PhysicsISC Class 12 · Chapter 7

Alternating Current ISC Class 12: Notes, Phasors and Numericals

Updated 14 October 2026 · Fact-checked

Alternating Current is Chapter 7 of ISC Class 12 Physics. It covers peak, mean and r.m.s. values, a.c. through a pure resistor, inductor and capacitor with phasor diagrams, inductive and capacitive reactance, the series LCR circuit (impedance, phase, resonance, Q factor), power and power factor, wattless current, LC oscillations and the a.c. generator.

What Alternating Current is about

Mains electricity is alternating: the voltage varies sinusoidally, $V = V_{0} sin ω t$ with $ω = 2 π f$ (in India f = 50 Hz). Because the average of a sine wave over a full cycle is zero, we describe a.c. by its r.m.s. value, $I_{r m s} = I_{0} / 2$ , the steady current that would produce the same heating, and by its mean value over half a cycle, $\frac{2 I _{0}}{π}$ . You must obtain both relations.

In a pure resistor current and voltage are in phase. In a pure inductor the current lags the voltage by $π /2$ , and the opposition is the inductive reactance $X_{L} = ω L$ . In a pure capacitor the current leads by $π /2$ , with capacitive reactance $X_{C} = \frac{1}{ω C}$ . The syllabus asks for phasor diagrams for each and graphs of $X_{L}$ and $X_{C}$ against frequency.

In a series LCR circuit the same current flows through all three; adding the voltage phasors gives $V^{2} = V_{R}^{2} + (V_{L} - V_{C})^{2}$ , so the impedance is $Z = R^{2} + (X_{L} - X_{C})^{2}$ and the phase angle satisfies $tan ϕ = \frac{X _{L} - X _{C}}{R}$ . At resonance $X_{L} = X_{C}$ , Z = R (minimum), current is maximum and $f_{0} = \frac{1}{2 π L C}$ . Bandwidth and Q factor are given without derivation.

Power in an a.c. circuit is $P = V_{r m s} I_{r m s} cos ϕ$ , where $cos ϕ = R / Z$ is the power factor. In a pure L or C circuit $cos ϕ = 0$ : the current is wattless, which is why a choke coil controls current with almost no power loss. An LC circuit oscillates at $ω_{0} = 1/ L C$ (qualitative). Finally, the a.c. generator produces $ε = N A B ω sin ω t$ ; you need its principle, construction, theory and the differences between a.c. and d.c.

This chapter is part of the 16 marks for magnetism, induction and a.c.; LR and LCR numericals, impedance graphs and identifying a component from V and I expressions have all appeared in recent papers.

Weightage: Alternating Current is part of the unit Magnetic Effects of Current and Magnetism; Electromagnetic Induction and Alternating Currents, which carries 16 of the 70 marks in the ISC Physics Paper 1 (Theory), shared with 3 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Physics syllabus, examination year 2027, distribution of marks

