜙 = = 1 ∴ 𝜙 = 45° or 𝜋/4 𝑋𝐿 = 2𝜋𝑣 𝐿 ⇒ 𝐿 = = = 0.16 𝐻
𝑅 2𝜋𝑣 2 × 3.14 × 50
268 (c) 278 (d)
(i) In a circuit having C alone, the voltage lags the The average power dissipation in the circuit is
𝜋 1
current by . 𝐸 𝐼 tan ϕ
2 2 0 0
(ii) In a circuit containing R and L, the voltage
𝜋
leads the current by .
2
279 (a)
(iii) In L – C circuit, the phase difference between
Final current in constant and L plays no role at
current and voltage can have any
𝜋 that instant. Therefore, 𝑖 = 𝐸/𝑅.
value between 0 to depending on the
2 280 (a)
values of L and C. A closed current carrying loop of any irregular
(iv) In a circuit containing L alone, the voltage shape and even not lying in a single plane, placed
𝜋
leads the current by . in a uniform magnetic field shall experience no
2
269 (d) net force. Therefore, force acting on the loop must
Induced emf 𝑒 = 𝐵𝑙𝑣 = 𝐵𝑊𝑣 be zero.
𝑒2 𝐵2 𝑊 2 𝑣 2 281 (b)
Power developed = 𝑅
= 𝑅 𝜇0 𝑁1 𝑁2 𝐴
As 𝑀 = ,
270 (b) 𝑙
In 𝑅𝐶 series circuit voltage across the capacitor ∴ 𝑀 can be increased by increasing the number
𝜋
leads the voltage across the resistance by 2 of turns in the coils.
282 (a)
271 (c)
𝑛𝑠 1 In L-C-R series circuit,
𝐸𝑆 = 𝐸 = × 240 = 12 V
𝑛𝑃 𝑃 200
𝑉 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2
272 (a)
𝐸0
The current is 𝐼 = = √(40)2 + (60 − 30)2
√𝑅 +𝜔2 𝐿2
2

=
4
= 0.8 A = √1600 + 900 = √2500 = 50V
√4 2 +(1000×3×10−3 )2 283 (c)
273 (b) For series L – C – R circuit
In parallel resonant circuit, resonance frequency
1
𝑉 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2
𝑓0 = 2𝜋
√𝐿𝐶 = √(80)2 + (40 − 100)2
1
= = 100 V
10×10−3
2𝜋√ ×0.04×10−6
𝜋2 285 (b)
104 1
= 2×0.2 = 25 kHz Power 𝑃 = 𝑉 𝐼 cos 𝜙
2 00
274 (d)

P a g e | 54
 22500 Charge on the capacitor 𝑞 = 𝐶𝑉
= 0.5 × 150 × 150 × cos 60° =
4 ⇒ 𝑞 = 0.5 × 10−6 × 6
= 5625 W 𝑞 = 3μ𝐶
286 (a) 290 (d)
If the capacitance is removed, it is an 𝐿 − 𝑅 circuit Current will be maximum in the condition of
𝜙 = 60° resonance so 𝑖max = = 𝐴
𝑉 𝑉
𝑋𝐿 𝑅 10
tan 𝜙 = = tan 60° = √3 1 2 1 𝐸 2
𝑅 Energy stored in the coil 𝑊𝐿 = 𝐿𝑖max = 𝐿( )
2 2 10
If inductance is removed, it is a capacitative 2
1 𝐸 1
circuit or 𝑅 − 𝐶 circuit. |𝜙| is the same = × 10−3 ( ) = × 10−5 𝐸 2 𝑗𝑜𝑢𝑙𝑒
1 2 100 2
∴ 𝐿𝜔 = 𝐶𝜔 This is a resonance circuit
∴ Energy stored in the capacitor
𝐸𝑟𝑚𝑠 1 1
𝑍 = 𝑅; 𝐼𝑟𝑚𝑠 = , 𝐸𝑟𝑚𝑠 = 200 𝑉 𝑊𝐶 = 𝐶𝐸 2 = × 2 × 10−6 𝐸 2 = 10−6 𝐸 2 𝑗𝑜𝑢𝑙𝑒
