JEE Physics Alternating Current MCQs Set 04

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Chapter-wise Objective Questions: Alternating Current

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Multiple Choice Questions

Question 1: The phase difference between current and voltage in an AC circuit is π/4 radian. If the frequency is 50 Hz, the phase difference is equivalent to a time difference of
(a) 0.78 s
(b) 15.7 ms
(c) 2.5 s
(d) 2.5 ms
Show Answer & Explanation

Answer: (d) 2.5 ms

Explanation:
1. \( \Delta t=\frac{\phi}{\omega}=\frac{\pi/4}{100\pi}=\frac1{400} \) s.
2. \( =2.5 \) ms.

Teacher's Note:
1. Time difference = phase / ω.
2. ω = 2π × 50 = 100π rad/s.

Question 2: The relation between an AC voltage source and time (SI units) is V = 120 sin(100πt) cos(100πt) volt. The peak voltage and frequency are respectively
(a) 120 V and 100 Hz
(b) \( \frac{120}{\sqrt2} \) V and 100 Hz
(c) 60 V and 200 Hz
(d) 60 V and 100 Hz
Show Answer & Explanation

Answer: (d) 60 V and 100 Hz

Explanation:
1. \( V=60\sin(200\pi t) \) using \( 2\sin\theta\cos\theta=\sin2\theta \).
2. Peak 60 V; \( \omega=200\pi \Rightarrow f=100 \) Hz.

Teacher's Note:
1. Convert the product into a single sine first.
2. f = ω/2π.

Question 3: For an alternating current I = I0 cos ωt, the rms value and peak value are respectively
(a) \( I_0,\ \frac{I_0}{\sqrt2} \)
(b) \( \frac{I_0}{\sqrt2},\ I_0 \)
(c) \( I_0,\ \frac{I_0}{2} \)
(d) \( 2I_0,\ \frac{I_0}{\sqrt2} \)
Show Answer & Explanation

Answer: (b) \( \frac{I_0}{\sqrt2},\ I_0 \)

Explanation:
1. Peak value is \( I_0 \); \( I_{rms}=\frac{I_0}{\sqrt2} \).

Teacher's Note:
1. The rms is smaller than the peak.
2. I_rms ≈ 0.707 I0.

Question 4: A resistance of 300 Ω and an inductance of 1/π henry are connected in series to an AC voltage of 20 V and 200 Hz frequency. The phase angle between the voltage and current is
(a) \( \tan^{-1}\frac43 \)
(b) \( \tan^{-1}\frac34 \)
(c) \( \tan^{-1}\frac32 \)
(d) \( \tan^{-1}\frac23 \)
Show Answer & Explanation

Answer: (a) \( \tan^{-1}\frac43 \)

Explanation:
1. \( X_L=2\pi fL=2\pi\times200\times\frac1\pi=400\,\Omega \).
2. \( \tan\phi=\frac{X_L}{R}=\frac{400}{300}=\frac43 \).

Teacher's Note:
1. tan φ = X/R for an RL circuit.
2. Voltage leads the current.

Question 5: The resistance that must be connected in series with an inductance of 0.2 H so that the phase difference between current and emf is 45° at 50 Hz is
(a) 6.28 Ω
(b) 62.8 Ω
(c) 628 Ω
(d) 31.4 Ω
Show Answer & Explanation

Answer: (b) 62.8 Ω

Explanation:
1. \( \phi=45^\circ\Rightarrow R=X_L=2\pi\times50\times0.2=62.8 \) Ω.

Teacher's Note:
1. tan 45° = 1.
2. R must equal the reactance.

Question 6: A series R-L-C circuit has its current plotted against frequency. If R decreases (L and C constant),
(a) only the maximum current increases
(b) the maximum current, the sharpness and the quality factor all increase
(c) the quality factor decreases
(d) the bandwidth increases
Show Answer & Explanation

Answer: (b) the maximum current, the sharpness and the quality factor all increase

Explanation:
1. At resonance \( I_m=\frac VR \) increases when R falls.
2. \( Q=\frac{\omega_0L}R \) increases, the curve is sharper and the bandwidth \( \frac{R}{L} \) decreases.

Teacher's Note:
1. Lower resistance means less damping.
2. Narrower bandwidth means better selectivity.

Question 7: An AC source of voltage V and frequency 50 Hz is connected to a 2 H inductor of negligible resistance, giving an rms current I. When the frequency is changed to 400 Hz with the same V, the current is
(a) 8I in phase with V
(b) 4I leading V by 90°
(c) I/4 lagging V by 90°
(d) I/8 lagging V by 90°
Show Answer & Explanation

Answer: (d) I/8 lagging V by 90°

Explanation:
1. \( I=\frac{V}{\omega L}\propto\frac1f \): frequency × 8 gives current I/8.
2. In a pure inductor the current lags the voltage by 90°.