Key concepts

Peak, mean and r.m.s. values: For $I = I_{0} sin ω t$ , the average over a full cycle is zero. Mean value over half a cycle: $I_{m e an} = \frac{1}{T /2} \int_{0}^{T /2} I_{0} sin ω t d t = \frac{2 I _{0}}{π} = 0.637 I_{0}$ . R.m.s. value: the steady current giving the same heat in a resistor. Heat in time T is $\int_{0}^{T} I^{2} R d t = I_{0}^{2} R \int_{0}^{T} sin^{2} ω t d t = \frac{I _{0}^{2} R T}{2}$ ; equating to $I_{r m s}^{2} R T$ gives $I_{r m s} = \frac{I _{0}}{2} = 0.707 I_{0}$ , and likewise $V_{r m s} = V_{0} / 2$ . A.c. meters read r.m.s. values; '230 V mains' is the r.m.s. value. Form factor $\frac{I _{r m s}}{I _{m e an}} = \frac{π}{2 2} = 1.11$ .; Worked example
The mains supply is 220 V (r.m.s.), 50 Hz. Find its peak value and its mean value over a half cycle.
Try it first, then show the solutionSolution
$V_{0} = 2 V_{r m s} = 1.414 \times 220 = 311 V$
$V_{m e an} = \frac{2 V _{0}}{π} = \frac{2 \times 311}{3.1416} = 198 V$
Answer: Peak 311 V; half-cycle mean 198 V
Pure R, L and C circuits: phase and reactance: Apply $V = V_{0} sin ω t$ . Resistor: $I = \frac{V _{0}}{R} sin ω t$ , in phase with V. Inductor: Kirchhoff's loop rule gives $V_{0} sin ω t = L \frac{d I}{d t}$ , so $I = I_{0} sin (ω t - π /2)$ : current lags voltage by $π /2$ , with $I_{0} = \frac{V _{0}}{ω L}$ ; inductive reactance $X_{L} = ω L = 2 π f L$ , proportional to f (a straight line through the origin on an $X_{L}$ -f graph). Capacitor: $q = C V_{0} sin ω t$ , $I = \frac{d q}{d t} = I_{0} sin (ω t + π /2)$ : current leads voltage by $π /2$ , with capacitive reactance $X_{C} = \frac{1}{ω C}$ , inversely proportional to f (a rectangular hyperbola). So a capacitor blocks d.c. (f = 0, $X_{C} \to \infty$ ) and an inductor passes d.c. freely ( $X_{L} = 0$ ). Reactances are measured in ohms.; Worked example
Find the reactance of (a) a 0.5 H inductor and (b) a 20 µF capacitor on a 50 Hz supply. Which would let more current through on 220 V?
Try it first, then show the solutionSolution
(a) $X_{L} = 2 π f L = 2 π \times 50 \times 0.5 = 157.1 Ω$
(b) $X_{C} = \frac{1}{2 π f C} = \frac{1}{2 π \times 50 \times 20 \times 1 0 ^{- 6}} = 159.2 Ω$
The inductor's reactance is slightly smaller, so it passes slightly more current ( $\frac{220}{157.1} = 1.40 A$ against $\frac{220}{159.2} = 1.38 A$ ).
Answer: (a) 157.1 Ω (b) 159.2 Ω; the inductor, marginally
Series LCR circuit: impedance and phase: The same current $I = I_{0} sin ω t$ flows through R, L and C. Phasors: $V_{R}$ in phase with I, $V_{L}$ ahead of I by $9 0^{\circ}$ , $V_{C}$ behind I by $9 0^{\circ}$ . $V_{L}$ and $V_{C}$ are opposite, so their resultant is $(V_{L} - V_{C})$ , perpendicular to $V_{R}$ . Hence $V^{2} = V_{R}^{2} + (V_{L} - V_{C})^{2}$ , and dividing by $I^{2}$ : $Z = R^{2} + (X_{L} - X_{C})^{2}$ . The voltage leads the current by φ where $tan ϕ = \frac{X _{L} - X _{C}}{R}$ , so $I = I_{0} sin (ω t - ϕ)$ with $I_{0} = V_{0} / Z$ . If $X_{L} > X_{C}$ the circuit is inductive (V leads I); if $X_{C} > X_{L}$ it is capacitive (I leads V). Special cases: RL circuit $Z = R^{2} + X_{L}^{2}$ ; RC circuit $Z = R^{2} + X_{C}^{2}$ . Never add R and X directly.; Worked example
A series circuit has R = 30 Ω, $X_{L} = 80 Ω$ and $X_{C} = 40 Ω$ across 200 V (r.m.s.). Find Z, the current, the phase angle and the voltage across each component.
Try it first, then show the solutionSolution
$Z = 3 0^{2} + (80 - 40)^{2} = 900 + 1600 = 50 Ω$ ; $I = \frac{200}{50} = 4 A$
$tan ϕ = \frac{40}{30} = 1.33$ , $ϕ = 5 3^{\circ}$ (voltage leads: inductive)
$V_{R} = 120 V$ , $V_{L} = 320 V$ , $V_{C} = 160 V$ ; check $12 0^{2} + (320 - 160)^{2} = 200 V$
Answer: Z = 50 Ω; 4 A; 53° (inductive); 120 V, 320 V, 160 V
Resonance, bandwidth and Q factor: As frequency rises, $X_{L}$ increases and $X_{C}$ decreases. At the resonant frequency they are equal: $ω_{0} L = \frac{1}{ω _{0} C}$ , so $ω_{0} = \frac{1}{L C}$ and $f_{0} = \frac{1}{2 π L C}$ . Then Z = R (minimum), the current is maximum ( $V / R$ ), V and I are in phase and the power factor is 1. The I-f graph is a peak at $f_{0}$ ; the Z-f graph is a dip to R at $f_{0}$ , rising on both sides. The bandwidth is the frequency range over which the power is at least half the maximum, and the Q factor $Q = \frac{ω _{0} L}{R} = \frac{1}{R} \frac{L}{C} = \frac{f _{0}}{ban d w i d t h}$ (no derivation) measures the sharpness of resonance. At resonance $V_{L} = V_{C} = Q \times V$ , which can be much larger than the source voltage. Used in tuning radio and TV receivers.; Worked example
A series circuit has L = 0.1 H, C = 10 µF and R = 10 Ω across a 20 V variable-frequency source. Find the resonant frequency, the current at resonance, the Q factor and the bandwidth.
Try it first, then show the solutionSolution
$f_{0} = \frac{1}{2 π 0.1 \times 1 0 ^{- 5}} = \frac{1}{2 π \times 1 0 ^{- 3}} = 159 H z$
$I = \frac{V}{R} = \frac{20}{10} = 2 A$
$Q = \frac{1}{R} \frac{L}{C} = \frac{1}{10} 1 0^{4} = 10$ ; bandwidth $= \frac{159}{10} = 15.9 H z$
Answer: 159 Hz; 2 A; Q = 10; bandwidth about 16 Hz
Power, power factor and wattless current: Average power over a cycle: $P = \frac{1}{2} V_{0} I_{0} cos ϕ = V_{r m s} I_{r m s} cos ϕ = I_{r m s}^{2} R$ . The power factor $cos ϕ = \frac{R}{Z}$ . Pure R: $cos ϕ = 1$ , all power used. Pure L or C: $ϕ = 9 0^{\circ}$ , $cos ϕ = 0$ , P = 0: the energy stored in a quarter cycle is returned in the next. The current then is wattless; more generally, the component $I_{r m s} sin ϕ$ (in quadrature with V) consumes no power. Choke coil: an inductor of large L and small R used to limit current (for example in a fluorescent lamp), wasting far less power than a series resistor would. A low power factor means a larger current (and line loss) for the same useful power.; Worked example
A lamp rated 50 V, 10 W is to run on a 200 V, 50 Hz supply. Find the inductance of a pure choke that makes it work normally, and the power that a series resistor would have wasted instead.
Try it first, then show the solutionSolution
Lamp: $I = \frac{10}{50} = 0.2 A$ , $R = \frac{50}{0.2} = 250 Ω$ . Needed $Z = \frac{200}{0.2} = 1000 Ω$ .
$X_{L} = 100 0^{2} - 25 0^{2} = 968 Ω$ ; $L = \frac{968}{2 π \times 50} = 3.08 H$
A resistor would need to drop 150 V at 0.2 A, wasting $150 \times 0.2 = 30 W$ ; the ideal choke wastes none.
Answer: L ≈ 3.1 H; a resistor would waste 30 W
LC oscillations (qualitative): A charged capacitor connected across an inductor discharges through it. The current builds up, storing energy in the inductor's magnetic field; when the capacitor is empty the current is maximum. The inductor then keeps the current going and charges the capacitor the other way. Energy shuttles between electric (capacitor) and magnetic (inductor) forms, like a mass on a spring, at angular frequency $ω_{0} = \frac{1}{L C}$ . In a real circuit resistance damps the oscillations. This natural frequency equals the resonant frequency of the LCR circuit.; Worked example
An LC circuit has L = 2 mH and C = 5 µF. Find its angular frequency and frequency of oscillation. What happens to the frequency if C is made four times larger?
Try it first, then show the solutionSolution
$ω_{0} = \frac{1}{2 \times 1 0 ^{- 3} \times 5 \times 1 0 ^{- 6}} = \frac{1}{1 0 ^{- 8}} = 1 0^{4} r a d s^{- 1}$
$f_{0} = \frac{1 0 ^{4}}{2 π} = 1592 H z$
$f \propto 1/ C$ : four times C halves f, to 796 Hz.
Answer: $1 0^{4} r a d s^{- 1}$ ; 1592 Hz; halved
The a.c. generator: Principle: electromagnetic induction. A coil of N turns and area A rotates with angular speed ω in a uniform field B (between the poles of a magnet). Construction: armature coil, field magnet, slip rings attached to the coil's ends, and carbon brushes pressing on them to connect to the external circuit. Theory: at time t the normal to the coil makes θ = ωt with B, so flux linkage is $N B A cos ω t$ and $ε = - \frac{d}{d t} (N B A cos ω t) = N B A ω sin ω t = ε_{0} sin ω t$ . The emf is maximum when the plane of the coil is parallel to B (flux zero but changing fastest) and zero when it is perpendicular. A.c. vs d.c.: a.c. reverses direction periodically and can be stepped up or down by transformers, so it is transmitted cheaply; d.c. has constant direction, is needed for electrolysis, charging batteries and electronics, and can be stored.; Worked example
A generator coil of 100 turns, area $0.02 m^{2}$ , rotates at 50 revolutions per second in a field of 0.5 T. Find the peak and r.m.s. emf.
Try it first, then show the solutionSolution
$ω = 2 π \times 50 = 314.2 r a d s^{- 1}$
$ε_{0} = N B A ω = 100 \times 0.5 \times 0.02 \times 314.2 = 314 V$
$ε_{r m s} = \frac{314}{2} = 222 V$
Answer: Peak 314 V; r.m.s. 222 V

