Class 12 Physics NCERT REPRINT 2026-27

Chapter 7: Alternating Current

Complete concise study notes, AC in R, L, C circuits, Phasors, Series LCR resonance, Power Factor, Wattless Current, Transformers, solved examples 7.1–7.10, and exercises 7.1–7.8.

01 / Exam-Focused Notes

Chapter 7: Alternating Current

Topics covered: AC applied to Pure Resistor, Inductor & Capacitor, Phasor Diagrams, RMS Values ($Irms = i_m/√2$), Inductive ($X_L = ω L$) and Capacitive Reactance ($X_C = 1/ω C$), Series LCR Circuit Impedance ($Z = √R^2 + (X_L - X_C)^2$), Resonance ($ω_0 = 1/√LC$), Quality Factor ($Q = ω_0 L/R$), Power Factor ($\cosφ$), Wattless Current, and Step-Up/Step-Down Transformers ($V_s/V_p = N_s/N_p$).

100% SYLLABUS
Chapter Navigation • NCERT Sections 7.1 – 7.8
7.1-7.2AC in Resistor & RMS 7.3-7.4Phasors & Inductor 7.5AC in Capacitor 7.6Series LCR Resonance
7.1 & 7.2
NCERT Sections

Introduction & AC Voltage Applied to a Resistor

AC Voltage & Pure Resistance

7.2.1 AC Circuit with Pure Resistor ($R$) & RMS Concept

  • AC Voltage: An alternating voltage varies sinusoidally with time: $$v(t) = v_m \sin(ω t)$$ where $v_m$ is voltage amplitude and $ω = 2πν$ is angular frequency.
  • Current through Resistor: Applying Kirchhoff's loop rule $v - i R = 0$: $$i(t) = v_mR\sin(ω t) = i_m \sin(ω t), \qquad i_m = v_mR$$ Phase Relation: Current $i(t)$ and voltage $v(t)$ are strictly in phase ($φ = 0^°$).
  • Root Mean Square (RMS) / Effective Values:
    Because average current over a complete cycle is zero ($\langle i \rangle = 0$), power dissipated depends on $i^2$: $$\langle \sin^2(ω t) \rangle = 12 \implies Pavg = 12 i_m^2 R = Irms^2 R$$
$$Irms = i_m√2 = 0.707\,i_m, \qquad Vrms = v_m√2 = 0.707\,v_m, \qquad P = Vrms Irms = Irms^2 R$$

Standard Line Voltage: Indian household supply of $220 V$ is an RMS value with peak value $v_m = √2 × 220 V ≈ 311 V$.

Example 7.1

Light Bulb on AC Supply — Peak Voltage & Current

A light bulb is rated at $100 W$ for a $220 V$ supply. Find: (a) resistance of the bulb, (b) peak voltage of the source, (c) rms current through the bulb.

(a) Resistance ($R$): $R = V^2P = (220 V)^2100 W = 484\,\Omega$.
(b) Peak Voltage ($v_m$): $v_m = √2 V = 1.414 × 220 V = 311 V$.
(c) RMS Current ($I$): $I = PV = 100 W220 V = 0.454 A$.
Answers: (a) $R = 484\,\Omega$, (b) $v_m = 311 V$, (c) $I = 0.454 A$.
7.3 & 7.4
NCERT Sections

Phasors & AC Voltage Applied to an Inductor

Pure Inductor & Inductive Reactance

7.4.1 AC Circuit with Pure Inductor ($L$) & Phasor Representation

  • Phasor: A vector rotating about origin at angular speed $ω$, whose projection on vertical axis represents instantaneous harmonic scalar quantities ($v(t)$ and $i(t)$).
  • Loop Equation for Pure Inductor: $$v - Ldidt = 0 \implies didt = v_mL\sin(ω t)$$
  • Integration for Current: $$i(t) = -v_mω L\cos(ω t) = i_m \sin(ω t - π2)$$ Phase Relation: Current lags voltage by $π/2$ rad ($90^°$) (or one-quarter cycle $T/4$).
  • Inductive Reactance ($X_L$): Opposition offered by inductor to alternating current:
$$X_L = ω L = 2πν L \quad (SI Unit: Ohm, \Omega), \qquad i_m = v_mX_L, \quad Irms = VrmsX_L$$
Fig 7.4: Phasor Relationships in Pure Resistor (In Phase), Pure Inductor (I Lags V by 90°), Pure Capacitor (I Leads V by 90°)
V I Pure Resistor (ϕ = 0°) V I Pure Inductor (I Lags 90°) V I Pure Capacitor (I Leads 90°)
Example 7.2

Pure Inductor in AC Circuit — Reactance and Current

A pure inductor of $25.0 mH$ is connected to a source of $220 V, 50 Hz$. Find inductive reactance and rms current in the circuit.

