JEE Physics Alternating Current MCQs Set 24

Multiple Choice Questions (MCQs) for JEE Physics: Alternating Current

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Multiple Choice Questions

Question 1: An AC voltage source of variable angular frequency \( \omega \) and fixed amplitude \( V_0 \) is connected in series with a capacitance \( C \) and an electric bulb of resistance \( R \) (inductance zero). When \( \omega \) 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
Show Answer & Explanation

Answer: (b) the bulb glows brighter

Explanation:
1. \( X_C=\dfrac{1}{\omega C} \) falls as \( \omega \) rises, so \( Z=\sqrt{R^2+X_C^2} \) falls, the current rises and the bulb glows brighter.

Teacher's Note:
1. Higher frequency means smaller capacitive reactance.

Question 2: In a series L-C-R circuit \( R=200\ \Omega \), and the voltage and the frequency of the main supply are 220 V and 50 Hz. On taking out the capacitor from the circuit the current lags behind the voltage by 30\(^\circ\). On taking out the inductor from the circuit the current leads the voltage by 30\(^\circ\). The power dissipated in the L-C-R circuit is
(a) 305 W
(b) 210 W
(c) zero
(d) 242 W
Show Answer & Explanation

Answer: (d) 242 W

Explanation:
1. Without C: \( \tan30^\circ=X_L/R \). Without L: \( \tan30^\circ=X_C/R \). So \( X_L=X_C \) and the circuit is when resonant.
2. \( Z=R \), so \( P=\dfrac{V^2}{R}=\dfrac{220^2}{200}=242 \) W.

Teacher's Note:
1. Equal reactances mean the resonant state, so the power is \( V^2/R \).

Question 3:
Assertion (A): 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.
Reason (R): Resultant emf in the circuit is given by \( E=\sqrt{V_R^2+(V_L-V_C)^2} \).
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( E=\sqrt{8^2+(16-10)^2}=\sqrt{64+36}=10 \) V, so A is true.
2. R is the correct phasor relation and gives A, so R explains A.

Teacher's Note:
1. Voltages across L and C are in antiphase, so subtract them before combining with \( V_R \).

Question 4: A capacitor and an inductance coil are connected in separate AC circuits with a bulb glowing in 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
Show Answer & Explanation

Answer: (d) a dielectric is introduced into the gap between the plates of the capacitor

Explanation:
1. Introducing a dielectric raises \( C \), lowers \( X_C \) and raises the current. The other three changes raise \( X_L \) or \( X_C \) and dim the bulb.

Teacher's Note:
1. Brighter means smaller reactance.

Question 5: 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 resistor are connected in series and then connected to the AC source of 150 V, 40 Hz. The current in the circuit becomes
(a) \( \dfrac{15}{\sqrt2} \) A
(b) 8 A
(c) 18 A
(d) 10 A
Show Answer & Explanation

Answer: (a) \( \dfrac{15}{\sqrt2} \) A

Explanation:
1. \( X_L(50)=100/8=12.5\ \Omega \), so \( X_L(40)=10\ \Omega \); \( R=100/10=10\ \Omega \).
2. \( Z=\sqrt{10^2+10^2}=10\sqrt2\ \Omega \), so \( I=\dfrac{150}{10\sqrt2}=\dfrac{15}{\sqrt2} \) A.

Teacher's Note:
1. Reactance is proportional to frequency.

Question 6: 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
Show Answer & Explanation

Answer: (c) 30 V

Explanation:
1. \( V_C=\sqrt{50^2-40^2}=30 \) V.

Teacher's Note:
1. Use \( V^2=V_R^2+V_C^2 \).

Question 7: 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
Show Answer & Explanation

Answer: (b) 23.4 A

Explanation:
1. \( X_L=2\pi\times50\times0.03=9.42\ \Omega \), so \( I=\dfrac{220}{9.42}=23.4 \) A.

Teacher's Note:
1. Find \( X_L \) first.

Question 8: A resistor of \( R=6\ \Omega \), an inductor of \( L=1 \) H and a capacitor of \( C=17.36\ \mu \)F are connected in series with an AC source. Find the Q-factor.
(a) 3.72
(b) 40
(c) 2.37
(d) 80
Show Answer & Explanation

Answer: (b) 40

Explanation:
1. \( Q=\dfrac1R\sqrt{\dfrac LC}=\dfrac16\sqrt{\dfrac{1}{17.36\times10^{-6}}}=\dfrac{240}{6}=40 \).

Teacher's Note:
1. Use \( Q=\dfrac1R\sqrt{L/C} \).

