JEE Physics Alternating Current MCQs Set 15

Physics Objective Questions and Answers: Alternating Current

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Multiple Choice Questions

Question 1: When 100 V dc is applied across a coil, a current of 1 A flows through it. When 100 V ac of 50 Hz is applied to the same coil, only 0.5 A flows. The resistance and inductance of the coil are
(a) \( 10\,\Omega,\ \dfrac{\sqrt3}{\pi} \) H
(b) \( 100\,\Omega,\ \dfrac{2\sqrt3}{\pi} \) H
(c) \( 100\,\Omega,\ \dfrac{\sqrt3}{\pi} \) H
(d) \( 100\,\Omega,\ \dfrac{3\sqrt3}{\pi} \) H
Show Answer & Explanation

Answer: (c) \( 100\,\Omega,\ \dfrac{\sqrt3}{\pi} \) H

Explanation:
1. Resistance from dc: \( R=100/1=100 \) Ω.
2. Impedance with ac: \( Z=100/0.5=200 \) Ω, so \( X_L=\sqrt{200^2-100^2}=100\sqrt3 \) Ω.
3. Inductance \( L=\dfrac{X_L}{2\pi f}=\dfrac{100\sqrt3}{100\pi}=\dfrac{\sqrt3}{\pi} \) H.

Teacher's Note:
1. DC gives only the resistance.
2. AC gives the impedance.

Question 2: In a series circuit, the applied voltage and current at any instant are \( E=141.4\cos(5000t-10^\circ) \) and \( I=5\cos(5000t-370^\circ) \). The inductance is 0.01 H. The capacitance is
(a) 6 μF
(b) \( 4\ \mu\text F \)
(c) 1 μF
(d) 5 μF
Show Answer & Explanation

Answer: (b) \( 4\ \mu\text F \)

Explanation:
1. The phases differ by \( 360^\circ \), i.e. they are equal: current and voltage are in phase, so the circuit is at resonance.
2. Then \( \omega^2LC=1 \): \( C=\dfrac{1}{(5000)^2\times0.01}=4\times10^{-6} \) F.

Teacher's Note:
1. Same phase means X_L = X_C.
2. Use ω = 5000 rad/s.

Question 3: The power factor of an L-R series circuit is 0.6 and that of a C-R series circuit is 0.5. If the elements (L, C and R) of the two circuits are joined in series, the power factor of the combination is 1. The ratio of the resistance in the L-R circuit to the resistance in the C-R circuit is
(a) \( \dfrac56 \)
(b) \( \dfrac65 \)
(c) \( \dfrac{4}{3\sqrt3} \)
(d) \( \dfrac{3\sqrt3}{4} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{3\sqrt3}{4} \)

Explanation:
1. L-R: \( \cos\phi=0.6\Rightarrow X_L/R_1=\tan\phi=4/3 \). C-R: \( \cos\phi=0.5\Rightarrow X_C/R_2=\sqrt3 \).
2. Power factor 1 in series needs \( X_L=X_C \): \( \dfrac43R_1=\sqrt3R_2 \).
3. So \( \dfrac{R_1}{R_2}=\dfrac{3\sqrt3}{4} \).

Teacher's Note:
1. Unity power factor means resonance.
2. Find each reactance in terms of its resistance.

Question 4: The potential difference and current in an ac circuit are \( V=5\cos\omega t \) volt and \( i=2\sin\omega t \) ampere. The average power dissipated in the circuit is
(a) 10 W
(b) 5 W
(c) 0 W
(d) 2.5 W
Show Answer & Explanation

Answer: (c) 0 W

Explanation:
1. The current and voltage differ in phase by 90° (sine and cosine).
2. Average power \( =V_{rms}I_{rms}\cos90^\circ=0 \).

Teacher's Note:
1. Zero phase-lag cosine gives full power; 90° gives zero.

Question 5: The power factor of a circuit is \( 1/\sqrt2 \). The circuit is driven by \( 2\sin100t \) volts and contains a 10 Ω resistor, a 0.1 H inductor and a capacitor in series. The capacitance is
(a) \( 500\ \mu\text F \)
(b) 300 μF
(c) 200 μF
(d) 400 μF
Show Answer & Explanation

Answer: (a) \( 500\ \mu\text F \)

Explanation:
1. \( X_L=100\times0.1=10 \) Ω. Power factor \( 1/\sqrt2 \) gives \( |X_L-X_C|=R=10 \).
2. So \( X_C=20 \) Ω (0 Ω is not possible), \( C=\dfrac{1}{100\times20}=500\ \mu F \).

Teacher's Note:
1. Equal reactance difference and resistance gives 45°.
2. A capacitor needs a positive reactance.

