Alternating Current — Previous Year Questions | eduPhysics
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Class XII · Chapter 07

Alternating Current — Exam Practice

Practice questions in CBSE, NEET, and JEE Main style covering RMS values, reactance, series LCR circuits, resonance, and transformers.

📝 10 questions 🎯 CBSE · NEET · JEE Main
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.

1CBSE Board Exam Style

CBSE · 1 Mark
Q1. The average value of alternating current over a full cycle is zero, but its RMS value is not. Why?
Answer AC is symmetric about zero over a full cycle, so the positive and negative halves cancel exactly, giving zero plain average. RMS involves squaring the current first — squaring makes every value positive, so there's nothing left to cancel, giving a genuinely nonzero result.
CBSE · 3 Marks
Q2. Derive an expression for the capacitive reactance of a capacitor connected to an AC source, and show that the current leads the voltage by 90°.
Answer For v = vmsin(ωt) applied across a capacitor, charge q = Cv = Cvmsin(ωt). Current i = dq/dt = Cvmω cos(ωt) = imsin(ωt + π/2), where im = vmωC.

Comparing to v = vmsin(ωt), current leads voltage by exactly π/2 (90°) — evident directly from the +π/2 phase in the current expression.

Capacitive reactance: XC = vm/im = 1/(ωC).
CBSE · 5 Marks
Q3. (a) Derive the expression for the impedance of a series LCR circuit, and state the condition for resonance. (b) A series LCR circuit has R = 10 Ω, L = 0.5 H, C = 8 μF, connected to a 200 V, 50 Hz AC supply. Find the impedance of the circuit.
Answer (a) Using a phasor diagram, the resistor's voltage is in phase with current, the inductor's voltage leads by 90°, and the capacitor's leads by −90° (i.e., lags by 90°). Adding these as phasors: V = √[(IR)² + (IXL − IXC)²], giving impedance Z = √[R² + (XL−XC)²].

Resonance occurs when XL = XC, minimizing Z to just R, and maximizing current.

(b) XL = 2πfL = 2π×50×0.5 ≈ 157.1 Ω. XC = 1/(2πfC) = 1/(2π×50×8×10⁻⁶) ≈ 397.9 Ω.
Z = √[10² + (157.1−397.9)²] = √[100 + 57,993] ≈ 241 Ω.

2NEET Style (Single-Correct MCQ)

NEET Style
Q4. The power factor of a circuit containing only a pure inductor connected to an AC source is:
1
✓ Zero
0.5
Infinite
Solution Power factor = cosφ. For a pure inductor, current lags voltage by exactly 90°, so cos90° = 0 — no average power is dissipated (wattless current).
NEET Style
Q5. At resonance in a series LCR circuit, the current in the circuit is:
Zero
✓ Maximum
Minimum but nonzero
Independent of R
Solution At resonance, Z = R (its minimum possible value, since XL=XC cancel), so current im = vm/Z is at its maximum.
NEET Style
Q6. The RMS value of AC is related to its peak value by:
I_rms = I_m
✓ I_rms = I_m/√2
I_rms = 2I_m
I_rms = I_m/2
Solution For a sinusoidal waveform, RMS value is always Im/√2 ≈ 0.707Im — this is what "220 V AC" actually refers to (the RMS value, not the peak).
NEET Style
Q7. In a step-up transformer:
N_s < N_p, and V_s < V_p
✓ N_s > N_p, and V_s > V_p
N_s = N_p
Current increases along with voltage
Solution More secondary turns than primary turns (Ns > Np) steps voltage up. Since an ideal transformer conserves power, current correspondingly decreases in the secondary as voltage rises.

3JEE Main Style

JEE Main Style · Numerical
Q8. Find the resonant frequency of a series LCR circuit with L = 2 H and C = 50 μF.
Solution ωr = 1/√(LC) = 1/√(2 × 50×10⁻⁶) = 1/√(1×10⁻⁴) = 1/0.01 = 100 rad/s.

fr = ωr/(2π) = 100/6.283 ≈ 15.9 Hz.
JEE Main Style · MCQ
Q9. Wattless (zero average power) current can flow in a circuit containing:
A pure resistor only
✓ A pure inductor or capacitor only
Any combination of R, L, C
A resistor and inductor together
Solution Only when the circuit is purely reactive (pure L or pure C, no R) does φ = ±90° exactly, giving cosφ = 0 and zero average power despite nonzero current flowing.
JEE Main Style · Numerical
Q10. A transformer has 500 turns in its primary coil and 2500 turns in its secondary. If the primary voltage is 220 V, find the secondary voltage.
Solution Vs/Vp = Ns/Np → Vs = 220 × (2500/500) = 220 × 5 = 1100 V.
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