Question 9: A series circuit has a resistance \( R=30\ \Omega \), an inductor of reactance \( X_L=25\ \Omega \) and a capacitor of reactance \( X_C=25\ \Omega \) connected to a 240 V source of negligible resistance. An ideal voltmeter is connected across the L-C combination and an ammeter is in series. The readings are respectively
(a) 0 V, 3 A
(b) 150 V, 3 A
(c) 150 V, 6 A
(d) 0 V, 8 A
Show Answer & Explanation

Answer: (d) 0 V, 8 A

Explanation:
1. \( X_L=X_C \), so the L-C combination has zero net reactance and the voltmeter reads 0 V.
2. \( Z=R=30\ \Omega \), so \( I=240/30=8 \) A.

Teacher's Note:
1. At the resonant state the voltages across L and C cancel.

Question 10: The values of L, C and R for a circuit are 1 H, 9 F and \( 3\ \Omega \). What is the quality factor for the circuit when resonant?
(a) 1
(b) 9
(c) \( \dfrac19 \)
(d) \( \dfrac13 \)
Show Answer & Explanation

Answer: (c) \( \dfrac19 \)

Explanation:
1. \( Q=\dfrac1R\sqrt{\dfrac LC}=\dfrac13\sqrt{\dfrac19}=\dfrac19 \).

Teacher's Note:
1. Substitute directly.

Question 11: 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 is
(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
Show Answer & Explanation

Answer: (d) decreases in the first circuit and increases in the other

Explanation:
1. \( X_L=\omega L \) rises with frequency so the current falls; \( X_C=\dfrac1{\omega C} \) falls so the current rises.

Teacher's Note:
1. Inductor opposes high frequency; capacitor passes it.

Question 12: In an AC circuit, the instantaneous values of emf and current are \( e=200\sin(314t) \) volt and \( i=\sin\left(314t+\dfrac\pi3\right) \) ampere. The average power consumed in watt is
(a) 200
(b) 100
(c) 50
(d) 25
Show Answer & Explanation

Answer: (c) 50

Explanation:
1. \( P=\tfrac12E_0I_0\cos\phi=\tfrac12\times200\times1\times\cos60^\circ=50 \) W.

Teacher's Note:
1. Phase difference is \( 60^\circ \).

Question 13: The impedance of an R-C circuit is \( Z_1 \) for frequency \( f \) and \( Z_2 \) for frequency \( 2f \). Then \( \dfrac{Z_1}{Z_2} \) is
(a) between 1 and 2
(b) 2
(c) between \( \tfrac12 \) and 1
(d) \( \tfrac12 \)
Show Answer & Explanation

Answer: (a) between 1 and 2

Explanation:
1. \( Z_1=\sqrt{R^2+X^2} \), \( Z_2=\sqrt{R^2+X^2/4} \). The ratio tends to 1 when \( X\ll R \) and to 2 when \( X\gg R \), so it lies between 1 and 2.

Teacher's Note:
1. Check the two limits.

Question 14: 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
(b) \( \dfrac{20}{\sqrt2} \) V
(c) \( 20\sqrt2 \) V
(d) \( \dfrac{10}{\sqrt2} \) V
Show Answer & Explanation

Answer: (d) \( \dfrac{10}{\sqrt2} \) V

Explanation:
1. \( V_L=V_C \) cancel, so the supply is \( V=V_R=10 \) V and \( X_C=R \).
2. After shorting L: \( Z=\sqrt2R \), \( I=\dfrac{10}{\sqrt2R} \), so \( V_C=IX_C=\dfrac{10}{\sqrt2} \) V.

Teacher's Note:
1. The supply voltage stays at 10 V.

Question 15: An AC voltage source has an output of \( \Delta V=(200\ \text{V})\sin2\pi ft \). This source is connected to a \( 100\ \Omega \) resistor. The rms current in the resistor is
(a) 1.41 A
(b) 2.41 A
(c) 3.41 A
(d) 0.71 A
Show Answer & Explanation

Answer: (a) 1.41 A

Explanation:
1. \( I_{rms}=\dfrac{200/\sqrt2}{100}=1.41 \) A.

Teacher's Note:
1. Divide the peak voltage by \( \sqrt2 \) first.

Question 16: 220 V, 50 Hz AC is applied to a resistor. The instantaneous value of voltage is
(a) \( 220\sqrt2\sin(100\pi t) \)
(b) \( 220\sin(100\pi t) \)
(c) \( 220\sqrt2\sin(50\pi t) \)
(d) \( 220\sin(50\pi t) \)
Show Answer & Explanation

Answer: (a) \( 220\sqrt2\sin(100\pi t) \)

Explanation:
1. \( V_0=220\sqrt2 \) V and \( \omega=2\pi\times50=100\pi \) rad/s.

Teacher's Note:
1. 220 V is the rms value.

