Total Questions: 395

JEE/NEET PHYSICS
ALTERNATING CURRENT

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√ 2 1
a) 𝐸 2 𝑅/ [𝑅 2 + (𝐿𝜔 − ) ] b) 𝐸 𝑅 + (𝐿𝜔 − 𝐶𝜔)
𝐶𝜔
𝑅
1 2 𝐸2𝑅
𝐸 2 [𝑅 2 + (𝐿𝜔 − 𝐶𝜔) ]
c) d) 1 2
𝑅 √𝑅 2 + (𝐿𝜔 − )
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

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 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
𝑉 = 50 sin(50𝑡)volt
𝜋
𝑖 = 50 sin (50𝑡 + 3 )mA
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

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 𝑉 𝑉𝑅1 𝑅2
c) 𝑅 at 𝑡 = 0 and at 𝑡 = ∞
2 √𝑅 2 + 𝑅 2
1 2
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
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°
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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
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 + (𝐿𝜔 − 𝐶𝜔)2 ]1/2 b) [𝑅 2 + (𝐿𝜔 − ) ]
𝐶𝜔
−1/2 1/2
1 2 1 2
c) [𝑅 2 + (𝐿𝜔 − ) ] d) 2
[(𝑅𝜔) + (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
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 a) 1/𝑒 𝐴 b) 𝑒 𝐴 c) 0.1 A d) 1 A
45. 𝑅
has the dimensions to
𝐿
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

   

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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) 2
b) c) d) 𝐸 ( )
𝑒𝑅 𝐸𝑅 𝐿 𝑅
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

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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.

C
R

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

a) b) 2 c) d)
5𝑑 + 3𝑣𝑡 2𝑑 − 3𝑑𝑣𝑡 − 9𝑣 2 𝑡 2 5𝑑 − 3𝑣𝑡 2𝑑2 + 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

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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
𝜇 𝑁 𝜇0 𝑁2
a) 0 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

Page|8
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 , as shown in the diagram. If the circuit
4
consists possibly only of R – C or R – L or L – C in series, find the relationship between the two elements

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 = 𝐸0 sin 𝜔𝑡
0 2
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

Page|9
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

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

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 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
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

P a g e | 11
 1
1/2
1 2 2 −2
a) [(𝑅𝜔)2 + (𝐿𝜔 − ) ] b) [𝑅 2 + (𝐿𝜔 − 1 ) ]
𝐶𝜔 𝐶𝜔
1/2
1 2
c) [𝑅 2 + (𝐿𝜔 − 𝐶𝜔)2 ]1/2 d) [𝑅 2 + (𝐿𝜔 − ) ]
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
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

P a g e | 12
 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
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

P a g e | 13
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
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

P a g e | 14
 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
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
P a g e | 15
 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
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

P a g e | 16
163. In an ac circuit, 𝑉 and 𝐼 are given by
𝜋
𝑉 = 100 sin (100 𝑡) 𝑣𝑜𝑙𝑡𝑠, 𝐼 = 100 sin (100𝑡 + ) 𝑚𝐴. The power dissipated in circuit is
3
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 m is kept inside a magnetic field which is normal to its plane. The magnetic field
2

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
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
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

P a g e | 17
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
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 . The emf developed across
−1

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

P a g e | 18
 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
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
P a g e | 19
 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
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

P a g e | 20
 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
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 + 3 ) amp. The average power consumed in watt is
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
1/2
𝑒 − 1 2
𝑒 −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

P a g e | 21
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
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 103 As−1 . The emf induced is
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

P a g e | 22
 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) b) c) d)
i i i i

   
236. The average power dissipated in a pure inductor of inductance 𝐿 when an ac current is passing through it,
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 c) d)
𝑉 (𝑅 + 𝜔 2 𝐿2 ) (𝑅 2 + 𝜔 2 𝐿2 ) 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

P a g e | 23
 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
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?

P a g e | 24
 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 = 𝐸0 sin 𝜔𝑡. The resulting current in the circuit is 𝐼 =
𝜋
𝐼0 sin (𝜔𝑡 − ). The power consumption in the circuit is given by
2
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 − ) and the ac potential is 𝑉 = 200 sin(100) 𝑣𝑜𝑙𝑡.
𝜋
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

P a g e | 25
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

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
P a g e | 26
 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 𝑅 + 𝐿2
2 𝑅 + 𝐿2
2

280. An irregular closed loop carrying a current has a shape such that the entire loop cannot lie in a single
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

P a g e | 27
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
√3 2
a) 2𝐴 b) 1𝐴 c) 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
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 1 ℎ𝑒𝑛𝑟𝑦 at 50 𝐻𝑧 frequency is
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

P a g e | 28
 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

c
Reactance

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
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 = ∞

P a g e | 29
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= s, the instantaneous value of the emf is
100
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
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

P a g e | 30
 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
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

P a g e | 31
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?

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
P a g e | 32
 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
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

P a g e | 33
 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
⃗ , figure. At the position 𝑀𝑁𝑄, the speed of the ring is 𝑣. The potential difference developed
induction 𝐵
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
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
P a g e | 34
357. A 𝐿𝐶𝑅 series 𝐴. 𝐶. circuit is tuned to resonance. The impedence of the circuit is now
2 1/2
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

a) b) c) d)

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

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

P a g e | 35
 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 | 36
Total Questions: 395

JEE/NEET PHYSICS
ALTERNATING CURRENT

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

P a g e | 37
321) a 322) d 323) d 324) b 349) d 350) b 351) a 352) b
325) c 326) c 327) d 328) a 353) d 354) d 355) c 356) b
329) c 330) b 331) c 332) d 357) a 358) c 359) d 360) b
333) a 334) c 335) d 336) c 361) d 362) c 363) c 364) c
337) a 338) c 339) c 340) c 365) c 366) d 367) a 368) d
341) a 342) d 343) c 344) a 369) a 370) c 371) d
345) a 346) d 347) d 348) b

P a g e | 38
Total Questions: 395

JEE/NEET PHYSICS
ALTERNATING CURRENT

: HINTS AND SOLUTIONS :

1 (b) or 0.5 =
200

𝑒 = 300√2 sin 𝜔𝑡 √𝑅2 +𝑋𝐿2

𝑒0 300√2 or 𝑅 2 + (2𝜋𝑓𝐿)2 = (400)2

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

P a g e | 39
 8𝐵𝑟𝑣
𝐼= 𝑅
13 (c)
1 2
𝑍 = √𝑅 2 + (2𝜋𝑓𝐿 − )
2𝜋𝑓𝐶
𝜋
From above equation at 𝑓 = 0, 𝑧 = ∞ 𝑋𝐶 = 𝑅 tan … (i)
1 3
When 𝑓 = (resonant frequency) ⇒ 𝑍 = 𝑅 𝑋𝐿 𝜋
2𝜋√𝐿𝐶
1
= tan
For 𝑓 > ⇒ 𝑍 starts increasing 𝑅 3
2𝜋√𝐿𝐶
𝑖. 𝑒., for frequency 0 − 𝑓𝑟 , 𝑍 decreases and for 𝑓𝑟 to
∞, 𝑍 increases
This is justified by graph 𝑐
14 (a)
1 1 1 𝜋
𝑋𝐶 = ⇒ = 𝑋𝐿 = 𝑅 tan … (ii)
2𝜋𝑣𝐶 1000 2𝜋 × 𝑣 × 5 × 10−6 3
100 Net impedance 𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2 = 𝑅
⇒𝑣= 𝑀𝐻𝑧
𝜋 𝑅
15 (a) Power factor cos 𝜙 = = 1
𝑍
𝑖0 5 21 (b)
𝑖𝑟𝑚𝑠 = = = 3.536 A 𝜔𝐿
√2 √2 From the relation, tan 𝜙 =
𝑅
16 (a) 1
Power factor cos 𝜙 =
𝑋𝐿 = 𝜔𝐿. √1+tan2 𝜙
𝑋 10 1
or 𝐿 = 𝐿 = = 0.5 H = 2
𝜔 20 √1+(𝜔𝐿)
17 (b) 𝑅
𝑅

When a circuit contains inductance only, then the = 2 2 2

√𝑅 +𝜔 𝐿
current lags behind the voltage by the phase 22 (a)
𝜋 1 1 1 2 𝐿
difference of 2 or 90°.
= + = 𝑜𝑟 𝐿𝑃 =
While in a purely capacitive circuit, the current 𝐿𝑝 𝐿 𝐿 𝐿 2
𝜋
leads the voltage by a phase angle of or 90°. Where L is inductance of each part
2 1.8 ×10−4
In a purely resistive circuit current is in phase = 2
= 0.9 × 10−4 H
with the applied voltage. 𝐿 0.9×10−4
∴ 𝐿𝑃 = = = 0.45 × 10−4H
2 2
18 (b) 6
At t = 0, inductor behaves like an infinite Resistance of each part, 𝑟 = 2 = 3Ω
1 1 1 2
resistance. So at Now, =3+3=3
𝑟𝑃
𝑉
𝑡 = 0, 𝑖 = ∴ 𝑟𝑃 = 3/2Ω
𝑅2 𝐿 0.45×10−4
And at 𝑡 = ∞, inductor behaves like a conducting Time constant of circuit = 𝑟𝑃 = 3/2
= 8A
𝑃
wire, 23 (b)
𝑉 𝑉(𝑅1 +𝑅2 )
𝑖= = The applied voltage is given by 𝑉 = √𝑉𝑅2 + 𝑉𝐿2
𝑅eq 𝑅1 𝑅2
19 (c) 𝑉 = √(200)2 + (150)2 = 250 𝑣𝑜𝑙𝑡
Hot wire ammeter reads 𝑟𝑚𝑠 value of current. 24 (d)
Hence its peak value = 𝑖𝑟𝑚𝑠 × √2 = 14.14 𝑎𝑚𝑝 The voltage across secondary in zero, as
20 (c) transformer does not work on DC supply.
25 (c)
𝐿𝑑𝑖 𝑑𝑖
𝑋𝐶 𝜋 From 𝑒 = 𝑑𝑡
, when 𝑑𝑡 = 1, 𝑒 = 𝐿
= tan
𝑅 3 26 (b)
𝐿2 𝑁22
=
𝐿1 𝑁12

P a g e | 40
 𝑁2 500 2 2 (30)2
𝑉𝑟𝑚𝑠
∴ 𝐿2 = 𝐿1 𝑁22 = 1.5 (100) = 375 mH 𝑃= = = 90 𝑊
1 𝑅 10
27 (b) 37 (c)
𝑅 𝑅 In series LCR, the impedance of the circuit is given
cos 𝜙 = =
𝑍 √𝑅 2 + 𝜔 2 𝐿2 by
12
= ⇒ cos 𝜙 𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2
√(12)2 + 4 × 𝜋 2 × (60)2 × (0.1)2 At resonance, 𝑋𝐿 = 𝑋𝐶
= 0.30 ∴𝑍=𝑅
28 (d) At resonance, the phase difference between the
Magnetic field intensity at a distance r from the current and voltage is 0°. Current is maximum at
straight wire carrying current is resonance
𝜇0 𝑖
𝐵=
2𝜋𝑟
39 (c)
𝐸𝑃 𝑖𝑃 1100×100
As area of loop, 𝐴 = 𝑎2 𝑒= 𝐸𝑠
= 220
= 500 A=0.5 kA
And magnetic flux, 𝜙 = 𝐵𝐴 40 (c)
𝜇0 𝑖𝑎2
∴ 𝜙= Impedance 𝑍 = √𝑅 2 + 𝑋 2 = √(8)2 + (6)2 = 10Ω
2𝜋𝑟
41 (b)
The induced emf in the loop is
𝑑𝜙 𝑑 𝜇0 𝑖 𝑎 2 𝑣 𝑍 = √𝑅 2 + 𝑋𝐿2 , 𝑋𝐿 = 𝜔𝐿 and 𝜔 = 2𝜋𝑓
𝑒 = | 𝑑𝑡 | = |𝑑𝑡 2𝜋𝑟
|
2 2 ∴ 𝑍 = √𝑅 2 + 4𝜋 2 𝑓 2 𝐿2
𝜇0 𝑖𝑎 𝑑𝑟 𝜇 𝑖𝑎 𝑣
𝑒= 2 𝜋 𝑟 2 𝑑𝑡
= 20 𝜋 𝑟2 43 (b)
Where 𝑣 =
𝑑𝑟
is velocity . 𝑉0 120
𝑑𝑡 𝑉𝑟𝑚𝑠 = = = 84.8 𝑉
29 (c) √2 1.414
𝜋 −𝜋 𝜋 44 (a)
Phase difference ∆𝜙 = 𝜙2 − 𝜙1 = 6 − ( 6 ) = 3 This is a combined example of growth and decay
30 (b) of current in an L – R circuit.
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,
𝑖0 sin 𝜙 = 8 𝐸
4 𝐼0 = 𝑅 = 1 A
Hence, tan 𝜙 = 5
(as inductor would be shorted in steady state)
34 (d)
After this decay of current starts in the circuit
In a purely inductive circuit, current is
𝜋 according to the equation 𝐼 = 𝐼0 𝑒 −𝑡/𝜏
𝑖 = 𝑖0 sin (𝜔𝑡 − 2 ) Where 𝜏 = 𝐿/𝑅.

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 2 or 90°.
35 (b) 𝐼 = 1 × 𝑒 −(1×1 0−3 )/(1 00×1 0−3 /1 00)
= (1/𝑒) A

P a g e | 41
45 (d) Energy stored in a inductor 𝐿 carrying
𝑅 𝑒/𝑖 1 1
= = = frequency. Current 𝑖 is 𝑈 = 2 𝐿 𝑖 2
𝐿 𝑒𝑑𝑡/𝑑𝑖 𝑑𝑡
46 (a) Rate at which energy is stored
𝑑𝑈 1 𝑑𝑖 𝑑𝑖
Phase difference relative to the current = = 2 𝐿 2𝑖 (𝑑𝑡) = 𝐿𝑖 (𝑑𝑡)
𝑑𝑡
𝜋 𝜋 𝑑𝑈
𝜙 = (314𝑡 − ) − (314 𝑡) = − At 𝑡 = 0, 𝑖 = 0, ∴ =0
6 6 𝑑𝑡
47 (b) At 𝑡 = ∞, 𝑖 = 𝑖0 (constant) , ∴
𝑑𝑖
=0
𝑑𝑡
At 𝑡 = 0, phase of the voltage is zero, while phase
𝜋 𝜋 56 (d)
of the current is − 2 , 𝑖. 𝑒., voltage leads by 2 1
Required time 𝑡 = 𝑇/4 = 4×50 = 5 × 10−3 𝑠𝑒𝑐
48 (d)
57 (a)
𝑍 2 = 𝑅 2 + (2𝜋𝑓𝐿)2
Maximum voltage is AC circuit
0.4 2
= (30)2 + (2𝜋 × 50 × ) 𝑉0 = 282 𝑉
𝜋
= (900 + 1600) = 2500 𝑉0 282
𝑉= =
or 𝑍 = 50 Ω √2 √2
𝑉 200 282 28200
Also, 𝐼 = 𝑍 = 50 = 4 A 𝑉= =
1.41 141
50 (b) 𝑉 = 200 𝑉
We know that Q - factor of series resonant circuit 58 (b)
is given as 1 1
𝜔𝑟 𝐿 𝑋𝐶 = = =∞
𝑄= 2𝜋𝑣 𝐶 0
𝑅
Here, 𝐿 = 8.1 mH, 𝐶 = 12.5 𝜇F, 𝑅 = 10Ω, 𝑓 = 59 (b)
500 Hz 𝑍 = √𝑅 2 + 𝑋𝐿2 = √102 + (2𝜋 × 60 × 2)2 = 753.7
𝜔𝑟 𝐿 2𝜋𝑓𝐿
∴ 𝑄= = 120
𝑅 𝑅 ∴𝑖= = 0.159 𝐴
2×𝜋×500×8.1×10 −3 8.1𝜋 753.7
= 10
= 10
= 2.5434 60 (b)
51 (b) An alternating current is one whose magnitude
1
For capacitive circuits 𝑋𝐶 = changes continuously with time between zero and
𝜔𝐶
𝑉 a maximum value and whose direction reverses
∴𝑖= = 𝑉𝜔𝐶 ⇒ 𝑖 ∝ 𝜔 periodically. The relation between frequency (𝑓)
𝑋𝐶
52 (b) and time (𝑇) is.
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
𝑖0 𝜏 (𝐸/𝑅)(𝐿/𝑅) 𝑒𝑙 1 1
𝑞 = = = 2 𝑇= = = 0.02 𝑠
𝑒 𝑒 𝑒𝑅 𝑓 50
54 (b) As is clear from the figure time taken to reach the
𝑃𝑖 = 240 × 0.7 = 168 W, 𝑃0 = 140 W maximum value is
𝑃0 140 𝑇 0.02
𝜂 = × 100 = × 100 ≈ 80% 4
= 4
= 0.005 s
𝑃𝑖 168
61 (b)
55 (c)