Inductive Reactance ($X_L$):
$X_L = 2πν L = 2 × 3.1416 × 50 Hz × (25.0 × 10-3 H) = 7.85\,\Omega$.
RMS Current ($I$):
$I = VX_L = 220 V7.85\,\Omega = 28.0 A$.
Answers: $X_L = 7.85\,\Omega$, $I = 28.0 A$.
7.5
NCERT Section

AC Voltage Applied to a Capacitor

Pure Capacitor & Capacitive Reactance

7.5.1 AC Circuit with Pure Capacitor ($C$) & DC vs AC Blocking

  • Capacitor Behavior: A capacitor blocks DC after charging ($X_C \to ∞$ for $ν = 0$). For AC, charge continuously flows back and forth to charge/discharge the plates.
  • Derivation of Current: $$q(t) = C v(t) = C v_m \sin(ω t) \implies i(t) = dqdt = ω C v_m \cos(ω t) = i_m \sin(ω t + π2)$$ Phase Relation: Current leads voltage by $π/2$ rad ($90^°$).
  • Capacitive Reactance ($X_C$): Opposition offered by capacitor to AC:
$$X_C = 1ω C = 12πν C \quad (SI Unit: Ohm, \Omega), \qquad i_m = v_mX_C, \quad Irms = VrmsX_C$$
  • Average Power: $p_C(t) = v_m i_m2\sin(2ω t) \implies Pavg = 0$. Average power absorbed over a full cycle is zero.
Example 7.3

Lamp in Series with Capacitor — DC vs AC

A lamp is connected in series with a capacitor. Predict observations for DC and AC connections. What happens in each case if capacitance $C$ is reduced?

DC Connection: Capacitor charges immediately and blocks DC. Lamp does not glow at all. Reducing $C$ makes no difference.
AC Connection: Capacitor offers finite reactance $X_C = 1/ω C$; AC current flows and lamp shines. If $C$ is reduced, $X_C$ increases $\implies$ circuit current decreases and lamp glows less brightly.
Example 7.4

Capacitor in AC Circuit — Reactance and Frequency Doubling

A $15.0\,μF$ capacitor is connected to a $220 V, 50 Hz$ source. Find capacitive reactance, rms and peak current. What happens if frequency is doubled?

Capacitive Reactance ($X_C$):
$X_C = 12πν C = 12 × 3.1416 × 50 × (15.0 × 10-6) = 212\,\Omega$.
RMS Current ($I$): $I = VX_C = 220 V212\,\Omega = 1.04 A$.
Peak Current ($i_m$): $i_m = √2 I = 1.414 × 1.04 A = 1.47 A$.
Frequency Doubled ($2ν$): $X_C$ is halved ($106\,\Omega$) and current is doubled ($2.08 A$).
Answers: $X_C = 212\,\Omega$, $I = 1.04 A$, $i_m = 1.47 A$.
Example 7.5

Bulb and Inductor with Iron Rod Insertion

A light bulb and an open-coil inductor are connected to an AC source. An iron rod is inserted into the inductor. Does the glow (a) increase, (b) decrease, (c) remain unchanged?

Reasoning: Inserting iron core increases permeability $μ_r \implies$ self-inductance $L = μ_r μ_0 n^2 A l$ increases $\implies$ inductive reactance $X_L = ω L$ increases $\implies$ total impedance $Z = √R^2 + X_L^2$ increases $\implies$ circuit current decreases and voltage across bulb drops.
Answer: (b) The glow of the light bulb decreases.
7.6
NCERT Section