𝑅 2 2
200𝑉 𝑊 𝐶 1
∴ 𝐼𝑟𝑚𝑠 = = 2𝐴 ∴ =
100Ω 𝑊𝐿 5
287 (d) 291 (c)
Resistance of the bulb 1
𝑉 2 (100)2 𝑋𝐿 = 2𝜋𝑣𝐿 = 2 × 𝜋 × 50 × = 100Ω
𝑅= = = 200 Ω 𝜋
𝑃 50 293 (c)
𝑉
Current through bulb(𝐼) = 𝑅 In series R – L – C circuit, the impedance of the
=
100
= 0.5 A circuit is given by
200
𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2
In a circuit containing inductive reactance (𝑋𝐿 )
1
and resistance (R), impedance (Z) of the circuit is Also, 𝑋𝐿 = 𝜔𝐿, 𝑋𝐶 = 𝜔𝐶
𝑍 = √𝑅 2 + 𝜔 2 𝐿2 …(i)
1 2
Here, 𝑍 =
200
= 400 Ω ∴ 𝑍 = √𝑅 2 + (𝜔𝐿 − )
0.5 𝜔𝐶
Now, 𝑋𝐿2
= 𝑍 − 𝑅22
Given, 𝑅 = 300Ω, 𝜔 = 1000 rads −1 , 𝐿 = 0.9 𝐻,
= (400)2 − (200)2 𝐶 = 20𝜇F = 2 × 10−6 F
(2𝜋𝑓𝐿)2 = 12 × 104
Hence, 𝑍 = √(300)2 + (1000 × 0.9 −
2√3×100
𝐿 = 2𝜋×50 1 2
)
2√3 1000×2×10−6
= 𝜋 = 1.1 H
= √90000 + (900 − 500)2
288 (b)
= √90000 + 160000
The Q- factor of series resonant circuit is given by
𝜔𝑟2 𝐿 = √250000 = 500 Ω
𝑄= 𝑅 294 (b)
It is evident from the relation that as R is As coil 𝐴 is moved closer to 𝐵, field due to
increased, Q – factor of the circuit is decreased. 𝐴 intercepting 𝐵 is increasing. Induced current in
289 (d) 𝐵 must oppose this increase. Hence the current in
In steady state current through the branch having 𝐵 must be anti-clock-wise.
capacitor is zero. 295 (d)
1 1 1 1 1
∴ = + + Reactance 𝑋 = 𝑋𝐿 − 𝑋𝐶 = 2𝜋𝑓𝐿 − 2𝜋𝑓𝐶
𝑅 1 2 3
296 (d)
1 6+3+2 Circuit is resonant.
=
𝑅 6 Hence supply voltage equals
6
𝑅 = 11 𝑉𝑅 = 10 V
As 𝑉 = 𝑖𝑅 Also, 𝑋𝐶 = 𝑅
6 As the voltage drops are equals across them when
∴ 6=𝑖× L is shortened
11
Current through the battery 𝑖 = 11A
𝑍 = √𝑅 2 + 𝑋𝐶2 = √2𝑅

P a g e | 55
 𝑉 be minimum which can be done, if
𝑉𝐶 = 𝑖𝑋𝐶 (∵ 𝑖 = )
2 𝑋𝐿 = 𝑋𝐶
𝑉 10
or 𝑉𝐶 = 𝑍 𝑋𝐶 = 2𝑅 𝑅 This happens in resonance state of the circuit
√
1
𝑉𝐶 =
10
V i.e, 𝜔𝐿 = 𝜔𝐶
√2
1
297 (c) or 𝐿 = 𝜔2 𝐶 ...(i)
Adff sdaf sdfsdf dsf Given, 𝜔 = 1000s −1 , 𝐶 = 10𝜇𝐹 = 10 × 10−6 F
298 (b) Hence, 𝐿 = (1000)2
1
= 0.1 H = 100 mH
𝑒𝑑𝑡 8×0.05 ×10×10−6
From 𝑒 = 𝐿𝑑𝐼/𝑑𝑡, 𝐿 = 𝑑𝐼
= 2
= 0.2H
299 (a) 307 (b)
Capacitance of wire 𝐸 = 𝐸0 cos 𝜔𝑡 = 10 cos(2𝜋 × 𝑓𝑡)
𝐶 = 0.014 × 10−6 × 200 1
= 10 cos (2𝜋 × 50 × 600)
= 2.8 × 10−6 𝐹 = 2.84𝜇F
𝜋 √3
For impedance of the circuit to be minimum = 10 cos ( ) = 10 × = 5√3 V
6 2
𝑋𝐿 = 𝑋𝐶 309 (d)
1 As explained in solution (1) for frequency 0 −
⇒ 2𝜋𝑣𝐿 =
2𝜋𝑣𝐶 𝑓𝑟 , 𝑍 decreases hence (𝑖 = 𝑉/𝑍) increases and for
1
𝐿= 2 2 frequency 𝑓𝑟 − ∞, 𝑍 increases hence 𝑖 decreases
