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Alternating Current

Chapter 7Notes + practice

CBSE Class 12 Physics · NCERT Physics Part-I

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Shishya's notes

What this chapter is about

This chapter introduces alternating current (AC), the form of electric current that reverses direction periodically. While earlier chapters dealt with direct current (DC) that flows steadily in one direction, AC is what powers homes, schools and industries across India. Understanding AC is essential because the electricity supplied by power stations is alternating, not direct.

You will learn how voltage and current vary sinusoidally with time, and how resistors, capacitors and inductors behave differently when connected to an AC source. The chapter develops the idea of impedance, phase difference, and resonance in circuits containing combinations of these elements. You will also study transformers, which change AC voltage levels and make long-distance power transmission practical.

After studying this chapter, you should be able to write expressions for instantaneous and RMS values of AC quantities, analyse series LCR circuits, calculate power in AC circuits, explain resonance, and describe how a transformer works.

Key ideas

  • An alternating current varies sinusoidally as i = i₀ sin(ωt), where i₀ is the peak (maximum) current and ω = 2πf is the angular frequency; voltage follows a similar form v = v₀ sin(ωt + φ), where φ is the phase angle.
  • The root-mean-square (RMS) value of an AC quantity is its effective value for power calculations: I_rms = i₀ / √2 and V_rms = v₀ / √2.
  • In a pure resistor, current and voltage are in phase; in a pure inductor, current lags voltage by 90°; in a pure capacitor, current leads voltage by 90°.
  • Inductive reactance is X_L = ωL and capacitive reactance is X_C = 1/(ωC); both have the unit ohm.
  • The impedance of a series LCR circuit is Z = √[R² + (X_L − X_C)²], and the phase angle φ satisfies tan φ = (X_L − X_C) / R.
  • Resonance occurs in a series LCR circuit when X_L = X_C, giving ω₀ = 1/√(LC); at resonance, impedance is minimum (equal to R) and current is maximum.
  • Average power in an AC circuit is P = V_rms I_rms cos φ, where cos φ is called the power factor.
  • A transformer changes AC voltage by electromagnetic induction; the voltage ratio equals the turns ratio: V_s / V_p = N_s / N_p.

Formulas and facts to remember

  1. Instantaneous current: i = i₀ sin(ωt); instantaneous voltage: v = v₀ sin(ωt + φ).
  2. RMS values: I_rms = i₀ / √2 ≈ 0.707 i₀; V_rms = v₀ / √2.
  3. Inductive reactance: X_L = ωL = 2πfL (ohm).
  4. Capacitive reactance: X_C = 1/(ωC) = 1/(2πfC) (ohm).
  5. Impedance of series LCR circuit: Z = √[R² + (X_L − X_C)²].
  6. Resonant angular frequency: ω₀ = 1/√(LC); resonant frequency f₀ = 1/(2π√(LC)).
  7. Average power: P = V_rms I_rms cos φ; for a pure resistor cos φ = 1, for pure L or pure C cos φ = 0.
  8. Transformer relation: V_s / V_p = N_s / N_p = I_p / I_s (ideal transformer, no energy loss).

Worked examples

Example 1: Finding RMS voltage and current

An AC source has a peak voltage of 311 V and a frequency of 50 Hz. It is connected to a 100 Ω resistor. Find the RMS voltage, RMS current and power dissipated.

Step 1: RMS voltage = v₀ / √2 = 311 / 1.414 ≈ 220 V.

Step 2: Since the circuit is purely resistive, Ohm's law applies with RMS values. I_rms = V_rms / R = 220 / 100 = 2.2 A.

Step 3: Power = V_rms × I_rms × cos φ. For a resistor, cos φ = 1. So P = 220 × 2.2 × 1 = 484 W.


Example 2: Series LCR circuit at resonance

A series circuit has R = 40 Ω, L = 5.0 H, and C = 80 μF. Find the resonant frequency and the current at resonance when the RMS supply voltage is 200 V.

Step 1: Resonant angular frequency ω₀ = 1/√(LC). Here L = 5.0 H, C = 80 × 10⁻⁶ F.

LC = 5.0 × 80 × 10⁻⁶ = 4.0 × 10⁻⁴ s².

ω₀ = 1/√(4.0 × 10⁻⁴) = 1 / 0.02 = 50 rad/s.

Step 2: Resonant frequency f₀ = ω₀ / (2π) = 50 / 6.28 ≈ 7.96 Hz.

Step 3: At resonance, X_L = X_C, so impedance Z = R = 40 Ω.

I_rms = V_rms / Z = 200 / 40 = 5.0 A.


Example 3: Transformer turns ratio

A step-down transformer is used to reduce 220 V mains supply to 22 V for a table lamp. If the primary coil has 2000 turns, how many turns does the secondary have?

Step 1: Use the turns-ratio relation: V_s / V_p = N_s / N_p.

22 / 220 = N_s / 2000.

Step 2: N_s = 2000 × (22 / 220) = 2000 × 0.1 = 200 turns.

Common mistakes

  • Confusing peak and RMS values → Always divide peak values by √2 to get RMS; household 220 V is already RMS.
  • Ignoring phase when adding voltages in LCR circuits → Voltages across L and C are 180° apart, so they subtract; add by phasor method, not arithmetic.
  • Thinking reactance is the same as resistance → Reactance depends on frequency; resistance does not.
  • Assuming power is V_rms × I_rms in all circuits → Multiply by cos φ; in a pure inductor or capacitor, average power is zero.
  • Using DC formulas directly for AC capacitors and inductors → Capacitors block DC but pass AC; inductors pass DC easily but oppose rapid AC changes.

Quick revision

  • AC quantities vary sinusoidally; RMS value = peak value / √2.
  • In a pure inductor current lags voltage by 90°; in a pure capacitor current leads by 90°.
  • Impedance Z = √[R² + (X_L − X_C)²]; minimum at resonance when X_L = X_C.
  • Power factor cos φ determines real power; purely reactive circuits consume zero average power.
  • Transformers change voltage level; step-up increases voltage, step-down decreases it; works only with AC.

Written by Shishya's AI on 26 Sept 2026 from the chapter's title and class level, in Shishya's own words — not a copy or summary of the textbook. Read the official chapter for the book's own text, activities and exercises.

Practice: 5 questions on Alternating Current

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These practice questions are Shishya's own, written by AI and answer-checked before they are shown. They are not taken from the NCERT book or any board paper.