Physics in real life: Alternating Current

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

R.m.s. and peak voltage

Suppose an electrician is choosing a capacitor for a fan regulator on the 230 V (r.m.s.) mains.

What is the highest voltage the capacitor must withstand in each cycle?

Try it first, then show the solutionSolution

$V_{0} = 2 \times 230$
$V_{0} = 325 V$

Answer: About 325 V, so a component rated for 230 V d.c. would not be enough; it needs a rating above the peak.

Memory hook: 230 on the label, 325 at the peak.

Resonance in tuning circuits

Imagine tuning an FM radio to a station broadcasting at 100 MHz; the tuning circuit is an LC circuit with a variable capacitor.

Suppose the coil has inductance 1 µH. What capacitance tunes it to 100 MHz?

Try it first, then show the solutionSolution

$f_{0} = \frac{1}{2 π L C} \Rightarrow C = \frac{1}{4 π ^{2} f _{0}^{2} L}$
$C = \frac{1}{4 π ^{2} \times ( 1 0 ^{8} ) ^{2} \times 1 0 ^{- 6}} = 2.53 \times 1 0^{- 12} F$

Answer: About 2.5 pF; turning the knob changes C and hence the station picked out.

Memory hook: Tune C until $X_{L} = X_{C}$ : that station's current peaks.

Power factor

Suppose a workshop motor draws 10 A from 230 V mains with a power factor of 0.7.

Find the real power used. What current would deliver the same power at a power factor of 1?

Try it first, then show the solutionSolution

$P = V_{r m s} I_{r m s} cos ϕ = 230 \times 10 \times 0.7 = 1610 W$
At $cos ϕ = 1$ : $I = \frac{1610}{230} = 7 A$

Answer: 1610 W; only 7 A would be needed at unity power factor, so the low power factor makes the wires carry 3 A extra and waste more heat.

Memory hook: Low cos φ: more amps for the same watts.

Topics in this chapter

Alternating Current formula sheet

Sinusoidal a.c.: $V = V_{0} sin ω t, ω = 2 π f = \frac{2 π}{T}$; Indian mains: f = 50 Hz.
R.m.s. values: $I_{r m s} = \frac{I _{0}}{2}, V_{r m s} = \frac{V _{0}}{2}$; Meters read r.m.s.
Mean value over half a cycle: $I_{m e an} = \frac{2 I _{0}}{π} = 0.637 I_{0}$; Full-cycle average is zero.
Inductive reactance: $X_{L} = ω L = 2 π f L$; I lags V by π/2 in pure L.
Capacitive reactance: $X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C}$; I leads V by π/2 in pure C.
Impedance of series LCR: $Z = R^{2} + (X_{L} - X_{C})^{2}$; RL: $R^{2} + X_{L}^{2}$ ; RC: $R^{2} + X_{C}^{2}$ .
Phase angle: $tan ϕ = \frac{X _{L} - X _{C}}{R} = \frac{V _{L} - V _{C}}{V _{R}}$; Positive: inductive (V leads).
Voltage phasors: $V^{2} = V_{R}^{2} + (V_{L} - V_{C})^{2}$; Add as vectors, not as numbers.
Resonance: $ω_{0} = \frac{1}{L C}, f_{0} = \frac{1}{2 π L C}$; Z = R, current maximum, cos φ = 1.
Q factor (no derivation): $Q = \frac{ω _{0} L}{R} = \frac{1}{R} \frac{L}{C} = \frac{f _{0}}{Δ f}$; Δf = bandwidth.
Average power: $P = V_{r m s} I_{r m s} cos ϕ = \frac{1}{2} V_{0} I_{0} cos ϕ = I_{r m s}^{2} R$; Zero in pure L or C.
Power factor: $cos ϕ = \frac{R}{Z}$; 1 for pure R and at resonance.
A.c. generator: $ε = N B A ω sin ω t, ε_{0} = N B A ω$; Maximum when the coil's plane is parallel to B.