P a g e | 42
 From 𝑒 = 𝐿𝑑𝐼/𝑑𝑡 ⇒ 𝑑𝐼 = 𝐿 = 1 = 1 As −1
𝑒 1
71 (d)
𝜙 𝜇0 𝑁1 𝐴𝑖 𝜇0 𝑁 2 𝑖
62 (a) 𝐵= = =
After time t, thickness of liquid will remain 𝐴 𝐿𝐴 𝐿
𝑑 72 (c)
( 3 − 𝑣𝑡). 𝐸0 𝑖0 𝑅
𝑃 = 𝐸𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙 = × ×
Now, time constant as function of time √2 √2 𝑍
𝜏𝑐 = 𝐶𝑅 𝐸0
𝑅 𝐸02 𝑅 𝐸0
⇒ × × ⇒𝑃=
𝜀0 (1). 𝑅 √2 𝑍√2 𝑍 2𝑍 2
= (Applying 𝐶 𝐸2
𝑑
(𝑑 − 3 + 𝑣𝑡) +
𝑑/3−𝑣𝑡 Given 𝑋𝐿 = 𝑅 so, 𝑍 = √2𝑅 ⇒ 𝑃 = 4𝑅0
2
73 (a)
𝜀0 𝐴 Since, current lags behind the voltage in phase by
= 𝑡)
𝑑−𝑡+𝑘 a constant angle, then circuit must contain R and
6𝜀0 𝑅 L.
=
5𝑑 + 3𝑣𝑡
63 (d)
When wire is thick, its resistance reduces.
Therefore, Joules’ heating loss is reduced.
64 (c)
We find that in R – L circuit, voltage leads the
Peak value = 220√2 = 311 𝑉 current by a phase angle 𝜙, where
65 (c) 𝐴𝐾 𝑂𝐿
𝜋 tan 𝜙 = 𝑂𝐴 = 𝑂𝐴
𝐼𝐿 lags behind 𝐼𝑅 by a phase of 2 , while 𝐼𝐶 leads by
𝑉 𝐼0 𝑋𝐿
𝜋 = 𝑉𝐿 =
a phase of 2 𝑅 𝐼0 𝑅
𝑋𝐿
66 (b) ∴ tan 𝜙 =
𝑅
Time constant of R – C circuit is 𝜏 = 𝑅𝐶 Y
Here effective resistance of the circuit L K
|

2𝑅×3𝑅 6𝑅
= =
2𝑅+3𝑅 5
6𝑅 6𝑅𝐶 VL
∴ 𝜏= ×𝐶 =
5 5
67 (b) )
|

X
O A
𝑀𝑑𝑖 𝜇0 𝑁1 𝑁2 𝐴 𝑑𝑖
𝑒= = ( )
𝑑𝑡 𝑙 𝑑𝑡 VR |

4𝜋×10−7 ×2000×300×1.2 ×10−3 (4)

= 0.3×0.25
Y'

−2
= 4.8 × 10 V
68 (a) 74 (d)
𝜋 Current will be max at first time when
𝑉 = 5 cos 𝜔𝑡 = 5 sin (𝜔𝑡 + ) and 𝑖 = 2 sin 𝜔𝑡
2
100𝜋𝑡 + 𝜋/3 = 𝜋/2 ⇒ 100 𝜋𝑡 = 𝜋/6 ⇒ 𝑡
Power = 𝑉𝑟.𝑚.𝑠. × 𝑖𝑟.𝑚.𝑠. × cos 𝜙 = 0
𝜋 𝜋 = 1/600 𝑠
[Since 𝜙 = 2 , therefore cos 𝜙 = cos 2 = 0]
75 (d)
69 (d) The current will lag behind the voltage when
𝑉 4 reactance of inductance is more than the
𝑖= = = 0.8 𝐴
𝑍 √42 + (1000 × 3 × 10−3 )2 reactance of condenser.
70 (a) Thus, 𝜔𝐿 > 𝜔𝐶 or 𝜔 >
1 1
√𝐿𝐶
𝑋𝐿 = 31Ω, 𝑋𝐶 = 25Ω, 𝑅 = 8Ω 1
or 𝑛 > or 𝑛 > 𝑛𝑟 where 𝑛𝑟 = resonant
Impedance of series 𝐿𝐶𝑅 is 2𝜋√𝐿𝐶

𝑍 = √(𝑅)2 + (𝑋𝐿 − 𝑋𝐶 )2 frequency

76 (b)
= √(8)2 + (31 − 25)2 = √64 + 36 = 10Ω 𝑑𝐼 𝑒𝑑𝑡 8(0.05)
𝑅 8 𝑒 = 𝐿 𝑑𝑡 𝐿 = = = 0.2 H
Power factor, cos 𝜙 = 𝑍 = 10 = 0.8 𝑑𝐼 (4−2)
77 (c)
P a g e | 43
 In 𝐿 − 𝑅 circuit, current at any time 𝑡 is given by ∴ 𝜔𝐶𝑅 = 1
𝐸 𝑅 𝐸 𝐸 𝑅 As 𝜔 = 100 rads−1
𝑖 = (1 − 𝑒 − 𝐿 𝑡 ) = − 𝑒 − 𝐿 𝑡 1
𝑅 𝑅 𝑅 The product of C – R should be 100 s −1.
𝑑𝑖 𝐸 −𝑅 𝑡 𝑅 𝐸 𝑅
= 𝑒 𝐿 ( ) = 𝑒 −𝐿 𝑡 85 (d)
𝑑𝑡 𝑅 𝐿 𝐿
𝑑𝑖 𝑅 Phase angle 𝜙 = 90°, so power 𝑃 = 𝑉𝑖 cos 𝜙 = 0
Induced emf = 𝐿 𝑑𝑡 = 𝐸𝑒 − 𝐿 𝑡
86 (d)
𝑅
1
From Eq. (i), 𝑖𝑅 = 𝐸 − 𝐸𝑒 − 𝐿 𝑡 Current lags the voltage if 𝜔𝐿 > 𝜔𝐶
Using Eq. (ii), 𝑖𝑅 = 𝐸 − 𝑒 𝑜𝑟 𝑒 = 𝐸 − 𝑖𝑅 1
Therefore, graph between 𝑒 and 𝑖 is a straight line 𝑓> ⇒ 𝑓 > 𝑓𝑟
2𝜋√𝐿𝐶
with negative slope and positive intercept. The 87 (c)
choice (c) is correct. 𝜔 120 × 7
𝑣= = = 19 𝐻𝑧
78 (c) 2𝜋 2 × 22
1 𝑞2 1 2 𝑞2 240
𝑈= = (𝑞0 𝑒 −𝑡/𝜏 ) = 0 𝑒 −2𝑡/𝜏 (where 𝜏 = 𝑉𝑟.𝑚.𝑠 = = 120√2 = 170 𝑉
2 𝐶 2𝐶 2𝐶
√2
𝐶𝑅)
88 (c)
𝑈 = 𝑈 𝑒−2𝑡/𝜏
𝑖 For an R – C circuit
1
𝑈 = 𝑈 𝑒−2𝑡1/𝜏
2 𝑖 𝑖 Applied voltage, 𝑉 = √𝑉𝑅2 + 𝑉𝐶2
1 −2𝑡1 /𝜏
2
=𝑒
𝜏 ∴ 50 = √(40)2 + 𝑉𝐶2
⇒ 𝑡1 = ln 2
2
Now 𝑞 = 𝑞 𝑒−𝑡/𝜏 ⇒ 𝑉𝐶 = 30 V
1
0 89 (d)
𝑞 0 = 𝑞 −𝑡 2 /𝜏 Power factor
2 0𝑒
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 = 900+1600 = 50 = 0.6
√
resistive circuit. For resistive circuit 𝜙 = 0° 90 (a)
80 (b) Yes, in 𝐴𝐶 if branch 𝐴𝐵 has 𝑅, 𝐵𝐶 has a capacitor
2 2
𝑍 = √𝑅 + 𝑋 = √4 + 3 = 52 2 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
𝑛𝑠 4200
𝐸𝑠 = 𝐸 = 2100 × 120 = 240V
𝑛𝑃 𝑃 91 (c)
𝑛𝑠 2100 At 𝐴 ∶ 𝑋𝐶 > 𝑋𝐿
𝑖𝑠 = 𝑖 = × 10 = 5 A
𝑛𝑃 𝑃 4200
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 = At any instant, the 𝑒𝑚𝑓 of the battery is equal to
𝑅
𝜋
1 the sum of potential drop on the resistor and the
𝜔𝐶
or tan = 𝑒𝑚𝑓 developed in the induction coil
4 𝑅

P a g e | 44
 Hence, 𝐸 = 𝑖𝑅 + 𝐿 𝑑𝑡
𝑑𝑖 1
⇒ 𝐶=
𝑑𝑖 𝜔(𝜔𝐿 − 𝑅)
240 = 15 × 10−3 × 4000 + 𝐿 1
= 2𝜋𝑓(2𝜋𝑓𝐿−𝑅)
𝑑𝑡
𝑑𝑖
Hence, 𝐿
𝑑𝑡
= 240 − 60 = 180 𝑉 100 (d)
93 (b) In an L – C – R circuit in resonance condition
𝐿/𝑅 represents time constant of R-L circuit. 𝑋𝐿 = 𝑋𝐶
0
Therefore, its dimensions are [𝑀 𝐿 𝑇 ].0 1 or 𝑋 𝐶 − 𝑋 𝐿 =0
94 (a) 102 (c)
𝑑𝑖
This is a parallel circuit, For oscillation, the energy As 𝑒 = 𝑀
𝑑𝑡
in 𝐿 and 𝐶 will be alternately maximum 10
95 (a) ∴ 30 × 103 = 3 × ,
𝑑𝑡
The current in a coil is given by 30 −3
𝑑= = 10 s
−𝑡/𝜏 30×103
𝑖 = 𝑖0 𝑒 103 (b)
𝑖0
Now, 𝑖 = 𝜂
𝑖𝑛 𝑡 = 𝑡0 Wheatstone bridge is balanced. Current through
𝑖0 𝐴𝐶 is zero. Effective resistance R of bridge is
∴ 𝜂
= 𝑖0 𝑒 −𝑡0 /𝜏 1 1 1 1
−𝑡0 /𝜏 −1 𝑅
= 6 + 6 = 3 , 𝑅 = 3Ω
𝑒 = 𝜂
Total resistance = 1 + 3 = 4Ω
Taking log of both sides,
𝑡0 Induced emf
− log 𝑒 𝑒 = −1 log 𝑒 𝜂
𝜏 𝑒 = 𝑖𝑅 = 𝐵𝑙𝑣
𝑡0
𝜏
= log 𝑒 𝜂 𝑖𝑅 1 × 10−3 × 4
∴ 𝑣= =
𝜏 = 𝑡0 / log 𝑒 𝜂 = 𝑡0 /In𝜂 𝐵𝑙 2 × 0.1
−2 −1
96 (b) = 2 × 10 ms
Since voltage is lagging behind the current, so 104 (b)
there must be no inductor in the box Motional emf across 𝑃𝑄
97 (b) 𝑉 = 𝐵𝑙𝑣 = 4(1) (2) = 8 volt
Average power dissipated in an AC circuit This is the potential to which the capacitor is
𝑃𝑎𝑣 = 𝑉rms 𝐼rms cos 𝜙 …(i) charged.
Where the term cos 𝜙 is known as power factor. As 𝑞 = 𝐶𝑉
Given, 𝑉rms = 100 V, 𝑅 = 100 Ω, ϕ = 30° ∴ 𝑞 = (10 × 10−6 )8 = 10−5 𝐶 = 80𝜇 𝐶
∴ 𝐼rms =
𝑉rms
=
100
=1A As magnetic force on electron in the conducting
𝑅 100
rod 𝑃𝑄 is towards 𝑄, therefore, 𝐴 is positively
Putting the values in Eq. (i), we get
charged and 𝐵 is negatively charged
𝑃𝑎𝑣 = 100 × 1 × cos 30°
𝑖𝑒, 𝑞𝐴 = +80𝜇 𝐶 and 𝑞𝐵 = −80 𝜇 𝐶
√3
= 100 2 105 (c)
= 50√3 = 86.6 W The DC generator must be mixed wound to with
98 (c) stand the load variation.
1 106 (a)
𝜔𝐿 − 𝜔𝐶
tan 𝜙 = For imparting max power
𝑅 1
𝜙 being the angle by which the current leads the 𝑋𝐿 = 𝑋𝐶 ⇒ 𝜔𝐿 =
𝜔𝐶
voltage. 1 1 1
Given, 𝜙 = 45° 𝐶= 2 = =
𝜔 𝐿 (2𝜋𝑓) × 𝐿 (100𝜋)2 × 10
2
1
𝜔𝐿 − 𝜔𝐶 = 1 × 10−6 = 1𝜇𝐹
∴ tan 45° =
𝑅 108 (c)
1
𝜔𝐿 − 𝜔𝐶
⇒ 1=
𝑅
1
⇒ 𝑅 = 𝜔𝐿 −
𝜔𝐶