Teacher's Note:
1. Inductive reactance is proportional to frequency.
2. Phase relation is unchanged.

Question 8: A capacitor of capacitance C has reactance X. If both the capacitance and the frequency are doubled, the reactance becomes
(a) 4X
(b) X/2
(c) X/4
(d) 2X
Show Answer & Explanation

Answer: (c) X/4

Explanation:
1. \( X=\frac1{2\pi fC} \), so doubling both f and C divides it by 4: X/4.

Teacher's Note:
1. Reactance is inversely proportional to both f and C.
2. Combine the two factors.

Question 9: In an L-R circuit with R = 10 Ω and L = 2 H, a 120 V, 60 Hz alternating voltage is applied. The current is
(a) 0.32 A
(b) 0.16 A
(c) 0.48 A
(d) 0.80 A
Show Answer & Explanation

Answer: (b) 0.16 A

Explanation:
1. \( X_L=2\pi\times60\times2=754 \) Ω, \( Z=\sqrt{10^2+754^2}\approx754 \) Ω.
2. \( I=\frac{120}{754}\approx0.16 \) A.

Teacher's Note:
1. R is negligible next to X_L.
2. I = V/Z.

Question 10: If the current through an inductor of inductance L is I = I0 sin ωt, the voltage across the inductor is
(a) \( I_0\omega L\sin(\omega t-\pi/2) \)
(b) \( I_0\omega L\sin(\omega t+\pi/2) \)
(c) \( I_0\omega L\sin(\omega t-\pi) \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( I_0\omega L\sin(\omega t+\pi/2) \)

Explanation:
1. \( V_L=L\frac{dI}{dt}=I_0\omega L\cos\omega t=I_0\omega L\sin(\omega t+\pi/2) \).

Teacher's Note:
1. Inductor voltage leads the current by 90°.
2. ELI: emf leads current in an inductor.

Question 11: A resistance of 200 Ω and an inductance of 1 H are connected in series with an AC source of frequency 200/2π Hz. The phase difference between V and I is
(a) 30°
(b) 60°
(c) 45°
(d) 90°
Show Answer & Explanation

Answer: (c) 45°

Explanation:
1. \( \omega=200 \), \( X_L=200\,\Omega=R \).
2. \( \tan\phi=1\Rightarrow\phi=45^\circ \).

Teacher's Note:
1. ω = 2πf = 200 rad/s.
2. R = X_L gives 45°.

Question 12: An alternating current of frequency 60 Hz flows through an inductance L = 1 mH, and the voltage drop across it is 0.6 V. The current (rms) is
(a) \( \frac1\pi \) A
(b) \( \frac5\pi \) A
(c) \( \frac{50}\pi \) A
(d) \( \frac{20}\pi \) A
Show Answer & Explanation

Answer: (b) \( \frac5\pi \) A

Explanation:
1. \( X_L=2\pi\times60\times10^{-3}=0.12\pi \) Ω.
2. \( I=\frac{0.6}{0.12\pi}=\frac5\pi \) A.

Teacher's Note:
1. I = V_L/X_L.
2. Convert mH to H.

Question 13: A series circuit has inductive reactance 200 Ω and capacitive reactance 600 Ω, with impedance Z1. If the values of the two reactances are interchanged, the impedance
(a) remains unchanged
(b) will increase
(c) will decrease
(d) information insufficient
Show Answer & Explanation

Answer: (a) remains unchanged

Explanation:
1. \( Z=\sqrt{R^2+(X_L-X_C)^2} \): the square makes the sign irrelevant.
2. The difference is ±400 Ω in both cases, so Z is the same.

Teacher's Note:
1. Only |X_L − X_C| matters.
2. The phase angle changes sign though.

Question 14: An alternating voltage is connected in series with a resistance r and an inductance L. The voltage drop across the resistance is 200 V and across the inductance is 150 V. The applied voltage is
(a) 350 V
(b) 250 V
(c) 500 V
(d) 300 V
Show Answer & Explanation

Answer: (b) 250 V

Explanation:
1. \( V=\sqrt{V_R^2+V_L^2}=\sqrt{200^2+150^2}=250 \) V.

Teacher's Note:
1. Quadrature addition again.
2. Not 350.

