By the end of this chapter you'll be able to…

  • 1Define RMS values and average AC power
  • 2Analyse purely resistive, inductive, and capacitive circuits
  • 3Find impedance and phase angle of an LCR series circuit
  • 4Determine resonance frequency and Q-factor
  • 5Explain power factor and transformer operation
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Why this chapter matters
AC powers our homes and industry. Understanding RMS values, reactance, the LCR circuit, resonance, power factor, and transformers explains why AC won the current wars and how electricity is generated, transmitted, and used efficiently.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Alternating Current

'DC goes one way — AC alternates. And that alternation makes ALL the difference for power transmission and circuit behaviour.'

1. Chapter Overview

Alternating Current (AC) is the form of electricity used in homes and industry. Topics include: RMS VALUE of AC, PURELY RESISTIVE/INDUCTIVE/CAPACITIVE circuits, the LCR SERIES CIRCUIT (and its IMPEDANCE), RESONANCE in AC circuits, POWER IN AC circuits (real power, reactive power, power factor), and the TRANSFORMER.


2. AC Voltage and Current

  • AC voltage: V = V₀ sin(ωt). V₀ = peak voltage.
  • AC current: I = I₀ sin(ωt + φ). φ = phase difference between V and I.

RMS Values

  • V_rms = V₀/√2. I_rms = I₀/√2.
  • 'RMS gives the DC-equivalent value — the AC voltage that produces the SAME heating effect as an equivalent DC voltage.'

Average Power

  • P_avg = V_rms × I_rms × cos φ. 'Only the component of current IN PHASE with voltage contributes to average power.'

3. Purely Resistive, Inductive, and Capacitive Circuits

Circuit TypeV and I PhaseImpedancePower FactorAverage Power
Resistive (R)In phase (φ=0°)R1 (unity)V_rms × I_rms
Inductive (L)I LAGS V by 90°X_L = ωL00
Capacitive (C)I LEADS V by 90°X_C = 1/ωC00
  • 'In pure L or C, average power is ZERO — energy is alternately stored and returned, never dissipated.'

4. LCR Series Circuit

Impedance (Z)

  • Z = √[R² + (X_L − X_C)²].
  • Phase angle: tan φ = (X_L − X_C)/R.

Resonance

  • Resonance condition: X_L = X_C ⇒ ωL = 1/ωC ⇒ ω₀ = 1/√(LC).
  • At resonance: Z = R (minimum). I₀ = V₀/R (maximum).
  • Sharpness of resonance: Q-factor = (ω₀L)/R = 1/(ω₀RC). 'Higher Q means SHARPER resonance — a more frequency-selective circuit.'
  • 'At resonance, an LCR circuit behaves as a PURE RESISTOR — voltage and current are in phase.'

Worked Example 1

Problem: An LCR circuit has R = 10 Ω, L = 0.1 H, C = 100 μF. Find the resonant frequency. Solution: ω₀ = 1/√(LC) = 1/√(0.1 × 100×10⁻⁶) = 1/√(10⁻⁵) = 316.2 rad/s. f₀ = ω₀/(2π) ≈ 50.3 Hz.


5. Power in AC Circuits

  • True power: P = V_rms I_rms cos φ. 'The power actually DISSIPATED.'
  • Apparent power: S = V_rms I_rms. 'The product of RMS V and I.'
  • Power factor: cos φ = R/Z. 'How much of the apparent power is actual power?'
  • 'A LOW power factor means HIGH current for the same real power — causing higher losses in transmission lines.'

Wattless Current

  • I_rms sin φ — component of current that does NOT contribute to power.

6. Transformer

  • Principle: Mutual induction between two coils wound on a common core.
  • Voltage ratio: V_s/V_p = N_s/N_p.
  • Current ratio: I_s/I_p = N_p/N_s (for ideal transformer).
  • Power: P_in = P_out (ideal). P_out = P_in − losses (real).

Types

TypeN_s/N_pV_sI_sUse
Step-upN_s > N_pV_s > V_pI_s < I_pPower transmission (increase voltage, decrease current → reduce I²R loss)
Step-downN_s < N_pV_s < V_pI_s > I_pHome appliances (reduce voltage from transmission lines)

Energy Losses in a Transformer

  • Copper loss: I²R in the windings.
  • Eddy current loss: Laminated core reduces this.
  • Hysteresis loss: Use soft iron core (low hysteresis loop area).
  • Flux leakage: Not all flux from primary reaches secondary.