P a g e | 45
 𝑡
1 2 − = In 3 − In 5
4
√ 2
𝑍 = 𝑅 +( )
2𝜋𝑣𝐶 or t =4(In 5 − In 3) = 2s.
114 (c)
1 𝜇0 𝑁 2 𝐴 𝜇0 𝜇𝑟 𝑁 2 𝐴
= √(3000)2 + 2 From 𝐿 = = ,
2.5 𝑙 𝑙
(2𝜋 × 50 × 𝜋
× 10−6 ) 1
When 𝜇 = 1000 and 𝑁 becomes 10
⇒ 𝑍 = √(3000)2 + (4000)2 = 5 × 103 Ω 1 2
𝑅 3000 ∴ 𝐿 become𝑠 1000 × (10) = 10 times
So power factor cos 𝜙 = 𝑍 = 5×103 = 0.6 and
𝑖𝑒, 𝐿 = 10 × 0.1 = 1H
power 115 (b)
2
𝑉𝑟𝑚𝑠 cos 𝜙 𝐸−𝑉
𝑃 = 𝑉𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙 = ⇒𝑃 From 𝑅 =
𝑍 𝑖
2 120−𝑉
(200) × 0.6 0.5 = 8
= = 4.8𝑊
5 × 103 𝑉 = 116 𝑉
110 (d) 117 (b)
Rise of current in L – R circuit is given by In non resonant circuits
1
Impedance 𝑍 = 2
, with rise in
√ 12 +(𝜔𝐶− 1 )
𝑅 𝜔𝐿

frequency 𝑍 decreases, 𝑖. 𝑒., current increases so

circuit behaves as capacitive circuit
118 (c)
Phase difference in R – L circuit,
𝑋𝐿
𝜙 = tan−1
𝑅
𝑋𝐿
or tan 45° = 𝑅
𝐼 = 𝐼0 (1 − 𝑒 −𝑡/𝜏 ) or 𝑋𝐿 = 𝑅
𝐸 5
Where, 𝐼0 = 𝑅 = 5 = 1 A 119 (d)
𝐿 10 At resonant frequency current in series 𝐿𝐶𝑅
Now, 𝜏 = = =2s
𝑅 5 circuit is maximum
After 2s, ie, at t = 2 s 120 (d)
Rise of current 𝐼 = (1 − 𝑒 −1 )A Magnetic field at the centre of primary coil 𝐵 =
111 (b) 𝜇0 𝑖1 /2𝑅1 . Considering it to be uniform, magnetic
As 𝑒 = 𝑀𝑑𝐼/𝑑𝑡, flux passing through secondary coil is
𝑒𝑑𝑡 15000×0.001
∴ 𝑀= =− = 5H 𝜇0 𝑖1
𝑑𝐼 3 𝜙 = 𝐵𝐴 = (𝜋𝑅22 )
112 (b) 2𝑅1
𝜙 𝜇 𝜋𝑅2
At resonance, 𝐿𝐶𝑅 circuit behaves as purely Now, 𝑀 = 𝑖 2 = 02𝑅 2
1 1
resistive circuit. For purely resistive circuit power
𝑅22
factor = 1 ∴ 𝑀∝
𝑅1
113 (a)
121 (c)
Voltage across the capacitors will increase from 0
Power remains constant in a ideal step down
to 10 V exponentially. The voltage at time t will be
transformer.
given by
122 (b)
𝑉 = 10(1 − 𝑒 −𝑡/𝜏𝐶 )
𝑉 = 50 × 2 sin 100𝜋𝑡 cos 100𝜋𝑡 = 50 sin 200𝜋𝑡
Here 𝜏𝐶 = 𝐶net 𝑅net = (1 × 106 )(4 × 10−6 ) = 4 s
⇒ 𝑉0 = 50 𝑣𝑜𝑙𝑡𝑠 and 𝑣 = 100𝐻𝑧
∴ 𝑉 = 10(1 − 𝑒 −𝑡/4 )
123 (b)
Substituting 𝑉 = 4 volt, we have,
Capacitive reactance is given by
4=10(1 − 𝑒 −𝑡/4 ) 1
−𝑡/4 3 𝑋𝐶 = 𝜔 𝐶
𝑒 = 0.6 =
5 Where C is capacitance and 𝜔 the angular
Taking log both sides we have, frequency (𝜔 = 2𝜋𝑓).

P a g e | 46
 1 ∵ (𝑋𝐶 ) >> (𝑋𝐿 )
∴ 𝑋𝐶 =
2𝜋𝑓𝐶 130 (a)
1 200 5 5
⇒ 𝑋𝐶 ∝ 𝑓 𝑖𝑟𝑚𝑠 = 280 = 7 𝐴. So 𝑖0 = 𝑖𝑟𝑚𝑠 × √2 = 7 × √2 ≈ 1𝐴
Hence, when frequency 𝑓 increases capacitive 132 (d)
reactance decreases. In series L – R circuit, impedance is given by
124 (a) 𝑍 = √𝑅 2 + 𝑋𝐿2
𝑅
Power factor= cos 𝜙 = 𝑍 Where R is the resistance and 𝑋𝐿 the inductive
12 4 reactance.
= 15 = 5 = 0.8
Given, 𝑅 = 8Ω, 𝑋𝐿 = 6Ω
125 (d)
1 1 ∴ 𝑍 = √(8)2 + (6)2
Given 𝜔𝐿 = 𝜔𝐶 ⇒ 𝜔2 = 𝐿𝐶
= √64 + 36
1 1
Or 𝜔 = = = 104 = √100 = 10 Ω
√10−3 ×10×10−6 √10−8
𝑋𝐿 = 𝜔𝐿 = 10 × 10 4 −3
= 10 Ω 134 (a)
126 (b) If the current is wattles then power is zero. Hence
𝐸𝑠 22 phase difference 𝜙 = 90°
𝑖𝑠 = 𝑍
= 220 = 0.1 A
135 (a)
127 (a)
In 𝐿𝐶𝑅 circuit; in the condition of resonance 𝑋𝐿 =
The instantaneous value of voltage is
𝑋𝐶 , 𝑖. 𝑒., circuit behaves as resistive circuit. In
𝐸 = 100 sin(100𝑡) 𝑉 … (i)
resistive circuit power factor is maximum
Compare it with 𝐸 = 𝐸0 sin(𝜔𝑡) 𝑉
136 (c)
We get 𝑇/2 𝑇/2
𝐸0 = 100𝑉, 𝜔 = 100𝑟𝑎𝑑𝑠 −1 ∫0 𝑖 𝑑𝑡 ∫0 𝐼0 sin(𝜔𝑡)𝑑𝑡
𝐼𝑎𝑣 = 𝑇/2
=
The rms value of voltage is ∫0 𝑑𝑡 𝑇/2
𝐸0 100 𝜔𝑇
𝐸𝑟𝑚𝑠 = = 𝑉 = 70.7𝑉 2𝐼0 − cos 𝜔𝑡 𝑇/2 2𝐼0 cos ( 2 ) cos 0°
√2 √2 = [ ] = [− + ]
The instantaneous value of current is 𝑇 𝜔 0 𝑇 𝜔 𝜔
𝜋
𝐼 = 100 sin (100𝑡 + ) 𝑚𝐴 2𝐼0 2𝐼0 2𝐼0
3 = [− cos 𝜋 + cos 0°] = [1 + 1] =
𝜔𝑇 2𝜋 𝜋
Compare it with
137 (c)
𝐼 = 𝐼0 sin(𝜔𝑡 + 𝜙) 1
At resonance, 𝜔𝐿 = 𝜔𝐶
We get
𝐼0 = 100𝑚𝐴, 𝜔 = 100rads−1 Current flowing through the circuit,
𝑉𝑅
The rms value of current is 𝐼= 𝑅
𝐼0 100 100
𝐼𝑟𝑚𝑠 = = 𝑚𝐴 = 70.7𝑚𝐴 = 1000
= 0.1 A
√2 √2
128 (a) So, voltage across L is given by
100 𝑉𝐿 = 𝐼𝑋𝐿 = 𝐼𝜔𝐿
Resistance , 𝑅 = 10
= 10 Ω 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 × = 10 Ω When the direction of current is reversed, moving
8𝜋
from 𝐵 to 𝐴.
2
Impedance of the circuit is 𝑍 = √𝑅 2 + 𝑋𝐿′ 𝑉𝐵 − 𝑉𝐴 = [5 × 10−3 (−103 ) + 15 + 1 × 5]
= √(10)2 + (10)2 = 15 volt
139 (b)
= 10√2 Ω
𝑉 150 15 The instantaneous voltage through the given
Current in the circuit is 𝑖 = 𝑍 = 10 = A
√2 √2 device
129 (a) 𝑒 = 80 sin 100𝜋𝑡
P a g e | 47
 Comparing the given instantaneous voltage with 𝐼0
= 𝐼0 (1 − 𝑒 −𝑅𝑡/𝐿 )
standard instantaneous voltage 2
𝑒 = 𝑒0 sin 𝜔𝑡. 1
= (1 − 𝑒 −𝑅𝑡/𝐿 )
We get 𝑒0 = 80 𝑉 2
1
Where 𝑒0 is the peak value of voltage 𝑒 −𝑅𝑡/𝐿 =
2
Impedance (𝑍) = 20Ω 𝑅𝑡
𝑒 =1n2
Peak value of current 𝐼0 = 𝑍0 𝐿

80
𝐿 3 00 × 1 0−3
= 20 = 4A ∴ 𝑡= 1𝑛2= × 0.6 93
2 2
Effective value of current (root mean square value = 1 50 × 0.6 93 × 1 0−3
of current). = 0.10395s = 0.1 s
𝐼𝑟𝑚𝑠 =
𝐼0 147 (d)
√2
4 The instantaneous values of emf and current in
= 2 = 2√2 = 2.828 A inductive circuit are given by 𝐸 = 𝐸0 sin 𝜔𝑡 and
√
140 (b) 𝜋
𝑖 = 𝑖0 sin (𝜔𝑡 − ) respectively
𝐸 𝑡 2
−
Charging current, 𝐼 = 𝑅 𝑒 𝑅𝐶 𝜋
So, 𝑃inst = 𝐸𝑖 = 𝐸0 sin 𝜔𝑡 × 𝑖0 sin (𝜔𝑡 − 2 )
Taking log both sides, 𝜋 𝜋
𝐸 𝑡 = 𝐸0 𝑖0 sin 𝜔𝑡 (sin 𝜔𝑡 cos − cos 𝜔𝑡 sin )
Log 𝐼 = log (𝑅) − 𝑅𝐶 2 2
When R is doubled, slope of curve increases. Also = 𝐸 𝑖
0 0 sin 𝜔𝑡 cos 𝜔𝑡
1
at t = 0, the current will be less. Graph Q = 𝐸0 𝑖0 sin 2𝜔𝑡 (sin 2𝜔𝑡 = 2 sin 𝜔𝑡 cos 𝜔𝑡)
2
represents the best.
Hence, angular frequency of instantaneous power
141 (b)
is 2𝜔
The coil has inductance 𝐿 besides the resistance 𝑅.
148 (c)
Hence for ac it’s effective resistance √𝑅 2 + 𝑋𝐿2 will 1
Resonance frequency = 2𝜋 𝐿𝐶 does not depend on
be larger than it’s resistance 𝑅 for dc √

142 (a) resistance

In 𝑅-𝐶 circuit current increases exponentially 149 (c)
𝑅 𝑅 5
with time, so correct graph will be (a) cos 𝜙 = = =
143 (d) 𝑍 √𝑅 2 + 𝜔 2 𝐿2 √25 + (50)2 × (0.1)2
5 1
𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2 = √(11)2 + (25 − 18)2 = = ⇒ 𝜙 = 𝜋/4
= 13 Ω √25 + 25 √2
260 151 (c)
Current 𝑖 = = 20 𝐴 𝑉 𝜔𝑁𝐵𝐴 (2𝜋𝑣)𝑁𝐵(𝜋𝑟 2 )
13
Amplitude of 𝑎𝑐 = 𝑖0 = 𝑅0 = 𝑅 =
144 (c) 𝑅
200 −2 2
Given 𝑋𝐿 = 𝑋𝐶 = 5Ω, this is the condition of 2𝜋 × 60 × 1 × 10 × 𝜋 × (0.3)
resonance. So 𝑉𝐿 = 𝑉𝐶 , so net voltage across 𝐿 and ⇒ 𝑖0 = = 6 𝑚𝐴
𝜋2
𝐶 combination will be zero 152 (c)
145 (c) In the condition of resonance
𝑛𝑠 𝐸𝑠 2400 𝑋𝐿 = 𝑋𝐶
= = = 20 1
𝑛𝑃 𝐸𝑃 120 or 𝜔𝐿 = 𝜔𝐶 …(i)
𝑛𝑠 = 20 𝑛𝑃 = 20 × 75 = 1500.
Since, resonant frequency remains unchanged,
146 (b)
So, √𝐿𝐶 =constant
The current at any instant is given by
L .R or LC = constant
∴ 𝐿1 𝐶1 = 𝐿2 𝐶2
⇒ 𝐿 × 𝐶 = 𝐿2 × 2 𝐶
𝐿
2V
⇒ 𝐿2 =
2
𝐼 = 𝐼0 (1 − 𝑒 −𝑅𝑡/𝐿
) 153 (b)

P a g e | 48
 This is because, when frequency 𝑣 is increased, 161 (d)
1 As the magnetic field directed into the paper is
the capacitive reactance 𝑋𝐶 = 2𝜋𝑣𝐶
decreases and
increasing at a constant rate, therefore, induced
hence the current through the bulb increases
current should produce a magnetic field directed
155 (a)
out of the paper. Thus current in both the loops
In the rotation of magnet, 𝑁 pole moves closer to
must be anti-clock-wise.
coil 𝐶𝐷 and 𝑆 pole moves closer to coil 𝐴𝐵. As per
Lenz’s law, 𝑁 pole should develop at the end
corresponding to 𝐶. Induced current flows from
𝐶 𝑡𝑜 𝐷. Again 𝑆 pole should develop at the end
corresponding to 𝐵. Therefore, induced current in
the coil flows from 𝐴 to 𝐵. As area of loop on right side is more, therefore,
156 (d) induced emf o right side of loop will be more
As a given pole (𝑁 or 𝑆) of suspended magnet compared to the emf induced on the left-side of
goes into the coil and comes out of its, current is the loop
𝑑𝜙 𝑑𝐵
induced in the coil in two opposite directions. [∴ 𝑒 = − 𝑑𝑡 = −𝐴 𝑑𝑡 ]
Therefore, galvanometer deflection goes to left 162 (b)
and right both. As amplitude of oscillation of Given, 𝐿 = 30 mH
magnet goes on decreasing, so does the amplitude 𝑉𝑟𝑚𝑠 = 220 V
of deflection. 𝑓 = 50𝐻𝑧
157 (b) Now, 𝑋𝐿 = 𝜔𝐿 = 2𝜋𝑓𝐿
As is for Fig. (i), steady state current for t = both = 2 × 3.14 × 50 × 30 × 10−3
the circuits is same. Therefore, = 9.42 Ω
𝑉 𝑉
= The rms current in the coil is
𝑅1 𝑅2
𝑉𝑟𝑚𝑠 220𝑉
Or 𝑅1 = 𝑅2 𝑖𝑟𝑚𝑠 = 𝑋𝐿
= 9.42 Ω=23.4 A
Again, from the same figure, we observe that 163 (c)
𝐿1 𝐿2 𝑃 = 𝑉𝑟.𝑚.𝑠. × 𝑖𝑟.𝑚.𝑠. × cos 𝜙
𝜏1 < 𝜏2 ∴ <
𝑅1 𝑅2 100 100 × 10−3 𝜋
As 𝑅1 = 𝑅2 , therefore, 𝐿1 < 𝐿2 . = × × cos
√2 √2 3
158 (b) 4
10 × 10 −3
1 10
𝑃 = 𝑉𝐼 = × = = 2.5 𝑤𝑎𝑡𝑡
2 2 4
550 164 (d)
𝐼= = 2.5 𝐴
220 Initial flux linked with inner coil when 𝑖 = 0 is
159 (a)
zero. Final flux linked with inner coil when 𝑖 =
Let the applied voltage be 𝑉 volt. 𝜇0 𝑖
R C 𝑖 𝑖𝑠 ( )𝜋 𝑎2
2𝜋𝑏
𝜇 𝑖
|
0
∴ Change in flux, 𝑑𝜙 = (2𝜋𝑏 ) 𝜋 𝑎2
V V
R C
𝑑𝜙
As 𝑑𝑞 =
𝑅
Here, 𝑉𝑅 = 12 𝑉, 𝑉𝐶 = 5 𝑉
∴ Total charge circulating the inner coil is
𝜇 𝑖 𝜋𝑎 2 𝜇0 𝑖 𝑎 2
∴ 𝑉= √𝑉𝑅2 + 𝑉𝐶2 = √(12)2 + (5)2 = (2 𝜋0 𝑏) =
𝑅 2𝑅𝑏
= √144 + 25 = √169 = 13V 165 (a)
160 (c) Induced emf produced in coil
𝑅𝑡 −𝑑𝜙 −𝑑
𝑖 = 𝑖0 (1 − 𝑒 − 𝐿 ) 𝑒= = 𝑑𝑡 (𝐵𝐴)
𝑑𝑡
𝑑𝐵 1
𝑑𝑖 𝑑 𝑑 𝑅𝑡 𝑖0 𝑅 −𝑅𝑡 ∴ |𝑒| = 𝐴 = 0.01 ×
⇒ = 𝑖0 − (𝑖0 𝑒 − 𝐿 ) = 0 + 𝑒 𝐿 𝑑𝑡 1 × 10−3
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝐿
|𝑒| = 10 V
Initially, t = 0
𝑑𝑖 𝑖0 × 𝑅 𝐸 5 Current produced in coil,
⇒ = = = = 2.5 A𝑠 −1 𝑖=
|𝑒|
=
10
=5A
𝑑𝑡 𝐿 𝐿 2 𝑅 2

