PhysicsCBSE Class 12 · Alternating Current

Series LCR Circuit and Resonance Class 12: Impedance and Power Factor

Updated 15 October 2026 · Fact-checked

In a series LCR circuit the voltages across R, L and C add as phasors, giving impedance Z = √(R² + (X_L − X_C)²) and tan φ = (X_L − X_C)/R. At resonance X_L = X_C, ω₀ = 1/√(LC), Z = R and the current is largest. Average power is VI cos φ, with power factor cos φ = R/Z.

Understand Series LCR Circuit and Resonance

A radio picks one station out of hundreds because a tuned circuit responds strongly to just one frequency. That circuit is a series LCR circuit near resonance. The 2026-27 syllabus treats it with phasors only.

Phasor diagram. The same current $i = i_{m} sin (ω t + ϕ)$ flows through R, L and C in series, so draw the current phasor as reference. Then:

$V_{R} = I R$ is along I.
$V_{L} = I X_{L}$ is 90° ahead of I.
$V_{C} = I X_{C}$ is 90° behind I.

$V_{L}$ and $V_{C}$ point in opposite directions, so they combine to $V_{L} - V_{C}$ . The source voltage is the phasor sum: $V^{2} = V_{R}^{2} + (V_{L} - V_{C})^{2}$ .

Impedance and phase. Dividing by I: $Z = R^{2} + (X_{L} - X_{C})^{2}$ , the impedance (in ohm), and $I = \frac{V}{Z}$ . The angle φ between the source voltage and the current is given by $tan ϕ = \frac{X _{L} - X _{C}}{R}$ . If $X_{L} > X_{C}$ the circuit is inductive (current lags); if $X_{C} > X_{L}$ it is capacitive (current leads). R, $X_{L}$ , $X_{C}$ and Z form a right-angled 'impedance triangle'.

Resonance. $X_{L} = ω L$ grows with frequency and $X_{C} = \frac{1}{ω C}$ falls. At one frequency they are equal: $ω_{0} L = \frac{1}{ω _{0} C}$ , so $ω_{0} = \frac{1}{L C}$ and $f_{0} = \frac{1}{2 π L C}$ . At resonance:

Z = R, its smallest value, so the current $I = \frac{V}{R}$ is largest.
φ = 0: current and voltage are in phase; the circuit behaves as a pure resistor.
$V_{L}$ and $V_{C}$ are equal and opposite. Each can be much larger than the source voltage (when $X_{L} ≫ R$ ), though they cancel each other.

The smaller R is, the taller and narrower the current peak; this is what makes tuning selective. (The quality factor and LC oscillations are not in the 2026-27 syllabus.) Resonance needs both L and C: an RL or RC circuit cannot resonate.

Power in an AC circuit. The average power is $P = V I cos ϕ$ (rms values). The factor $cos ϕ$ is the power factor:

Pure R: φ = 0, cos φ = 1, $P = V I$ .
Pure L or C: φ = ±90°, cos φ = 0, P = 0.
Series LCR: $cos ϕ = \frac{R}{Z}$ , so $P = I^{2} Z \cdot \frac{R}{Z} = I^{2} R$ : only the resistor uses power on average.
At resonance, cos φ = 1.

Wattless current. Resolve the current into $I cos ϕ$ (in phase with V) and $I sin ϕ$ (90° out of phase). The second part contributes no average power; it is called the wattless current. A low power factor means a large current for the same useful power, so more heat is wasted in transmission lines ( $I^{2} R$ losses). That is why industries are asked to keep their power factor close to 1, for example by adding capacitors to inductive loads such as motors.

Series LCR Circuit and Resonance: formulas and key facts

Impedance: $Z = R^{2} + (X_{L} - X_{C})^{2}$; I = V/Z (rms or peak).
Phase angle: $tan ϕ = \frac{X _{L} - X _{C}}{R}$; φ > 0: inductive (I lags); φ < 0: capacitive (I leads).
Voltage phasor sum: $V = V_{R}^{2} + (V_{L} - V_{C})^{2}$; Never add V_R, V_L and V_C arithmetically.
Resonant frequency: $ω_{0} = \frac{1}{L C}, f_{0} = \frac{1}{2 π L C}$; X_L = X_C, Z = R, current largest, φ = 0.
Average power and power factor: $P = V I cos ϕ, cos ϕ = \frac{R}{Z}$; Equivalently P = I²R.
Wattless current: $I_{w a ttl ess} = I sin ϕ$; Contributes no average power.