Alternating Current: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Phasor diagrams for pure R, L, C and series LCR

Explaining phase relations; deriving Z and tan φ for an LCR circuit.

Draw

Pure R: I and V phasors along the same line. Pure L: V phasor 90° ahead (anticlockwise) of I. Pure C: V phasor 90° behind I.
LCR: draw I along the x-axis; $V_{R}$ along I; $V_{L}$ straight up; $V_{C}$ straight down (shorter if inductive).
Draw $(V_{L} - V_{C})$ up from the tip of $V_{R}$ , and the resultant V from the origin; mark φ between V and I.

Label

I
$V_{R}$
$V_{L}$
$V_{C}$
$V_{L} - V_{C}$
V
φ

Watch out: Phasors rotate anticlockwise; 'ahead' means further anticlockwise. Mark φ between V and I, not between V and $V_{L}$ .

Graphs: $X_{L}$ , $X_{C}$ , Z and I against frequency

Named in the syllabus; examiners penalise unmarked $f_{0}$ , swapped or unlabelled axes, and curves touching the axis.

Draw

$X_{L}$ -f: straight line through the origin. $X_{C}$ -f: falling curve (hyperbola) approaching but never touching the f-axis.
Z-f for series LCR: large at low f, dips to a minimum equal to R at $f_{0}$ , rises again; the minimum stays above the f-axis.
I-f: a peak at $f_{0}$ (sharper for smaller R), falling on both sides.

Label

axes with symbols and units
$f_{0}$
Z = R at $f_{0}$
$I_{ma x} = V / R$

Watch out: Draw two I-f curves for different R to show that smaller R gives a sharper peak (higher Q).

A.c. generator

'Draw a labelled diagram of an a.c. generator and explain its working'.

Draw

Two magnet poles N and S facing each other with field lines between them.
A rectangular coil ABCD between the poles, on an axle, with arrows showing rotation.
Two slip rings on the axle, each connected to one end of the coil, with carbon brushes touching them and leads to an external resistor or bulb.

Label

N, S field magnet
armature coil ABCD
axle and rotation
slip rings
carbon brushes
external circuit

Watch out: Slip rings (not a split-ring commutator) give a.c.; say this in the explanation.

How the board asks Alternating Current

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Peak, mean and r.m.s.	Find peak from r.m.s. or the reverse; derive $I_{r m s} = I_{0} / 2$ ; read values from $I = I_{0} sin ω t$ .	Multiply r.m.s. by $2$ for peak; for frequency use $f = ω /2 π$ , not ω.
Identify the component from V and I	Given $E = E_{0} sin ω t$ and $I = I_{0} sin (ω t \pm ϕ)$ , identify the unknown element and find Z, power factor and f.	Current lags: inductive; leads: capacitive. $Z = E_{0} / I_{0}$ ; $cos ϕ$ is the power factor; $R = Z cos ϕ$ .
LR, RC and LCR numericals (3-5 marks)	Reactance, impedance, current, phase angle, voltage across each element.	Find $X_{L}$ and $X_{C}$ first, then Z by the square-root formula; never add R and X directly.
Resonance and graphs	Resonant frequency, current at resonance, Q and bandwidth; sketch Z-f or I-f.	$f_{0} = 1/2 π L C$ ; at resonance Z = R; mark $f_{0}$ on the graph.
Power and power factor	Average power, wattless current, choke coil versus resistor, why power factor matters.	$P = V_{r m s} I_{r m s} cos ϕ$ ; $cos ϕ = R / Z$ ; pure L or C gives zero power.
A.c. generator and a.c. vs d.c.	Principle, labelled diagram, derivation of $ε = N B A ω sin ω t$ , two differences between a.c. and d.c.	Flux linkage $N B A cos ω t$ , differentiate; differences in direction, transformability and uses.

Tricks to remember Alternating Current

Short memory aids. Say each one aloud once, then use it on a question the same day.

ELI the ICE man
In an inductor (L), E (voltage) comes before I: ELI. In a capacitor (C), I comes before E: ICE.
Use it when: Phase of current in L and C circuits.
X_L likes high f, X_C likes low f
Inductors choke high frequencies; capacitors block low ones (d.c. totally).
Use it when: Reactance-frequency graphs and reasoning.
Root-two up, root-two down
Peak = r.m.s. × √2; r.m.s. = peak ÷ √2. 220 ↔ 311, 230 ↔ 325.
Use it when: Avoiding the reversed formula examiners report.
Pythagoras for impedance
Z is the hypotenuse; R and $(X_{L} - X_{C})$ are the other two sides; cos φ = R/Z is adjacent over hypotenuse.
Use it when: Z, power factor and phase in one triangle.
Resonance: everything cancels except R
At $f_{0}$ , L and C cancel each other: Z = R, I max, cos φ = 1.
Use it when: Resonance questions.
Wattless is ninety
90° between V and I means zero power: cos 90° = 0.
Use it when: Pure L or C power questions.

Alternating Current: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Phase in a pure inductor · 1 mark · MCQBoard level

In a purely inductive a.c. circuit, the current:
(a) is in phase with the voltage
(b) leads the voltage by $π /2$
(c) lags the voltage by $π /2$
(d) lags the voltage by π

Try it first, then show the solutionSolution

$V = L \frac{d I}{d t}$ gives $I = I_{0} sin (ω t - π /2)$ : the current lags.