Question 17: An L-C circuit contains a 20 mH inductor and a 50 \( \mu \)F 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
Show Answer & Explanation

Answer: (b) \( t=1.57 \) ms

Explanation:
1. \( T=2\pi\sqrt{LC}=2\pi\sqrt{20\times10^{-3}\times50\times10^{-6}}=6.28 \) ms. The energy is wholly magnetic first at \( T/4=1.57 \) ms.

Teacher's Note:
1. The capacitor is fully discharged after a quarter period.

Question 18: 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 circuit will be
(a) 10 V
(b) 25 V
(c) 5 V
(d) 20 V
Show Answer & Explanation

Answer: (c) 5 V

Explanation:
1. \( V=\sqrt{5^2+(10-10)^2}=5 \) V.

Teacher's Note:
1. \( V_L \) and \( V_C \) cancel.

Question 19: What is the average power dissipation in an ideal capacitor in AC circuit?
(a) \( 2CV^2 \)
(b) \( \tfrac12CV^2 \)
(c) zero
(d) \( CV^2 \)
Show Answer & Explanation

Answer: (c) zero

Explanation:
1. The current leads the voltage by \( 90^\circ \), so \( \cos\phi=0 \) and the average power is zero.

Teacher's Note:
1. An ideal capacitor stores and returns energy.

Question 20: 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
Show Answer & Explanation

Answer: (b) 2.5 ms

Explanation:
1. \( I_0\sin\omega t=\dfrac{I_0}{\sqrt2} \Rightarrow \omega t=\dfrac\pi4 \), so \( t=\dfrac T8=\dfrac{20}{8}=2.5 \) ms.

Teacher's Note:
1. \( T=20 \) ms.

Question 21: In a 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
Show Answer & Explanation

Answer: (b) two capacitors and an inductor

Explanation:
1. A Colpitts oscillator uses a tank circuit with two capacitors and one inductor.

Teacher's Note:
1. Hartley uses two inductors.

Question 22: 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 \( 45^\circ \). The value of C is
(a) \( \dfrac{1}{2\pi f(2\pi fL+R)} \)
(b) \( \dfrac{1}{\pi f(2\pi fL+R)} \)
(c) \( \dfrac{1}{2\pi f(2\pi fL-R)} \)
(d) \( \dfrac{1}{\pi f(2\pi fL-R)} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{1}{2\pi f(2\pi fL+R)} \)

Explanation:
1. \( \tan45^\circ=\dfrac{X_C-X_L}{R}=1 \Rightarrow \dfrac{1}{2\pi fC}=2\pi fL+R \), so \( C=\dfrac{1}{2\pi f(2\pi fL+R)} \).

Teacher's Note:
1. Leading current means capacitive net reactance.

Question 23: What is the value of inductance L for which the current is a maximum in a series L-C-R circuit with \( C=10\ \mu \)F and \( \omega=1000\ \text{s}^{-1} \)?
(a) 100 mH
(b) 1 mH
(c) cannot be calculated unless R is known
(d) 10 mH
Show Answer & Explanation

Answer: (a) 100 mH

Explanation:
1. Current is maximum when resonant: \( L=\dfrac{1}{\omega^2C}=\dfrac{1}{10^6\times10^{-5}}=0.1 \) H \( =100 \) mH.

Teacher's Note:
1. R does not affect the resonant condition.

Question 24: A series L-C-R circuit has voltages \( V_R=80 \) V, \( V_L=40 \) V and \( V_C=100 \) V. The value of the alternating emf E of the source is
(a) 220 V
(b) 140 V
(c) 100 V
(d) 20 V
Show Answer & Explanation

Answer: (c) 100 V

Explanation:
1. \( E=\sqrt{80^2+(100-40)^2}=\sqrt{10000}=100 \) V.

Teacher's Note:
1. Subtract \( V_L \) from \( V_C \) before squaring.

Question 25: The resonant frequency of the tank circuit of an oscillator when \( L=\dfrac{10}{\pi^2} \) mH and \( C=0.04\ \mu \)F are connected in parallel is
(a) 250 kHz
(b) 25 kHz
(c) 2.5 kHz
(d) 25 MHz
Show Answer & Explanation

Answer: (b) 25 kHz

Explanation:
1. \( f=\dfrac{1}{2\pi\sqrt{LC}} \), \( LC=\dfrac{10^{-2}}{\pi^2}\times4\times10^{-8}=\dfrac{4\times10^{-10}}{\pi^2} \), so \( \sqrt{LC}=\dfrac{2\times10^{-5}}{\pi} \) and \( f=25 \) kHz.

Teacher's Note:
1. Work out \( \sqrt{LC} \) first.

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