4𝜋 𝑣 𝐶
1
= 4(3.14)2 ×(5×103 )2 ×2.8×10−6 310 (a)
At resonance, 𝑉𝐿 and 𝑉𝐶 are equal in magnitude
= 0.35 × 10−3 H = 0.35 mH
but have phase difference of 180° relative to each
300 (c)
other
At resonance 𝑋𝐿 = 𝑋𝐶
∴ 𝑉𝐿𝐶 = 𝑉𝐿 − 𝑉𝐶 = 0
303 (c)
Hence, voltmeter 𝑉2 read 0 volt
As 𝐵0 = 𝜇0 𝑛𝑖, therefore 𝐵0 does not depend upon
311 (b)
radius (𝑟) of the solenoid.
When 𝐶 is removed circuit becomes 𝑅𝐿 circuit
304 (c)
hence
Here, 𝐿 = 25 𝑚𝐻 = 25 × 10−3 𝐻 𝜋 𝑋𝐿
𝑣 = 50 𝐻𝑧, 𝑉𝑟𝑚𝑠 = 220 𝑉 tan = … (i)
3 𝑅
The inductive reactance is When 𝐿 is removed circuit becomes 𝑅𝐶 circuit
22 hence
𝑋𝐿 = 2𝜋𝑣 𝐿 = 2 × × 50 × 25 × 10−3 Ω
7 𝜋 𝑋𝐶
The 𝑟𝑚𝑠 current in the circuit is tan = … (ii)
3 𝑅
𝑉𝑟𝑚𝑠 220 From equation (i) and (ii) we obtain 𝑋𝐿 = 𝑋𝐶 .
𝐼𝑟𝑚𝑠 = = 22
𝑋𝐿 2 × 7 × 50 × 25 × 10−3 This is the condition of resonance and in
7 × 1000 resonance 𝑍 = 𝑅 = 100Ω
= 𝐴 = 28 𝐴
2 × 5 × 25 312 (c)
305 (b) 𝑉𝐿 = 46 𝑣𝑜𝑙𝑡𝑠, 𝑉𝐶 = 40 𝑣𝑜𝑙𝑡𝑠, 𝑉𝑅 = 8 𝑣𝑜𝑙𝑡𝑠
To decrease current in an AC circuit, choke coil is E.M.F. of source 𝑉 = √82 + (46 − 40)2 = 10 𝑣𝑜𝑙𝑡𝑠
used. The choke coil has high inductance and 313 (b)
negligible resistance, so that the energy loss in the 1
Resonance frequency, 𝜔 = 𝐿𝐶
circuit is negligible. √
1
Hence, 𝑋𝐿 >> 𝑅 =
√8×10−3 ×20×10−6
306 (a) 1 104
= −4 = 4
Current in a L-C-R series circuit, 4×10
𝑉 = 2500 rad 𝑠 −1
𝑖= 2 2 𝑉 220
√𝑅 +(𝑋𝐿 −𝑋𝐶 )
Amplitude of current, 𝐼0 = 𝑅 = 44 = 5 A
Where 𝑉 is rms value of current, R is resistance,
314 (a)
𝑋𝐿 is inductive reactance and 𝑋𝐶 is capacitive
reactance. 1 𝑇
𝑉𝑟𝑚𝑠 = √ ∫ 102 𝑑𝑡 = 10 𝑉
For current to be maximum, denominator should 𝑇 0

P a g e | 56
315 (c) Impedance,
𝑉0 423
Effective voltage 𝑉𝑟.𝑚.𝑠. = = = 300 𝑉 𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2
√2 √2
316 (b) ∴ 10 = √(102 + (𝑋𝐿 − 𝑋𝐶 )2
Given, 𝐿 = 20 mH = 20 × 10−3 H ⇒ 100 = 100 + (𝑋𝐿 − 𝑋𝐶 )2
𝐶 = 50𝜇F = 50 × 10−6 F ⇒ 𝑋𝐿 − 𝑋𝐶 = 0
For LC circuit the frequency , …(i)
1 Let 𝜙 is the phase difference between current and
𝑓=
2𝜋√𝐿𝐶
1 voltage
or 𝑇 = 2𝜋√𝐿𝐶 (∵ 𝑇 = 𝑓) 𝑋𝐿 −𝑋𝐶
tan 𝜙 = 𝑅
𝑇
At time 𝑡 = , energy stored is completely 0
4
∴ tan 𝜙 =
magnetic. 𝑅
𝑇 ⇒ 𝜙=0 [From Eq.(i)]
The time, 𝑡 =
4 322 (d)
2𝜋√𝐿𝐶
𝑡= In an L – C – R series AC circuit, the voltage across
4
2𝜋√20×10−3 ×50×10−6 inductor L leads the current by 90° and the
or 𝑡= 4 voltage across capacitor C lags behind the current
3.14√10−6
or 𝑡= by 90°.