Answer: (c)

Example 2: Power in a pure inductor · 1 mark · Assertion-reasonBoard level

Assertion: The average power consumed by a pure inductor over a complete cycle is zero.
Reason: In a pure inductor, the phase difference between voltage and current is $π /2$ .
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

$P = V_{r m s} I_{r m s} cos ϕ$ with $ϕ = π /2$ gives zero; the reason explains the assertion.

Answer: (a)

Example 3: Peak and mean from r.m.s. · 2 marks · Short answerHarder board level

An a.c. ammeter reads 4 A. Find the peak value of the current and its mean value over a half cycle. [2]

Try it first, then show the solutionSolution

Ammeter reads r.m.s.: $I_{0} = 2 \times 4 = 5.66 A$
$I_{m e an} = \frac{2 I _{0}}{π} = \frac{2 \times 5.66}{3.1416} = 3.60 A$

Answer: Peak 5.66 A; half-cycle mean 3.60 A

Example 4: Derivation: r.m.s. value · 3 marks · Short answerHarder board level

Define the r.m.s. value of an alternating current and show that $I_{r m s} = \frac{I _{0}}{2}$ for a sinusoidal current. [3]

Try it first, then show the solutionSolution

R.m.s. value: the steady (d.c.) current that produces the same heat in a resistor in the same time as the a.c. does.
Heat in one period: $H = \int_{0}^{T} I_{0}^{2} sin^{2} ω t R d t = I_{0}^{2} R \int_{0}^{T} \frac{1 - c o s 2 ω t}{2} d t = \frac{I _{0}^{2} R T}{2}$ (the cosine term integrates to zero).
Equate to $I_{r m s}^{2} R T$ : $I_{r m s}^{2} = \frac{I _{0}^{2}}{2} \Rightarrow I_{r m s} = \frac{I _{0}}{2} = 0.707 I_{0}$ .

Answer: $I_{r m s} = I_{0} / 2$

Example 5: A coil with inductance and resistance · 3 marks · Short answerHarder board level

A coil of inductance 0.4 H and resistance 100 Ω is connected to a 220 V, 50 Hz supply. Find (a) its reactance, (b) its impedance, (c) the current, and (d) the power factor. [3]

Try it first, then show the solutionSolution

(a) $X_{L} = 2 π \times 50 \times 0.4 = 125.7 Ω$
(b) $Z = 10 0^{2} + 125. 7^{2} = 160.6 Ω$ ; (c) $I = \frac{220}{160.6} = 1.37 A$
(d) $cos ϕ = \frac{R}{Z} = \frac{100}{160.6} = 0.62$ (current lags by about 51.5°)

Answer: (a) 125.7 Ω (b) 160.6 Ω (c) 1.37 A (d) 0.62

Example 6: LCR circuit: derivation and numerical · 5 marks · Long answerHarder board level

(a) Using a phasor diagram, obtain expressions for the impedance of a series LCR circuit and the phase angle between voltage and current. [3]
(b) R = 30 Ω, $X_{L} = 80 Ω$ and $X_{C} = 40 Ω$ are in series across 200 V (r.m.s.). Find the current and the average power. [2]

Try it first, then show the solutionSolution

(a) Same current I in all. Phasors: $V_{R} = I R$ along I; $V_{L} = I X_{L}$ 90° ahead; $V_{C} = I X_{C}$ 90° behind.
$V_{L}$ and $V_{C}$ are opposite, giving $(V_{L} - V_{C})$ perpendicular to $V_{R}$ : $V^{2} = V_{R}^{2} + (V_{L} - V_{C})^{2} = I^{2} [R^{2} + (X_{L} - X_{C})^{2}]$ .
$Z = \frac{V}{I} = R^{2} + (X_{L} - X_{C})^{2}$ and $tan ϕ = \frac{X _{L} - X _{C}}{R}$ .
(b) $Z = 3 0^{2} + 4 0^{2} = 50 Ω$ ; $I = 200/50 = 4 A$ .
$P = V_{r m s} I_{r m s} cos ϕ = 200 \times 4 \times \frac{30}{50} = 480 W$ (check: $I^{2} R = 16 \times 30 = 480 W$ ).

Answer: (b) 4 A; 480 W

Example 7: Above and below resonance · 3 marks · Short answerHOTS / competency

A series circuit has L = 0.1 H, C = 10 µF and R = 10 Ω. (a) Find its resonant frequency. (b) Is the circuit inductive or capacitive at 100 Hz? (c) Sketch Z against f and mark the minimum. [3]

Try it first, then show the solutionSolution

(a) $f_{0} = \frac{1}{2 π 1 0 ^{- 6}} = 159 H z$ .
(b) At 100 Hz: $X_{L} = 2 π \times 100 \times 0.1 = 62.8 Ω$ , $X_{C} = \frac{1}{2 π \times 100 \times 1 0 ^{- 5}} = 159 Ω$ . $X_{C} > X_{L}$ , so the circuit is capacitive (current leads).
(c) Z falls from very large values at low f to a minimum of 10 Ω (= R) at 159 Hz, then rises; the minimum does not touch the f-axis.

Answer: (a) 159 Hz (b) capacitive (c) Z minimum = 10 Ω at 159 Hz

Example 8: A capacitor at two frequencies · 2 marks · Short answerHOTS / competency

Find the reactance of a 100 µF capacitor at 50 Hz and at 5 kHz. Use the answers to explain why a capacitor 'blocks d.c. but passes a.c.' [2]

Try it first, then show the solutionSolution

$X_{C} = \frac{1}{2 π f C}$ : at 50 Hz, $\frac{1}{2 π \times 50 \times 1 0 ^{- 4}} = 31.8 Ω$ ; at 5 kHz, $0.318 Ω$ .
$X_{C} \propto 1/ f$ : high-frequency a.c. meets little opposition, while for d.c. (f = 0) $X_{C}$ is infinite, so no steady current flows.

