PhysicsCBSE Class 12 · Alternating Current

AC Through Resistor, Inductor and Capacitor Class 12: Reactance and Phase

Updated 15 October 2026 · Fact-checked

For AC, rms values are peak ÷ √2. In a resistor, current and voltage are in phase. In an inductor, current lags the voltage by 90° and is limited by the reactance X_L = ωL; in a capacitor, current leads by 90° with reactance X_C = 1/ωC. Pure L and C draw no average power.

Watch the dashed line at the voltage peak: current peaks there for R, later for L (lags), and earlier for C (leads).

Illustration drawn for QuestionBuddy.

Understand AC Through Resistor, Inductor and Capacitor

The mains supply reverses direction 100 times a second (50 Hz). A heater does not care about the direction, but an inductor and a capacitor respond very differently to AC than to DC. This topic looks at each element on its own.

Alternating voltage and rms values. An AC source gives $v = v_{m} sin ω t$ , where $v_{m}$ is the peak value and $ω = 2 π f$ . The current averaged over a cycle is zero, so we use the root mean square value, which gives the same heating as a steady current: $I = \frac{i _{m}}{2} \approx 0.707 i_{m}$ and $V = \frac{v _{m}}{2}$ . AC meters read rms values; '220 V mains' means $V_{r m s} = 220$ V, peak about 311 V.

AC through a resistor. $i = \frac{v _{m}}{R} sin ω t$ : current and voltage are in phase, reaching peaks together. Average power $P = I^{2} R = V I$ . In a phasor diagram, the current and voltage phasors point the same way.

AC through an inductor. For a pure inductor, the source voltage equals $L \frac{d i}{d t}$ , so $\frac{d i}{d t} = \frac{v _{m}}{L} sin ω t$ . Integrating: $i = - \frac{v _{m}}{ω L} cos ω t = i_{m} sin (ω t - \frac{π}{2})$ , with $i_{m} = \frac{v _{m}}{ω L}$ . So:

The current lags the voltage by $\frac{π}{2}$ (90°).
The opposition is the inductive reactance $X_{L} = ω L = 2 π f L$ (in ohm); $i_{m} = \frac{v _{m}}{X _{L}}$ .
$X_{L} \propto f$ : an inductor passes DC (f = 0) freely but opposes high frequencies.
Average power over a cycle is zero: energy stored in the magnetic field in one quarter cycle is returned in the next.

AC through a capacitor. For a capacitor, $q = C v$ , so $i = \frac{d q}{d t} = C \frac{d v}{d t} = ω C v_{m} cos ω t = i_{m} sin (ω t + \frac{π}{2})$ , with $i_{m} = ω C v_{m}$ . So:

The current leads the voltage by $\frac{π}{2}$ .
The capacitive reactance is $X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C}$ (in ohm); $i_{m} = \frac{v _{m}}{X _{C}}$ .
$X_{C} \propto \frac{1}{f}$ : a capacitor blocks DC (infinite reactance) but passes high frequencies easily.
Average power is again zero: energy goes into the electric field and comes back.

Phasors. A phasor is a rotating arrow of length equal to the peak value, turning anticlockwise at $ω$ ; its projection on the vertical axis gives the instantaneous value. For R the current phasor is along the voltage phasor; for L it is 90° behind; for C it is 90° ahead. Phasors make it easy to add voltages of different phase, which is needed for the series LCR circuit.

Lamp tests. A bulb in series with an inductor glows less than with a resistor-free DC path, and dims further if an iron rod is pushed into the coil (L and $X_{L}$ rise). A bulb in series with a capacitor glows brighter if the frequency is raised or a bigger capacitor is used ( $X_{C}$ falls).