2
3.14×10−3
or 𝑡=
2
−3
or 𝑡 = 1.57 × 10 s = 1.57 ms
317 (c)
Ignoring mutual induction, resultant, inductance
𝐿′ = 𝐿1 + 𝐿2
=𝐿 + 𝐿 = 2𝐿
319 (a)
In an L – C circuit the impedance of circuit is
𝑍 = 𝑋𝐿 − 𝑋𝐶 Hence, the voltage across 𝐿 − 𝐶 combination will
When𝑋𝐿 = 𝑋𝐶 , then Z = 0. In this situation the be zero.
amplitude of current in the circuit would be 323 (d)
infinite. It will be condition of electrical resonance 𝜇0 𝑁 2 𝐴 𝜇0 𝑁 2 (𝜋𝑟 2 )
𝐿= =
and frequency is given by 𝑙 𝑙
4𝜋×10−7 ×(500)2 ×𝜋(0.025)2
1 =
𝑓= 1
2𝜋√𝐿𝐶 = 4 × 10 × 107 × (500)2 × (0.025)2
1
=
2×3.14×√10×10−3 ×0.25×10−6
= 6.25 × 10−4H
= 3184.7 cycle s −1 325 (c)
velocity The power loss in AC circuit will be minimum, if
Also frequency = wavelength
resistance is low. In inductance power loss is zero.
𝑐 3×108
⇒ 𝜆 = 𝑓 = 3184.7 It applies to high as well as low inductances.
⇒ 𝜆 = 9.42 × 104 m 326 (c)
320 (a) Inductive reactance 𝑋𝐿 = 𝜔𝐿
As the inductors are in parallel, therefore, induced = 2𝜋𝑣𝐿
emf across the two inductors is the same 𝑖𝑒, = 2𝜋 × 50 × 1
𝑒1 = 𝑒2 = 100𝜋
𝑑𝑖 𝑑𝑖 327 (d)
𝐿1 ( 1 ) = 𝐿2 ( 2 )
𝑑𝑡 𝑑𝑡 A series resonance circuit admits maximum
Integrating both sides w.r.t. 𝑡, we get current, as
𝐿1 𝑖1 = 𝐿2 𝑖2 𝑃 = 𝑖2𝑅
𝑖1 𝐿2 So, power dissipated is maximum at resonance.
∴ =
𝑖2 𝐿1 So, frequency of the source at which maximum
321 (a) power is dissipated in the circuit is

P a g e | 57
 𝑣 = 2𝜋
1 335 (d)
√𝐿𝐶
1 Given, the frequency of alternating current 𝑓
=
2×3.14√25×10−3 ×400×10−6 = 50 Hz
1
= = 50.3 Hz The time for alternating current to become its rms
2×3.14√10−5
value from zero.