Answer: 31.8 Ω and 0.318 Ω; $X_{C} \to \infty$ as f → 0

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Identifying a hidden component

Suppose a sealed box contains a resistor in series with one unknown component X. When an alternating emf $E = 100 sin (100 π t)$ volts is applied, the current is $I = 2 sin (100 π t + π /4)$ amperes.

(i) · 1 mark
Is X an inductor or a capacitor? Give a reason.
Answer
Answer: A capacitor.
The current leads the emf by π/4; a leading current means the circuit is capacitive.
(ii) · 1 mark
Find the frequency of the source and the r.m.s. current.
Answer
Answer: 50 Hz; 1.41 A
$ω = 100 π$ , so $f = ω /2 π = 50 H z$ . $I_{r m s} = 2/ 2 = 1.41 A$ .
(iii) · 1 mark
Find the impedance and the power factor.
Answer
Answer: 50 Ω; 0.707
$Z = E_{0} / I_{0} = 100/2 = 50 Ω$ ; $cos ϕ = cos 4 5^{\circ} = 0.707$ .
(iv) · 2 marks
Find R and the capacitance of X.
Answer
Answer: R = 35.4 Ω; C ≈ 90 µF
$R = Z cos ϕ = 35.4 Ω$ ; $X_{C} = Z sin ϕ = 35.4 Ω$ ; $C = \frac{1}{ω X _{C}} = \frac{1}{314.2 \times 35.4} = 9.0 \times 1 0^{- 5} F$ .

Sweeping through resonance

Suppose a student connects L = 0.1 H, C = 10 µF and R = 10 Ω in series to a 20 V (r.m.s.) signal generator and slowly raises the frequency, noting the current.

(i) · 1 mark
At what frequency does the current peak, and what is the peak current?
Answer
Answer: 159 Hz; 2 A
$f_{0} = \frac{1}{2 π L C} = 159 H z$ ; at resonance Z = R, so $I = 20/10 = 2 A$ .
(ii) · 2 marks
Find the voltage across the inductor at resonance. Why can it exceed the supply voltage?
Answer
Answer: 200 V
$X_{L} = 2 π f_{0} L = 100 Ω$ , so $V_{L} = 2 \times 100 = 200 V$ . $V_{L}$ and $V_{C}$ are equal and opposite in phase, so they cancel in the total; each can be Q = 10 times the supply.
(iii) · 1 mark
Find the Q factor and bandwidth.
Answer
Answer: Q = 10; about 16 Hz
$Q = \frac{ω _{0} L}{R} = \frac{100}{10} = 10$ ; bandwidth $= f_{0} / Q = 15.9 H z$ .
(iv) · 1 mark
How would the I-f curve change if R were increased to 40 Ω?
Answer
Answer: Same f₀ but a lower (0.5 A) and broader peak.
$f_{0}$ depends only on L and C; peak current V/R falls to 0.5 A and Q falls to 2.5, so the curve is flatter and wider.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

An a.c. ammeter reads 4 A. Find the peak value of the current and its mean value over half a cycle. (2 marks)

Step	Marks
$I_{0} = 2 I_{r m s} = 5.66 A$	1
$I_{m e an} = 2 I_{0} / π = 3.60 A$	1

A coil of inductance 0.4 H and resistance 100 Ω is connected to 220 V, 50 Hz. Find its reactance, impedance and the current. (3 marks)

Step	Marks
$X_{L} = 2 π f L = 125.7 Ω$	1
$Z = R^{2} + X_{L}^{2} = 160.6 Ω$	1
$I = V / Z = 1.37 A$	1

Describe the principle and construction of an a.c. generator and derive the expression for the induced emf. A coil of 100 turns, area 0.02 m², rotates at 50 rev/s in 0.5 T. Find the peak emf. (5 marks)

Step	Marks
Principle: electromagnetic induction (changing flux linkage of a rotating coil)	1
Labelled diagram: field magnet, armature coil, slip rings, brushes	1
Flux linkage $N ϕ = N B A cos ω t$	1
$ε = - \frac{d}{d t} (N B A cos ω t) = N B A ω sin ω t$	1
$ε_{0} = 100 \times 0.5 \times 0.02 \times 314.2 = 314 V$	1

Where students lose marks in Alternating Current

The slips examiners see most often, and what to write instead.

Mistake: Adding R and $X_{L}$ instead of using the square root
CISCE examiners' comments, 2024: for a coil with L and R on a.c., candidates treated it as an LCR circuit, gave the reactance as the resistance, added $X_{L} + R$ , used V = IR instead of V = IZ, or took Z = R as if at resonance.
Wrong working
$Z = R + X_{L} = 100 + 125.7 = 225.7 Ω$
Right working
$Z = R^{2} + X_{L}^{2} = 160.6 Ω$ ; $I = V / Z$ .
Do this: Draw the impedance triangle; R and X are perpendicular.
Usually costs: 1-2 marks.
Mistake: Reversing the peak-r.m.s. relation
CISCE examiners' comments, 2024: asked for the peak value from an r.m.s. of 5 A, candidates used $I_{0} = I_{r m s} / 2$ .
Wrong working
$I_{0} = 5/ 2 = 3.54 A$
Right working
$I_{0} = 2 \times 5 = 7.07 A$ (peak is always larger).
Do this: Sanity check: peak > r.m.s. > half-cycle mean.
Usually costs: 1 mark.
Mistake: Poor Z-f graphs
CISCE examiners' comments, 2024: for the Z-f graph of an LCR circuit, $f_{0}$ was not marked, axes were swapped or unlabelled, the shape was wrong, or the curve touched the f-axis.
Wrong working
A V-shaped curve dipping to zero.
Right working
A U-shaped curve with its minimum at Z = R above the axis, at the marked $f_{0}$ .
Do this: Label Z and f, mark $f_{0}$ and R, keep the minimum above the axis.
Usually costs: 1-2 marks.
Mistake: Using ω as the frequency and the phase angle as the power factor
CISCE examiners' comments, 2025: given E and I as sine functions, candidates could not identify the component, gave ω as the frequency, wrote the phase-angle formula for the power factor and miscalculated Z.
Wrong working
From $E = 5 sin (314 t)$ : f = 314 Hz; power factor = π/3.
Right working
$f = 314/2 π = 50 H z$ ; power factor $= cos (π /3) = 0.5$ ; current lagging means an inductor.
Do this: Read $ω$ , then divide by 2π; the power factor is the cosine of the phase angle.
Usually costs: 1 mark per part.
Mistake: Incomplete explanation of transformer currents
CISCE examiners' comments, 2025: explaining why the secondary current of a step-down transformer is larger, candidates only said the secondary has fewer turns.
Wrong working
Because the secondary has fewer turns.
Right working
For an ideal transformer input power = output power, $V_{p} I_{p} = V_{s} I_{s}$ ; since $V_{s} < V_{p}$ , $I_{s} > I_{p}$ .
Do this: Always bring in power conservation.
Usually costs: 1 mark.
Mistake: Naming a 'step transformer' or 'rectifier' for a charger
CISCE examiners' comments, 2024: asked which transformer a phone charger uses, candidates wrote 'rectifier' or 'step transformer'.
Wrong working
Rectifier.
Right working
A step-down transformer (230 V to a few volts), followed by a rectifier.
Do this: Give the full term: step-up or step-down.
Usually costs: 1 mark.

