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NCERT Exercises and Solutions: Alternating Current

🎓 Class 12 Physics CBSE Theory Ch 7 – Alternating Current ⏱ ~8 min
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NCERT Exercises and Solutions: Alternating Current

Chapter 7 — Summary

  • Sinusoidal AC: \(v = v_m\sin\omega t\); RMS values \(V = v_m/\sqrt{2}\), \(I = i_m/\sqrt{2}\).
  • Pure R: V and I in phase, P = VI = I²R.
  • Pure L: I lags V by 90°; reactance XL = ωL; P = 0.
  • Pure C: I leads V by 90°; reactance XC = 1/(ωC); P = 0.
  • Series LCR: \(Z = \sqrt{R^2 + (X_L - X_C)^2}\); \(\tan\phi = (X_L - X_C)/R\).
  • Resonance: \(\omega_0 = 1/\sqrt{LC}\); at resonance Z = R, I is max.
  • Q-factor: \(Q = \omega_0 L/R = (1/R)\sqrt{L/C}\); bandwidth = R/L = ω₀/Q.
  • Average AC power: \(P = V_{rms}I_{rms}\cos\phi\); power factor cos φ = R/Z.
  • LC oscillations: natural frequency ω₀ = 1/√(LC), energy oscillates between L and C.
  • Transformer (ideal): Vs/Vp = Ns/Np; VpIp = VsIs.
QuantitySymbolFormulaSI Unit
Peak / amplitudevm, im—V, A
RMS valueV, Ivm/√2V, A
Inductive reactanceXLωLΩ
Capacitive reactanceXC1/(ωC)Ω
ImpedanceZ√[R² + (XL−XC)²]Ω
Power factorcos φR/Z—
Resonance frequencyω₀, f₀1/√(LC), 1/(2π√(LC))rad/s, Hz
Q-factorQ(1/R)√(L/C)—

NCERT Exercises — Worked Solutions

Exercise 7.1

A 100 Ω resistor is connected to a 220 V, 50 Hz AC supply. (a) What is the RMS current in the circuit? (b) What is the net power consumed over a full cycle?

(a) I = V/R = 220/100 = 2.20 A.

(b) P = VI = 220 × 2.20 = 484 W (since the resistor is in phase, cos φ = 1).

Exercise 7.2

(a) The peak voltage of an AC supply is 300 V. What is the RMS voltage? (b) The RMS value of current in an AC circuit is 10 A. What is the peak current?

(a) V = vm/√2 = 300/1.414 = 212 V.

(b) im = I·√2 = 10 × 1.414 = 14.14 A.

Exercise 7.3

A 44 mH inductor is connected to a 220 V, 50 Hz AC supply. Determine the RMS value of the current.

XL = 2π fL = 2π × 50 × 0.044 = 13.82 Ω.

I = V/XL = 220/13.82 = 15.92 A.

Exercise 7.4

A 60 μF capacitor is connected to a 110 V, 60 Hz AC supply. Determine the RMS value of the current.

XC = 1/(2π × 60 × 60 × 10⁻⁶) = 44.21 Ω.

I = 110/44.21 = 2.49 A.

Exercise 7.5

For the circuits in Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle?

Both circuits contain only a pure reactance (no R). The phase angle is ±90°, so cos φ = 0. Net power absorbed = 0 W in both cases.

Exercise 7.6

Obtain the resonant frequency ωr of a series LCR circuit with L = 2.0 H, C = 32 μF, R = 10 Ω. What is the Q-value of this circuit?

ωr = 1/√(LC) = 1/√(2.0 × 32 × 10⁻⁶) = 1/0.008 = 125 rad/s.

Q = ωrL/R = 125 × 2/10 = 25.

Exercise 7.7

A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit?

ω = 1/√(LC) = 1/√(27 × 10⁻³ × 30 × 10⁻⁶) = 1/√(8.1 × 10⁻⁷) = 1.11 × 10³ rad/s.

Exercise 7.8

Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?

U = q²/2C = (6 × 10⁻³)² / (2 × 30 × 10⁻⁶) = 36 × 10⁻⁶ / 60 × 10⁻⁶ = 0.6 J.

For an ideal (lossless) LC circuit total energy is conserved: U = 0.6 J at all later times.

Exercise 7.9

A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a 200 V (rms) AC source. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

At resonance Z = R = 20 Ω, cos φ = 1.

I = V/R = 200/20 = 10 A.

P = VI cos φ = 200 × 10 × 1 = 2000 W.

Exercise 7.10

A radio can tune over the frequency range of a portion of MW broadcast band: 800 kHz to 1200 kHz. If its LC circuit has an effective inductance of 200 μH, what must be the range of its variable capacitor?

From ω0 = 1/√(LC): C = 1/(ω²L) = 1/(4π² f² L).

For f = 1200 kHz: Cmin = 1/(4π² × (1.2 × 10⁶)² × 200 × 10⁻⁶) = 88 pF.

For f = 800 kHz: Cmax = 1/(4π² × (8 × 10⁵)² × 200 × 10⁻⁶) = 198 pF.

Variable capacitor range: 88 pF to 198 pF.

