Physics · Electromagnetic induction and AC · chapter 7 of 14

Class 12 Physics Chapter 7: Alternating Current MCQs with answers

10 multiple-choice questions on Alternating Current, each with its answer and the reason, written from the NCERT chapter. For the CBSE board, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
An ac voltmeter connected across a laboratory socket reads 230 V. During each cycle the voltage between the socket's terminals varies between
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Answer: (a) 325 V and −325 V

The meter reads the rms value V = 230 V. The peak is vm = √2 V = 1.414 × 230 ≈ 325 V, and the sinusoidal voltage swings from +325 V to −325 V. Taking the reading as the peak gives ±230 V; dividing by √2 instead of multiplying gives ±163 V; an ac voltage is not one-signed.

Question 2
The current in a 50 Ω heating element is i = 2.0 sin(100πt) A. The average power dissipated in the element is
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Answer: (b) 100 W

p̄ = ½ im² R = ½ × (2.0)² × 50 = 100 W. Using im² R without the ½ gives 200 W; taking I = im/2 gives 50 W; the average current is zero but i² is never negative, so the power is not zero.

Question 3
An inductor of negligible resistance has a reactance of 30 Ω at 50 Hz. Its reactance at 150 Hz is
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Answer: (c) 90 Ω

XL = 2πνL is directly proportional to ν: XL = 30 × (150/50) = 90 Ω. 10 Ω treats XL as inversely proportional to ν; 270 Ω squares the ratio.

Question 4
A capacitor has a reactance of 120 Ω at 60 Hz. At 240 Hz its reactance is
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Answer: (d) 30 Ω

XC = 1/(2πνC) is inversely proportional to ν: XC = 120 × (60/240) = 30 Ω. 480 Ω treats it as proportional to ν; 7.5 Ω squares the ratio.

Question 5
An ac voltage is applied across an ideal inductor. The current in the inductor
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Answer: (a) lags the voltage by π/2

For a pure inductor i = im sin(ωt − π/2): the current reaches its maximum a quarter period after the voltage, so it lags by π/2. A lead of π/2 is the capacitor's case; in phase is the resistor's.

Question 6
An inductor of inductance (0.50/π) H and negligible resistance is connected to a 200 V (rms) source whose frequency can be changed. At what frequency does it draw an rms current of 2.0 A?
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Answer: (b) 100 Hz

XL = V/I = 200/2.0 = 100 Ω. XL = 2πνL gives ν = 100/(2π × 0.50/π) = 100/1.0 = 100 Hz. 628 is ω = XL/L in rad s⁻¹, not ν; 141 Hz takes 2.0 A as the peak current; 71 Hz takes 200 V as the peak voltage.

Question 7
An ideal capacitor is connected to a 220 V (rms), 50 Hz supply. The current in it reaches a peak value of 3.11 A. The capacitance is about
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Answer: (c) 31.8 μF

im = ωCvm with vm = √2 × 220 = 311 V, so C = im/(ωvm) = 3.11/(2π × 50 × 311) ≈ 3.18 × 10⁻⁵ F = 31.8 μF. 45.0 μF pairs the peak current with the rms voltage; 22.5 μF pairs the rms current with the peak voltage; 200 μF uses ν for ω, dropping 2π.

Question 8
Two identical bulbs P and Q are connected across the same ac source, P in series with a capacitor and Q in series with an inductor of negligible resistance. The rms voltage of the source stays fixed while its frequency is raised. Then
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Answer: (b) P glows more brightly and Q less brightly

XC = 1/ωC falls as the frequency rises, so the current through P rises; XL = ωL rises, so the current through Q falls. P brightens and Q dims. (a) and (c) treat both reactances alike; (d) swaps how XL and XC depend on frequency.

Question 9
In a series LCR circuit R = 30 Ω, XL = 70 Ω and XC = 30 Ω. The impedance of the circuit is
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Answer: (c) 50 Ω

Z = √[R² + (XL − XC)²] = √[30² + 40²] = 50 Ω. Adding the reactances, √(30² + 100²), gives 104 Ω; adding all three gives 130 Ω; XL − XC alone gives 40 Ω.

Question 10
In a series LCR circuit R = 40 Ω, XL = 10 Ω and XC = 50 Ω. The current in the circuit
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Answer: (d) leads the source voltage by 45°

tan φ = (XC − XL)/R = (50 − 10)/40 = 1, so φ = +45°. φ is positive because XC > XL: the circuit is predominantly capacitive and the current leads.

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