Practice questions on Alternating Current

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The peak value of an alternating voltage of r.m.s. value 220 V is about:
(a) 156 V
(b) 220 V
(c) 311 V
(d) 440 V
Hint
Multiply by √2.
Answer
(c) 311 V
Q2 · 1 mark · MCQBoard level
The capacitive reactance of a capacitor:
(a) increases with frequency
(b) decreases with frequency
(c) is independent of frequency
(d) is zero for d.c.
Hint
$X_{C} = 1/2 π f C$ .
Answer
(b)
Q3 · 1 mark · MCQBoard level
At resonance in a series LCR circuit:
(a) Z is maximum
(b) current is minimum
(c) Z = R and current is maximum
(d) power factor is zero
Hint
$X_{L} = X_{C}$ .
Answer
(c)
Q4 · 1 mark · Very short answerBoard level
What is meant by wattless current?
Hint
Phase difference 90°.
Answer
Current (or its component) that is 90° out of phase with the voltage, so the average power consumed is zero.
Q5 · 2 marks · Short answerBoard level
Find the reactances of a 0.5 H inductor and a 20 µF capacitor at 50 Hz. If both are connected in series with a small resistor, is the circuit nearly at resonance?
Hint
Compare $X_{L}$ and $X_{C}$ .
Answer
$X_{L} = 157.1 Ω$ , $X_{C} = 159.2 Ω$ ; nearly equal, so the circuit is close to resonance (exact $f_{0} = 50.3 H z$ ).
Q6 · 2 marks · Short answerHarder board level
A series circuit has R = 60 Ω, $X_{L} = 120 Ω$ , $X_{C} = 40 Ω$ on 220 V (r.m.s.). Find Z, the power factor and the average power.
Hint
$Z = R^{2} + (X_{L} - X_{C})^{2}$ .
Answer
Z = 100 Ω; cos φ = 0.6; I = 2.2 A; P = 220 × 2.2 × 0.6 = 290.4 W.
Q7 · 2 marks · Short answerHarder board level
A radio's tuning coil has inductance 1 µH. What capacitance tunes it to 100 MHz?
Hint
$C = 1/4 π^{2} f^{2} L$ .
Answer
$C = 2.53 \times 1 0^{- 12} F$ (about 2.5 pF)
Q8 · 2 marks · Short answerHOTS / competency
A 50 V, 10 W lamp is to be run from 200 V, 50 Hz mains. Compare using a series resistor with using a pure choke coil.
Hint
Find the current, then the required R or L.
Answer
I = 0.2 A. Resistor: 150 V/0.2 A = 750 Ω, wasting 30 W. Choke: lamp R = 250 Ω, Z = 1000 Ω, $X_{L} = 968 Ω$ , L = 3.08 H, wasting (ideally) no power since current through L is wattless.
Q9 · 3 marks · Short answerHarder board level
A generator coil of 100 turns, area $0.02 m^{2}$ , rotates at 50 rev/s in a 0.5 T field. Find the peak and r.m.s. emf. What happens to each if the speed is doubled?
Hint
$ε_{0} = N B A ω$ .
Answer
Peak 314 V, r.m.s. 222 V. Doubling the speed doubles ω, so both emfs double (628 V, 444 V) and the frequency doubles to 100 Hz.
Q10 · 3 marks · Short answerHarder board level
An LC circuit has L = 2 mH and C = 5 µF. Find the frequency of oscillation and describe how energy is exchanged during one cycle.
Hint
$ω_{0} = 1/ L C$ .
Answer
$ω_{0} = 1 0^{4} r a d s^{- 1}$ , f = 1592 Hz. Energy starts in the capacitor's electric field, moves to the inductor's magnetic field as current builds (capacitor empty, current maximum), then back to the capacitor with reversed polarity, and so on twice per cycle.
Q11 · 3 marks · Short answerHarder board level
For L = 0.1 H, C = 10 µF and R = 10 Ω in series, find the resonant frequency, Q factor and bandwidth. What does a high Q mean for a radio receiver?
Hint
$Q = \frac{1}{R} L / C$ .
Answer
$f_{0} = 159 H z$ ; Q = 10; bandwidth 15.9 Hz. High Q means a sharp, narrow resonance, so the receiver selects one station and rejects neighbouring frequencies.
Q12 · 5 marks · Long answerHarder board level
(a) State the principle of an a.c. generator and draw a labelled diagram. [2]
(b) Derive $ε = N B A ω sin ω t$ . [2]
(c) Give one difference between a.c. and d.c. and one advantage of a.c. [1]
Hint
Flux linkage changes as the coil turns.
Answer
(a) Electromagnetic induction: a coil rotating in a magnetic field has a changing flux linkage, so an emf is induced; diagram with magnet, coil, slip rings and brushes. (b) $N ϕ = N B A cos ω t$ ; $ε = - d (N ϕ) / d t = N B A ω sin ω t$ . (c) A.c. reverses direction periodically while d.c. flows one way; a.c. can be stepped up or down by transformers for efficient transmission.
Q13 · 2 marks · Short answerHarder board level
Find the form factor ( $I_{r m s} / I_{m e an}$ , using the half-cycle mean) of a sinusoidal current.
Hint
Divide the two expressions.
Answer
$\frac{I _{0} / 2}{2 I _{0} / π} = \frac{π}{2 2} = 1.11$