328 (a) 𝑇
As initially charge is maximum, 𝑡=4
1
𝑞 = 𝑞0 cos 𝜔𝑡 or 𝑡 = 4𝑓
𝑑𝑞 1
⇒ 𝑖= = −𝜔𝑞0 sin 𝜔𝑡 or 𝑡 = 200 s
𝑑𝑡
1 2 𝑞2 or 𝑡 = 5 ms
Given 𝐿𝑖 =
2 2𝐶
1 2
(𝑞0 cos 𝜔𝑡)2
⇒ 𝐿 (𝜔𝑞0 sin 𝜔𝑡) =
2 2𝐶
1
𝜔 = 𝐿𝐶
√
∴ tan 𝜔𝑡 = 1
𝜋
𝜔𝑡 = tan 4
𝜋 𝜋
𝑡 = 4𝜔 = 4 √𝐿𝐶
330 (b)
(1) For time interval 0 < 𝑡 < 𝑇/2
𝐼 = 𝑘𝑡, where 𝑘 is the slope 336 (c)
For inductor as we know, induced voltage 𝑉 = The full cycle of alternating current consists of
𝑑𝑖
−𝐿 𝑑𝑡 two half cycles. For one – half, current is positive
⇒ 𝑉1 = −𝐾𝐿 and for second half, current is negative. Therefore,
𝑇 for an AC cycle, the net value of current average
(2) For time interval < 𝑡 < 𝑇
2 out to zero. While the DC ammeter, read the
𝐼 = −𝐾𝑡 ⇒ 𝑉2 = 𝐾𝐿 average value. Hence, the alternating current
331 (c) cannot DC measured by DC ammeter.
Average power in AC circuits is given by 𝑃 = 337 (a)
𝑉rms 𝐼rms cos 𝜙 for pure capacitive circuit 𝜙 = Current will be maximum at the condition of
90° so, 𝑃 = 0. 1
resonance. So resonant frequency 𝜔0 = =
332 (d) √𝐿𝐶
𝑉 𝑉 1
𝐼𝐿 = 𝑋 and 𝐼𝐶 = 𝑋 √0.5×8×10−6
𝐿 𝐶
1 = 500 𝑟𝑎𝑑/𝑠
i.e, 𝐼𝐿 ∝ 𝜔 338 (c)
and 𝐼𝐶 ∝ 𝜔 Input power 𝑃1 = 220 × 1.5 = 330 W
∴ With increase in 𝜔, 𝐼𝐿 decreases 3 2
while 𝐼𝐶 increases. Loss of power 𝑖 2 𝑅 = (2) × 20 = 45 W
333 (a) Output power, 𝑃0 = 330 − 45 = 285 W
𝑃0 285
Geometric length of a magnet is 6/5 times its ∴ Peak emf induced, 𝑉0 = = = 190 V
𝑖 1.5
magnetic length. 339 (c)
∴ Geometric length =6/5×10=12 cm
𝑉0 770√2
334 (c) 𝑉𝑟𝑚𝑠 = = ≈ 500 V
√2 2
The current in the circuit
𝑉𝑅 340 (c)
𝑖= 𝑅 In L – R circuit, impedance
100
= 1000
=0.1 A 𝑍 = √𝑅 2 + 𝑋𝐿2
At resonance, Here, 𝑋𝐿 = 𝜔𝐿 = 2𝜋 𝑓𝐿
𝑖
𝑉𝐿 = 𝑉𝐶 = 𝑖𝑋𝐶 = ∴ 𝑍 = √𝑅 2 + 4𝜋 2 𝑓 2 𝐿2
𝜔𝐶
0.1 341 (a)
= 200×2×10−6 = 250 V
Given, 𝑉𝐶 = 3𝑉𝑅 = 3(𝑉 − 𝑉𝐶 )

P a g e | 58
 Here, V is the applied potential. Across L, 𝑉𝐿 = 𝑖𝑋𝐿 = 2 × 15 = 30 V
3 Across C, 𝑉𝐶 = 𝑖𝑋𝐶 = 2 × 11 = 22 V
∴ 𝑉𝐶 = 4
V
3 So, potential difference across series combination
Or 𝑉(1 − 𝑒 −𝑡/𝜏𝐶 ) = V