P a g e | 49
 Heat evolved = 𝑖 2 𝑅𝑡 Capacitive reactance, 𝑋𝐶 = 14Ω
= (5)2 × (2) × 1 × 10−3 = 0.05 J The impedance of the series 𝐿𝐶𝑅 circuit is
166 (c) 𝑍 = √𝑅 2 + (𝑋𝐶 + 𝑋𝐿 )2 = √(3)2 + (14 − 10)2
Power = Rate of work done in one complete cycle. 𝑍 = 5Ω
𝑊
or 𝑃𝑎𝑣 = 𝑇
174 (d)
(𝐸 𝐼 cos 𝜙)𝑇/2 In purely inductive circuit voltage leads the
or 𝑃𝑎𝑣 = 0 0 𝑇
current by 90°
𝐸 𝐼 cos 𝜙
or 𝑃𝑎𝑣 = 0 0 175 (a)
2
Where cos 𝜙 is called the power factor of an AC 1 𝐿
𝑄 factor is given by 𝑅 √𝐶
circuit.
167 (c) So, for large quality factor the inductance should
Terminal velocity of the rod is attained when be large and resistance and capacitance must be
magnetic force on the rod (𝐵𝑖𝑙) balances the small
component of weight of the rod(mg sin θ) , figure. 176 (b)
true power
As, power factor = apparent power
= cos 𝜙
in
mg s 𝑅
Fm =
√𝑅2 +(𝑋𝐿 −𝑋𝐶 )2
𝑅
R ∴ power factor= cos 𝜙 =
𝑍
In a non-inductive circuit, 𝑋𝐿 = 𝑋𝐶
𝑅 𝑅
∴ Power factor = cos 𝜙 = = =1
𝑖𝑒 , 𝐵𝑖𝑙 = mg sin θ √𝑅 2 𝑅
𝑒 ∴ 𝜙 = 0°
𝐵 ( ) 𝑙 = mg sin θ
𝑅 This is the maximum value of power factor. In a
𝐵𝑙
(𝑒) = mg sin θ pure inductor or an ideal capacitor
𝑅
𝐵𝑙 𝜙 = 90°
(𝐵𝑙𝑣𝑟 ) = mg sin θ
𝑅
mg 𝑅 sin θ ∴ Power factor= cos 𝜙 = cos 90° = 0
𝑟𝑇 = 𝐵2 𝑙2 Average power consumed in a pure inductor or
168 (a) ideal capacitor
1 1 1 𝑃 = 𝐸𝑣 . 𝐼𝑣 cos 90° = zero
𝑋𝐶 = ⇒𝐶= =
2𝜋𝑣𝐶 2𝜋𝑣𝑋𝐶 2 × 𝜋 × 400 × 25 Therefore, current through pure L or pure C,
𝜋
= 50 𝜇𝐹 which consumes no power for its maintenance in
170 (c) the circuit is called ideal current or wattles
Here, 𝑀 = 2H, 𝑑𝜙 = 4 Wb, 𝑑𝑡 = 10 s current.
As 𝜙 = 𝑀 𝑖 177 (d)
𝑑𝜙 = 𝑀 𝑑𝑖 Potential difference across the capacitor = emf
𝑑𝜙 4 induced across 𝐻𝐸 = 𝐵𝑙𝑣 which is constant.
Or 𝑑𝑖 = 𝑀
=2=2A
Therefore, charge stored in the capacitor is
Also, 𝑑𝜙 = 𝑀 (𝑑𝑖) = 2(1) constant. Hence current in the circuit 𝐻𝐾𝐷𝐸 is
= 2 Wb zero.
171 (a) 178 (b)
𝑋𝐿 √3 𝑅 𝑒 = −𝐿𝑑𝐼/𝑑𝑡 = −5 × (−2) = +10 V
tan 𝜙 = = = √3 ⇒ 𝜙 = 60° = 𝜋/3
𝑅 𝑅 179 (b)
172 (c) (Rated voltage)2
𝑇 (1/50) 𝜋 1 Resistance of a bulb = Rated power
Time difference = 2𝜋
×𝜙 = 2𝜋
×4 = 400
𝑠 =
(220)2
2.5𝑚-𝑠 = = 484 Ω
100
173 (a)
Peak voltage of the source, 𝑉0 = 220√2 𝑉 = 311 𝑉
Here, Resistance, 𝑅 = 3Ω
180 (a)
Inductive reactance, 𝑋𝐿 = 10Ω

P a g e | 50
 𝜇0 𝑁1 𝑁2 𝐴
As 𝑀 = , therefore, 𝑀 becomes 4 times
𝑙
189 (d)
181 (a)
Impedance of 𝐿𝐶𝑅 circuit will be minimum at 𝑍 = √(𝑅)2 + (𝑋𝐿 − 𝑋𝐶 )2 ;
resonant frequency so 𝑅 = 10Ω, 𝑋𝐿 = 𝜔𝐿 = 2000 × 5 × 10−3 = 10Ω
1 1 1 1
𝑉0 = = 𝑋𝐶 = = = 10Ω, 𝑖. 𝑒. , 𝑍
2𝜋√𝐿𝐶 2𝜋√1 × 10−3 × 0.1 × 10−6 𝜔𝐶 2000 × 50 × 10−6
10 5 = 10Ω
= 𝐻𝑧 𝑉0 20
2𝜋 Maximum current 𝑖0 = 𝑍
= 10 = 2𝐴
182 (b) Hence 𝑖𝑟𝑚𝑠 =
2
= 1.4 𝐴 and 𝑉𝑟𝑚𝑠 = 4 × 1.41 =
√2
Here, resistance of rod = 2Ω. 𝑟 = 0.1 m, 𝐵 =
5.64 𝑉
50 T, along 𝑧 − axis 𝜔 = 20 rads −1 .
191 (d)
Potential difference between centre of the ring
and the rim is 𝑉 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2
1 1
𝑉 = 𝐵𝜔𝑟 2 = × 50 × 20 × (0.1)2 = 5 V ∵ 𝑉𝑅 = 𝑉 ∵ 𝑉𝐿 = 𝑉𝐶
2 2
The equivalent circuit of the arrangement is ∴ Reading of voltmeter = 220𝑉
shown in figure 𝐸
Reading of ammeter 𝐼𝑟𝑚𝑠 = 𝑟𝑚𝑠 𝑍
A 220
= = 2.2𝐴
10 Ω 100
10 Ω
192 (a)
O
Motion emf induced in the connector
10 Ω
𝑒 = 𝐵𝑙𝑣 = 2(1)(2) = 4 V
This acts as a cell of emf 4 V and internal
B
5V resistance 2 Ω. 6Ω and 3 Ω resistors are in
5V 10 Ω parallel.
5Ω 1 1 1 1+2 3 1
∴ 𝑅𝑃
=6+3= 6
=6=2
5V 10 Ω 𝑅𝑃 = 2 Ω
10 Ω
10 Ω 4V 4V
Current through external resistance, 6Ω 3Ω 2Ω
𝐸 5 1 2Ω
𝑖= = = A 2Ω
𝑅+𝑟 10+5 3
183 (c)
𝜔𝐿 2𝜋 × 50 × 0.21 ∴ Current through the connector (𝑖)
tan 𝜙 = = = 5.5 ⇒ 𝜙 = 80° 𝐸 4
𝑅 12 =𝑅 = 2+2 = 1 A.
𝑃 +𝑟
185 (d)
5
Magnetic force on the connector
𝑒 = 𝐿𝑑𝑖/𝑑𝑡 = 4 × 1/1500 = 30000V=30kV = 𝐵𝑖𝑙 = (1)(1) = 2 N
187 (a) Therefore, to keep the connector moving with a
50 constant velocity, a force of 2 N has to be applied to
𝑋𝐿 = 2𝜋𝑓𝐿 = 2𝜋 ( ) × 1 = 100Ω
𝜋 the right side.
1 193 (c)
𝑋𝐶 =
2𝜋𝑓𝐶 Heat produced by ac = 3 × Heat produced by dc
1 2
∴ 𝑖𝑟𝑚𝑠 𝑅𝑡 = 3 × 𝑖 2 𝑅𝑡 ⇒ 𝑖𝑟𝑚𝑠
2
= 3 × 22
= 50
2𝜋 ( 𝜋 ) × 20 × 10−6 ⇒ 𝑖𝑟𝑚𝑠 = 2√3 = 3.46 𝐴
= 500 Ω 194 (d)
1
Impedence Z = √(𝑅)2 + (𝑋𝑐 − 𝑋𝐿 )2 Brightness ∝ 𝑃consumed ∝ 𝑅. For bulb, 𝑅𝑎𝑐 = 𝑅𝑑𝑐 ,
= √(300)2 + (400)2 so brightness will be equal in both the cases
195 (c)
= 500 Ω 𝐸 = 141 sin 628𝑡

P a g e | 51
 𝐸0 Also 𝑍 2 = 𝑅 2 + 4𝜋 2 𝑣 2 𝐿2
∴ 𝐸𝑟𝑚𝑠 =
√2 ⇒ 50 = (5)2 + 4(3.14)2 𝑣 2 (10 × 10−3 )2 ⇒ 𝑣
141
= 1.41 = 100 V = 80 𝐻𝑧
𝜔 203 (d)
and 𝑣=
2𝜋 The voltage 𝑉𝐿 and 𝑉𝐶 are equal and opposite so,
628
= 2×3.14 = 100 Hz voltmeter reading will be zero.
196 (c) Also, 𝑅 = 30 Ω, 𝑋𝐿 = 𝑋𝐶 = 25 Ω
𝑉
Here, 𝑅 = 10 Ω. As is known, So, 𝑖=
𝑑𝜙 √𝑅2 +(𝑋𝐿 −𝑋𝐶 )2
|𝑑𝑞| = = |𝑖 𝑑𝑡| = area under 𝑖 − 𝑡 graphs. 𝑉 240
𝑅 = 𝑅 = 30 = 8A
𝑑𝜙 (4)(0.1)
∴ = = 0.2 204 (c)
𝑅 2
𝑑𝜙 = 0.2 𝑅 = 0.2 × 10 = 2 Wb If 𝜔 = 50 × 2𝜋 then 𝜔𝐿 = 20Ω
197 (a) If 𝜔′ = 100 × 2𝜋 then 𝜔′ 𝐿 = 40Ω
200 200 200
From 𝜙 = 𝑀𝑖 𝐼= = =
𝑀1 𝜙 10−3 ×200 10 𝑍 √𝑅 2 + (𝜔 ′ 𝐿)2 √(30)2 + (40)2
𝑀2
= 𝜙1 = 0.8×10−3 ×400 = 16 = 0.625
2 𝐼 =4𝐴
198 (a) 205 (a)
Here, 𝑉𝑟𝑚𝑠 = 220V, v = 50 Hz Eddy currents are produced when a metal is kept
Peak value of voltage 𝑉0 = √2 𝑉rms = 220√2 V in a varying magnetic field.
∴The instantaneous value of voltage is 206 (c)
𝑉 = 𝑉0 sin 2𝜋𝑣𝑡 = 220 √2 sin 2𝜋 × 50𝑡 After the current in the inductor reaches its
= 220√2 sin 100𝜋𝑡 maximum value 𝐼0 , it falls from 𝐼0 to zero. The
1
199 (d) energy 𝐿𝐼02 supplied by the source during build
2
𝑀𝑑𝐼 20
𝑒= 𝑑𝑡
= 0.09 × 0.006 = 300V up of current is returned back to the source
200 (a) during the fall of current.
Current through the bulb 𝑖 = 𝑉 = 10 = 6𝐴
𝑃 60 Thus, net power supplied by the source in a
60W, 10V
complete cycle is zero.
i
L 207 (a)
1 1
10 V VL 𝑓= ⇒𝑓∝
i 2𝜋√𝐿𝐶 √𝐶
208 (a)
100V, 50Hz
The same emf is induced in all the three rings
because emf is only due to linear motion and does
𝑉 = √𝑉𝑅2 + 𝑉𝐿2 not depend on spin.
(100)2 = (10)2 + 𝑉𝐿2 209 (b)
𝑑𝑖
⇒ 𝑉𝐿 = 99.5 𝑉𝑜𝑙𝑡 As 𝑒 = 𝑀
𝑑𝑡
Also 𝑉𝐿 = 𝑖𝑋𝐿 = 𝑖 × (2𝜋𝑣𝐿) 𝑒 25×10−3
∴ 𝑀= = = 1.67 × 10−3 H
⇒ 99.5 = 6 × 2 × 3.14 × 50 × 𝐿 𝑑𝑖/𝑑𝑡 15.0

⇒ 𝐿 = 0.052 𝐻 As 𝜙 = 𝑀𝑖
201 (b) ∴ 𝜙 = 1.67 × 10−3 × 3.6 = 6 × 10−3 Wb
𝑑𝐼 (2 − 3) = 6 m Wb
𝐿 = ? 𝑒 = 5 𝑉, = = −103 As−1 210 (b)
𝑑𝑡 10−3
As 𝑒 = −𝐿 𝑑𝑡
𝑑𝐼 In Colpitt oscillator two capacitors are placed
5
across a common inductor and the centre of the
∴ 5 = −𝐿(−103 ), 𝐿 = 103H = 5mH two capacitors is tapped.
202 (c) 211 (d)
𝑉2 (10)2 In an AC circuit, the coil of high inductance and
With dc : 𝑃 = 𝑅
⇒ 𝑅= 20
= 5Ω;
2
𝑉𝑟𝑚𝑠 𝑅 (10)2 ×5
negligible resistance used to control current, is
With ac : 𝑃 = ⇒ 𝑍2 = = 50 Ω2
𝑍2 10