Exercise 7.11

Figure shows a series LCR circuit connected to a variable-frequency 230 V source. L = 5.0 H, C = 80 μF, R = 40 Ω. (a) Find source frequency at which current amplitude is maximum. (b) Find that maximum value. (c) Find RMS potential drops across each element at resonance. (d) Find Q-value of the circuit.

(a) ω0 = 1/√(LC) = 1/√(5 × 80 × 10⁻⁶) = 1/0.02 = 50 rad/s; f₀ = 7.96 Hz.

(b) At resonance Z = R = 40 Ω. im = vm/R = (√2 × 230)/40 = 8.13 A.

(c) I = 230/40 = 5.75 A. VR = IR = 230 V. VL = IXL = 5.75 × 50 × 5 = 1437.5 V. VC = IXC = 5.75 × 1/(50 × 80×10⁻⁶) = 5.75 × 250 = 1437.5 V. (VL and VC equal in magnitude but 180° out of phase; they cancel.)

(d) Q = ω0L/R = 50 × 5 / 40 = 6.25.

Exercise 7.12

An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. The energy is shared equally between the inductor and the capacitor. Find the time t (in terms of the period T) when this first happens, and find the natural frequency f.

ω0 = 1/√(20 × 10⁻³ × 50 × 10⁻⁶) = 1/√(10⁻⁶) = 1000 rad/s ⇒ f = 159 Hz.

q = q0 cos ω0t. Energy in C is q²/2C; energy in L is total minus this. They are equal when q² = q₀²/2, i.e. cos²ω₀t = 1/2 ⇒ cos ω₀t = 1/√2 ⇒ ω₀t = π/4 ⇒ t = T/8 = (1/8)(2π/ω₀) = π/(4 × 1000) = 7.85 × 10⁻⁴ s.

Exercise 7.13 — Transformer step-down

A power transmission line feeds input power at 2300 V to a step-down transformer with its primary winding having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V?

Ns/Np = Vs/Vp = 230/2300 = 1/10 ⇒ Ns = 4000/10 = 400 turns.

Quick AC Calculator

Enter Vrms, R, L, C and frequency. The calculator returns Z, I, phase angle, P and the resonance frequency.

XL—
XC—
Impedance Z—
RMS current I—
Phase angle φ—
Power factor cos φ—
Average power P—
Resonance f₀—

Competency-Based Questions L1L2L3L4L5

A FM radio receiver has an LC tank with L = 0.1 μH. The station broadcasts at 98.5 MHz.

1. What value of C tunes the receiver to this station? L3

C = 1/(4π²f²L) = 1/(4π² × (98.5 × 10⁶)² × 10⁻⁷) = 26.1 pF.

2. Define RMS value of an alternating current. L1

The RMS value of an AC current is the steady DC current that would dissipate the same average power in the same resistor. For sinusoidal current I = im/√2.

3. Why is high-Q tuning preferred for FM reception? L4

High Q means a narrow resonance peak ⇒ the receiver rejects neighbouring stations and ambient noise, capturing only the desired carrier.

4. Distinguish wattless current from working current. L2

Working current (I cos φ) is in phase with voltage and dissipates real power. Wattless current (I sin φ) is 90° out of phase and contributes zero average power.

5. An electricity board penalises industries with cos φ < 0.85. Critique the reasoning behind this policy. L5

Low power factor means large reactive (wattless) current in the transmission line. The line still has to be sized to carry this current, and copper losses (I²R) are large even though no useful power is delivered. The penalty pushes industries to install capacitor banks, improving the grid's efficiency.

Assertion-Reason Questions

Assertion: A choke coil is preferred over a resistor for limiting AC current.

Reason: A choke dissipates almost no power compared to a resistor.

(A). Both true and R explains A. The inductor's average power is zero (XL stores and returns energy) while a resistor would waste energy as heat.

Assertion: At resonance in a series LCR circuit, the voltage across L is exactly opposite in phase to the voltage across C.

Reason: VL leads I by 90° while VC lags I by 90°.

(A). Both true; R explains A. The total phase difference between VL and VC is 180°.

Assertion: Step-up transformer increases the voltage but decreases the current.

Reason: Energy conservation requires Vp Ip = Vs Is for an ideal transformer.

(A). Both true; R explains A.

Frequently Asked Questions - NCERT Exercises and Solutions: Alternating Current

What are the key NCERT exercise types in Chapter 7 Alternating Current?
NCERT Class 12 Physics Chapter 7 Alternating Current exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Alternating Current?
For numerical problems in NCERT Class 12 Physics Chapter 7 Alternating Current: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 7?
From NCERT Class 12 Physics Chapter 7 (Alternating Current), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 12 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula. For Chapter 7 Alternating Current problems, write the dimensions of each quantity, substitute, and simplify. If the dimensions match, the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 7 exercises?
Common mistakes in NCERT Class 12 Physics Chapter 7 Alternating Current exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for charges/currents/vectors, (3) forgetting to consider all field/force contributions, (4) algebra errors during derivations, (5) misreading the problem. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Physics Chapter 7 Alternating Current solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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Physics Class 12 Part I – NCERT (2025-26)
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