AC Through Resistor, Inductor and Capacitor: formulas and key facts

Rms values: $I = \frac{i _{m}}{2}, V = \frac{v _{m}}{2}$; AC meters read rms values.
Resistor: $i = \frac{v _{m}}{R} sin ω t, P = I^{2} R$; In phase.
Inductive reactance: $X_{L} = ω L = 2 π f L$; Current lags voltage by π/2; average power 0.
Capacitive reactance: $X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C}$; Current leads voltage by π/2; average power 0.
Current through L or C alone: $I = \frac{V}{X _{L}} or \frac{V}{X _{C}}$; Use rms with rms, or peak with peak.
Instantaneous currents: $i_{L} = i_{m} sin (ω t - \frac{π}{2}), i_{C} = i_{m} sin (ω t + \frac{π}{2})$; For v = v_m sin ωt.

How to solve AC Through Resistor, Inductor and Capacitor questions

Use this method for AC questions on a single resistor, inductor or capacitor.

1Read the source: from $v = v_{m} sin ω t$ get $v_{m}$ and ω; $V_{r m s} = \frac{v _{m}}{2}$ , $f = \frac{ω}{2 π}$ .
2Find the opposition: R, $X_{L} = ω L$ or $X_{C} = \frac{1}{ω C}$ (C in farad, L in henry).
3Current: $I = \frac{V}{X}$ , keeping rms with rms and peak with peak.
4Phase: in phase for R, lagging by 90° for L, leading by 90° for C. Write the expression for i(t).
5Power: $I^{2} R$ for R; zero average power for pure L or C.
6For frequency changes, use $X_{L} \propto f$ and $X_{C} \propto \frac{1}{f}$ .
7State units: Ω for reactance, A for current, W for power.

AC Through Resistor, Inductor and Capacitor in real life

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.

Rms and peak values of mains

Suppose your home supply is 230 V rms at 50 Hz and an electric kettle rated 2300 W is connected to it.

(a) Find the peak voltage of the supply. (b) Find the rms and peak current in the kettle. (c) Why must the insulation of the wiring be rated for more than 230 V?

Try it first, then show the solutionSolution

$V_{m} = 2 \times 230 \approx 325$ V.
$I = \frac{P}{V} = \frac{2300}{230} = 10$ A; $i_{m} = 2 \times 10 \approx 14.1$ A.
The voltage actually reaches about 325 V twice in every cycle, so the insulation must withstand the peak, not the rms value.

Answer: (a) about 325 V; (b) 10 A rms and about 14.1 A peak; (c) the voltage peaks at about 325 V.

Memory hook: Meters show rms; insulation feels the peak.

Capacitor in an AC circuit

Imagine a 2.5 μF capacitor connected directly across a 230 V, 50 Hz supply (as a test, with no other components).

(a) Find its reactance. (b) Find the rms current. (c) How much average power does it take from the supply? Explain.

Try it first, then show the solutionSolution

$X_{C} = \frac{1}{2 π f C} = \frac{1}{2 π \times 50 \times 2.5 \times 1 0 ^{- 6}} \approx 1273$ Ω.
$I = \frac{230}{1273} \approx 0.18$ A.
Zero on average: the current leads the voltage by 90°, so the energy stored in the capacitor's field each quarter cycle is returned in the next.

Answer: (a) about 1273 Ω; (b) about 0.18 A; (c) zero average power.

Memory hook: A capacitor borrows energy and gives it back: current flows, but no power is used on average.

Tricks to remember AC Through Resistor, Inductor and Capacitor

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

ELI the ICE man
ELI: in an inductor (L), E (voltage) comes before I. ICE: in a capacitor (C), I comes before E.
Use it when: Phase relations.
L loves low, C loves high
Inductors pass low frequencies (X_L = ωL small); capacitors pass high frequencies (X_C = 1/ωC small).
Use it when: Frequency-dependence questions.
Root two between rms and peak
Peak = √2 × rms; 220 V rms is about 311 V peak.
Use it when: Converting AC values.
90° means no power
Pure L or C: voltage and current 90° apart, so cos 90° = 0 average power.
Use it when: Wattless current questions.

