Class 12 Physics NCERT Solutions

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Alternating Current Class 12 Physics NCERT Solutions

The complete NCERT exercise solutions for Chapter 7, Alternating Current — 8 questions from 7.1 to 7.8, each worked through step by step in the CBSE marking pattern. AC voltage applied to resistors, inductors and capacitors, phasors, LCR circuits, resonance, power factor and the transformer.

Class:12Subject:PhysicsChapter:7

4 Key Formulas25 Practice MCQs

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 12 Physics Chapter 7?

Chapter 7 carries 8 exercise questions, numbered 7.1 to 7.8. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Table of Contents9 sections

Revision & Practice

Chapter Overview

Alternating Current runs on the rms toolkit: V_rms = V₀/√2, X_L = ωL, X_C = 1/(ωC), Z = √(R² + (X_L – X_C)²), resonance at ω = 1/√(LC), and average power P = V_rms I_rms cosφ. Two recurring traps: a pure L or pure C circuit absorbs zero average power, and at resonance the huge voltage drops across L and C are exactly opposite and cancel. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Write which quantity is asked — rms or peak — before substituting; mixing them is the classic slip. Reactance is frequency-dependent, so recompute X_L and X_C for the actual frequency before touching LCR formulas. 'Amplitude of current' means the peak value I₀ = V₀/Z. At resonance Z = R, the current and voltage are in phase, and X_L = X_C, which is why the LC drop vanishes even though each inductor/capacitor drop alone can exceed the supply voltage.

$Ex 7.1 — rms current and power in a resistive ac circuit$
$Ex 7.2 — Peak ⇌ rms conversions$
$Ex 7.3 — rms current through an inductor$
$Ex 7.4 — rms current through a capacitor$
$Ex 7.5 — Why L and C circuits absorb no net power$
$Ex 7.6 — Angular frequency of LC oscillations$
$Ex 7.7 — Power at resonance in a series LCR circuit$
$Ex 7.8 — Series LCR: resonance, impedance, amplitude, drops$

NCERT Exercise 7.1 — rms Current and Power in a Resistive ac Circuit

1Exercise question

Step-by-step solution

1(a) I_rms = V_rms/R = 220/100 = 2.2 A.
2(b) For a pure resistor the power factor is 1, so net power over a full cycle is P = V_rms I_rms = 220 × 2.2 = 484 W.
3(Check: P = V²/R = 220²/100 = 484 W, the same value.)

Final answer

(a) I_rms = 2.2 A. (b) Net power = 484 W.

NCERT Exercise 7.2 — Peak ⇌ rms Conversions

1Exercise question

Step-by-step solution

1(a) V_rms = V₀/√2 = 300/1.414 = 212 V.
2(b) I₀ = √2 · I_rms = 1.414 × 10 = 14.1 A.

Final answer

(a) V_rms ≈ 212 V. (b) I₀ ≈ 14.1 A.

NCERT Exercise 7.3 — rms Current Through an Inductor

1Exercise question

Step-by-step solution

1X_L = 2πfL = 2π × 50 × 44 × 10⁻³ = 13.82 Ω.
2I_rms = V_rms/X_L = 220/13.82 = 15.9 A ≈ 16 A.

Final answer

I_rms ≈ 15.9 A ≈ 16 A.

NCERT Exercise 7.4 — rms Current Through a Capacitor

1Exercise question

Step-by-step solution

1X_C = 1/(2πfC) = 1/(2π × 60 × 60 × 10⁻⁶).
2X_C = 1/(2.26 × 10⁻²) = 44.2 Ω.
3I_rms = V_rms/X_C = 110/44.2 = 2.49 A ≈ 2.5 A.

Final answer

I_rms ≈ 2.5 A.

NCERT Exercise 7.5 — Why L and C Circuits Absorb No Net Power

1Exercise question

Step-by-step solution

1Average power in any ac circuit is P = V_rms I_rms cosφ, where φ is the phase angle between voltage and current.
2In a pure inductor the current lags the voltage by 90° (φ = 90°); in a pure capacitor it leads by 90°.
3cos90° = 0, hence the net power absorbed over a complete cycle is zero for both the inductor circuit (Exercise 7.3) and the capacitor circuit (Exercise 7.4).
4Physically: whatever energy is delivered to the field during one quarter-cycle is fully returned to the supply during the next, so no energy is dissipated on average.