Question 6: An LCR series circuit with 100 Ω resistance is connected to a 200 V ac source of angular frequency 300 rad/s. When only the capacitance is removed, the current lags the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. The current and the power dissipated in the LCR circuit are
(a) 2 A, 300 W
(b) 3 A, 100 W
(c) 1 A, 400 W
(d) 2 A, 400 W
Show Answer & Explanation

Answer: (d) 2 A, 400 W

Explanation:
1. Without C: \( \tan60^\circ=X_L/R \), so \( X_L=100\sqrt3 \). Without L: \( X_C=100\sqrt3 \).
2. So \( X_L=X_C \) and the circuit is at resonance: \( I=200/100=2 \) A.
3. Power \( =I^2R=4\times100=400 \) W.

Teacher's Note:
1. Equal angles mean equal reactances.
2. At resonance Z = R.

Question 7: A current of 4 A flows in a coil connected to a 12 V dc source. When the same coil is connected to a 12 V, 50 rad/s ac source, a current of 2.4 A flows. The inductance of the coil, and the power developed if a 2500 μF capacitor is connected in series with the coil, are
(a) 0.5 H, 17 W
(b) 2 H, 5.9 W
(c) 0.08 H, 17.28 W
(d) 1 H, 5.28 W
Show Answer & Explanation

Answer: (c) 0.08 H, 17.28 W

Explanation:
1. Resistance \( R=12/4=3 \) Ω; impedance \( Z=12/2.4=5 \) Ω so \( X_L=4 \) Ω and \( L=4/50=0.08 \) H.
2. With the capacitor: \( X_C=\dfrac{1}{50\times2500\times10^{-6}}=8 \) Ω; \( Z=\sqrt{3^2+(4-8)^2}=5 \) Ω; \( I=2.4 \) A.
3. Power \( =I^2R=2.4^2\times3=17.28 \) W.

Teacher's Note:
1. DC gives R; AC gives Z.
2. Power is I²R.

Question 8: An alternating voltage \( V=200\sin(100\pi t) \) volt is applied. Which set of statements is correct? (1) The peak value is 200 V. (2) The rms value is 220 V. (3) The rms value is \( 100\sqrt2 \) V. (4) The frequency is 50 Hz.
(a) (2) and (3) only
(b) (1), (3) and (4) only
(c) (1) and (4) only
(d) (1), (2) and (4) only
Show Answer & Explanation

Answer: (b) (1), (3) and (4) only

Explanation:
1. Peak = 200 V (1 true).
2. rms = \( 200/\sqrt2=100\sqrt2\approx141 \) V (3 true, 2 false).
3. Frequency \( f=\omega/2\pi=50 \) Hz (4 true).

Teacher's Note:
1. Read V₀ and ω from the equation.
2. rms = peak/√2.

Question 9: A 50 Ω electric heater is connected to a 100 V (rms), 60 Hz ac sinusoidal supply. Which of the statements are correct? (1) The peak value of the voltage is 100 V. (2) The peak value of the current is \( 2\sqrt2 \) A. (3) The rms value of the voltage is 100 V. (4) The rms value of the current is 2 A.
(a) (1), (2) and (3) only
(b) (1) and (4) only
(c) (3) and (4) only
(d) (2), (3) and (4) only
Show Answer & Explanation

Answer: (d) (2), (3) and (4) only

Explanation:
1. The quoted 100 V is the rms value, so (3) is true and the peak is \( 100\sqrt2\ne100 \) V, making (1) false.
2. Rms current \( =100/50=2 \) A (4 true); peak current \( 2\sqrt2 \) A (2 true).

Teacher's Note:
1. Supply ratings are rms values.
2. Peak = √2 × rms.

Question 10: In a certain series LCR circuit connected to an ac source, \( X_L>X_C \). Which of the following will increase the current in the circuit? (1) slightly increasing the capacitance; (2) slightly increasing the inductance; (3) slightly decreasing the capacitance; (4) slightly decreasing the inductance.
(a) (1) and (4) only
(b) (3) and (4) only
(c) (1) and (2) only
(d) (2) and (3) only
Show Answer & Explanation

Answer: (b) (3) and (4) only

Explanation:
1. The net reactance is \( X_L-X_C>0 \); current increases if it is reduced.
2. Decreasing \( C \) increases \( X_C=1/\omega C \) and decreasing \( L \) decreases \( X_L \): both reduce the net reactance.

Teacher's Note:
1. Move towards resonance.
2. XL > XC means the circuit is inductive.