P a g e | 52
 called the choke coil. The power factor of such a 𝑉0 200
𝐼0 = = =2A
coil is given by 𝑅 100
𝑅 𝐼0
Cos 𝜙 = 𝐼𝑟𝑚𝑠 = = 1.414 A
√𝑅2 +𝜔2 𝐿2
𝑅 √2
≈ 𝜔𝐿
(as 𝑅 << 𝜔𝐿) 219 (b)
As 𝑅 << 𝜔𝐿, cos 𝜙 is very small. Thus, the power The phase angle (𝜃) between I and V is given by
𝑋 −𝑋
absorbed by the coil is very small. The only loss of tan 𝜃 = 𝐿 𝑅 𝐶 …(i)
energy is due to hysteresis in the iron core, which Where, 𝑋𝐿 = 2𝜋𝑓𝐿
is much less than the loss of energy in the 200
resistance that can also reduce the current if = 2𝜋 × 50 × [ × 10−3 ]
𝜋
placed instead of the choke coil. = 20 Ω
212 (d) 1
𝑋𝐿 =
𝑒0 = 𝑁𝐴𝐵𝜔. When 𝐵 and 𝜔 are doubled, 𝑒0 2𝜋𝑓𝐶
becomes 4 times. 1×𝜋
=
213 (c) 2𝜋 × 50 × 10−3
200 1 =10 Ω
𝑉rms = , 𝐼rms =
√2 √2 and 𝑅 = 10 Ω
200 1 𝜋 Substituting values of 𝑋𝐿 , 𝑋𝐶 and R in eq.(i), we
∴ 𝑃 = 𝑉rms 𝐼rms cos 𝜙 = cos = 50W
√2 √2 3 get
214 (b) 20−10
tan 𝜃 = 10 = 1
Here, 𝑖 = 𝑖0 at 𝑡 = ∞. Let 𝑖 be the current at 𝑡 = 1 𝜋
s ⇒ tan 𝜃 = tan
4
𝑅
− 𝑡 𝜋
From 𝑖 = 𝑖0 (1 − 𝑒 𝐿 ) ∴ 𝜃=
4
10 𝜋
= 𝑖 (1 − 𝑒 − 5 ×1 ) = 𝑖 (1 − )
1 The phase angle of the circuit is 4 .
0 0 𝑒2
𝑖0 𝑒 2 220 (b)
∴ = 2 𝑛𝑃 1
𝑖 𝑒 −1 =
215 (c) 𝑛𝑠 20
𝑖𝑃 𝑛
In a series L – C – R circuit, potential difference As 𝑖𝑠
= 𝑛𝑠
𝑃
leads the current by an angle ϕ(let). 𝑖𝑃
𝑋𝐿 −𝑋𝐶 ∴ = 20 ∶ 1
ϕ = tan−1 ( 𝑅
) 𝑖𝑠
𝜔𝐿 −
1 221 (b)
or ϕ = tan−1 ( 𝑅
𝜔𝐶
) 1 1
𝑃 = 𝑉0 𝑖0 cos 𝜙 ⇒ 1000 = × 200 × 𝑖0 cos 60°
1 2 2
At resonance, 𝑋𝐿 = 𝑋𝐶 , 𝑖𝑒, 𝜔𝐿 = 𝜔𝐶 𝑖0 20
⇒ 𝑖0 = 20 𝐴 ⇒ 𝑖𝑟𝑚𝑠 = = = 10√2𝐴
Hence, ϕ = tan−1 (0) = 0 √2 √2
Therefore, phase difference between current and 222 (c)
voltage at resonance is zero. 𝜔𝐿 1 1
𝑄= = × ×𝐿
216 (b) 𝑅 𝑅 √𝐿𝐶
The average power output of the emf source is 1 𝐿
1
= √
𝑅 𝐶
𝑃= |𝑉 ||𝐼 | cos 𝜃
2 0 0
1 1 1
Since, 𝑉0 = 𝐼0 𝑅 = 3 × √9 = 9
1
∴ 𝑃 = 𝑅|𝐼0 |2 223 (c)
2
Given, 𝑉𝑅 = 5 V, 𝑉𝐿 = 10 V and 𝑉𝐶 = 10 V
It is clear that only the resistor dissipates energy
In the L – C – R circuit, the AC voltage applied to
in the circuit. The inductor and capacitor both
the circuit will be
store energy but they eventually return it to the
circuit without dissipation. 𝑉 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2
217 (a) = √(5)2 + (10 − 10)2 = 5V

P a g e | 53
 VL Y 232 (c)
V
Resistances of both the bulbs are
V L - VC 𝑉 2 2202
𝑅1 = =
𝑃1 25
2
𝑉 2202
𝑅2 = =
𝑃2 100
IV X
VR ∴ 𝑅2 > 𝑅2
Hence 25 𝑊 bulb will fuse
233 (b)
𝑉−𝐸 𝑉 − 𝐸 220 − 80
VC 𝑖= ,𝑅 = = = 5.6 Ω
𝑅 𝑖 25
234 (c)
224 (b)
The current in 𝐿𝐶𝑅 circuit becomes maximum at
The impedance of R – C circuit for frequency 𝑓1 is
series resonance condition. At this point the total
1
𝑍1 = √𝑅 2 + 2 2 2 reactance of the circuit is zero. That means the
4𝜋 𝑓 𝐶
reactance of inductance becomes equal and
The impedance of R – C circuit for frequency 2𝑓 is
opposite to the reactance by the capacitor
1
𝑍2 = √𝑅 2 + 4𝜋2 (2𝑓2 )𝐶 2 235 (b)
For anti-resonant circuit current is minimum at
2 1
or 𝑍2 = √𝑅 + 16𝜋2 𝑓2 𝐶 2 resonant frequency and at frequencies other than
1 resonant frequency current rises with frequency
𝑍12 𝑅2 + 2 2 2
Then, = 2
4𝜋 𝑓 𝐶
1 237 (b)
𝑍22 𝑅 +
16𝜋2 𝑓2 𝐶2 𝑉 𝑅 𝑉 2𝑅 𝑉 2𝑅
1+
1 𝑃 = 𝑉𝑖 cos 𝜙 = 𝑉 ( ) ( ) = =
𝑍12 4𝜋2 𝑓2 𝐶2 𝑅2 𝑍 𝑍 𝑍2 (𝑅 2 + 𝜔 2 𝐿2 )
Or = 1
𝑍22 1+
16𝜋2 𝑓2 𝑅2 𝐶2 238 (b)
𝑍1
Values are greater than 1 then 𝑍 = lies between 1 Moving from 𝐴 to 𝐵.
2
𝑉𝐵 − 𝑉𝐴 = [5 × 10−3 (−103 ) + 15 + 1 × 5]
and 2.
= 15 volt
225 (c)
1 239 (d)
𝑓 = 2𝜋 𝐿𝐶 For a series L – C – R
√
1 circuit at resonance
∴ 𝑓=
2 × 3.14√0.5 × 10 × 20 × 10−3 −6 Phase difference, 𝜙 = 0°
1 Power factor= cos 𝜙 = 1
= ≈ 1600 Hz
2 × 3.14 × 10−4 241 (c)
226 (c) Resonance frequency in 𝑟𝑎𝑑𝑖𝑎𝑛/𝑠𝑒𝑐𝑜𝑛𝑑 is
200
𝑋𝐿 2𝜋𝑣𝐿 2𝜋 × 2𝜋 × 1 1 1
tan 𝜙 = = = =1⇒𝜙 𝜔= = = 500 𝑟𝑎𝑑/𝑠𝑒𝑐
𝑅 𝑅 200 √𝐿𝐶 √8 × 0.5 × 10−6
= 45° 242 (b)
227 (a) 1 1
𝑑𝐼 𝑋𝐶 = =
𝑒 = 𝐿 𝑑𝑡 = 2 × 10 = 2V−3 𝜔𝐶 2𝜋𝑓𝐶
1
228 (d) 𝑖. 𝑒. , 𝑋𝐶 ∝
2 𝑓
𝑉𝑟𝑚𝑠 𝑅
𝑃𝑎𝑣 = 243 (b)
𝑍2
𝑉𝑟𝑚𝑠
229 (b) 𝑍 = √𝑅 2 + 𝑋𝐶2 ∶ 𝐼𝑟𝑚𝑠 = 2
: 𝑃 = 𝐼𝑟𝑚𝑠 𝑅
𝑉0 = 𝑖0 𝑍 ⇒ 200 = 100 𝑍 ⇒ 𝑍 = 2Ω 𝑍
1
Also 𝑍 2 = 𝑅 2 + 𝑋𝐿2 ⇒ (2)2 = (1)2 + 𝑋𝐿2 ⇒ 𝑋𝐿 = Where 𝑋𝐶 = 𝜔𝐶
√3Ω As 𝜔 is increased, 𝑋𝐶 will decrease or 𝑍 will
231 (b) decrease. Hence 𝐼𝑟𝑚𝑠 𝑜𝑟 𝑃 will increase.
1
Power loss ∝ (Voltage)2 Therefore, bulb glows brighter.
Hence the correct option is (b).
P a g e | 54
244 (d) 𝑋𝐿 = 𝑋𝐶
𝑡 10 −2 4 4𝑅 4 110 440
𝐶= 𝑅
= 103
= 10 𝐹 = 10 𝜇 F ∴ 𝑋𝐶 = 3
=3 × 3
= 9
Ω [From Eq.
245 (c) (i)]
1 440
2𝜋𝑣 = 377 ⇒ 𝑣 = 60.03 𝐻𝑧 = Ω
2𝜋𝑓𝐶 9
247 (c) 9
1 𝐶 = 2×3.14×60×440
𝑓 = 2𝜋
√𝐿𝐶
1
= 54 𝜇F
or 𝑓∝ 251 (b)
√𝐶
When capacitor C is replaced by another capacitor Capacitive reactance( 𝑋𝐶 ) is given by
C’ of dielectric constant K, then 1
𝑋𝐶 = 𝜔𝐶
𝐶 ′ = 𝐾𝐶
Where 𝜔 is angular frequency and C the
𝑓′ 𝐶 capacitance.
∴ =√
𝑓 𝐶′ Also, 𝜔 = 2𝜋𝑓,where 𝑓 is frequency.
In a DC circuit 𝑓 = 0 ∴ 𝜔 = 0
125000−25000 𝐶
or =√
125000 𝐾𝐶 𝑋 = =∞
1
𝐶 0
100 1
or = 252 (c)
125 √𝐾
125 2
As continuous flow of DC do not take place
or 𝐾=( ) = 1.56 through a capacitor, Therefore resistance of the
100
248 (a) circuit
𝐸𝑆 𝑖𝑃 𝐸𝑆 4.6 R = 1+0.5 = 1.5
= ⇒ 𝑖𝑃 = × 𝑖𝑆 = × 5 = 0.1 A
𝐸𝑃 𝑖𝑆 𝐸𝑃 230
Current with circuit
Frequency is not affected by transformer. 𝐸
249 (a) 𝐸 = 𝑅′
2 4
2 2 2√2 = 1.5 = 3 A
𝑉𝑎𝑣 = 𝑉0 = × (𝑉𝑟𝑚𝑠 × √2) = . 𝑉𝑟𝑚𝑠
𝜋 𝜋 𝜋 Potential difference across capacitor =Potential
2√2 difference across 1 Ω resistor
= × 220 = 198 𝑉 4 4
𝜋 = 3 × 1 = 3V
250 (b)
Ist case From formula ∴ Charge on capacitor 𝑞 = 𝐶𝑉
4
𝑉2 = 1 × 3 = 1.33 μF
𝑅= 𝑃
110×110 110 253 (a)
= 330 = Ω 1
3 Frequency of 𝐿𝐶 oscillation = 2𝜋
√𝐿𝐶
Since, current lags the voltage thus, the circuit
contains resistance and inductance. 𝑓1 1 𝐿2 𝐶2 1/2
⇒ = √𝐿 𝐶 = ( )
Power factorcos 𝜙 = 0.6 𝑓2 √𝐿1 𝐶1 2 2 𝐿1 𝐶1
𝑅
= 0.6 2𝐿 × 4𝐶 1/2
√𝑅2 +𝑋𝐿2 =( ) = (8)1/2
𝐿×𝐶
𝑓 𝑓 𝑓
𝑅 2 ∴ 𝑓1 = 2√2 ⇒ 𝑓2 = 2 12 or, 𝑓2 = 2 2 [∵ 𝑓1 = 𝑓]
⇒ 𝑅 2 + 𝑋𝐿2 = ( ) 2 √ √
0.6 254 (a)
2
𝑅
⇒ 𝑋𝐿2 = − 𝑅2 From 𝑒 = 𝐿 𝑑𝐼/𝑑𝑡
(0.6)2 𝑑𝐼 𝑒 90
= 𝐿 = 0.2 = 450 As −1
𝑅 2 × 0.64 𝑑𝑡
⇒ 𝑋𝐿2 = 255 (d)
0.36
∴
0.8 𝑅
𝑋𝐿 = 0.6 = 3
4𝑅
…(i) 𝑉 = 120 sin 100𝜋𝑡 cos 100𝜋𝑡 ⇒ 𝑉 = 60 sin 200𝜋𝑡
𝑉max = 60𝑉 and 𝑣 = 100 𝐻𝑧
IInd case
256 (c)
Now cos 𝜙 = 1 (given)
Power = Rate of work done in one complete cycle.
Therefore, circuit is purely resistive, ie, it contains 𝑊
or 𝑃𝑎𝑣 =
only resistance. This is the condition of resonance 𝑇
(𝐸 𝐼 cos 𝜙)𝑇/2
in which or 𝑃𝑎𝑣 = 0 0
𝑇

P a g e | 55
 𝐸0 𝐼0 cos 𝜙 The maximum current is
or 𝑃𝑎𝑣 = 2
Where cos 𝜙 is called the power factor of an AC 𝐼𝑚 = √2 𝐼𝑟𝑚𝑠 = √2(10) = 10√2 𝐴
circuit. Maximum potential difference is 𝑉𝑚 = 𝐼𝑚 𝑅
257 (b) = 10√2 × 12 = 169.68 𝑉
The impedance (Z) of an R – L – C series circuit is 266 (b)
given by 𝑍 = √𝑅 2 + 𝑋𝐶2
1 2
𝑍 = √𝑅 2 + (𝜔𝐿 − 𝜔𝐶 ) 1 2
= √(𝑅)2 + (𝜔𝐶 )
As frequency of alternating emf applied to the
circuit is increased, 𝑋𝐿 goes on increasing and 𝑋𝐶 In case (b) capacitance (c) will be more.
goes on decreasing. Therefore, impedance Z will be less. Hence
For a particular value of 𝜔 = (𝜔𝑟 say) current will be more.
𝑋𝐿 = 𝑋𝐶 267 (c)
1 𝑋𝐿
ie, 𝜔𝑟 𝐿 = tan 𝜙 = = 1 ∴ 𝜙 = 45° or 𝜋/4
𝜔𝑟 𝐶 𝑅
1 268 (c)
or 𝜔𝑟 = 𝐿𝐶
√ (i) In a circuit having C alone, the voltage lags the
1 𝜋
or 2𝜋𝑣𝑟 = 𝐿𝐶 current by 2 .
√
1
or 𝑣𝑟 = (ii) In a circuit containing R and L, the voltage
2𝜋√𝐿𝐶 𝜋
1 leads the current by .
2
∴ 𝑣=
2 × 3.14 × √5 × 80 × 10−6 (iii) In L – C circuit, the phase difference between
1 current and voltage can have any
=
2×3.14×√(400×10−6 ) 𝜋
1 value between 0 to 2 depending on the
= 2×3.14×2×10−2 values of L and C.
100
= 3.14×4 (iv) In a circuit containing L alone, the voltage
25 25 𝜋
= = 𝜋 Hz leads the current by .
3.15 2
259 (b) 269 (d)
The time taken by AC in reaching from zero to Induced emf 𝑒 = 𝐵𝑙𝑣 = 𝐵𝑊𝑣
𝑒2 𝐵2 𝑊 2 𝑣 2
maximum value is Power developed = =
𝑇 1 𝑅 𝑅
𝑡 = 4 = 4𝑓 270 (b)
=
1
= 5 × 10−3 s In 𝑅𝐶 series circuit voltage across the capacitor
50×4 𝜋
leads the voltage across the resistance by 2
260 (b)
For given circuit current is lagging the voltage by 271 (c)
𝑛 1
𝜋/2, so circuit is purely inductive and there is no 𝐸𝑆 = 𝑛 𝑠 𝐸𝑃 = 200 × 240 = 12 V
𝑃
power consumption in the circuit. The work done 272 (a)
by battery is stored as magnetic energy in the 𝐸
The current is 𝐼 = 2 0 2 2
inductor. √𝑅 +𝜔 𝐿
4
262 (a) = 2 = 0.8 A
√4 +(1000×3×10−3 )2
In a pure inductor (zero resistance), voltage leads 273 (b)
the current by 90° 𝑖. 𝑒. , 𝜋/2 In parallel resonant circuit, resonance frequency
263 (d) 1
𝑓0 = 2𝜋 𝐿𝐶
𝑃 = 𝑉𝑖 cos 𝜙 √
𝜋 1
Phase difference 𝜙 = ⇒ 𝑃 = zero =
2 10×10−3
2𝜋√ ×0.04×10−6
𝜋2
264 (c)
104
1
We have 𝑋𝐶 = 𝐶×2𝜋𝑓 and 𝑋𝐿 = 𝐿 × 2𝜋𝑓 = 2×0.2 = 25 kHz
274 (d)
265 (c)
Here 𝐼𝑟𝑚𝑠 = 10 𝐴, 𝑅 = 12 Ω

