Class 12 Physics Chapter 7: Alternating Current

CBSE 2026–27 | NCERT-aligned free notes, diagrams, examples and practice

Alternating current changes magnitude and direction periodically. The chapter develops RMS and average values, AC through resistors, inductors and capacitors, phasors, series LCR circuits, resonance, power and transformers.

1. AC Voltage and Current

A sinusoidal voltage may be written v = V_m sinωt. The angular frequency is ω = 2πf. For a pure resistor, current is in phase with voltage.

2. RMS Value

For a sinusoidal current, I_rms = I_m/√2 and V_rms = V_m/√2. RMS values are used for practical AC power calculations.

3. AC Through R, L and C

4. Series LCR Circuit

Impedance Z = √[R² + (X_L − X_C)²]. Current I_rms = V_rms/Z. The phase angle satisfies tanφ = (X_L − X_C)/R.

5. Resonance

For a series LCR circuit, resonance occurs at ω₀ = 1/√LC. At resonance, impedance is minimum and current is maximum for the ideal series circuit. The circuit behaves as resistive with φ = 0.

6. AC Power

Average power P = V_rmsI_rmscosφ. The factor cosφ is the power factor. A purely inductive or purely capacitive ideal circuit has zero average power.

7. Transformer

An ideal transformer uses mutual induction to change AC voltage. V_s/V_p = N_s/N_p. A step-up transformer has more secondary turns; a step-down transformer has fewer.

Worked Example

For a 230 V RMS supply, peak voltage is V_m = √2×230 ≈ 325 V.

Common Exam Traps

• Do not confuse peak and RMS values.
• Inductor: current lags; capacitor: current leads.
• At series resonance, X_L equals X_C, not both equal zero.
• Power factor is cosφ, not φ itself.

Practice Questions

1. Explain RMS value and derive the relation with peak value.
2. Compare AC behaviour of R, L and C.
3. Derive impedance of a series LCR circuit.
4. Explain resonance and its condition.
5. Describe transformer operation and voltage ratio.

📌 Concept Diagram: Series LCR Circuit

Real-Life Connection

AC circuits are not only about calculations. The phase relation between voltage and current affects power transfer, motor operation and the efficiency of electrical systems. A low power factor means more current is required for the same useful power.

Application Questions

1. Why does a capacitor offer lower reactance at higher frequency?
2. Explain why current becomes maximum at series resonance.
3. A circuit has V_rms = 100 V, R = 20 Ω and is at resonance. Find the current.
4. Why cannot an ordinary transformer operate directly with steady DC?

AC Circuit Visuals & Numericals

Series LCR Circuit

Inductive reactance X_L=ωL and capacitive reactance X_C=1/(ωC). At resonance X_L=X_C, so the impedance of a series LCR circuit is minimum and current is maximum.

Worked Numerical

If R=10 Ω, X_L=8 Ω and X_C=4 Ω, Z=√(100+16)=√116≈10.77 Ω.

Transformer

For an ideal transformer, V_s/V_p=N_s/N_p. A step-up transformer has more secondary turns; a step-down transformer has fewer.

Power Factor

For an AC circuit, average power is P=V_rmsI_rmscosφ. Power factor indicates how effectively apparent power is converted into useful average power.

Practice

1. Calculate rms values.
2. Find impedance of a series LCR circuit.
3. Determine resonance frequency.
4. Solve transformer turns-ratio problems.
5. Explain the role of power factor.

Common Mistakes to Avoid

• Confusing rms and peak values.
• Using resistance instead of impedance in an AC circuit.
• Forgetting the resonance condition X_L=X_C.