Question 15: The self-inductance of the motor of an electric fan is 10 H. To impart maximum power at 50 Hz it should be connected to a capacitance of
(a) \( 2\times10^{-6} \) F
(b) \( 3\times10^{-6} \) F
(c) \( 10^{-4} \) F
(d) \( 10^{-6} \) F
Show Answer & Explanation

Answer: (d) \( 10^{-6} \) F

Explanation:
1. Maximum power means resonance: \( \omega^2LC=1 \).
2. \( C=\frac1{\omega^2L}=\frac1{(100\pi)^2\times10}\approx1.0\times10^{-6} \) F.

Teacher's Note:
1. Maximum power occurs when X_L = X_C.
2. ω = 2π × 50 = 100π.

Question 16: In a series L-C-R circuit, the voltage across the resistor, inductor and capacitor are 1 V, 3 V and 2 V respectively (rms). At the instant the source voltage is V = V0 cos ωt, the current in the circuit is
(a) \( I_0\cos(\omega t+\frac\pi4) \)
(b) \( I_0\cos(\omega t-\frac\pi4) \)
(c) \( I_0\cos(\omega t+\frac\pi3) \)
(d) \( I_0\cos(\omega t-\frac\pi3) \)
Show Answer & Explanation

Answer: (b) \( I_0\cos(\omega t-\frac\pi4) \)

Explanation:
1. \( \tan\phi=\frac{V_L-V_C}{V_R}=\frac{3-2}{1}=1\Rightarrow\phi=\frac\pi4 \).
2. V_L > V_C, so the circuit is inductive and the current lags the voltage by π/4.

Teacher's Note:
1. Net reactive voltage = V_L − V_C.
2. Inductive circuit: current lags.

Question 17: In a series LCR circuit, the voltages across the resistance, inductance and capacitance are 40 V, 30 V and 60 V. The voltage across the source is
(a) 130 V
(b) 100 V
(c) 70 V
(d) 50 V
Show Answer & Explanation

Answer: (d) 50 V

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

Teacher's Note:
1. V_L and V_C oppose each other.
2. Subtract before squaring.

Question 18: An AC circuit draws 5 A at 160 V and the power consumption is 600 W. The power factor is
(a) 1
(b) 0.75
(c) 0.50
(d) zero
Show Answer & Explanation

Answer: (b) 0.75

Explanation:
1. Apparent power \( =160\times5=800 \) VA.
2. \( \cos\phi=\frac{600}{800}=0.75 \).

Teacher's Note:
1. Power factor = real power/apparent power.
2. Take rms values.

Question 19: For a series LCR circuit, the power loss at resonance is
(a) \( \frac{V^2}{\omega L-\frac1{\omega C}} \)
(b) \( I^2L\omega \)
(c) \( I^2R \)
(d) \( \frac{V^2}{C\omega} \)
Show Answer & Explanation

Answer: (c) \( I^2R \)

Explanation:
1. Only the resistance dissipates power (average power in L and C is zero).
2. So the power loss is I²R (with rms current).

Teacher's Note:
1. Reactive elements do not consume average power.
2. Option (a) would be infinite at resonance.

Question 20: An AC supply gives 30 V (rms) across a 10 Ω resistance. The power dissipated in it is
(a) \( 90\sqrt2 \) W
(b) 90 W
(c) \( 45\sqrt2 \) W
(d) 45 W
Show Answer & Explanation

Answer: (b) 90 W

Explanation:
1. \( P=\frac{V_{rms}^2}{R}=\frac{900}{10}=90 \) W.

Teacher's Note:
1. The given voltage is already rms.
2. No √2 is needed.

Question 21: For a series circuit with R = 1100 Ω and L = 3.5 H connected to a 50 Hz source, the power factor is
(a) 0
(b) 1/2
(c) \( 1/\sqrt2 \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( 1/\sqrt2 \)

Explanation:
1. \( X_L=2\pi\times50\times3.5\approx1100 \) Ω = R.
2. \( \cos\phi=\frac RZ=\frac{R}{\sqrt{2}R}=\frac1{\sqrt2} \).

Teacher's Note:
1. R = X_L gives 45°.
2. cos 45° = 1/√2.

Question 22: The power factor of an L-R circuit is
(a) \( \frac{\omega L}{R} \)
(b) \( \frac{R}{\sqrt{(\omega L)^2+R^2}} \)
(c) \( \omega LR \)
(d) \( \sqrt{\omega LR} \)
Show Answer & Explanation

Answer: (b) \( \frac{R}{\sqrt{(\omega L)^2+R^2}} \)

Explanation:
1. \( \cos\phi=\frac RZ=\frac{R}{\sqrt{R^2+\omega^2L^2}} \).

Teacher's Note:
1. Power factor = resistance/impedance.
2. It is at most 1.

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