P a g e | 56
 22500
(𝑇/2)𝑉02 + 0 𝑉0 = 0.5 × 150 × 150 × cos 60° =
𝑉𝑟𝑚𝑠 =√ = 4
𝑇 √2 = 5625 W
275 (a) 286 (a)
The voltage across a 𝐿 − 𝑅 combination is given If the capacitance is removed, it is an 𝐿 − 𝑅 circuit
by 𝜙 = 60°
2 2 2 𝑋𝐿
𝑉 = 𝑉𝑅 + 𝑉𝐿
tan 𝜙 = = tan 60° = √3
2 2
𝑅
𝑉𝐿 = √𝑉 − 𝑉𝑅 = √400 − 144 = √256 = 16 𝑣𝑜𝑙𝑡 If inductance is removed, it is a capacitative
276 (d) circuit or 𝑅 − 𝐶 circuit. |𝜙| is the same
1
The average power of L – C – R circuit ∴ 𝐿𝜔 = 𝐶𝜔 This is a resonance circuit
𝑃𝑎𝑣 = 𝑉𝑟𝑚𝑠 . 𝑖𝑟𝑚𝑠 cos 𝜙 𝐸𝑟𝑚𝑠
Hence, the average power depends upon current, 𝑍 = 𝑅; 𝐼𝑟𝑚𝑠 = , 𝐸𝑟𝑚𝑠 = 200 𝑉
𝑅
emf and phase difference. 200𝑉
∴ 𝐼𝑟𝑚𝑠 = = 2𝐴
277 (d) 100Ω
𝑋𝐿 50 287 (d)
𝑋𝐿 = 2𝜋𝑣 𝐿 ⇒ 𝐿 = = = 0.16 𝐻
2𝜋𝑣 2 × 3.14 × 50 Resistance of the bulb
278 (d) 𝑉 2 (100)2
𝑅= = = 200 Ω
The average power dissipation in the circuit is 𝑃 50
1 𝑉
𝐸0 𝐼0 tan ϕ Current through bulb(𝐼) = 𝑅
2
100
= 200 = 0.5 A

279 (a) In a circuit containing inductive reactance (𝑋𝐿 )

Final current in constant and L plays no role at that and resistance (R), impedance (Z) of the circuit is
instant. Therefore, 𝑖 = 𝐸/𝑅. 𝑍 = √𝑅 2 + 𝜔 2 𝐿2 …(i)
200
280 (a) Here, 𝑍 =
0.5
= 400 Ω
A closed current carrying loop of any irregular Now, 𝑋𝐿2
= 𝑍 − 𝑅22

shape and even not lying in a single plane, placed = (400)2 − (200)2
in a uniform magnetic field shall experience no (2𝜋𝑓𝐿)2 = 12 × 104
net force. Therefore, force acting on the loop must 2√3×100
𝐿=
be zero. 2𝜋×50
2√3
281 (b) = 𝜋 = 1.1 H
𝜇0 𝑁1 𝑁2 𝐴
As 𝑀 = 𝑙
, 288 (b)
∴ 𝑀 can be increased by increasing the number The Q- factor of series resonant circuit is given by
of turns in the coils. 𝜔𝑟2 𝐿
𝑄= 𝑅
282 (a)
It is evident from the relation that as R is
In L-C-R series circuit,
increased, Q – factor of the circuit is decreased.
𝑉 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2 289 (d)
In steady state current through the branch having
= √(40)2 + (60 − 30)2 capacitor is zero.
= √1600 + 900 = √2500 = 50V 1 1 1 1
∴ = + +
283 (c) 𝑅 1 2 3
For series L – C – R circuit
𝑉 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2 1 6+3+2
=
𝑅 6
= √(80)2 + (40 − 100)2 6
𝑅 = 11
= 100 V
285 (b) As 𝑉 = 𝑖𝑅
1 6
Power 𝑃 = 𝑉 𝐼 cos 𝜙 ∴ 6=𝑖×
2 00 11
Current through the battery 𝑖 = 11A
P a g e | 57
 Charge on the capacitor 𝑞 = 𝐶𝑉
𝑍 = √𝑅 2 + 𝑋𝐶2 = √2𝑅
⇒ 𝑞 = 0.5 × 10−6 × 6
𝑞 = 3μ𝐶 𝑉
𝑉𝐶 = 𝑖𝑋𝐶 (∵ 𝑖 = )
290 (d) 2
𝑉 10
Current will be maximum in the condition of or 𝑉𝐶 = 𝑍 𝑋𝐶 = 2𝑅 𝑅
√
𝑉 𝑉 10
resonance so 𝑖max = = 𝐴 𝑉𝐶 = V
𝑅 10 √2
𝐸 2
2 1
Energy stored in the coil 𝑊𝐿 = 2 𝐿𝑖max
1
= 2 𝐿 (10) 297 (c)
Adff sdaf sdfsdf dsf
1 𝐸2 1
= × 10−3 ( ) = × 10−5 𝐸 2 𝑗𝑜𝑢𝑙𝑒 298 (b)
2 100 2 𝑒𝑑𝑡 8×0.05
∴ Energy stored in the capacitor From 𝑒 = 𝐿𝑑𝐼/𝑑𝑡, 𝐿 = 𝑑𝐼
= 2
= 0.2H
1 1 299 (a)
𝑊𝐶 = 𝐶𝐸 2 = × 2 × 10−6 𝐸 2 = 10−6 𝐸 2 𝑗𝑜𝑢𝑙𝑒
2 2 Capacitance of wire
𝑊𝐶 1 𝐶 = 0.014 × 10−6 × 200
∴ =
𝑊𝐿 5 = 2.8 × 10−6 𝐹 = 2.84𝜇F
291 (c) For impedance of the circuit to be minimum
1 𝑋𝐿 = 𝑋𝐶
𝑋𝐿 = 2𝜋𝑣𝐿 = 2 × 𝜋 × 50 × = 100Ω
𝜋 1
293 (c) ⇒ 2𝜋𝑣𝐿 =
2𝜋𝑣𝐶
In series R – L – C circuit, the impedance of the 1
𝐿 = 4𝜋2 𝑣 2 𝐶
circuit is given by 1
= 4(3.14)2 ×(5×103 )2 ×2.8×10−6
𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2
Also, 𝑋𝐿 = 𝜔𝐿, 𝑋𝐶 = 𝜔𝐶
1 = 0.35 × 10−3 H = 0.35 mH
300 (c)
2 At resonance 𝑋𝐿 = 𝑋𝐶
1
∴ 𝑍 = √𝑅 2 + (𝜔𝐿 − ) 303 (c)
𝜔𝐶
Given, 𝑅 = 300Ω, 𝜔 = 1000 rads −1 , 𝐿 = 0.9 𝐻, As 𝐵0 = 𝜇0 𝑛𝑖, therefore 𝐵0 does not depend upon
𝐶 = 20𝜇F = 2 × 10−6 F radius (𝑟) of the solenoid.
304 (c)
Hence, 𝑍 = √(300)2 + (1000 × 0.9 −
Here, 𝐿 = 25 𝑚𝐻 = 25 × 10−3 𝐻
1 2
) 𝑣 = 50 𝐻𝑧, 𝑉𝑟𝑚𝑠 = 220 𝑉
1000×2×10−6
The inductive reactance is
= √90000 + (900 − 500)2 22
= √90000 + 160000 𝑋𝐿 = 2𝜋𝑣 𝐿 = 2 × × 50 × 25 × 10−3 Ω
7
= √250000 = 500 Ω The 𝑟𝑚𝑠 current in the circuit is
294 (b) 𝑉𝑟𝑚𝑠 220
𝐼𝑟𝑚𝑠 = = 22
As coil 𝐴 is moved closer to 𝐵, field due to 𝑋𝐿 2 × × 50 × 25 × 10−3
7
𝐴 intercepting 𝐵 is increasing. Induced current in 7 × 1000
𝐵 must oppose this increase. Hence the current in = 𝐴 = 28 𝐴
2 × 5 × 25
𝐵 must be anti-clock-wise. 305 (b)
295 (d) To decrease current in an AC circuit, choke coil is
1
Reactance 𝑋 = 𝑋𝐿 − 𝑋𝐶 = 2𝜋𝑓𝐿 − 2𝜋𝑓𝐶 used. The choke coil has high inductance and
negligible resistance, so that the energy loss in the
296 (d)
circuit is negligible.
Circuit is resonant.
Hence, 𝑋𝐿 >> 𝑅
Hence supply voltage equals
306 (a)
𝑉𝑅 = 10 V
Current in a L-C-R series circuit,
Also, 𝑋𝐶 = 𝑅 𝑉
As the voltage drops are equals across them when 𝑖= 2
√𝑅 +(𝑋𝐿 −𝑋𝐶 )2
L is shortened

P a g e | 58
 Where 𝑉 is rms value of current, R is resistance, 𝑉 220
Amplitude of current, 𝐼0 = 𝑅 = 44
= 5A
𝑋𝐿 is inductive reactance and 𝑋𝐶 is capacitive
314 (a)
reactance.
For current to be maximum, denominator should 1 𝑇
𝑉𝑟𝑚𝑠 = √ ∫ 102 𝑑𝑡 = 10 𝑉
be minimum which can be done, if 𝑇 0
𝑋𝐿 = 𝑋𝐶 315 (c)
This happens in resonance state of the circuit Effective voltage 𝑉𝑟.𝑚.𝑠. =
𝑉0
=
423
= 300 𝑉
1 √2 √2
i.e, 𝜔𝐿 =
𝜔𝐶 316 (b)
1
or 𝐿 = 𝜔2 𝐶 ...(i) Given, 𝐿 = 20 mH = 20 × 10−3 H
Given, 𝜔 = 1000s −1
, 𝐶 = 10𝜇𝐹 = 10 × 10−6 F 𝐶 = 50𝜇F = 50 × 10−6 F
Hence, 𝐿 = (1000)2
1
= 0.1 H = 100 mH For LC circuit the frequency ,
×10×10−6 1
𝑓 = 2𝜋
√𝐿𝐶
1
307 (b) or 𝑇 = 2𝜋√𝐿𝐶 (∵ 𝑇 = )
𝑓
𝐸 = 𝐸0 cos 𝜔𝑡 = 10 cos(2𝜋 × 𝑓𝑡) 𝑇
1
At time 𝑡 = 4, energy stored is completely
= 10 cos (2𝜋 × 50 × 600)
magnetic.
𝜋 √3 𝑇
= 10 cos ( 6 ) = 10 × 2 = 5√3 V The time, 𝑡 = 4
309 (d) 2𝜋√𝐿𝐶
𝑡= 4
As explained in solution (1) for frequency 0 −
2𝜋√20×10−3 ×50×10−6
𝑓𝑟 , 𝑍 decreases hence (𝑖 = 𝑉/𝑍) increases and for or 𝑡= 4
frequency 𝑓𝑟 − ∞, 𝑍 increases hence 𝑖 decreases 3.14√10−6
or 𝑡=
310 (a) 2
3.14×10−3
At resonance, 𝑉𝐿 and 𝑉𝐶 are equal in magnitude or 𝑡=
2
but have phase difference of 180° relative to each or 𝑡 = 1.57 × 10−3 s = 1.57 ms
other 317 (c)
∴ 𝑉𝐿𝐶 = 𝑉𝐿 − 𝑉𝐶 = 0 Ignoring mutual induction, resultant, inductance
Hence, voltmeter 𝑉2 read 0 volt 𝐿′ = 𝐿1 + 𝐿2
311 (b) =𝐿 + 𝐿 = 2𝐿
When 𝐶 is removed circuit becomes 𝑅𝐿 circuit 319 (a)
hence In an L – C circuit the impedance of circuit is
𝜋 𝑋𝐿
tan = … (i) 𝑍 = 𝑋𝐿 − 𝑋𝐶
3 𝑅 When𝑋𝐿 = 𝑋𝐶 , then Z = 0. In this situation the
When 𝐿 is removed circuit becomes 𝑅𝐶 circuit
amplitude of current in the circuit would be
hence
𝜋 𝑋𝐶 infinite. It will be condition of electrical resonance
tan = … (ii) and frequency is given by
3 𝑅
1
From equation (i) and (ii) we obtain 𝑋𝐿 = 𝑋𝐶 . 𝑓=
This is the condition of resonance and in 2𝜋√𝐿𝐶
1
resonance 𝑍 = 𝑅 = 100Ω =
2×3.14×√10×10−3 ×0.25×10−6
312 (c) = 3184.7 cycle s −1
𝑉𝐿 = 46 𝑣𝑜𝑙𝑡𝑠, 𝑉𝐶 = 40 𝑣𝑜𝑙𝑡𝑠, 𝑉𝑅 = 8 𝑣𝑜𝑙𝑡𝑠 Also frequency =
velocity
wavelength
E.M.F. of source 𝑉 = √82 + (46 − 40)2 = 10 𝑣𝑜𝑙𝑡𝑠 𝑐 3×108
313 (b) ⇒ 𝜆= =
𝑓 3184.7
1
Resonance frequency, 𝜔 = ⇒ 𝜆 = 9.42 × 104 m
√𝐿𝐶
1 320 (a)
= −3 −6
√8×10 ×20×10 As the inductors are in parallel, therefore, induced
1 104
= 4×10−4 = 4 emf across the two inductors is the same 𝑖𝑒,
= 2500 rad 𝑠 −1 𝑒1 = 𝑒2

