Last Updated: May 2026
CBSE Class 12 Physics Chapter 7 — Alternating Current is one of the highest-weight chapters in the Class 12 Physics NCERT (Part 1, Chapter 7), carrying roughly 6-8 marks in the CBSE Board examination 2027 and forming the backbone of the entire Electromagnetic Induction → AC → EM Waves cluster. This chapter introduces sinusoidal voltage and current, RMS values, the behaviour of AC circuits with R, L and C — individually and in series LCR — resonance, power dissipation in AC, the wattless current, transformers and the AC generator.
This guide gives you the full NCERT chapter summary, every important formula in a single reference table, fully worked NCERT in-text and exercise solutions for the most-asked problems, and 10 boards-pattern MCQs at the end. Use this with our Class 12 Physics Crash Course for full chapter coverage.
Chapter 7 — Quick Map
| Section (NCERT) | Topic | Boards-asked formats |
|---|---|---|
| 7.1 | Introduction to AC | 1-mark conceptual MCQ |
| 7.2 | AC Voltage applied to a Resistor | 2-mark derivation |
| 7.3 | Representation of AC by rotating phasor | 1-mark MCQ |
| 7.4 | AC Voltage applied to an Inductor | 3-mark derivation: I = (V0/ωL) sin(ωt – π/2) |
| 7.5 | AC Voltage applied to a Capacitor | 3-mark derivation |
| 7.6 | AC Voltage applied to series LCR circuit (incl. resonance) | 5-mark long-answer with phasor diagram |
| 7.7 | Power in AC circuit; Wattless Current | 3-mark numerical |
| 7.8 | LC Oscillations (qualitative) | 1-2 mark conceptual |
| 7.9 | Transformers | 3-5 mark derivation + working principle |
1. Alternating Current and Voltage
An alternating voltage is one whose magnitude varies sinusoidally with time and reverses its direction periodically. Mathematically:
V(t) = V0 sin(ωt), where V0 is the peak value and ω = 2πf is the angular frequency.
In India the household supply is 220 V (RMS) at 50 Hz. The peak voltage is therefore V0 = Vrms × √2 ≈ 311 V.
2. RMS (Root Mean Square) Value
The RMS value of an AC current is the equivalent DC current that would produce the same heating effect in a resistor over the same time. For a sinusoidal current:
Irms = I0 / √2 ≈ 0.707 I0
Similarly, Vrms = V0 / √2.
Why RMS, not average? The average of sin(ωt) over a full cycle is zero, but the average of sin²(ωt) is ½. Hence the meaningful effective measure of AC is the RMS value.
3. AC Through a Resistor
For V = V0 sin(ωt) applied to a pure resistor R:
I = (V0/R) sin(ωt)
Current and voltage are in phase. Average power dissipated: P = Vrms·Irms = I²rms·R.
4. AC Through an Inductor
For a pure inductor L with V = V0 sin(ωt):
I = (V0/ωL) sin(ωt – π/2)
Current lags voltage by π/2 (90°). Inductive reactance: XL = ωL = 2πfL (in ohms). Average power = 0 over a full cycle (purely reactive).
5. AC Through a Capacitor
For a pure capacitor C with V = V0 sin(ωt):
I = (V0·ωC) sin(ωt + π/2)
Current leads voltage by π/2. Capacitive reactance: XC = 1/(ωC) = 1/(2πfC). Average power = 0.
6. Series LCR Circuit
Impedance: Z = √(R² + (XL − XC)²)
Phase angle: tan φ = (XL − XC) / R
If XL > XC: circuit is inductive, current lags voltage. If XC > XL: capacitive, current leads. If XL = XC: resonance.
Resonance
At resonance: ω0 = 1/√(LC) and Z = R (minimum). Current is maximum: Imax = Vrms/R.
Quality factor (Q-factor): Q = (1/R)√(L/C) = ω0L/R = 1/(ω0RC). High Q means a sharper resonance (used in radio tuning circuits).
7. Power in AC Circuit — Wattless Current
Average power: Pavg = Vrms·Irms·cos φ, where cos φ is the power factor.
For pure R: cos φ = 1. For pure L or C: cos φ = 0 → no average power consumed; this current is called wattless current.
8. Transformer
A transformer is a static device that transfers electrical power between two circuits at different voltage levels through electromagnetic induction. It works only on AC.
Turns ratio: Vs/Vp = Ns/Np = Ip/Is (for an ideal transformer).
- Step-up transformer: Ns > Np, Vs > Vp.
- Step-down transformer: Ns < Np, Vs < Vp.
- Energy losses: copper losses (I²R in windings), iron losses (eddy currents — minimised by lamination — and hysteresis), flux leakage, humming losses.
Long-distance transmission uses high-voltage AC because power loss = I²R; doubling voltage at the same power means halving current, reducing loss to 1/4.
Key Formula Sheet
| Quantity | Formula |
|---|---|
| Peak vs RMS voltage | Vrms = V0/√2 |
| Peak vs RMS current | Irms = I0/√2 |
| Inductive reactance | XL = 2πfL |
| Capacitive reactance | XC = 1/(2πfC) |
| Impedance (LCR series) | Z = √(R² + (XL − XC)²) |
| Phase angle | tan φ = (XL − XC)/R |
| Resonant frequency | f0 = 1/(2π√(LC)) |
| Quality factor | Q = (1/R)√(L/C) |
| Average power | P = Vrms·Irms·cos φ |
| Power factor | cos φ = R/Z |
| Transformer ratio | Vs/Vp = Ns/Np = Ip/Is |
NCERT Solutions (Selected Important Questions)
Q 7.1: A 100 Ω resistor is connected to a 220 V, 50 Hz AC supply. (a) Find Irms. (b) Find net power consumed.