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define r.m.s. value of a.c.Show answer
The steady current that produces the same heat in a resistor in the same time; $I_{r m s} = I_{0} / 2$ .
2.Mean value of a.c. over half a cycle?Show answer
$2 I_{0} / π = 0.637 I_{0}$ .
3.Inductive and capacitive reactance?Show answer
$X_{L} = ω L$ , $X_{C} = 1/ ω C$ .
4.Impedance of a series LCR circuit?Show answer
$Z = R^{2} + (X_{L} - X_{C})^{2}$ .
5.Condition and frequency of resonance?Show answer
$X_{L} = X_{C}$ ; $f_{0} = \frac{1}{2 π L C}$ .
6.Define power factor.Show answer
$cos ϕ = R / Z$ , the ratio of true power to apparent power ( $V_{r m s} I_{r m s}$ ).
7.What is a choke coil?Show answer
An inductor of high inductance and low resistance used to limit a.c. with very little power loss.
8.Emf of an a.c. generator?Show answer
$ε = N B A ω sin ω t$ .

Best order to study it

Learn $V = V_{0} sin ω t$ ; derive the r.m.s. and half-cycle mean values.
Study pure R, L and C: phase, reactance, phasor diagrams and reactance-frequency graphs.
Derive Z and tan φ for the series LCR circuit; practise RL and RC special cases.
Learn resonance, the Z-f and I-f graphs, Q factor and bandwidth.
Learn power, power factor, wattless current and the choke coil; then LC oscillations.
Finish with the a.c. generator and the differences between a.c. and d.c.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you derive $I_{r m s} = I_{0} / 2$ and $I_{m e an} = 2 I_{0} / π$ ?
Can you state the phase relation in pure R, L and C and draw phasors?
Can you sketch $X_{L}$ -f and $X_{C}$ -f?
Can you derive Z and tan φ for an LCR circuit and solve an LR numerical?
Can you identify L or C from V and I expressions and find f, Z and power factor?
Can you find $f_{0}$ , Q and bandwidth and draw Z-f and I-f with $f_{0}$ marked?
Can you explain wattless current and why a choke coil is used?
Can you derive $ε = N B A ω sin ω t$ and draw the generator?

Alternating Current: frequently asked questions

Do I need to derive the Q factor or bandwidth?

No. The ISC 2027 syllabus lists bandwidth and Q factor with 'no derivation'. Use Q = \omega_0L/R = f_0/\text{bandwidth} and know what a high Q means.

Are LC oscillations examined numerically?

The syllabus lists LC oscillations as qualitative, with \omega_0 = 1/\sqrt{LC}. You may be asked the frequency or to describe the energy exchange, but not the differential equation.

Which derivations are important?

R.m.s. and mean values, the phase relations for pure R, L and C, the impedance and phase angle of a series LCR circuit using phasors, average power, and the emf of an a.c. generator.

Why is the full-cycle average of a.c. zero but the r.m.s. value not?

The current is positive for half a cycle and negative for the other half, so it averages to zero. Squaring makes every value positive, so the mean of I^2 is I_0^2/2, and its square root is not zero.

How do I tell whether an LCR circuit is inductive or capacitive?

Compare X_L and X_C at the given frequency: if X_L > X_C (above resonance) it is inductive and the current lags; if X_C > X_L (below resonance) it is capacitive and the current leads.

How many marks does this chapter carry?

It shares 16 of the 70 theory marks with Moving Charges and Magnetism, Magnetism and Matter and Electromagnetic Induction. Recent papers set 5-mark questions on LR circuits and on identifying a component from V and I.

Based on: CISCE ISC 2027 Physics syllabus (revised, uploaded February 2026); CISCE ISC 2027 Physics specimen Paper 1 (Theory) question paper and answer key; CISCE Analysis of Pupil Performance, ISC Physics, 2025; CISCE Analysis of Pupil Performance, ISC Physics, 2024

Alternating Current ISC Class 12: Notes, Phasors and Numericals

What Alternating Current is about

Key concepts

Physics in real life: Alternating Current

Topics in this chapter

Alternating Current formula sheet

Alternating Current: diagrams to practise

Phasor diagrams for pure R, L, C and series LCR

Graphs: XL​, XC​, Z and I against frequency

A.c. generator

How the board asks Alternating Current

Tricks to remember Alternating Current

Alternating Current: board-level solved questions

Case-study practice (Section D style)

Identifying a hidden component

Sweeping through resonance

Where the marks are

Where students lose marks in Alternating Current

Practice questions on Alternating Current

Test your recall

Best order to study it

Quick revision checklist

Alternating Current: frequently asked questions

Do I need to derive the Q factor or bandwidth?

Are LC oscillations examined numerically?

Which derivations are important?

Why is the full-cycle average of a.c. zero but the r.m.s. value not?

How do I tell whether an LCR circuit is inductive or capacitive?

How many marks does this chapter carry?

Graphs: $X_{L}$ , $X_{C}$ , Z and I against frequency