P a g e | 59
 𝐿1 (
𝑑𝑖1
)
𝑑𝑖2
= 𝐿2 ( ) 327 (d)
𝑑𝑡 𝑑𝑡
A series resonance circuit admits maximum
Integrating both sides w.r.t. 𝑡, we get
current, as
𝐿1 𝑖1 = 𝐿2 𝑖2
𝑖1 𝐿2 𝑃 = 𝑖2𝑅
∴ = 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
Impedance, 1
𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2 𝑣 = 2𝜋
√𝐿𝐶
1
∴ 10 = √(102
+ (𝑋𝐿 − 𝑋𝐶 )2 =
2×3.14√25×10−3 ×400×10−6
2 1
⇒ 100 = 100 + (𝑋𝐿 − 𝑋𝐶 ) = = 50.3 Hz
2×3.14√10−5
⇒ 𝑋𝐿 − 𝑋𝐶 = 0
328 (a)
…(i)
Let 𝜙 is the phase difference between current and As initially charge is maximum,
𝑞 = 𝑞0 cos 𝜔𝑡
voltage
𝑋 −𝑋 𝑑𝑞
tan 𝜙 = 𝐿 𝐶 ⇒ 𝑖= = −𝜔𝑞0 sin 𝜔𝑡
𝑅 𝑑𝑡
0 1 2 𝑞2
∴ tan 𝜙 = Given 2
𝐿𝑖 = 2𝐶
𝑅
⇒ 𝜙=0 [From Eq.(i)] 1 2
(𝑞0 cos 𝜔𝑡)2
⇒ 𝐿 (𝜔𝑞0 sin 𝜔𝑡) =
322 (d) 2 2𝐶
1
In an L – C – R series AC circuit, the voltage across 𝜔 = 𝐿𝐶
√
inductor L leads the current by 90° and the ∴ tan 𝜔𝑡 = 1
voltage across capacitor C lags behind the current 𝜋
𝜔𝑡 = tan 4
by 90°. 𝜋 𝜋
𝑡 = 4𝜔 = 4 √𝐿𝐶
330 (b)
(1) For time interval 0 < 𝑡 < 𝑇/2
𝐼 = 𝑘𝑡, where 𝑘 is the slope
For inductor as we know, induced voltage 𝑉 =
𝑑𝑖
−𝐿
𝑑𝑡
⇒ 𝑉1 = −𝐾𝐿
𝑇
(2) For time interval 2 < 𝑡 < 𝑇
Hence, the voltage across 𝐿 − 𝐶 combination will 𝐼 = −𝐾𝑡 ⇒ 𝑉2 = 𝐾𝐿
be zero. 331 (c)
323 (d) Average power in AC circuits is given by 𝑃 =
𝜇0 𝑁 2 𝐴 𝜇0 𝑁 2 (𝜋𝑟 2 ) 𝑉rms 𝐼rms cos 𝜙 for pure capacitive circuit 𝜙 =
𝐿= = 90° so, 𝑃 = 0.
𝑙 𝑙
=
4𝜋×10−7 ×(500)2 ×𝜋(0.025)2 332 (d)
1 𝑉 𝑉
= 4 × 10 × 10 × 7 (500)2× (0.025)2 𝐼𝐿 = 𝑋 and 𝐼𝐶 = 𝑋
𝐿 𝐶
1
= 6.25 × 10−4H i.e, 𝐼𝐿 ∝ 𝜔
325 (c) and 𝐼𝐶 ∝ 𝜔
The power loss in AC circuit will be minimum, if ∴ With increase in 𝜔, 𝐼𝐿 decreases
resistance is low. In inductance power loss is zero. while 𝐼𝐶 increases.
It applies to high as well as low inductances. 333 (a)
326 (c) Geometric length of a magnet is 6/5 times its
Inductive reactance 𝑋𝐿 = 𝜔𝐿 magnetic length.
= 2𝜋𝑣𝐿 ∴ Geometric length =6/5×10=12 cm
= 2𝜋 × 50 × 1 334 (c)
= 100𝜋
P a g e | 60
 The current in the circuit 𝑉0 770√2
𝑉𝑅 𝑉𝑟𝑚𝑠 = = ≈ 500 V
𝑖= √2 2
𝑅
100 340 (c)
= =0.1 A
1000 In L – R circuit, impedance
At resonance,
𝑖 𝑍 = √𝑅 2 + 𝑋𝐿2
𝑉𝐿 = 𝑉𝐶 = 𝑖𝑋𝐶 =
𝜔𝐶 Here, 𝑋𝐿 = 𝜔𝐿 = 2𝜋 𝑓𝐿
0.1
= 200×2×10−6
= 250 V ∴ 𝑍 = √𝑅 2 + 4𝜋 2 𝑓 2 𝐿2
335 (d) 341 (a)
Given, the frequency of alternating current 𝑓 Given, 𝑉𝐶 = 3𝑉𝑅 = 3(𝑉 − 𝑉𝐶 )
= 50 Hz Here, V is the applied potential.
3
The time for alternating current to become its rms ∴ 𝑉𝐶 = 4 V
value from zero. 3
𝑇 Or 𝑉(1 − 𝑒 −𝑡/𝜏𝐶 ) = V
𝑡=4 4
1
or
1
𝑡 = 4𝑓 ∴ 𝑒 −𝑡/𝜏𝐶 = … (𝑖)
4
1 Here, 𝜏𝐶 = 𝐶𝑅 = 10s
or 𝑡 = 200 s
Substituting this value of 𝜏𝐶 in Eq.(i) and solving
or 𝑡 = 5 ms for t, we get
𝑡 = 1 3.8 6s
342 (d)
𝐹 12
𝑙1 = = =6𝐴
𝑅1 2
𝑑𝑙2
𝐸=𝐿 𝑑𝑡
+ 𝑅2 × 𝑙2
𝐼2 = 𝐼0 (1 − 𝑒 −𝑡/𝑡𝑐 )
𝐸 12
⇒ 𝐼0 = = = 6A
𝑅2 2
𝐿 400 × 10−3
𝑡𝑐 = = = 0.2
336 (c) 𝑅 2
The full cycle of alternating current consists of 𝐼2 = 6(1 − 𝑒 −𝑡/0.2 )
two half cycles. For one – half, current is positive Potential drop across
and for second half, current is negative. Therefore, 𝐿 = 𝐸 − 𝑅2 𝐿2 = 12 − 2 × 6(1 − 𝑒 −𝑏𝑡 ) = 12𝑒 −5𝑡
for an AC cycle, the net value of current average 343 (c)
out to zero. While the DC ammeter, read the Reactance of capacitor or capacitive reactance is
average value. Hence, the alternating current denoted by 𝑋𝐶 , given by
cannot DC measured by DC ammeter. 1
𝑋𝐶 = 𝜔𝐶
337 (a)
Given, 𝜔 = 50 rad s −1 , 𝐶 = 50𝜇F = 50 × 10−6 F
Current will be maximum at the condition of 1
1 ∴ 𝑋𝐶 =
resonance. So resonant frequency 𝜔0 = 𝐿𝐶 = 50 × 50 × 10−6
√
1 From Ohm’s law, current flowing through the
√0.5×8×10−6 circuit is given by
= 500 𝑟𝑎𝑑/𝑠 𝑉
𝐼rms = 𝑋rms
338 (c) 𝐶
𝑉rms 220
Input power 𝑃1 = 220 × 1.5 = 330 W = = −6
1/𝜔𝐶 1/50×50×10
2 3 2
Loss of power 𝑖 𝑅 = (2) × 20 = 45 W ⇒ 𝐼rms = 220 × 50 × 50 × 10−6
Output power, 𝑃0 = 330 − 45 = 285 W = 55 × 10−2 A= 0.55 A
𝑃0 285 344 (a)
∴ Peak emf induced, 𝑉0 = 𝑖
= 1.5
= 190 V
In an ideal choke, ratio of its inductance L to its
339 (c)
DC resistance R is infinity.
345 (a)

P a g e | 61
 Given, 𝑅 = 3Ω, 𝑋𝐿 = 15 Ω, 𝑋𝐶 = 11 Ω increase and hence, impedance of the circuit will
𝑉rms = 10 volt decrease. Thus, current and hence brightness of
∴ Current through the circuit the bulb increase.
𝑖= 2
𝑉rms 354 (d)
√𝑅 +(𝑋𝐿 −𝑋𝐶 )2
10
When resonance occurs emf E and current i are in
= phase. In this case, the impedance is minimum
√(3)2 +(15−11)2
10 10
= 9+16 = 5 = 2A and current is maximum. At resonance inductive
√ reactance is equal to capacitive reactance
Since, L, C and R are connected in series
𝑋𝐿 = 𝑋𝐶
combination then potential difference across R is
355 (c)
𝑉𝑅 = 𝑖 × 𝑅 = 2 × 3 = 6 V
The circuit element connected to the AC source
Across L, 𝑉𝐿 = 𝑖𝑋𝐿 = 2 × 15 = 30 V
will be pure resistor. In pure resistive AC circuit,
Across C, 𝑉𝐶 = 𝑖𝑋𝐶 = 2 × 11 = 22 V
voltage and current are in the same phase
So, potential difference across series combination
356 (b)
of L and C, 1
Frequency = 2𝜋 𝐿𝐶
= 𝑉𝐿 − 𝑉𝐶 √
= 30 − 22 = 8 V So the combination which represents dimension
1
346 (d) of frequency is = (𝐿𝐶)−1/2
√𝐿𝐶
At resonance net voltage across 𝐿 and 𝐶 is zero 357 (a)
347 (d) In 𝐿𝐶𝑅 series circuit, impedance 𝑍 of the circuit is
∵ 𝑃 = 𝑉𝑖 cos 𝜙 , ∴ 𝑃 ∝ cos 𝜙 given by
348 (b)
𝑑𝑖 𝑑
𝑍 = √(𝑅)2 + (𝑋𝐿 − 𝑋𝐶 )2 where 𝑋𝐿 = 𝜔𝐿, 𝑋𝐶 =
As 𝑒 = 𝑀 𝑑𝑡 = 𝑀 𝑑𝑡 (𝑖0 sin 𝜔𝑡) 1/𝜔𝐶
∴ 𝑒 = 𝑀𝑖0 cos 𝜔𝑡(𝜔) At resonance 𝑋𝐿 = 𝑋𝐶 ∴ 𝑍 = 𝑅
𝑒max = 𝑀𝑖0 × 1 × 𝜔 358 (c)
= 0.005 × 10 × 10𝜋 = 5𝜋 𝑋𝐿 = 2𝜋𝑓
349 (d) ⇒ 𝑋𝐿 ∝ 𝑓
The emf induced in a conductor does not depend 1 1
⇒ ∝
o its shape, but only on its end points, 𝑀 and 𝑄 in 𝑋𝐿 𝑓
this case. Thus the conductor is equivalent to an 1
𝑖. 𝑒., graph between 𝑋 and 𝑓 will be a hyperbola
𝐿
imaginary straight conductor of 𝑙 = 𝑀𝑄 = 2𝑅.
359 (d)
Therefore, potential difference developed across
For current to be maximum , 𝑋𝐿 = 𝑋𝐶
the ring =𝐵𝑙𝑣 = 𝐵(2𝑅)𝑣. And the direction of
Hence, resonant frequency
induced current is from 𝑄 to 𝑀. Therefore, 𝑄 is at
1 1 1 103
higher potential. 𝑓= = = =
2𝜋 √𝐿𝐶 √0.5 × 8 × 10−6 4𝜋
350 (b)
𝑒 𝑒𝑑𝑡 5×10−3 But angular frequency
𝐿 = 𝑑𝑖/𝑑𝑡 = 𝑑𝑖 = (3−2) H = 5 mH 𝜔 = 2𝜋𝑓
2𝜋×1000
351 (a) 𝜔= = 500 Hz
4𝜋
As 𝑀 ∝ 𝑁1 𝑁2 , therefore, M remains the same. 360 (b)
352 (b) The average power consumed in an AC circuit is
For purely capacitive circuit 𝑒 = 𝑒0 sin 𝜔𝑡 given by
𝜋
𝑖 = 𝑖0 sin (𝜔𝑡 + 2 ) , 𝑖. 𝑒., current is ahead of emf by 𝑃=
𝑉0 𝐼0
cos 𝜙
𝜋 2
2 Where 𝜙 is phase angle and 𝑉0 𝑎𝑛𝑑 𝐼0 the peak
353 (d) value of voltage and current.
𝑉 𝜋
Current, 𝑖 = Given, 𝑉0 = 200𝑉, 𝐼0 = 2 𝐴, 𝜙 = 3 .
√𝑅2 +𝑋𝐶2 200×2 𝜋
𝑃= 2
cos 3
If a dielectric is introduced into the gap between 200×2 1
the plates of capacitor. Its capacitance will = 2 ×2= 100 W

P a g e | 62
361 (d) 367 (a)
The given circuit is under resonance as 𝑋𝐿 = 𝑋/𝐶 For purely 𝐿-circuit 𝑃 = 0
Hence, power dissipated in the circuit is 369 (a)
𝑉2 𝑅 = 6 + 4 = 10Ω
𝑃= 𝑅
= 242 W
𝑋𝐿 = 𝜔𝐿 = 2000 × 5 × 10−3 = 10Ω
362 (c)
1 1
𝑍 = √𝑅 2 + (2𝜋𝑣𝐿)2 𝑋𝐶 = = = 10Ω
𝜔𝐶 2000 × 50 × 10−6
= √(40)2 + 4𝜋 2 × (50)2 × (95.5 × 10−3 )2 ∴ 𝑍 = √𝑅 2 + (𝑋𝐿 − 𝑋𝐶 )2 = 10Ω
= 50 𝑜ℎ𝑚 𝑉0 20
Amplitude of current = 𝑖0 = 𝑍
= 10 = 2𝐴
363 (c)
370 (c)
90
𝑖𝐿 = = 3𝐴 Impedance, 𝑍 = √(𝑋𝐿 ~ 𝑋𝐶 )2 + 𝑅 2
30
90
𝑖𝐶 = = 4.5 𝐴
20
Net current through circuit 𝑖 = 𝑖𝐶 − 𝑖𝐿 = 1.5 𝐴
𝑉 90
∴𝑍= = = 60Ω
𝑖 1.5
364 (c)
1 2
∴ 𝜙 = 𝑀𝑖 𝑍 = √(𝜔𝐿~ ) + 𝑅2
𝜙 0.4 𝜔𝐶
∴ 𝑀= 𝑖
= 2
= 0.2 H
Inductive reaction
365 (c)
𝑋𝐿 = 𝜔𝐿 = 70 × 103 × 100 × 10−6 = 7Ω
𝑉 120
𝑖= = = 0.455 𝐴 Capacitance reactance
𝑋𝐿 2 × 3.14 × 60 × 0.7 1 1
366 (d) 𝑋𝐶 = =
𝜔𝐶 70 × 10 × 1 × 10−6
3
𝑋𝐿 2𝜋𝑣𝐿 2𝜋 × 50 × 𝐿 100
tan 𝜙 = = ⇒ tan 30° = = [𝑋𝐶 > 𝑋𝐿 ]
𝑅 𝑅 𝜋√3 7
= 0.01 𝐻 Hence, circuit behaves like an 𝑅-𝐶 circuit

P a g e | 63
Session: 2023-24 Total Questions: 395

JEE/NEET PHYSICS
7.ALTERNATING CURRENT

Assertion - Reasoning Type

This section contain(s) 0 questions numbered 1 to 0. Each question contains STATEMENT 1(Assertion) and
STATEMENT 2(Reason). Each question has the 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

a) Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1

b) Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1

c) Statement 1 is True, Statement 2 is False

d) Statement 1 is False, Statement 2 is True

Statement 1: In a series R-L-C circuit the voltages across resistor, inductor and capacitor are 8 V, 16 V
and 10 V respectively. The resultant emf in the circuit is 10 V.
Statement 2: Resultant emf of the circuit is given by the relation. 𝐸 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2

Statement 1: The capacitor blocks DC but allows AC.

Statement 2: The capacitor offers infinite resistance to DC.

Statement 1: The alternating current lags behind the 𝑒.m.f. by a phase angle of 𝜋/2, when ac flows
through an inductor
Statement 2: The inductive reactance increases as the frequency of ac source decreases

Statement 1: In series 𝐿 − 𝐶 − 𝑅 circuit resonance can take place

Statement 2: Reasonance takes place if inductive and capacitive reactance are equal and opposite.

Statement 1: The mutual inductance of two coils is doubled if the self-inductance of the primary or
secondary coil is doubled.
Statement 2: Mutual inductance is proportional to the self-inductance of primary and secondary coils.

Statement 1: The average value of alternating emf is 63.69% of the peak value.

P a g e | 64
 Statement 2: The rms value of alternating emf is 70.72% of peak value.

Statement 1: An inductance and a resistance are connected in series with an ac circuit. In this circuit
the current and the potential difference across the resistance lags behind potential
difference across the inductance by an angle 𝜋/2
Statement 2: In 𝐿𝑅 circuit voltage leads the current by phase angle which depends on the value of
inductance and resistance both
8

Statement 1: Making or breaking of current in a coil produces no momentary current in the

neighbouring coil of another circuit
Statement 2: At the time of making or breaking of current changes.

Statement 1: Inductance coil are made of copper.

Statement 2: Induced current is more in wire having less resistance

10

Statement 1: The divisions are equally marked on the scale of ac ammeter

Statement 2: Heat produced is directly proportional to the current

11

Statement 1: Two identical heaters are connected to two different sources one DC and other AC having
same potential difference across their terminals. The heat produced in heater supplied
with AC source is greater.
Statement 2: The et impedance of a AC source is greater than resistance.

