NCERT Solutions for Class 12 Physics Chapter 7: Alternating Current

Concept and context. An alternating voltage v = v_m sin ωt applied across a resistor R drives a current i = i_m sin ωt with i_m = v_m/R: voltage and current are in phase (NCERT Fig. 7.1 and 7.2). The instantaneous power p = i²R = i_m²R sin²ωt is always positive and its average over a cycle is P = ½ i_m²R. Because ac values change continuously, they are expressed through the root-mean-square (rms) values, defined so that the ac produces the same average heating as a dc of the same value: I_rms = i_m/√2 ≈ 0.707 i_m and V_rms = v_m/√2. With rms values the power takes the dc form P = I_rms² R = V_rms I_rms = V_rms²/R. Unless stated otherwise, a quoted ac voltage (such as 220 V) is the rms value.

Given: R = 100 Ω; V_rms = 220 V; ν = 50 Hz (not needed for a resistor — a resistor’s opposition does not depend on frequency).

(a) rms current

1. Ohm’s law for rms values: I_rms = V_rms/R = 220/100 = 2.2 A.
2. (Peak values for completeness: v_m = √2 × 220 = 311 V; i_m = √2 × 2.2 = 3.11 A.)

(b) Net power consumed over a full cycle

1. Average power P = I_rms² R = (2.2)² × 100 = 4.84 × 100 = 484 W.
2. Check: P = V_rms I_rms = 220 × 2.2 = 484 W ✓ and P = V²/R = 220²/100 = 484 W ✓.
3. Explanation: the instantaneous power p = i_m² R sin²ωt varies between 0 and i_m²R = 968 W (twice the average), and its average over a full cycle is half of the peak: ½ i_m² R = ½ × (3.11)² × 100 = 484 W.

I_rms = 2.2 A ; P_avg = 484 W

• Why use rms values? The simple average of a sinusoidal current over a cycle is zero, but the heating effect (∝ i²) is not zero; the rms value is the square root of the mean of i². One ampere of rms ac heats a resistor exactly as much as one ampere of dc.
• Frequency independence: for a pure resistor the current does not depend on the frequency; the 50 Hz is irrelevant here (but is needed for inductors and capacitors).

Conclusion: the rms current is 2.2 A and the net (average) power consumed over a full cycle is 484 W.

Concept and context. For a sinusoidal alternating quantity v = v_m sin ωt, the rms value is the square root of the average of v² over a cycle. Since the average of sin²ωt over a full cycle is ½, v²_avg = v_m²/2 and so V_rms = v_m/√2 = 0.707 v_m; similarly I_rms = i_m/√2 (NCERT Fig. 7.3 shows how the rms current is related to the peak current: i² has a mean value i_m²/2). Hence peak = √2 × rms ≈ 1.414 × rms. The rms value is the DC equivalent for heating: an ac with rms value V produces the same average heat in a resistor as a dc voltage V.

(a) rms voltage from the peak voltage 300 V

1. V_rms = v_m/√2 = 300/1.414.
2. V_rms ≈ 212 V (212.1 V).

(b) Peak current from the rms current 10 A

1. i_m = √2 I_rms = 1.414 × 10.
2. i_m ≈ 14.1 A (14.14 A).

V_rms = v_m/√2 = 212 V ; i_m = √2 I_rms = 14.1 A

• Derivation: I_rms² = ⟨i²⟩ = i_m²⟨sin²ωt⟩ = i_m²/2 ⇒ I_rms = i_m/√2.
• Practical significance: a household supply quoted as 230 V (rms) swings between ±325 V; insulation and capacitors must be rated for the peak value, while heating and power calculations use the rms value.
• The relation applies to sinusoidal waveforms only (a square wave has rms = peak).

Conclusion: (a) the rms voltage is ≈ 212 V; (b) the peak current is ≈ 14.1 A.

Concept and context. When an ac voltage v = v_m sin ωt is applied across a pure inductor L, the changing current induces a back emf, and Kirchhoff’s loop rule gives v = L di/dt. Integrating: i = −(v_m/ωL) cos ωt = i_m sin(ωt − π/2). Thus the current amplitude is i_m = v_m/X_L with the inductive reactance X_L = ωL = 2πνL (unit ohm), and the current lags the voltage by π/2 (phasor diagram, NCERT Fig. 7.4 and 7.6). Reactance is the opposition of the inductor to ac; it grows with frequency (for dc, ω = 0 and X_L = 0). The same relation holds for rms values: I_rms = V_rms/X_L.