Solution:
(a) Irms = Vrms/R = 220/100 = 2.2 A.
(b) P = Vrms·Irms = 220 × 2.2 = 484 W.
Q 7.2: Peak voltage of an AC supply is 300 V. What is the RMS voltage?
Vrms = V0/√2 = 300/1.414 ≈ 212.1 V.
Q 7.3: The RMS value of current in an AC circuit is 10 A. What is the peak current?
I0 = √2 × 10 ≈ 14.14 A.
Q 7.4: A 44 mH inductor is connected to a 220 V, 50 Hz supply. Find the RMS current.
XL = 2πfL = 2 × 3.14 × 50 × 44 × 10−3 ≈ 13.82 Ω.
Irms = 220/13.82 ≈ 15.92 A.
Q 7.5: A 60 μF capacitor is connected to a 110 V, 60 Hz supply. Find the RMS current.
XC = 1/(2π × 60 × 60 × 10−6) ≈ 44.21 Ω.
Irms = 110/44.21 ≈ 2.49 A.
Q 7.7: A series LCR circuit has L = 5 H, C = 80 μF, R = 40 Ω, Vrms = 230 V. Find resonance frequency.
ω0 = 1/√(LC) = 1/√(5 × 80 × 10−6) = 1/√(4 × 10−4) = 1/(0.02) = 50 rad/s.
f0 = ω0/2π ≈ 7.96 Hz.
Q 7.8: At resonance for the same circuit (Q 7.7), find Irms and amplitude of voltage across L and C.
At resonance Z = R = 40 Ω. Irms = 230/40 = 5.75 A.
VL = I × XL = 5.75 × (50 × 5) = 1437.5 V (note this is much greater than supply — resonance voltage magnification).
VC = same magnitude as VL (180° opposite); they cancel in vector sum.
Concept Highlights to Remember
- RMS values of household 220 V AC mean peak ≈ 311 V — wires must be insulated for the peak, not the RMS.
- Inductors oppose change in current (Lenz’s Law); capacitors oppose change in voltage.
- “Choke coil” is a high-inductance, low-resistance coil used in fluorescent tube lights to control current without dissipating much power.
- Transformer cores are made of soft iron and laminated to reduce eddy current loss.
- Step-up at the generating station, step-down at the substation — basis of national grid.
10 Boards-Pattern MCQs — CBSE Class 12 Physics Chapter 7 (2027)
- The RMS value of an alternating current of peak value 10 A is approximately:
(a) 5 A (b) 7.07 A (c) 10 A (d) 14.14 A
Ans: (b) 7.07 A - The reactance of a capacitor is inversely proportional to:
(a) Frequency only (b) Capacitance only (c) Both frequency and capacitance (d) Voltage
Ans: (c) - In a purely inductive AC circuit, the current:
(a) Leads the voltage by π/2 (b) Lags the voltage by π/2 (c) Is in phase with voltage (d) Lags by π
Ans: (b) - At resonance in a series LCR circuit:
(a) Z is maximum (b) Current is minimum (c) Z = R (d) Power factor is zero
Ans: (c) - The power factor of a pure capacitor is:
(a) 0 (b) 0.5 (c) 1 (d) Infinity
Ans: (a) - Transformer works on the principle of:
(a) Self-induction (b) Mutual induction (c) Faraday’s electrolysis law (d) Coulomb’s law
Ans: (b) - In a step-down transformer:
(a) Ns > Np (b) Vs > Vp (c) Is > Ip (d) Power is increased
Ans: (c) - The Q-factor of a series LCR circuit is given by:
(a) ω0L/R (b) R/ω0L (c) ω0RC (d) RC/ω0
Ans: (a) - An AC source of 220 V, 50 Hz is connected to a 100 Ω resistor. The peak current is approximately:
(a) 2.2 A (b) 3.11 A (c) 4.4 A (d) 5.0 A
Ans: (b) — I0 = √2 × 2.2 ≈ 3.11 A - Eddy current losses in a transformer core are reduced by:
(a) Using thick solid iron core (b) Using laminated core (c) Increasing turns ratio (d) Using copper core
Ans: (b)
Frequently Asked Questions
Q1. What is the weightage of Chapter 7 (AC) in the CBSE Class 12 Physics Board exam 2027?
A: Chapter 7 along with Chapter 6 (Electromagnetic Induction) and Chapter 8 (EM Waves) forms the unit “Electromagnetic Induction and Alternating Currents”, which carries approximately 8 marks. Chapter 7 alone typically contributes 5-6 marks split across one MCQ, one short-answer derivation and one numerical or long-answer.
Q2. Which derivations are most important?
A: (i) AC through an inductor — show I = (V0/ωL) sin(ωt − π/2). (ii) Series LCR circuit phasor analysis leading to Z = √(R² + (XL − XC)²). (iii) Working principle of a transformer with the turns ratio. (iv) Resonance condition and the formula for f0.
Q3. Are LC oscillations (Section 7.8) asked in numerical form?
A: Rarely. Boards usually frame Section 7.8 as a 1-2 mark conceptual MCQ comparing LC oscillations to a mechanical mass-spring system. The numerical depth on LC is reserved for JEE — boards prefer the analogy.
Q4. What is the difference between average power and apparent power?
A: Apparent power = Vrms × Irms (in volt-amperes, VA). Average (true) power = Vrms × Irms × cos φ (in watts, W). Their ratio is the power factor cos φ.
Q5. Why is power loss in long-distance AC transmission less at higher voltages?
A: Power loss = I²R. For a fixed power P = VI, doubling V halves I, so loss falls to one-quarter. That is why grid transmission uses 132 kV / 220 kV / 400 kV / 765 kV step-up at the generating station.
Master every derivation and numerical of Chapter 7 with structured weekly tests at Ready For Boards Class 12 Physics Course.