AC Through Resistor, Inductor and Capacitor: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Rms and peak values in a resistor · 2 marks · Very short answerBoard level

A 44 Ω heater is connected to a 220 V, 50 Hz supply. Find (a) the peak voltage, (b) the rms and peak current, and (c) the average power.

Try it first, then show the solutionSolution

(a) $v_{m} = 2 \times 220 \approx 311$ V. (½)
(b) $I = \frac{220}{44} = 5$ A; $i_{m} = 52 \approx 7.07$ A. (1)
(c) $P = I^{2} R = 25 \times 44 = 1100$ W. (½)

Answer: (a) about 311 V; (b) 5 A rms, about 7.07 A peak; (c) 1100 W.

Example 2: Derivation: AC through a pure inductor · 3 marks · Short answerBoard level

An AC voltage $v = v_{m} sin ω t$ is applied to a pure inductor L. Derive the expression for the current, show that it lags the voltage by π/2, and define inductive reactance. Draw the phasor diagram.

Try it first, then show the solutionSolution

The back emf balances the source: $v_{m} sin ω t = L \frac{d i}{d t}$ , so $\frac{d i}{d t} = \frac{v _{m}}{L} sin ω t$ . (1)
Integrating: $i = - \frac{v _{m}}{ω L} cos ω t = i_{m} sin (ω t - \frac{π}{2})$ with $i_{m} = \frac{v _{m}}{ω L}$ : the current lags by π/2. (1)
Inductive reactance $X_{L} = ω L$ , so $i_{m} = \frac{v _{m}}{X _{L}}$ . Phasor diagram: V phasor at angle ωt, I phasor 90° behind it (clockwise). (1)

Answer: $i = \frac{v _{m}}{ω L} sin (ω t - \frac{π}{2})$ ; current lags by 90°; $X_{L} = ω L$ .

Example 3: Capacitor at two frequencies · 3 marks · Short answerHarder board level

A 40 μF capacitor is connected to a 220 V supply. Find its reactance and the rms current at (a) 50 Hz and (b) 100 Hz. (c) What would happen on a 220 V DC supply?

Try it first, then show the solutionSolution

(a) $X_{C} = \frac{1}{2 π \times 50 \times 40 \times 1 0 ^{- 6}} \approx 79.6$ Ω; $I = \frac{220}{79.6} \approx 2.76$ A. (1)
(b) Doubling f halves $X_{C}$ to about 39.8 Ω, so I doubles to about 5.53 A. (1)
(c) For DC, f = 0 and $X_{C} \to \infty$ : after a brief charging current, no current flows. (1)

Answer: (a) about 79.6 Ω and 2.76 A; (b) about 39.8 Ω and 5.53 A; (c) no steady current.

Example 4: Inductor on the mains · 3 marks · Short answerHarder board level

A pure inductor of 0.2 H is connected to a 220 V, 50 Hz supply. Find (a) its reactance, (b) the rms and peak currents, and (c) the average power drawn.

Try it first, then show the solutionSolution

(a) $X_{L} = 2 π \times 50 \times 0.2 \approx 62.8$ Ω. (1)
(b) $I = \frac{220}{62.8} \approx 3.50$ A; $i_{m} = 2 \times 3.50 \approx 4.95$ A. (1)
(c) Zero: the current lags by 90°, so the energy taken in one quarter cycle is returned in the next. (1)

Answer: (a) about 62.8 Ω; (b) about 3.50 A rms and 4.95 A peak; (c) zero.