12

Statement 1: When capacitive reactance is smaller than the inductive reactance in 𝐿𝐶𝑅 circuit, 𝑒.m.f.
leads the current
Statement 2: The phase angle is the angle between the alternating 𝑒.m.f. and alternating current of the
circuit
13

Statement 1: If the frequency of alternating current in an ac circuit consisting of an inductance coil is

increased then current gets decreased
Statement 2: The current is inversely proportional to frequency of alternating current

14

Statement 1: If a variable frequency AC source is connected to a capacitor, the displacement current in

it increases with increase in frequency.
Statement 2: The increase in frequency results in an increase of impedance.

15

Statement 1: A capacitor of suitable capacitance can be used in an ac circuit in place of the choke coil

P a g e | 65
 Statement 2: A capacitor blocks dc and allows ac only

16

Statement 1: Average value of ac over a complete cycle is always zero

Statement 2: Average value of ac is always defined over half cycle

17

Statement 1: Ac is more dangerous than dc

Statement 2: Frequency of ac is dangerous for human body

18

Statement 1: In a series 𝑅 − 𝐿 − 𝐶 circuit the voltage across resistor, inductor and capacitor are 8V,
16V and 10V respectively. The resultant emf in the circuit is 10.
Statement 2: Resultant emf of the circuit is given by the relation 𝐸 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2

19

Statement 1: Capacitor serves as a block for dc and offers an easy path to ac

Statement 2: Capacitive reactance is inversely proportional to frequency

20

Statement 1: The armature current in DC motor maximum when the motor has just started.

Statement 2: Armature current is given by 𝑖 = 𝐸−𝑒

Where 𝑒 = the back emf and 𝑅𝑎 = resistance of
𝑅𝑎
armature.
21

Statement 1: The energy stored in the inductor of 2 H, when a current of 10 A flows through it is 100 J.

Statement 2: Energy stored in an inductor is directly proportional to its inductance.

22

Statement 1: In a series 𝑅 − 𝐿 − 𝐶 circuit the voltage across resistor, inductor and capacitor are 8 V, 16
V and 10 V respectively. The resultant emf the circuit is 10 V
Statement 2: Resultant emf of the circuit is given by the relation 𝐸 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2

23

Statement 1: For an electric lamp connected in series with a variable capacitor and ac source, its
brightness increases with increase in capacitance
Statement 2: Capacitive reactance decreases with increase in capacitance of capacitor

24

Statement 1: A sinusoidal AC current flows through a resistance 𝑅. If the peak current is 𝐼0 , then the
𝑅𝐼02
power dissipated is 2
.
Statement 2: For purely resistive circuit, power factor cos ϕ = 1.

P a g e | 66
Session: 2023-24 Total Questions: 395

JEE/NEET PHYSICS
7.ALTERNATING CURRENT

: ANSWER KEY :
1) a 2) a 3) c 4) a 17) a 18) a 19) a 20) b
5) c 6) b 7) b 8) d 21) b 22) a 23) a 24) b
9) b 10) d 11) a 12) b
13) a 14) c 15) b 16) b

P a g e | 67
Session: 2023-24 Total Questions: 395

JEE/NEET PHYSICS
7.ALTERNATING CURRENT

: HINTS AND SOLUTIONS :

1 (a) In other words we can say that at resonant
The resultant emf in the L – C – R circuit is given frequency (𝑓0 = 2𝜋
1
) resonance can take place.
√𝐿𝐶
by

𝐸 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2 5 (c)

It two coil of inductance 𝐿1 and 𝐿2 are joined
together, then their mutual inductance is given by
𝐸 = √(8)2 + (16 − 10)2
𝐸 = √64 + 36 ⇒ 𝐸 = 10V 𝑀 = 𝑘√𝐿1 𝐿2

2 (a) It is clear from the relation, if self-inductance of

From the formula the capacitive reactance is primary and secondary coil are doubled the
mutual inductance of the coils will be doubled.
1 1
𝑋𝐶 = =
𝜔𝐶 2𝜋 𝑓𝐶 6 (b)
The average value of alternating emf 𝐸 =
Where 𝜔 represents angular frequency, 𝐶 for
𝐸0 sin 𝜔𝑡 over a positive half cycle is
capacitance and 𝑓 represent the frequency.
2𝐸0
The frequency of DC is zero hence capacitive 𝐸av half cycle = = 0.6369
𝜋
reactance becomes infinity. When the same
capacitor is connected with AC, capacitive Percentage of 𝐸av = 0.6369 × 100 = 63.69%
reactance is not equal to infinity hence capacitor
allows AC to pass through it, and offers infinite Also rms value of emf is given as
resistance to DC. 𝐸0
𝐸rms = = 0.7072
3 (c) √2
When ac flows through an inductor, current lags Percentage 𝐸rms = 0.7072 × 100 = 70.72%
behind the 𝑒𝑚𝑓 by phase of 𝜋/2. Inductive
reactance, 𝑋𝐿 = 𝜔𝐿 = 𝜋. 2𝑓. 𝐿, so when frequency 7 (b)
increases correspondingly inductive reactance As both the inductive and resistance are joined in
also increases series, hence current through both will be same.
4 (a) But in case of resistance, both the current and
Resonance in 𝐿 − 𝐶 − 𝑅 series circuit takes place potential vary simultaneously, hence they are in
when inductive reactance and capacitive same phase. In case of an inductance when
reactance are equal and opposite 𝑖𝑒, current is zero, potential difference across it is
maximum and when current reaches maximum
𝑋𝐿 = 𝑋𝐶 (at 𝜔 𝑡 = 𝜋/2), potential difference across it
becomes zero, 𝑖. 𝑒., potential difference leads the
Or 𝜔0 𝐿 = 𝜔0 𝐶
current by 𝜋/2 or current lags behind the
1 1 potential difference by 𝜋/2. Phase angle in case of
Or 𝜔0 = or 𝑓0 =
√𝐶 2𝜋√𝐿𝐶 𝜔𝐿
𝐿𝑅 circuit is given as 𝜙 = tan−1 ( 𝑅 )

P a g e | 68
 VL inductive coil increases. Current (𝑋𝐿 − 𝜔𝐿 =
2𝜋𝑓𝐿) in the circuit containing inductor is given
V 𝑉 𝑉
by 𝐼 = 𝑋 = 2𝜋𝑓𝐿. As inductive resistance of the
𝐿

 inductor increases, current in the circuit

I
VR decreases
8 (d) 14 (c)
Before making current in a coil, the current is zero The impedance of a capacitor is given by
and before breaking the current is maximum. In 1
𝑋𝐶 =
other words, it is constant in both the cases. 2𝜋𝑓𝐶
Obviously, on making or breaking the current in a 1
circuit, the current starts changing. The changing ⇒ 𝑋𝐶 ∝ 𝑓
current produces changing magnetic field, which
in turn produces induced current in the Thus, the impedance of the capacitor decreases
neighbouring coil of the circuit. with increase in the frequency 𝑓 of a source.

9 (b) Current now increase in the capacitor with

Since, copper consists of a very small ohmic decrease of impedance or resistance. This is equal
resistance so, inductance coils are made of to displacement current between the plates of
copper. A large induced current is produced in capacitor.
such an inductance due to change in flux, which
15 (b)
offers a pretty opposition to the flow of current.
We can use a capacitor of suitable capacitance as a
10 (d) choke coil because average power consumed per
An ac ammeter is constructed on the basis of cycle in an ideal capacitor is zero. Therefore, like a
heating effect of the electric current. Since heat choke coil, a condenser can reduce ac without
produced varies as square of current (𝐻 = 𝐼 2 𝑅), power dissipation
therefore the division marked on the scale of ac 16 (b)
ammeter are not equally spaced The mean average value of alternating current (or
11 (a) 𝑒mf) during half cycle is given by 𝐼𝑚 = 0.636 𝐼0
For the case of DC, the frequency is zero and the (or 𝐸𝑚 = 0.636𝐸0 )
net impedance is equal to the resistance. During the next half cycle, the mean value of ac
will be equal in magnitude but opposite in
For the case of AC, the impedance of the AC circuit direction
is given by For this reason the average value of ac over a
complete cycle is always zero. So the average
𝑍 = √𝑅 2 + 𝜔 2 𝐿2 value is always defined over a half cycle of ac
17 (a)
Where 𝑅 = resistance, 𝜔 = angular frequency and
It is because for a given value of voltage, peak
𝐿 = inductance.
value of ac voltage is greater than the dc voltage
12 (b) For 𝑒. 𝑔., In case of 220 𝑉 ac, 𝑉0 = 220√2 = 311 𝑉
The phase angle for the 𝐿𝐶𝑅 circuit is given by ∴ 311 𝑉 will cause more harm to the human body
𝑋𝐿 − 𝑋𝐶 𝜔𝐿 − 1/𝜔𝐶 than 220 𝑉 dc
tan 𝜙 = =
𝑅 𝑅 18 (a)
Where 𝑋𝐿 , 𝑋𝐶 are inductive reactance and The resultant emf in the 𝐿 − 𝐶 − 𝑅 circuit is given
capacitive reactance respectively. When 𝑋𝐿 > 𝑋𝐶 by
then tan 𝜙 is positive, 𝑖. 𝑒. , 𝜙 is positive (between
0 and 𝜋/2). Hence 𝑒mf leads the current 𝐸 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2
13 (a)
When frequency of alternating current is
increasing, the effective resistance of the 𝐸 = √(8)2 + (16 − 10)2

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 = √64 + 36 𝐸 = √𝑉𝑅2 + (𝑉𝐿 − 𝑉𝐶 )2

𝐸 = 10 V 𝐸 = √(8)2 + (16 − 10)2

19 (a)
= √64 + 36
The capacitive reactance of capacitor is given by
1 1 𝐸 = 10 V
𝑋𝐶 = =
𝜔𝐶 2𝜋𝑓𝐶
So this is infinite for dc (𝑓 = 0) and has a very 23 (a)
1
small value of ac. Therefore a capacitor blocks dc Capacitive reactance 𝑋𝐶 = . When capacitance
𝜔𝐶
20 (b) (𝐶) increases, the capacitive reactance decreases.
From the relation Due to decrease in its values, the current in the
𝐸−𝑒
𝑖= 𝑅𝑎 circuit will increase (𝐼 =
𝐸
) and hence
√𝑅2 +𝑋𝐶2
When the motor is started 𝑒 = 0 brightness of source (or electric lamp) will also
𝐸 increase
Hence, Eq (i) becomes 𝑖star𝑡 = 𝑅
𝑎 24 (b)
Power dissipated is given by
It is obvious that current is maximum when the
motor has just started. 𝑃 = 𝐸rms 𝐼rms cos ϕ

21 (b) We know that for purely resistance circuit power

The energy stored in the inductor is given by 𝑅 𝑅
factor cos ϕ = 𝑍 = 𝑅 = 1
1 1
𝑉= 𝐿𝑖02 = × 2 × (10)2 = 100 J Hence, 𝑃 = 𝐸rms × 𝐼rms
2 2

It is obvious that energy stored is directly = (𝑅𝐼rms ) × 𝐼rms

proportional to its inductance.
= 𝑅 (𝐼rms )2
22 (a)
2
The resultant emf in the L-C-R circuit is given by 𝐼0 𝑅𝐼02
= 𝑅( ) =
√2 2

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Session: 2023-24 Total Questions: 395

JEE/NEET PHYSICS
7.ALTERNATING CURRENT

Matrix-Match Type

This section contain(s) 0 question(s). Each question contains Statements given in 2 columns which have to be
matched. Statements (A, B, C, D) in columns I have to be matched with Statements (p, q, r, s) in columns II.

1. You are given many resistances, capacitors and inductors. These are connected to a variable DC voltage
source (the first two circuits) or an AC voltage source of 50 𝐻𝑧 frequency (the next three circuits) in
different ways as shown in Column-II. When a current I (steady state for DC or 𝑟𝑚𝑠 for AC) flows through
the circuit, the corresponding voltage 𝑉1 and 𝑉2 (indicated in circuits) are related as shown in Column-I.
Match the two
Column-I Column- II

(A) 𝐼 ≠ 0, 𝑉1 is proportional to 𝐼 (p)

(B) 𝐼 ≠ 0, 𝑉2 > 𝑉1 (q)

(C) 𝑉1 = 0, 𝑉2 = 𝑉 (r)

(D) 𝐼 ≠ 0, 𝑉2 is proportional to 𝐼 (s)

(t)

CODES :

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 A B C D

a) r,s,t q,r,s,t p,q q,r,s,t

b) q,r,s,t p,q r,s,t q,r

c) p,q r,s,t q,r,s,t s,t

d) r,s,t p,q q,r,s,t q,r,s,t

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Session: 2023-24 Total Questions: 395

JEE/NEET PHYSICS
7.ALTERNATING CURRENT

: ANSWER KEY :

1) a

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Session: 2023-24 Total Questions: 395

JEE/NEET PHYSICS
7.ALTERNATING CURRENT

: HINTS AND SOLUTIONS :

1 (a) So, of 𝑟 ⇒ A, B, D

As per given conditions, there will be no steady

state in circuit ‘p’, so it should not be considered
in options of ‘C’

(s)
Here, 𝑉1 = 𝐼𝑋𝐿 , where 𝑋𝐿 = 6𝜋 × 10−1 Ω
Also, 𝑉2 = 𝐼𝑋𝐶 , where
104
(p) 𝑋𝐶 =
As 𝐼 is steady state current 3𝜋
So, 𝑉2 > 𝑉1
𝑉1 = 0; 𝐼 = 0
𝑉1 ∝ 𝐼
Hence, 𝑉2 = 𝑉
𝑉2 ∝ 𝐼
So, of 𝑝 ⇒ 𝐶
So, of s ⇒ A, B, D

(q) (t)
In the steady state; Here, 𝑉1 = 𝐼𝑅, where
𝑑𝐼
𝑉1 = 0 as 𝑑𝑡 = 0 𝑅 = 1000Ω,
∴ 𝑉2 = 𝑉 = 𝐼𝑅 or 104
𝑋𝐶 = Ω
𝑉2 ∝ 𝐼 3𝜋
and 𝑉2 > 𝑉1 𝑉2 = 𝐼𝑋𝐶 ,
So, of 𝑞 ⇒ 𝐵, 𝐶, 𝐷 So, 𝑉2 > 𝑉1 and 𝑉1 ∝ 𝐼; 𝑉2 ∝ 𝐼
So, of t ⇒ A, B, D
Note: for circuit ‘p’
𝐿𝑑𝑖 𝑞 𝑑𝑖 𝑑2 𝑖
𝑉− 𝑑𝑡
− 𝐶 = 0 or 𝐶𝑉 = 𝐶𝐿 𝑑𝑡 + 𝑞 or 0 = 𝐿𝐶 𝑑𝑡 2 +
𝑑𝑞
𝑑𝑡
(r) 𝑑2 𝑖 1 𝑑𝑞
or =−
𝑑𝑡 2 𝐿𝐶 𝑑𝑡
Inductive reactance 1
𝑋𝐿 = 𝜔𝐿 So, 𝑖 = 𝑖0 sin ( 𝐿𝐶 𝑡 + 𝜙0 )
√
𝑋𝐿 = 6𝜋 × 10−1 Ω and resistance As per given conditions, there will be no steady
= 𝑅 = 2Ω state in circuit ‘p’. So it should not be considered
So, 𝑉1 = 𝐼𝑋𝐿 and 𝑉2 = 𝐼𝑅 in options of ‘C’
Hence, 𝑉2 > 𝑉1 ∴ (A) → r,s,t; (B) → q,r,s,t; (C) → p,q; (D) → q,r,s,t

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