Series LCR Circuit & Resonance

Phasor Solution & Impedance Diagram

7.6.1 Series LCR Impedance ($Z$), Phase Angle ($φ$) & Resonance Condition

  • Kirchhoff Loop Equation: $$Ldidt + i R + qC = v_m \sin(ω t)$$
  • Phasor Addition & Impedance ($Z$):
    Voltages $V_L$ and $V_C$ are collinear and opposite in phase ($π$). Resultant reactive voltage is $(V_L - V_C)$ or $(V_C - V_L)$. $$v_m^2 = vRm^2 + (vCm - vLm)^2 = i_m^2 [R^2 + (X_C - X_L)^2]$$
$$Z = √R^2 + (X_C - X_L)^2 = √R^2 + (1ω C - ω L)^2, \qquad i_m = v_mZ, \quad I = VZ$$
Fig 7.15: Series LCR Resonance Curve & Bandwidth (Sharpness / Q-Factor)
Frequency ω → RMS Current I ω₀ = 1 / √(LC) I_max = V / R (Small R) Large R I_max / √2 (Half-Power) Quality Factor Q = ω₀ L / R = 1 / (R √(C/L))
  • Phase Difference ($φ$): $$\tanφ = X_C - X_LR = vCm - vLmvRm$$
    • If $X_C > X_L$: $φ > 0 \implies$ Circuit is predominantly capacitive (current leads voltage).
    • If $X_L > X_C$: $φ < 0 \implies$ Circuit is predominantly inductive (current lags voltage).
    • If $X_L = X_C$: $φ = 0 \implies$ Circuit is purely resistive (Resonance).
  • Series LCR Resonance:
    Occurs when $X_L = X_C \implies ω_0 L = 1ω_0 C$:
$$ω_0 = 1√LC, \qquad ν_0 = 12π√LC, \qquad Zmin = R, \qquad imax = v_mR$$
02 / Interactive Assessment

Concept Check & Practice Quiz

15 targeted MCQs covering RMS values, pure R/L/C circuits, series LCR resonance, power factor, wattless current, and transformers with in-depth explanations.

15 MCQS
0 / 15 Answered
Score: 0 (0%)
03 / Textbook Solutions

NCERT Exercises 7.1 – 7.8

Stepwise solutions for textbook exercise questions in NCERT Class 12 Physics Chapter 7 Alternating Current Reprint 2026-27.

8 QUESTIONS
Ex 7.1 Resistor on AC Supply

7.1 A $100\,\Omega$ resistor is connected to a $220 V, 50 Hz$ AC supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?

(a) $I = 2.2 A$ • (b) $P = 484 W$.
(a) RMS Current ($I$): $$I = VR = 220 V100\,\Omega = 2.2 A$$
(b) Net Power Consumed ($P$): $$P = V I = 220 V × 2.2 A = 484 W \quad (or P = I^2 R = (2.2)^2 × 100 = 484 W)$$
Ex 7.2 RMS and Peak Voltage/Current Relations

7.2 (a) The peak voltage of an AC supply is $300 V$. What is the rms voltage?
(b) The rms value of current in an AC circuit is $10 A$. What is the peak current?

(a) $Vrms = 212.1 V$ • (b) $i_m = 14.14 A$.
(a) RMS Voltage ($V$): $$V = v_m√2 = 300 V1.414 = 212.1 V$$
(b) Peak Current ($i_m$): $$i_m = √2 I = 1.414 × 10 A = 14.14 A$$
Ex 7.3 Pure Inductor Current

7.3 A $44 mH$ inductor is connected to $220 V, 50 Hz$ AC supply. Determine the rms value of the current in the circuit.

RMS Current $I = 15.9 A$.
Inductive Reactance ($X_L$): $$X_L = 2πν L = 2 × 227 × 50 s-1 × (44 × 10-3 H) = 447 × 2.2 = 13.83\,\Omega$$
RMS Current ($I$): $$I = VX_L = 220 V13.83\,\Omega = 15.9 A$$
Ex 7.4 Pure Capacitor Current

7.4 A $60\,μF$ capacitor is connected to a $110 V, 60 Hz$ AC supply. Determine the rms value of the current in the circuit.

RMS Current $I = 2.49 A$.
Capacitive Reactance ($X_C$): $$X_C = 12πν C = 12 × 3.1416 × 60 × (60 × 10-6) = 10^622619.5 = 44.21\,\Omega$$
RMS Current ($I$): $$I = VX_C = 110 V44.21\,\Omega = 2.49 A$$
Ex 7.5 Power Dissipation in Pure L and C

7.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle? Explain your answer.

