Physics · Electromagnetic induction and AC · chapter 7 of 14

Class 12 Physics Chapter 7: Alternating Current case study question with answers

One case-based question of the kind the board asks: a passage and a figure, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks

In a physics laboratory Ananya builds the series circuit shown, with a resistor R = 20 Ω, an inductor of negligible resistance L = 0.25 H and a capacitor C = 16 μF. A signal generator keeps the rms voltage at 10 V while she changes its angular frequency ω. She records the rms current and notes how it rises to a maximum at one value of ω and falls on either side of it.

Part 1 · 1 mark

The angular frequency at which the current is maximum is

(a) 500 rad s⁻¹(b) 2.5 × 10⁵ rad s⁻¹(c) 79.6 rad s⁻¹(d) 1000 rad s⁻¹

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Answer: (a) 500 rad s⁻¹

ω₀ = 1/√(LC) = 1/√(0.25 × 16 × 10⁻⁶) = 1/(2.0 × 10⁻³) = 500 rad s⁻¹. 2.5 × 10⁵ omits the square root; 79.6 is ν₀ in hertz; 1000 doubles ω₀.

Part 2 · 1 mark

The maximum rms current she records is

(a) 0.50 A(b) 0.71 A(c) 0.35 A(d) 2.0 A

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Answer: (a) 0.50 A

At resonance Z = R, so I = 10/20 = 0.50 A. 0.71 A is the peak current; 0.35 A divides by √2 again; 2.0 A inverts V/R.

Part 3 · 2 marks

Find the rms current at ω = 1000 rad s⁻¹ and state whether it leads or lags the source voltage.

Answer and marking scheme

X_L = 1000 × 0.25 = 250 Ω, X_C = 1/(1000 × 16 × 10⁻⁶) = 62.5 Ω. Z = √[20² + (250 − 62.5)²] ≈ 189 Ω, I = 10/189 ≈ 0.053 A. X_L > X_C: the circuit is inductive and the current lags.

X_L = 250 Ω, X_C = 62.5 Ω, Z ≈ 189 Ω (1)
I ≈ 0.053 A, lags (1)

Find the power factor of the circuit at ω = 250 rad s⁻¹ and state whether the current leads or lags the source voltage.

Answer and marking scheme

X_L = 250 × 0.25 = 62.5 Ω, X_C = 1/(250 × 16 × 10⁻⁶) = 250 Ω. Z = √[20² + (62.5 − 250)²] ≈ 189 Ω, cos φ = R/Z = 20/189 ≈ 0.11. X_C > X_L: the circuit is capacitive and the current leads.

X_L = 62.5 Ω, X_C = 250 Ω, Z ≈ 189 Ω (1)
cos φ ≈ 0.11, current leads (1)

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