Given: L = 44 mH = 44 × 10⁻³ H; V_rms = 220 V; ν = 50 Hz.

1. Inductive reactance: X_L = ωL = 2πνL = 2 × 3.14 × 50 × 44 × 10⁻³ = 13.82 Ω.
2. rms current: I_rms = V_rms/X_L = 220/13.82 = 15.9 A.
3. Peak current (extra): i_m = √2 × 15.9 = 22.5 A.

X_L = 13.8 Ω ; I_rms ≈ 15.9 A

• Contrast with Example 7.2 in the text (25 mH inductor: X_L = 7.85 Ω, I = 28 A): a larger inductance gives a larger reactance and a smaller current.
• Frequency dependence: at 100 Hz the reactance doubles and the current halves (7.95 A); at dc the inductor (ignoring its resistance) would carry an enormous current — that is why ac “chokes” are used.
• Power: the average power absorbed is zero (see Exercise 7.5), so although 15.9 A flows, no net energy is dissipated — the inductor merely stores and returns energy.

Conclusion: the rms value of the current in the circuit is ≈ 15.9 A, lagging the voltage by π/2.

Concept and context. When an ac voltage v = v_m sin ωt is applied across a capacitor C, the charge q = Cv varies with time and the current i = dq/dt = ωC v_m cos ωt = i_m sin(ωt + π/2), where i_m = v_m/X_C and the capacitive reactance X_C = 1/ωC = 1/(2πνC). The current leads the voltage by π/2 (NCERT Fig. 7.7, 7.8). The capacitive reactance is inversely proportional to the frequency and the capacitance: for dc (ω = 0) X_C is infinite — a capacitor blocks dc — while at high frequency it offers little opposition. For rms values I_rms = V_rms/X_C.

Given: C = 60 μF = 60 × 10⁻⁶ F; V_rms = 110 V; ν = 60 Hz.

1. Capacitive reactance: X_C = 1/(2πνC) = 1/(2 × 3.14 × 60 × 60 × 10⁻⁶) = 1/(2.26 × 10⁻²) = 44.2 Ω.
2. rms current: I_rms = V_rms/X_C = 110/44.2 = 2.49 A.
3. Peak current (extra): i_m = √2 × 2.49 = 3.52 A.

X_C = 44.2 Ω ; I_rms ≈ 2.49 A

• Contrast with the inductor (Exercise 7.3): X_C decreases with frequency (∝ 1/ν) while X_L increases (∝ ν). At twice the frequency this current would be doubled (4.98 A) — compare NCERT Example 7.4.
• Why the current exists even though the dielectric is an insulator: the capacitor is repeatedly charged and discharged; the “current” is the charging/discharging current in the circuit, not a current through the dielectric.
• Average power absorbed over a cycle is zero (Exercise 7.5).

Conclusion: the rms value of the current in the circuit is ≈ 2.49 A, leading the voltage by π/2.

Concept and context. The instantaneous power delivered by an ac source to a circuit is p = v i. Over one full cycle the average power is P = V_rms I_rms cos φ, where φ is the phase difference between the voltage and the current and cos φ is the power factor. In a pure inductor the current lags the voltage by π/2 and in a pure capacitor it leads by π/2; in both cases φ = ±90° and cos φ = 0. Hence no net power is absorbed by an ideal inductor or capacitor over a complete cycle — such elements store energy in one part of the cycle and return it to the source in another. Only a resistor dissipates energy (as heat).

(a) Inductor (Exercise 7.3)

1. v = v_m sin ωt and i = i_m sin(ωt − π/2) = −i_m cos ωt.
2. Instantaneous power p = v i = −v_m i_m sin ωt cos ωt = −(v_m i_m/2) sin 2ωt.
3. Average over a complete cycle: ⟨sin 2ωt⟩ = 0 ⇒ P = 0. Equivalent: P = V_rms I_rms cos(π/2) = 0.

(b) Capacitor (Exercise 7.4)

1. v = v_m sin ωt and i = i_m sin(ωt + π/2) = i_m cos ωt.
2. p = v i = (v_m i_m/2) sin 2ωt ⇒ average over a cycle = 0 ⇒ P = 0. Equivalent: P = V_rms I_rms cos(−π/2) = 0.