Net Power Absorbed $= 0 W$ in both circuits.
Explanation:
The average power in any AC circuit is $P = V I \cosφ$.
  • In a pure inductor (Exercise 7.3), current lags voltage by $π/2 \implies φ = -90^° \implies \cosφ = 0 \implies P = 0$. Energy stored during one quarter-cycle in magnetic field is completely returned to source in the next quarter-cycle.
  • In a pure capacitor (Exercise 7.4), current leads voltage by $π/2 \implies φ = +90^° \implies \cosφ = 0 \implies P = 0$. Energy stored in electric field during charging is completely returned during discharging.
Ex 7.6 LC Natural Oscillation Frequency

7.6 A charged $30\,μF$ capacitor is connected to a $27 mH$ inductor. What is the angular frequency of free oscillations of the circuit?

Angular Frequency $ω_0 = 1.11 × 10^3 rad s-1$.
Formula: $$ω_0 = 1√LC$$
Given: $C = 30\,μF = 30 × 10-6 F$, $L = 27 mH = 27 × 10-3 H$.
Calculation: $$LC = (27 × 10-3) × (30 × 10-6) = 810 × 10-9 = 8.1 × 10-7 s^2$$ $$ω_0 = 1√8.1 × 10-7 = 19.0 × 10-4 = 1111 rad s-1 ≈ 1.11 × 10^3 rad s-1$$
Ex 7.7 Resonant Power in Series LCR

7.7 A series LCR circuit with $R = 20\,\Omega, L = 1.5 H$ and $C = 35\,μF$ is connected to a variable-frequency $200 V$ AC supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Average Power Transferred $P = 2000 W = 2.0 kW$.
Resonant Condition: At natural frequency (resonance), $X_L = X_C$, so impedance is purely resistive: $Z = R = 20\,\Omega$ and $\cosφ = 1$.
Calculation: $$P = V^2R = (200 V)^220\,\Omega = 4000020 = 2000 W = 2.0 kW$$
Ex 7.8 Detailed Resonant LCR Circuit Parameters

7.8 A series LCR circuit is connected to a variable frequency $230 V$ source with $L = 5.0 H, C = 80\,μF, R = 40\,\Omega$.
(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

(a) $ω_0 = 50 rad s-1$ ($ν_0 = 7.96 Hz$) • (b) $Z = 40\,\Omega, i_m = 8.13 A$ • (c) $V_R = 230 V, V_L = 1437.5 V, V_C = 1437.5 V, VLC = 0 V$.
(a) Resonant Frequency ($ω_0$):
$$ω_0 = 1√LC = 1√5.0 × (80 × 10-6) = 1√400 × 10-6 = 120 × 10-3 = 50 rad s-1$$ $ν_0 = ω_02π = 502 × 3.1416 = 7.96 Hz$.
(b) Impedance & Current Amplitude:
At resonance, $Z = R = 40\,\Omega$.
RMS current $I = VR = 230 V40\,\Omega = 5.75 A$.
Current amplitude $i_m = √2 I = 1.414 × 5.75 A = 8.13 A$.
(c) RMS Potential Drops:
$X_L = ω_0 L = 50 × 5.0 = 250\,\Omega$.
$X_C = 1ω_0 C = 150 × (80 × 10-6) = 250\,\Omega$.
  • $V_R = I R = 5.75 × 40 = 230 V$ (equals source voltage).
  • $V_L = I X_L = 5.75 × 250 = 1437.5 V$.
  • $V_C = I X_C = 5.75 × 250 = 1437.5 V$.
Because $V_L$ and $V_C$ are $180^°$ out of phase, the net voltage drop across LC is: $$VLC = |V_L - V_C| = 1437.5 V - 1437.5 V = 0 V$$
04 / Rapid Reference

Chapter Summary & Points to Ponder

8 core summary principles, Physical Quantities dimension and unit table, and all 11 official NCERT Points to Ponder for Chapter 7.

100% SYLLABUS

RMS & Peak Relations

I = i_m / √2 • V = v_m / √2

Effective DC equivalent for Joule heating; $220 V$ RMS has $311 V$ peak voltage.

Pure Resistor ($R$)

φ = 0 • P = I² R = V I

Voltage and current are in phase; maximum power dissipation occurs through resistor.

Pure Inductor ($L$)

X_L = ωL • Current lags by π/2 • P = 0

Reactance proportional to frequency; average power absorbed over a cycle is zero.