7. Comparison Table: DC vs AC

FeatureDCAC
DirectionConstant (one direction)Reverses periodically
Frequency050 Hz (India) / 60 Hz (USA)
GenerationBattery, cellGenerator, alternator
TransmissionHigh loss over distanceCan be stepped up/down — LOW loss
CapacitorBlocks DCPasses AC (impedance depends on frequency)
InductorZero impedance (steady state)Impedance X_L = ωL

8. Common Mistakes

  1. RMS vs peak: V_rms = V₀/√2. I_rms = I₀/√2. Many students use these interchangeably.
  2. Phase in LCR: tan φ = (X_L − X_C)/R. If X_L > X_C, φ > 0 (voltage leads). If X_C > X_L, φ < 0 (current leads).
  3. Resonance misunderstanding: At resonance, Z = R is MINIMUM, and I is MAXIMUM. NOT the other way around.
  4. Transformer equations: V_s/V_p = N_s/N_p = I_p/I_s. Note the INVERSE relationship for current.

9. CBSE Exam Focus

  1. RMS value — definition and calculation for sinusoidal waveforms
  2. Phasor diagrams — R, L, C, and LCR series circuits
  3. Impedance and phase angle — Z = √(R² + (XL − XC)²), tan φ = (XL − XC)/R
  4. Resonance — ω₀ = 1/√(LC), Q-factor
  5. Power in AC — P = V_rms I_rms cos φ, power factor
  6. Transformer — voltage/current ratios, step-up/step-down, losses

10. Self-Test

Q1: A 220 V AC supply has a peak voltage of? A1: V₀ = V_rms × √2 = 220 × 1.414 = 311 V.

Q2: An LCR circuit has R = 20 Ω, XL = 30 Ω, XC = 10 Ω. Find impedance and phase angle. A2: Z = √(20² + (30−10)²) = √(400 + 400) = √800 = 28.28 Ω. tan φ = (30−10)/20 = 20/20 = 1 ⇒ φ = 45° (voltage LEADS current).

Q3: A 0.1 H inductor and 50 μF capacitor are in series. Find the resonant frequency. A3: ω₀ = 1/√(0.1 × 50×10⁻⁶) = 1/√(5×10⁻⁶) = 447.2 rad/s. f₀ = 447.2/2π ≈ 71.2 Hz.

Q4: A step-down transformer converts 220 V to 11 V. If the primary has 1000 turns, find secondary turns. A4: N_s/N_p = V_s/V_p ⇒ N_s/1000 = 11/220 = 0.05 ⇒ N_s = 50 turns.

Q5: In an LCR circuit, V = 200 V, I = 5 A, and the power consumed is 800 W. Find the power factor. A5: P = V_rms I_rms cos φ ⇒ 800 = 200×5×cos φ ⇒ cos φ = 800/1000 = 0.8 (lagging or leading).


11. Conclusion

AC power is the BACKBONE of modern electrical systems:

  • RMS: 'The DC-equivalent — what you actually pay for on your electricity bill.'
  • REACTANCE: 'Inductors and capacitors have frequency-dependent resistance — they STORE and RELEASE energy.'
  • RESONANCE: 'The magic frequency where L and C cancel — current MAXIMISES, impedance MINIMISES.'
  • POWER FACTOR: 'How efficiently you use the current drawn from the supply.'
  • TRANSFORMER: 'The reason AC won over DC — voltage can be transformed UP for efficient transmission.'

'AC is the universal language of power distribution — its ability to transform voltage makes it indispensable.'