4 of L and C,
1 = 𝑉𝐿 − 𝑉𝐶
∴ 𝑒 −𝑡/𝜏𝐶 = … (𝑖)
4 = 30 − 22 = 8 V
Here, 𝜏𝐶 = 𝐶𝑅 = 10s
346 (d)
Substituting this value of 𝜏𝐶 in Eq.(i) and solving
At resonance net voltage across 𝐿 and 𝐶 is zero
for t, we get
347 (d)
𝑡 = 1 3.8 6s
∵ 𝑃 = 𝑉𝑖 cos 𝜙 , ∴ 𝑃 ∝ cos 𝜙
342 (d)
348 (b)
𝐹 12 𝑑𝑖 𝑑
𝑙1 = = =6𝐴 As 𝑒 = 𝑀 𝑑𝑡 = 𝑀 𝑑𝑡 (𝑖0 sin 𝜔𝑡)
𝑅1 2
𝐸=𝐿
𝑑𝑙2
+ 𝑅2 × 𝑙2 ∴ 𝑒 = 𝑀𝑖0 cos 𝜔𝑡(𝜔)
𝑑𝑡
𝑒max = 𝑀𝑖0 × 1 × 𝜔
𝐼2 = 𝐼0 (1 − 𝑒 −𝑡/𝑡𝑐 )
= 0.005 × 10 × 10𝜋 = 5𝜋
𝐸 12
⇒ 𝐼0 = = = 6A 349 (d)
𝑅2 2
The emf induced in a conductor does not depend
𝐿 400 × 10−3
𝑡𝑐 = = = 0.2 o its shape, but only on its end points, 𝑀 and 𝑄 in
𝑅 2
this case. Thus the conductor is equivalent to an
𝐼2 = 6(1 − 𝑒 −𝑡/0.2 )
imaginary straight conductor of 𝑙 = 𝑀𝑄 = 2𝑅.
Potential drop across
Therefore, potential difference developed across
𝐿 = 𝐸 − 𝑅2 𝐿2 = 12 − 2 × 6(1 − 𝑒 −𝑏𝑡 ) = 12𝑒 −5𝑡
the ring =𝐵𝑙𝑣 = 𝐵(2𝑅)𝑣. And the direction of
343 (c)
induced current is from 𝑄 to 𝑀. Therefore, 𝑄 is at
Reactance of capacitor or capacitive reactance is
higher potential.
denoted by 𝑋𝐶 , given by
1
350 (b)
𝑋𝐶 = 𝜔𝐶 𝑒 𝑒𝑑𝑡 5×10−3
𝐿= 𝑑𝑖/𝑑𝑡
= 𝑑𝑖
= (3−2)
H= 5 mH
Given, 𝜔 = 50 rad s −1 , 𝐶 = 50𝜇F = 50 × 10−6 F
1 351 (a)
∴ 𝑋𝐶 = As 𝑀 ∝ 𝑁1 𝑁2 , therefore, M remains the same.
50 × 50 × 10−6
From Ohm’s law, current flowing through the 352 (b)
circuit is given by For purely capacitive circuit 𝑒 = 𝑒0 sin 𝜔𝑡
𝑉rms 𝜋
𝐼rms = 𝑖 = 𝑖0 sin (𝜔𝑡 + 2 ) , 𝑖. 𝑒., current is ahead of emf by
𝑋𝐶
𝜋
𝑉rms 220
= 1/𝜔𝐶
= 1/50×50×10−6 2
353 (d)
⇒ 𝐼rms = 220 × 50 × 50 × 10−6 𝑉
= 55 × 10−2 A= 0.55 A Current, 𝑖 =
√𝑅2 +𝑋𝐶2
344 (a)
In an ideal choke, ratio of its inductance L to its If a dielectric is introduced into the gap between
DC resistance R is infinity. the plates of capacitor. Its capacitance will
345 (a) increase and hence, impedance of the circuit will
Given, 𝑅 = 3Ω, 𝑋𝐿 = 15 Ω, 𝑋𝐶 = 11 Ω decrease. Thus, current and hence brightness of
𝑉rms = 10 volt the bulb increase.