Question 11: In an LCR circuit with \( C=\dfrac1\pi\ \mu\text F \) and \( L=\dfrac1\pi \) H connected to a supply of constant rms voltage \( V \) and variable frequency, at resonance which statements are correct? (1) The current is \( V/R \). (2) The resonance frequency is 500 Hz. (3) The voltages across the capacitor and the inductor are 180° out of phase. (4) The impedance is \( \sqrt{R^2+(2/\pi)^2} \).
(a) (1), (2) and (3) only
(b) (1) and (2) only
(c) (1) and (4) only
(d) (2), (3) and (4) only
Show Answer & Explanation

Answer: (a) (1), (2) and (3) only

Explanation:
1. \( f_0=\dfrac1{2\pi\sqrt{LC}}=\dfrac{1}{2\pi\sqrt{10^{-6}/\pi^2}}=500 \) Hz (2 true).
2. At resonance \( X_L=X_C \), so \( Z=R \) and \( I=V/R \) (1 true, 4 false).
3. The capacitor and inductor voltages are always in antiphase (3 true).

Teacher's Note:
1. At resonance Z is minimum and equals R.
2. VL and VC are 180° apart at all frequencies.

Question 12: An alternating current is given by \( I=I_0+I_1\sin\omega t \). Its rms value is
(a) \( \sqrt{\dfrac{I_0^2+I_1^2}{2}} \)
(b) \( \sqrt{I_0^2+\dfrac{I_1^2}{2}} \)
(c) \( \sqrt{I_0^2+I_1^2} \)
(d) \( I_0+\dfrac{I_1}{\sqrt2} \)
Show Answer & Explanation

Answer: (b) \( \sqrt{I_0^2+\dfrac{I_1^2}{2}} \)

Explanation:
1. \( \langle I^2\rangle=I_0^2+2I_0I_1\langle\sin\omega t\rangle+I_1^2\langle\sin^2\omega t\rangle=I_0^2+\dfrac{I_1^2}2 \).
2. Take the square root.

Teacher's Note:
1. Average the square, not the current.
2. The cross term averages to zero.

Question 13: An alternating current is given by \( i=i_1\cos\omega t+i_2\sin\omega t \). Its rms value is
(a) \( \dfrac{i_1^2+i_2^2}{2} \)
(b) \( \sqrt{i_1^2+i_2^2} \)
(c) \( \dfrac{i_1+i_2}{\sqrt2} \)
(d) \( \sqrt{\dfrac{i_1^2+i_2^2}{2}} \)
Show Answer & Explanation

Answer: (d) \( \sqrt{\dfrac{i_1^2+i_2^2}{2}} \)

Explanation:
1. \( i=\sqrt{i_1^2+i_2^2}\,\sin(\omega t+\varphi) \), so the peak is \( \sqrt{i_1^2+i_2^2} \).
2. Rms \( =\text{peak}/\sqrt2=\sqrt{\dfrac{i_1^2+i_2^2}{2}} \).

Teacher's Note:
1. Combine sine and cosine into one sinusoid.
2. rms = peak/√2.

Question 14: A 50 W, 100 V lamp is to be connected to an ac mains of 200 V, 50 Hz. The capacitance that must be put in series with the lamp is nearly
(a) \( 9.2\ \mu\text F \)
(b) \( 92\ \mu\text F \)
(c) \( 18.4\ \mu\text F \)
(d) \( 4.6\ \mu\text F \)
Show Answer & Explanation

Answer: (a) \( 9.2\ \mu\text F \)

Explanation:
1. Lamp: \( I=P/V=0.5 \) A, \( R=V^2/P=200 \) Ω.
2. Needed impedance \( Z=200/0.5=400 \) Ω, so \( X_C=\sqrt{400^2-200^2}=346.4 \) Ω.
3. Capacitance \( C=\dfrac{1}{2\pi\times50\times346.4}\approx9.2\times10^{-6} \) F.

Teacher's Note:
1. The lamp must still carry its rated current.
2. A capacitor drops voltage without dissipating power.

Question 15: An inductance should be connected in series with a capacitor of 5 μF, a resistance of 10 Ω and an ac source of 50 Hz so that the power factor of the circuit is unity. The inductance is nearly
(a) 4.05 H
(b) 2.03 H
(c) 0.2 H
(d) 1.01 H
Show Answer & Explanation

Answer: (b) 2.03 H

Explanation:
1. Unity power factor means resonance: \( \omega^2LC=1 \).
2. \( L=\dfrac{1}{(100\pi)^2\times5\times10^{-6}}\approx2.03 \) H.

Teacher's Note:
1. The resistance does not affect the resonance condition.
2. ω = 2πf = 100π rad/s.