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

RMS values
V_rms = V0/sqrt(2); I_rms = I0/sqrt(2)
DC-equivalent heating values.
Impedance of LCR series
Z = sqrt(R^2 + (XL - XC)^2); tan(phi) = (XL - XC)/R
XL = omega L, XC = 1/(omega C).
Resonance
omega0 = 1/sqrt(LC); Q = omega0 L/R
At resonance Z = R is minimum and current is maximum.
Power and transformer
P = V_rms I_rms cos(phi); Vs/Vp = Ns/Np = Ip/Is
Power factor cos(phi) = R/Z.
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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Confusing RMS and peak values
RMS = peak/sqrt(2); use RMS for power and heating calculations.
WATCH OUT
Thinking resonance gives maximum impedance
At resonance in a series LCR circuit, impedance is minimum (Z = R) and current is maximum.
WATCH OUT
Assuming pure L or C dissipates power
Average power in a pure inductor or capacitor is zero; energy is stored and returned.
WATCH OUT
Getting the transformer current ratio backwards
Voltage ratio Vs/Vp = Ns/Np, but the current ratio is inverse: Is/Ip = Np/Ns.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· RMS
A 220 V AC supply has what peak voltage?
Show solution
V0 = V_rms x sqrt(2) = 220 x 1.414 = 311 V.
Q2MEDIUM· Impedance
An LCR circuit has R = 20, XL = 30, XC = 10 ohm. Find impedance and phase angle.
Show solution
Z = sqrt(20^2 + (30-10)^2) = sqrt(800) = 28.28 ohm. tan(phi) = 20/20 = 1, so phi = 45 degrees (voltage leads).
Q3MEDIUM· Resonance
A 0.1 H inductor and 50 uF capacitor are in series. Find the resonant frequency.
Show solution
omega0 = 1/sqrt(0.1 x 50e-6) = 1/sqrt(5e-6) = 447.2 rad/s. f0 = 447.2/(2 pi) = 71.2 Hz.
Q4EASY· Transformer
A step-down transformer converts 220 V to 11 V with 1000 primary turns. Find the secondary turns.
Show solution
Ns/Np = Vs/Vp = 11/220 = 0.05, so Ns = 50 turns.
Q5MEDIUM· Power Factor
In an LCR circuit V = 200 V, I = 5 A, power consumed = 800 W. Find the power factor.
Show solution
P = V I cos(phi): 800 = 200 x 5 x cos(phi), so cos(phi) = 0.8.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • AC: V = V0 sin(omega t); V_rms = V0/sqrt(2).
  • Resistor: V and I in phase; inductor: I lags 90; capacitor: I leads 90.
  • Reactances XL = omega L, XC = 1/(omega C); pure L/C dissipate no power.
  • LCR series: Z = sqrt(R^2 + (XL - XC)^2); tan(phi) = (XL - XC)/R.
  • Resonance: omega0 = 1/sqrt(LC); Z = R minimum, I maximum; Q = omega0 L/R.
  • Power P = V_rms I_rms cos(phi); power factor cos(phi) = R/Z.
  • Transformer: Vs/Vp = Ns/Np = Ip/Is; losses are copper, eddy, hysteresis, flux leakage.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6-8 marks across the chapter

Question typeMarks eachTypical countWhat it tests
LCR / resonance3-51Impedance, phase, resonance, Q-factor
Transformer31Voltage/current ratios and losses
RMS / power factor2-31RMS values and power calculations
Prep strategy
  • Always distinguish RMS from peak values
  • Draw phasor diagrams for LCR circuits
  • Memorise resonance and Q-factor formulas
  • Use inverse current ratio for transformers

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Power transmission

Transformers step voltage up and down to transmit AC efficiently over long distances.

Radio tuning

LCR resonance selects the desired station frequency in radios and TVs.

Household electricity

RMS values describe the 220 V supply that powers homes and appliances.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

  1. Convert between peak and RMS correctly
  2. Use phasor diagrams to find phase and impedance
  3. Apply resonance condition XL = XC
  4. Use transformer ratios with inverse current relation

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

  • Analyse bandwidth and sharpness of resonance via the Q-factor.
  • Solve parallel LCR (rejector) circuits and their resonance behaviour.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 Physics examHigh
JEE Main and Advanced (AC Circuits)High
NEET PhysicsMedium

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

In a pure inductor the current lags the voltage by 90 degrees, and in a pure capacitor it leads by 90 degrees. Average power is V_rms I_rms cos(phi), and with phi = 90 degrees, cos(phi) = 0, so the average power is zero. Physically, during one part of the cycle the inductor or capacitor absorbs energy from the source (storing it in its magnetic or electric field), and during the next part it returns that energy to the source. No net energy is dissipated, unlike in a resistor.

AC can be easily stepped up to very high voltage using transformers, which rely on mutual induction and only work with changing (alternating) current. Transmitting power at high voltage means low current for the same power, and since transmission losses are I^2 R, reducing the current dramatically reduces energy lost as heat in the lines. At the destination, the voltage is stepped down again for safe use. This easy voltage transformation, impossible with simple DC, is why AC dominates power distribution.
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Last reviewed on 30 May 2026. Written and reviewed by subject-matter experts — read about our process.
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