∴ Current through the circuit 354 (d)
𝑉rms When resonance occurs emf E and current i are in
𝑖=
√𝑅2 +(𝑋𝐿 −𝑋𝐶 )2 phase. In this case, the impedance is minimum
10
= and current is maximum. At resonance inductive
√(3)2 +(15−11)2
10 10 reactance is equal to capacitive reactance
= = 5 = 2A 𝑋𝐿 = 𝑋𝐶
√9+16
Since, L, C and R are connected in series 355 (c)
combination then potential difference across R is The circuit element connected to the AC source
𝑉𝑅 = 𝑖 × 𝑅 = 2 × 3 = 6 V will be pure resistor. In pure resistive AC circuit,
P a g e | 59
 voltage and current are in the same phase 90
𝑖𝐶 = = 4.5 𝐴
356 (b) 20
Frequency = 2𝜋
1 Net current through circuit 𝑖 = 𝑖𝐶 − 𝑖𝐿 = 1.5 𝐴
√𝐿𝐶 𝑉 90
So the combination which represents dimension ∴𝑍= = = 60Ω
𝑖 1.5
1
of frequency is = (𝐿𝐶)−1/2 364 (c)
√𝐿𝐶
357 (a) ∴ 𝜙 = 𝑀𝑖
𝜙 0.4
In 𝐿𝐶𝑅 series circuit, impedance 𝑍 of the circuit is ∴ 𝑀= 𝑖
= 2
= 0.2 H
given by 365 (c)
2 2
𝑍 = √(𝑅) + (𝑋𝐿 − 𝑋𝐶 ) where 𝑋𝐿 = 𝜔𝐿, 𝑋𝐶 = 𝑉 120
𝑖= = = 0.455 𝐴
1/𝜔𝐶 𝑋𝐿 2 × 3.14 × 60 × 0.7
At resonance 𝑋𝐿 = 𝑋𝐶 ∴ 𝑍 = 𝑅 366 (d)
358 (c) 𝑋𝐿 2𝜋𝑣𝐿 2𝜋 × 50 × 𝐿
tan 𝜙 = = ⇒ tan 30° =
𝑋𝐿 = 2𝜋𝑓 𝑅 𝑅 𝜋√3
⇒ 𝑋𝐿 ∝ 𝑓 = 0.01 𝐻
1 1 367 (a)
⇒ ∝
𝑋𝐿 𝑓 For purely 𝐿-circuit 𝑃 = 0
1
𝑖. 𝑒., graph between and 𝑓 will be a hyperbola 369 (a)
𝑋𝐿
𝑅 = 6 + 4 = 10Ω
359 (d)
𝑋𝐿 = 𝜔𝐿 = 2000 × 5 × 10−3 = 10Ω
For current to be maximum , 𝑋𝐿 = 𝑋𝐶 1 1
Hence, resonant frequency 𝑋𝐶 = = = 10Ω
𝜔𝐶 2000 × 50 × 10−6
1 1 1 103
𝑓= = = = ∴ 𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2 = 10Ω
2𝜋 √𝐿𝐶 √0.5 × 8 × 10−6 4𝜋 𝑉0 20
Amplitude of current = 𝑖0 = = = 2𝐴
But angular frequency 𝑍 10
𝜔 = 2𝜋𝑓 370 (c)
𝜔=
2𝜋×1000
= 500 Hz Impedance, 𝑍 = √(𝑋𝐿 ~ 𝑋𝐶 )2 + 𝑅 2
4𝜋
360 (b)
The average power consumed in an AC circuit is
given by
𝑉0 𝐼0
𝑃= cos 𝜙
2
Where 𝜙 is phase angle and 𝑉0 𝑎𝑛𝑑 𝐼0 the peak
1 2
value of voltage and current. 𝑍 = √(𝜔𝐿~ ) + 𝑅2
𝜋 𝜔𝐶
Given, 𝑉0 = 200𝑉, 𝐼0 = 2 𝐴, 𝜙 = .
3
200×2 𝜋
Inductive reaction
𝑃 = 2 cos 3 𝑋𝐿 = 𝜔𝐿 = 70 × 103 × 100 × 10−6 = 7Ω
200×2 1
= 2 ×2= 100 W Capacitance reactance
1 1
361 (d) 𝑋𝐶 = =
The given circuit is under resonance as 𝑋𝐿 = 𝑋/𝐶 𝜔𝐶 70 × 10 × 1 × 10−6
3
100
Hence, power dissipated in the circuit is = [𝑋𝐶 > 𝑋𝐿 ]
7
𝑉2
𝑃= = 242 W Hence, circuit behaves like an 𝑅-𝐶 circuit
𝑅
362 (c)
𝑍 = √𝑅 2 + (2𝜋𝑣𝐿)2
= √(40)2 + 4𝜋 2 × (50)2 × (95.5 × 10−3 )2
= 50 𝑜ℎ𝑚
363 (c)
90
𝑖𝐿 = = 3𝐴
30

P a g e | 60