Final answer

Net power = 0 in each circuit, because current and voltage are 90° out of phase (cosφ = 0).

NCERT Exercise 7.6 — Angular Frequency of LC Oscillations

1Exercise question

Step-by-step solution

1ω = 1/√(LC).
2ω = 1/√(27 × 10⁻³ × 30 × 10⁻⁶) = 1/√(8.1 × 10⁻⁷).
3√(8.1 × 10⁻⁷) = 9.0 × 10⁻⁴, so ω = 1.11 × 10³ rad s⁻¹.

Final answer

ω ≈ 1.1 × 10³ rad s⁻¹.

NCERT Exercise 7.7 — Power at Resonance in a Series LCR Circuit

1Exercise question

Step-by-step solution

1At resonance the inductive and capacitive reactances cancel, so Z = R = 20 Ω and the circuit is purely resistive (cosφ = 1).
2P = V_rms²/R = 200²/20.
3P = 40000/20 = 2000 W.

Final answer

Average power at resonance = 2000 W.

NCERT Exercise 7.8 — Series LCR: Resonance, Impedance, Amplitude, Drops

1Exercise question

Step-by-step solution

1(a) Resonance: ω = 1/√(LC) = 1/√(5.0 × 80 × 10⁻⁶) = 1/√(4.0 × 10⁻⁴) = 1/2.0 × 10⁻² = 50 rad s⁻¹.
2Source frequency f = ω/2π = 50/(2π) = 7.96 Hz ≈ 8 Hz.
3(b) At resonance Z = R = 40 Ω.
4Amplitude of current (peak): I₀ = V₀/Z, with V₀ = √2 × 230 = 325.3 V; I₀ = 325.3/40 = 8.13 A ≈ 8.1 A.
5(c) rms current: I_rms = 230/40 = 5.75 A.
6Drop across R: V_R = I_rms R = 5.75 × 40 = 230 V.
7At resonance X_L = ωL = 50 × 5.0 = 250 Ω and X_C = 1/(ωC) = 1/(50 × 80 × 10⁻⁶) = 250 Ω.
8Drop across L: V_L = I_rms X_L = 5.75 × 250 = 1437 V ≈ 1.4 × 10³ V.
9Drop across C: V_C = I_rms X_C = 5.75 × 250 = 1437 V ≈ 1.4 × 10³ V.
10V_L and V_C are in antiphase, so the net drop across the LC series combination is V_L – V_C = 0 at resonance.

Final answer

(a) f ≈ 8 Hz. (b) Z = 40 Ω; current amplitude ≈ 8.1 A. (c) V_R = 230 V, V_L = V_C ≈ 1.4 kV; LC combination drop = 0.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

RMS value of an AC

V_{r m s} = \frac{V _{0}}{2}

Impedance of an LCR circuit

Z = R^{2} + (ω L - \frac{1}{ω C})^{2}

Resonant frequency

ω_{0} = \frac{1}{L C}, f_{0} = \frac{1}{2 π L C}

Power in an AC circuit

P = V_{r m s} I_{r m s} cos ϕ

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Resonance is the condition X_L = X_C, at which the circuit's impedance is purely resistive and the current is maximum — not the condition of maximum voltage, which is what is sometimes misremembered.
Only the resistive part of the circuit averages power over a cycle; an ideal inductor or capacitor consumes net zero power, which is why the power factor multiplies through.

FAQ

Frequently asked questions

How many questions are in NCERT Class 12 Physics Chapter 7 (Alternating Current)?

There are 8 exercise questions in this chapter, numbered 7.1 to 7.8. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Alternating Current Class 12 Physics?

The formulas this chapter's questions actually turn on are: RMS value of an AC, Impedance of an LCR circuit, Resonant frequency, Power in an AC circuit. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Alternating Current important for JEE Main and NEET?

Very important — LCR resonance, the power factor and transformer questions are standard in every board paper and in JEE Main and NEET, and they carry marks in the numerical section.

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