Alternating Current – Class 12 Physics Chapter 7 Notes & NCERT PDF

Introduction

The electricity in our homes is not steady dc. It is an alternating voltage that varies like a sine function, reversing direction 100 times each second at 50 Hz. AC is preferred because transformers can easily and efficiently raise or lower ac voltages, which makes long-distance transmission economical. Nikola Tesla and George Westinghouse championed ac over Edison's dc. In this chapter you will see how resistors, inductors and capacitors behave with ac, what phasors, impedance and resonance are, why the power factor matters, and how a transformer works. Tuning a radio to your favourite station is an everyday use of resonance in an ac circuit.

Key Concepts

7.2 AC Voltage Applied to a Resistor

With v = v_m sin ωt, the current is i = i_m sin ωt with i_m = v_m/R: voltage and current are in phase. The average current over a cycle is zero, but the average power is not: p̄ = ½i_m²R. The rms (effective) values are I = i_m/√2 = 0.707 i_m and V = v_m/√2 = 0.707 v_m, giving P = I²R = V²/R = VI, just like dc. Household 220 V is an rms value; its peak is √2 × 220 ≈ 311 V.

7.3 Phasors

A phasor is a vector rotating anticlockwise with angular speed ω. Its length is the peak value and its vertical projection gives the instantaneous value. Phasors are a tool for adding sinusoidal quantities; voltage and current are still scalars.

7.4 AC Voltage Applied to an Inductor

L di/dt = v_m sin ωt gives i = i_m sin(ωt − π/2). The current lags the voltage by π/2 (a quarter cycle). Inductive reactance X_L = ωL, so i_m = v_m/X_L. X_L rises with frequency. Average power over a cycle is zero.

7.5 AC Voltage Applied to a Capacitor

q = Cv gives i = ωCv_m cos ωt = i_m sin(ωt + π/2). The current leads the voltage by π/2. Capacitive reactance X_C = 1/ωC, falling with frequency; a capacitor blocks dc (ω = 0). Average power is again zero.

7.6 Series LCR Circuit

The voltages across R, L and C are v_R (in phase with i), v_L (leads i by π/2) and v_C (lags i by π/2). Adding the phasors: v_m² = v_Rm² + (v_Cm − v_Lm)², so i_m = v_m/Z, where the impedance is Z = √[R² + (X_C − X_L)²] and the phase angle is tan φ = (X_C − X_L)/R. If X_C > X_L the circuit is capacitive (current leads); if X_L > X_C it is inductive (current lags).

7.6.2 Resonance: when X_L = X_C, Z = R is minimum and current is maximum. This occurs at ω₀ = 1/√(LC). The book's curve for L = 1.00 mH and C = 1.00 nF peaks at ω₀ = 1.00 × 10⁶ rad/s, and is sharper for R = 100 Ω than for R = 200 Ω. Resonance needs both L and C; there is no resonance in an RL or RC circuit. Radio tuning adjusts C so that ω₀ matches the station frequency. A metal detector at an airport also works on resonance: metal changes the inductance, shifting resonance and triggering an alarm.

7.7 Power in an AC Circuit: the Power Factor

Average power P = VI cos φ = I²Z cos φ. The factor cos φ = R/Z is the power factor.

A low power factor means a larger current for the same power and hence larger I²R transmission losses; a capacitor in parallel with an inductive load improves it.

7.8 Transformers

A transformer uses mutual induction between a primary (N_p turns) and a secondary (N_s turns) wound on a laminated iron core. For an ideal transformer, V_s/V_p = N_s/N_p and, since V_pI_p = V_sI_s, I_s/I_p = N_p/N_s. Step-up: N_s > N_p; step-down: N_s < N_p. Energy losses: flux leakage, resistance of windings (reduced with thick wire), eddy currents (reduced by laminating the core) and hysteresis (reduced by a low-loss core material). Power stations step up voltage to cut I²R loss in transmission, then substations step it down to about 240 V for homes.

Worked Examples

Example 1 (Book Example 7.1). A 100 W, 220 V bulb: R = V²/P = 220²/100 = 484 Ω; peak voltage v_m = √2 × 220 ≈ 311 V; I = P/V = 100/220 ≈ 0.454 A.

Example 2 (Book Example 7.4). A 15.0 µF capacitor on 220 V, 50 Hz: X_C = 1/(2π × 50 × 15 × 10⁻⁶) ≈ 212 Ω; I = 220/212 ≈ 1.04 A; i_m = 1.41 × 1.04 ≈ 1.47 A. Doubling the frequency halves X_C and doubles the current.

Example 3 (Book Example 7.6). R = 200 Ω and C = 15.0 µF in series on 220 V, 50 Hz. Z = √(200² + 212.3²) ≈ 291.7 Ω, I ≈ 0.755 A, V_R ≈ 151 V, V_C ≈ 160.3 V. Note that 151 + 160.3 ≠ 220; the phasor sum √(151² + 160.3²) = 220 V.

Example 4 (Book Examples 7.8 and 7.9). v_m = 283 V, 50 Hz, R = 3 Ω, L = 25.48 mH, C = 796 µF. X_L = 8 Ω, X_C = 4 Ω, Z = √(3² + 4²) = 5 Ω; φ = tan⁻¹(−4/3) = −53.1° (current lags). V = 283/√2 ≈ 200 V, I = 40 A, P = I²R = 4800 W, power factor = 3/5 = 0.6. At resonance, ω₀ = 1/√(LC) ≈ 222.1 rad/s, ν ≈ 35.4 Hz, Z = R = 3 Ω, I = 200/3 ≈ 66.7 A and P ≈ 13.35 kW.

Example 5 (Transformer). N_p = 100, N_s = 200, input 220 V at 10 A: output 440 V at 5.0 A (ideal).

Formula and Key-Facts Sheet

Board Exam Focus

Common questions: define rms value; derive the phase relation and reactance for L and C; draw the phasor diagram and derive Z for series LCR; resonance condition and graph; power factor cases; transformer principle, equations and losses; numericals from Exercises 7.1–7.8.

Answer outline: "Derive the impedance of a series LCR circuit." Draw I along the x-axis; V_R along I, V_L up (π/2 ahead), V_C down. Resultant V² = V_R² + (V_C − V_L)². Substitute V_R = IR, V_L = IX_L, V_C = IX_C: V = I√[R² + (X_C − X_L)²], so Z = √[R² + (X_C − X_L)²], tan φ = (X_C − X_L)/R.

Answer outline: Exercise 7.8. L = 5.0 H, C = 80 µF, R = 40 Ω, 230 V. ω₀ = 1/√(5 × 80 × 10⁻⁶) = 50 rad/s (ν ≈ 7.96 Hz). Z = R = 40 Ω, I = 230/40 = 5.75 A (amplitude 8.13 A). V_R = 230 V; V_L = Iω₀L = 1437.5 V; V_C = I/(ω₀C) = 1437.5 V. V_L and V_C are opposite in phase, so the LC drop is zero.

Answer outline: Exercises 7.3 and 7.4. 44 mH on 220 V, 50 Hz: X_L ≈ 13.8 Ω, I ≈ 15.9 A. 60 µF on 110 V, 60 Hz: I = VωC ≈ 2.49 A. Net power in each is zero because φ = π/2 (Exercise 7.5).

Key Terms

Common Mistakes