Example 5: Lamp in series with a coil · 5 marks · Long answerHOTS / competency

A lamp of resistance 50 Ω is connected in series with a coil of negligible resistance and inductance 0.1 H across a 220 V, 50 Hz supply.
(a) Find the current and the power used by the lamp. (2)
(b) An iron rod is pushed into the coil, raising its inductance to 0.5 H. Find the new current and lamp power. (2)
(c) What would you see if the AC supply were replaced by a 220 V DC supply? (1)

Try it first, then show the solutionSolution

(a) $X_{L} = 2 π \times 50 \times 0.1 \approx 31.4$ Ω; the voltages across R and L are 90° apart, so the total opposition is $5 0^{2} + 31. 4^{2} \approx 59.0$ Ω. (1)
$I \approx \frac{220}{59.0} \approx 3.73$ A; lamp power $I^{2} R \approx 694$ W. (1)
(b) $X_{L} \approx 157$ Ω, opposition $\approx 5 0^{2} + 15 7^{2} \approx 165$ Ω, $I \approx 1.33$ A. (1)
Lamp power $\approx 1.3 3^{2} \times 50 \approx 89$ W: the lamp dims sharply. (1)
(c) For DC the inductor has no reactance (f = 0), so I = 220/50 = 4.4 A and the lamp glows at full brightness; the iron rod makes no difference to the steady current. (1)

Answer: (a) about 3.73 A and 694 W; (b) about 1.33 A and 89 W, the lamp dims; (c) the lamp glows brightly and the rod has no effect.

Example 6: Reading an AC equation · 4 marks · Case-basedHarder board level

Read the passage and answer.

An AC source gives $v = 311 sin (100 π t)$ volt.
(a) Find the rms voltage and the frequency. (1)
(b) Find the voltage at t = 1/600 s. (1)
(c) How long after t = 0 does the voltage first reach its peak? (1)
(d) This source is connected across a pure capacitor. Write the expression for the current if its peak is 2 A. (1)

Try it first, then show the solutionSolution

(a) $V = \frac{311}{2} \approx 220$ V; $f = \frac{100 π}{2 π} = 50$ Hz. (1)
(b) $v = 311 sin \frac{π}{6} = 155.5$ V. (1)
(c) Peak at $ω t = \frac{π}{2}$ : $t = \frac{T}{4} = \frac{0.02}{4} = 5$ ms. (1)
(d) The current leads by π/2: $i = 2 sin (100 π t + \frac{π}{2}) = 2 cos (100 π t)$ ampere. (1)

Answer: (a) about 220 V, 50 Hz; (b) 155.5 V; (c) 5 ms; (d) i = 2 cos(100πt) A.

Where students lose marks in AC Through Resistor, Inductor and Capacitor

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

Mistake: Mixing rms and peak values in one equation.
Not noticing which the question gives.
Wrong working
i_m = V_rms/X_L
Right working
I_rms = V_rms/X_L, or i_m = v_m/X_L
Do this: Label every value as rms or peak before substituting.
Usually costs: 1 mark.
Mistake: Getting lead and lag the wrong way round.
Memory slip.
Do this: Use 'ELI the ICE man': voltage leads in an inductor, current leads in a capacitor.
Usually costs: 1 mark.
Mistake: Using f instead of ω in reactance.
Writing X_L = fL.
Wrong working
X_L = 50 × 0.2 = 10 Ω
Right working
X_L = 2π × 50 × 0.2 ≈ 62.8 Ω
Do this: Always write ω = 2πf.
Usually costs: The answer mark.
Mistake: Leaving capacitance in μF in X_C.
Forgetting 1 μF = 10⁻⁶ F.
Do this: Convert to farad first.
Usually costs: The answer mark.
Mistake: Saying an inductor uses power because current flows.
Thinking any current means power.
Do this: Average power = VI cos φ; for a pure L or C, φ = 90° and the average power is zero.
Usually costs: 1 mark.

Exam tips

In derivations, draw the v and i graphs on the same axes and the phasor diagram; both earn marks.
State the phase relation in words and as an equation.
Reactance and phase questions were set in 2023 (55/5/1 Q27, identifying elements from the phase of the current) and 2024 (55/1/1 Q7 on capacitive reactance, and Q32(a), which includes a lamp run on AC through a series coil).
When the question gives an equation like v = 311 sin 100πt, write v_m, ω, f and V_rms first.
Keep this topic and the series LCR topic linked: reactances and phases are the building blocks of impedance.