Question 16: A circuit takes a current of 3 A at a power factor of 0.6 lagging when connected to a 115 V, 50 Hz supply. Another circuit takes a current of 5 A at a power factor of 0.07 leading from the same supply. If the two circuits are connected in series across a 230 V, 50 Hz supply, the current, power consumed and power factor are nearly
(a) 8.9 A, 1.0 kW, 0.07
(b) 8.9 A, 1.96 kW, 0.95
(c) 5 A, 1.15 kW, 0.5
(d) 3 A, 0.5 kW, 0.6
Show Answer & Explanation

Answer: (b) 8.9 A, 1.96 kW, 0.95

Explanation:
1. Circuit 1: \( Z_1=38.3 \) Ω, \( R_1=23 \) Ω, \( X_{L}=30.7 \) Ω. Circuit 2: \( Z_2=23 \) Ω, \( R_2=1.61 \) Ω, \( X_C=22.9 \) Ω.
2. Series: \( R=24.6,\ X=30.7-22.9=7.7 \), \( Z=25.8 \) Ω.
3. Current \( 230/25.8=8.9 \) A; power \( I^2R=1.96 \) kW; power factor \( R/Z=0.95 \).

Teacher's Note:
1. Resolve each circuit into R and X.
2. Reactances of L and C subtract.

Question 17: A coil of inductance 300 mH and resistance 2 Ω is connected to a source of voltage 2 V. The current reaches half of its steady-state value in
(a) 0.05 s
(b) 0.15 s
(c) 0.1 s
(d) 0.3 s
Show Answer & Explanation

Answer: (c) 0.1 s

Explanation:
1. \( \tau=L/R=0.15 \) s. \( 1-e^{-t/\tau}=\tfrac12 \) gives \( t=\tau\ln2=0.104\approx0.1 \) s.

Teacher's Note:
1. Half the final value is reached after 0.69 time constants.

Question 18: The self-inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitor of
(a) 8 μF
(b) 4 μF
(c) 2 μF
(d) \( 1\ \mu\text F \)
Show Answer & Explanation

Answer: (d) \( 1\ \mu\text F \)

Explanation:
1. Maximum power at resonance: \( C=\dfrac{1}{\omega^2L}=\dfrac{1}{(100\pi)^2\times10}\approx10^{-6} \) F.

Teacher's Note:
1. Resonance maximises the current and hence the power.

Question 19: A circuit has a resistance of 12 Ω and an impedance of 15 Ω. The power factor of the circuit is
(a) 0.8
(b) 0.125
(c) 1.25
(d) 0.4
Show Answer & Explanation

Answer: (a) 0.8

Explanation:
1. \( \cos\phi=R/Z=12/15=0.8 \).

Teacher's Note:
1. Power factor cannot exceed 1.

Question 20: An inductor (\( L=100 \) mH), a resistor (\( R=100\,\Omega \)) and a battery (\( E=100 \) V) are in series. After a long time the battery is disconnected after short-circuiting points A and B (so that the current continues through L and R). The current 1 ms after the short circuit is
(a) \( e \) A
(b) \( 1/e \) A
(c) 1 A
(d) 0.1 A
Show Answer & Explanation

Answer: (b) \( 1/e \) A

Explanation:
1. Steady current \( I_0=100/100=1 \) A; \( \tau=L/R=1 \) ms.
2. Decay: \( i=I_0e^{-t/\tau}=e^{-1} \) A at \( t=1 \) ms.

Teacher's Note:
1. τ = L/R = 1 ms here.
2. Decay gives 36.8% in one time constant.

Question 21: In a series resonant LCR circuit the voltage across R is 100 V, with \( R=1\,\text{k}\Omega \) and \( C=2\ \mu\text F \). The resonant angular frequency is 200 rad/s. At resonance, the voltage across L is
(a) 100 V
(b) 25 V
(c) \( 2.5\times10^{-2} \) V
(d) 250 V
Show Answer & Explanation

Answer: (d) 250 V

Explanation:
1. Current \( I=100/1000=0.1 \) A.
2. At resonance \( X_L=X_C=\dfrac{1}{\omega C}=\dfrac{1}{200\times2\times10^{-6}}=2500 \) Ω.
3. So \( V_L=IX_L=0.1\times2500=250 \) V.

Teacher's Note:
1. Voltage across L can exceed the source voltage at resonance.

Question 22: In an ac circuit the applied voltage is \( E=E_0\sin\omega t \) and the resulting current is \( I=I_0\sin(\omega t-\pi/2) \). The power consumption in the circuit is
(a) \( \dfrac{E_0I_0}{2} \)
(b) zero
(c) \( \dfrac{E_0I_0}{\sqrt2} \)
(d) \( \sqrt2E_0I_0 \)
Show Answer & Explanation

Answer: (b) zero

Explanation:
1. The phase difference is 90°, so \( P=\dfrac{E_0I_0}{2}\cos90^\circ=0 \).

Teacher's Note:
1. Zero power factor means no power is consumed.

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Multiple Choice Questions (MCQs) for JEE Physics Alternating Current

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