Explanation in terms of energy

• Inductor: during the quarter cycle in which the current grows, the source supplies energy ½Li² that is stored in the magnetic field of the inductor (p positive); in the next quarter cycle, as the current falls, the field collapses and returns the same energy to the source (p negative). Over a full cycle the energy absorbed equals the energy returned.
• Capacitor: while the voltage rises, the capacitor stores energy ½Cv² in its electric field (p positive); as the voltage falls it discharges and returns this energy to the source (p negative).
• Result: the net power absorbed over a complete cycle is zero in both circuits, even though currents of 15.9 A (Exercise 7.3) and 2.49 A (Exercise 7.4) flow; such currents are called wattless currents. (In real inductors the resistance of the winding gives a small non-zero power loss, and real capacitors have a small dielectric loss.)

Conclusion: the net power absorbed over a complete cycle is zero in both the inductor circuit (Exercise 7.3) and the capacitor circuit (Exercise 7.4), because the current and voltage are 90° out of phase (cos φ = 0) and energy is merely exchanged between the source and the field.

Concept and context. When a charged capacitor is connected across an inductor, the capacitor discharges through the inductor; the current builds up and the inductor’s magnetic field stores energy; when the capacitor is fully discharged the inductor’s induced emf keeps the current flowing and recharges the capacitor with the opposite polarity, and the process repeats. The circuit therefore executes free (natural) electrical oscillations, with energy swinging between the electric field of the capacitor (q²/2C) and the magnetic field of the inductor (½Li²) — exactly like a mass on a spring exchanging potential and kinetic energy. Applying Kirchhoff’s loop rule: L d²q/dt² + q/C = 0, which is the equation of simple harmonic motion with ω₀ = 1/√(LC). (This is also the resonant frequency of a series LCR circuit, at which X_L = X_C.)

Given: C = 30 μF = 30 × 10⁻⁶ F; L = 27 mH = 27 × 10⁻³ H.

1. Formula: ω₀ = 1/√(LC).
2. LC = 27 × 10⁻³ × 30 × 10⁻⁶ = 810 × 10⁻⁹ = 8.1 × 10⁻⁷ s².
3. √(LC) = √(8.1 × 10⁻⁷) = 9.0 × 10⁻⁴ s.
4. ω₀ = 1/(9.0 × 10⁻⁴) = 1.11 × 10³ rad s⁻¹ (1111 rad s⁻¹).

ω₀ = 1/√(LC) = 1.1 × 10³ rad s⁻¹

• Damping: a real circuit has some resistance, so the oscillations die out (damped); for an ideal LC circuit the amplitude remains constant.
• Analogy: L ↔ mass, 1/C ↔ spring constant, q ↔ displacement, i ↔ velocity.
• Application: the same expression gives the resonant frequency used in radio tuning circuits (vary C to change ω₀).

Conclusion: the angular frequency of the free oscillations of the LC circuit is ≈ 1.1 × 10³ rad s⁻¹ (frequency about 177 Hz).

Concept and context. In a series LCR circuit driven by v = v_m sin ωt, the current amplitude is i_m = v_m/Z with impedance Z = √[R² + (ωL − 1/ωC)²]. When the supply frequency equals the natural frequency ω₀ = 1/√(LC), the inductive and capacitive reactances are equal (X_L = X_C), their voltages (opposite in phase) cancel, the impedance is minimum, Z = R, the current is maximum, i_m = v_m/R, and the current is in phase with the voltage (φ = 0, power factor cos φ = 1). This is resonance (NCERT Figs 7.12–7.14). The average power transferred to the circuit is P = V_rms I_rms cos φ = I_rms² R, all dissipated in the resistor.

Given: R = 20 Ω; L = 1.5 H; C = 35 μF = 35 × 10⁻⁶ F; V_rms = 200 V; at resonance ω = ω₀.