Practice questions on AC Through Resistor, Inductor and Capacitor

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
In a pure inductor connected to an AC source, 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
Hint
ELI.
Answer
(c) lags the voltage by π/2.
Q2 · 1 mark · Very short answerBoard level
The current in a circuit is $i = 10 sin (100 π t)$ A. Find its rms value and frequency.
Hint
i_m/√2 and ω/2π.
Answer
$I = \frac{10}{2} \approx 7.07$ A; f = 50 Hz.
Q3 · 2 marks · Very short answerHarder board level
Find the reactance of a 0.5 H inductor at 50 Hz. At what frequency would its reactance be 1000 Ω?
Hint
X_L = 2πfL.
Answer
$X_{L} = 2 π \times 50 \times 0.5 \approx 157$ Ω. For 1000 Ω: $f = \frac{1000}{2 π \times 0.5} \approx 318$ Hz.
Q4 · 1 mark · Very short answerHarder board level
A 10 μF capacitor is connected to a 220 V, 50 Hz supply. Find the rms current.
Hint
I = VωC.
Answer
$I = 220 \times 2 π \times 50 \times 1 0^{- 5} \approx 0.69$ A.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A capacitor blocks direct current but allows alternating current to flow in the circuit.
Reason (R): The capacitive reactance 1/(ωC) is infinite for ω = 0 and finite for ω > 0.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
DC has zero frequency.
Answer
(a).
Q6 · 2 marks · Very short answerHOTS / competency
Two identical lamps are connected to the same AC supply, one in series with a capacitor and the other in series with an inductor. The supply frequency is then increased. What happens to the brightness of each lamp? Explain.
Hint
X_C ∝ 1/f, X_L ∝ f.
Answer
The lamp with the capacitor gets brighter, because X_C falls and more current flows. The lamp with the inductor gets dimmer, because X_L rises and less current flows.
Q7 · 3 marks · Short answerHOTS / competency
Show that the average power over a full cycle in a pure inductor is zero.
Hint
Multiply v and i and average.
Answer
With $v = v_{m} sin ω t$ and $i = - i_{m} cos ω t$ , $p = v i = - v_{m} i_{m} sin ω t cos ω t = - \frac{v _{m} i _{m}}{2} sin 2 ω t$ . The average of $sin 2 ω t$ over a full cycle is zero, so the average power is zero: energy stored in the field during one quarter cycle is returned in the next.

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.Relation between rms and peak current?Show answer
I = i_m/√2.
2.Inductive reactance?Show answer
X_L = ωL.
3.Capacitive reactance?Show answer
X_C = 1/(ωC).
4.Phase of current in a pure capacitor?Show answer
Leads the voltage by π/2.
5.Average power in a pure inductor?Show answer
Zero.
6.How does X_C change if frequency doubles?Show answer
It halves.

AC Through Resistor, Inductor and Capacitor: frequently asked questions

Why do we use rms values for AC?

The average of a sinusoidal current over a cycle is zero, which says nothing about its heating effect. The rms value is the steady current that produces the same heating in a resistor, so it is the useful measure.

Why does a capacitor block DC?

Once charged to the DC voltage, no further charge can flow, so the steady current is zero. In terms of reactance, X_C = 1/(ωC) is infinite when ω = 0.

Why does the current lag in an inductor?

The inductor's back emf opposes changes in current, so the current builds up only after the voltage has been applied. For a sinusoidal voltage this delay is exactly a quarter cycle (90°).

What is the unit of reactance?

Ohm. Like resistance, it is the ratio of voltage to current, but it causes no energy loss on average.

What does an iron core do to a coil in an AC circuit?

It increases the coil's inductance, and so its reactance X_L = ωL. A lamp in series with the coil then gets dimmer.