Pure Capacitor ($C$)

X_C = 1/(ωC) • Current leads by π/2 • P = 0

Blocks DC ($ν = 0$); offers finite reactance to AC; average power is zero.

Series LCR Impedance

Z = √[R² + (X_C − X_L)²]

Phase angle: $\tanφ = (X_C - X_L)/R$; power factor $\cosφ = R/Z$.

Series LCR Resonance

ω₀ = 1/√(LC) • Z_min = R • I_max = V/R

Requires both $L$ and $C$; voltages across $L$ and $C$ cancel completely ($VLC = 0$).

Power & Wattless Current

P = V I cosφ • I_q = I sinφ

Wattless current consumes no energy ($\cosφ = 0$ in pure $L$ or $C$).

Transformer Relations

V_s / V_p = N_s / N_p = I_p / I_s

Step-up ($N_s > N_p$): increases voltage, cuts transmission current & line losses.

NCERT Official Table

Physical Quantities, Symbols, Dimensions & Units

Physical Quantity Symbol Dimensions SI Unit Remarks / Formulas
RMS Voltage V $[M L^2 T-3 A-1]$ V (volt) $V = v_m / √2 = 0.707\,v_m$
RMS Current I $[A]$ A (ampere) $I = i_m / √2 = 0.707\,i_m$
Inductive Reactance X_L $[M L^2 T-3 A-2]$ Ω (ohm) $X_L = ω L = 2πν L$
Capacitive Reactance X_C $[M L^2 T-3 A-2]$ Ω (ohm) $X_C = 1/(ω C) = 1/(2πν C)$
Impedance Z $[M L^2 T-3 A-2]$ Ω (ohm) $Z = √R^2 + (X_C - X_L)^2$
Resonant Frequency ω₀, ν₀ $[T-1]$ rad s$-1$ / Hz $ω_0 = 1/√LC, \quad ν_0 = 1/(2π√LC)$
Quality Factor Q $[M^0 L^0 T^0 A^0]$ Dimensionless $Q = ω_0 LR = 1ω_0 C R = 1R√LC$
Power Factor cosφ $[M^0 L^0 T^0 A^0]$ Dimensionless $\cosφ = R/Z = P/(V I)$
NCERT Official

Points to Ponder (11 Principles)

  1. RMS Specification: Standard values given for AC voltage or current (e.g. $240 V$ wall socket) always refer to RMS values. The amplitude is $v_m = √2 × 240 V ≈ 340 V$.
  2. Power Rating: The power rating of an AC appliance refers to its time-averaged power.
  3. Non-Negative Power: The net power consumed in an AC circuit is never negative.
  4. Definition of AC Ampere: Because AC alternates direction, the AC ampere cannot be defined via magnetic attraction between wires (which averages to zero). It is defined through Joule heating: $1 A$ RMS is the alternating current that generates the same average heat as $1 A$ of steady DC current in the same resistance.
  5. Vectorial Addition of Voltages: In AC circuits, voltages cannot be added algebraically. For an RC series circuit, $Vtotal = √V_R^2 + V_C^2 \ne V_R + V_C$ because $V_C$ lags $V_R$ by $90^°$.
  6. Phasors as Analytical Tools: Phasors are rotating vectors used as graphical tools to combine sinusoidally oscillating scalars; voltages and currents remain scalar physical quantities.
  7. Absence of Power Loss in Pure L and C: Pure inductances and capacitances do not dissipate energy. The only element that dissipates electrical energy in an AC circuit is resistance $R$.
  8. Requirement for Resonance: Series resonance requires the simultaneous presence of both $L$ and $C$ to allow reactive voltage cancellation. A pure RL or RC circuit cannot exhibit resonance.
  9. Power Factor Significance: The power factor $\cosφ = R/Z$ measures how effectively the circuit converts apparent power $VI$ into true useful power.
  10. Motors vs Generators: Motors convert electrical energy into mechanical energy; generators convert mechanical energy into electrical energy.
  11. Transformers and Energy Conservation: A step-up transformer increases voltage ($V_s > V_p$) but reduces current ($I_s < I_p$) by the same proportion, preserving energy conservation ($V_p I_p = V_s I_s$).
05 / Practice Tests

3-Tier Practice Tests

Level 1 (Foundation), Level 2 (Application), and Level 3 (Challenge). Select options and submit to evaluate your preparation.

24 QUESTIONS
Level 1 Active • 8 Questions