1. Natural frequency: ω₀ = 1/√(LC) = 1/√(1.5 × 35 × 10⁻⁶) = 1/√(5.25 × 10⁻⁵) = 1/(7.25 × 10⁻³) = 138 rad s⁻¹ (ν₀ = 22 Hz).
2. Check reactances: X_L = ω₀L = 138 × 1.5 = 207 Ω; X_C = 1/(ω₀C) = 1/(138 × 35 × 10⁻⁶) = 207 Ω — equal, so Z = R = 20 Ω.
3. Current at resonance: I_rms = V_rms/Z = 200/20 = 10 A.
4. Average power: P = I_rms² R = (10)² × 20 = 2000 W (2 kW). Also P = V_rms I_rms cos φ = 200 × 10 × 1 = 2000 W ✓ (cos φ = R/Z = 1).

P_avg = V²/R = 200²/20 = 2000 W

• Remarks: although the reactances (207 Ω each) are large, their voltages (10 A × 207 Ω = 2070 V each!) are 180° out of phase and cancel, so the net voltage across L and C together is zero and the full source voltage 200 V appears across R. These large voltages across L and C at resonance (greater than the applied voltage) are why resonant circuits can produce high voltage “magnification” (quality factor).
• Away from resonance Z > R, the current falls and the power factor drops below 1, so less power is transferred.

Conclusion: at the natural frequency the impedance equals R, the rms current is 10 A, and the average power transferred to the circuit in one cycle is 2000 W (2 kW).

Concept and context. A series LCR circuit driven by an ac source has impedance Z = √[R² + (X_L − X_C)²], with X_L = ωL and X_C = 1/ωC. Resonance occurs at the frequency ω₀ for which X_L = X_C: ω₀ = 1/√(LC) (ν₀ = ω₀/2π). At resonance the impedance is minimum (= R), the current amplitude is maximum (= v_m/R) and is in phase with the source voltage; the voltages across the inductor and the capacitor are equal in magnitude and opposite in phase (180° apart), so they cancel in the series combination and the entire source voltage appears across the resistor. Resonance requires both L and C in the circuit.

Given: L = 5.0 H; C = 80 μF = 80 × 10⁻⁶ F; R = 40 Ω; V_rms = 230 V.

(a) Resonant frequency

1. ω₀ = 1/√(LC) = 1/√(5.0 × 80 × 10⁻⁶) = 1/√(4.0 × 10⁻⁴) = 1/(2.0 × 10⁻²) = 50 rad s⁻¹.
2. ν₀ = ω₀/2π = 50/(2 × 3.14) = 7.96 Hz ≈ 8 Hz.

(b) Impedance and amplitude of current at resonance

1. At resonance X_L = X_C so Z = √[R² + 0] = R = 40 Ω.
2. Peak voltage v_m = √2 × 230 = 325.3 V.
3. Amplitude of current i_m = v_m/Z = 325.3/40 = 8.13 A (rms current I_rms = 230/40 = 5.75 A).

(c) rms potential drops across the three elements

1. Resistor: V_R = I_rms R = 5.75 × 40 = 230 V (the entire source voltage).
2. Inductor: X_L = ω₀L = 50 × 5.0 = 250 Ω; V_L = I_rms X_L = 5.75 × 250 = 1437.5 V ≈ 1.44 kV.
3. Capacitor: X_C = 1/(ω₀C) = 1/(50 × 80 × 10⁻⁶) = 250 Ω; V_C = I_rms X_C = 5.75 × 250 = 1437.5 V.
4. Potential drop across the LC combination: in a series LC combination the voltage across L leads the current by π/2 and that across C lags it by π/2, so they are 180° out of phase and V_LC = |V_L − V_C| = 1437.5 − 1437.5 = 0. (Equivalently: the impedance of the LC combination is |X_L − X_C| = 0 and V_LC = I × 0 = 0.)

V_R = 230 V = source ; V_L = V_C = 1437.5 V, V_LC = 0

• Surprising fact: the voltages across L and C (1.44 kV) are about 6 times larger than the source voltage (230 V) — they do not violate energy conservation since they are opposite in phase and cancel; this “voltage magnification” at resonance is exploited in radio tuning but is also a hazard (insulation must withstand it).
• Check by phasors: the source voltage equals the vector sum √[V_R² + (V_L − V_C)²] = √[230² + 0²] = 230 V ✓.

Conclusion: (a) resonance occurs at ω₀ = 50 rad s⁻¹ (ν₀ ≈ 8 Hz); (b) Z = R = 40 Ω and the current amplitude is 8.13 A; (c) V_R = 230 V, V_L = V_C = 1437.5 V, and the potential drop across the LC combination is zero at resonance.