How to solve Series LCR Circuit and Resonance questions

Use this method for series LCR numericals.

1Find ω (= 2πf), then $X_{L} = ω L$ and $X_{C} = \frac{1}{ω C}$ .
2Find $Z = R^{2} + (X_{L} - X_{C})^{2}$ and $I = \frac{V}{Z}$ .
3Find φ from $tan ϕ = \frac{X _{L} - X _{C}}{R}$ or $cos ϕ = \frac{R}{Z}$ ; say whether the circuit is inductive or capacitive.
4Voltages: $V_{R} = I R$ , $V_{L} = I X_{L}$ , $V_{C} = I X_{C}$ ; check that $V_{R}^{2} + (V_{L} - V_{C})^{2} = V$ .
5Power: $P = V I cos ϕ = I^{2} R$ .
6For resonance, set $X_{L} = X_{C}$ : $ω_{0} = \frac{1}{L C}$ ; then Z = R and I = V/R.
7To make a circuit resonate at a given frequency, solve $C = \frac{1}{ω ^{2} L}$ or $L = \frac{1}{ω ^{2} C}$ .

Series LCR Circuit and Resonance 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.

Resonance in radio tuning

Suppose a radio's tuning circuit has a fixed 250 μH coil and a variable capacitor, and it must be tuned across stations from 600 kHz to 1500 kHz.

(a) Find the capacitance needed for 600 kHz and for 1500 kHz. (b) Why does the radio pick out one station at a time?

Try it first, then show the solutionSolution

$C = \frac{1}{( 2 π f ) ^{2} L}$ . At 600 kHz: $C = \frac{1}{( 2 π \times 6 \times 1 0 ^{5} ) ^{2} \times 2.5 \times 1 0 ^{- 4}} \approx 281$ pF.
At 1500 kHz: $C \approx 45$ pF (the frequency is 2.5 times, so C is $\frac{1}{6.25}$ times).
At its resonant frequency the circuit's impedance is least and the current from that station's signal is largest; other frequencies meet a much larger impedance and give tiny currents.

Answer: (a) about 281 pF and 45 pF; (b) only the station at the resonant frequency produces a large current.

Memory hook: Tuning = choosing C so that X_L = X_C at the station you want.

Why power factor matters

Imagine a workshop that uses 10 kW of power at 400 V with a power factor of 0.6. The electricity board asks it to raise the power factor to 0.9.

(a) Find the current drawn at each power factor. (b) By what factor does the heating loss in the supply cables fall? (c) Suggest how the power factor can be raised.

Try it first, then show the solutionSolution

$I = \frac{P}{V c o s ϕ}$ : at 0.6, $I = \frac{1 0 ^{4}}{240} \approx 41.7$ A; at 0.9, $I = \frac{1 0 ^{4}}{360} \approx 27.8$ A.
Cable loss ∝ I²: ratio $= (\frac{27.8}{41.7})^{2} = (\frac{0.6}{0.9})^{2} \approx 0.44$ .
Motors are inductive; adding suitable capacitors reduces $X_{L} - X_{C}$ , bringing φ towards zero.

Answer: (a) about 41.7 A and 27.8 A; (b) to about 44% of the original loss; (c) add capacitors to cancel part of the inductive reactance.

Memory hook: Low power factor: more current for the same useful power, and more heat in the wires.

Tricks to remember Series LCR Circuit and Resonance

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

L up, C down, R across
In the phasor diagram draw V_R along I, V_L straight up and V_C straight down.
Use it when: Drawing the phasor diagram quickly.
Pythagoras for AC
Z² = R² + (X_L − X_C)²: a right triangle with R as base and the net reactance as height.
Use it when: Impedance and phase angle.
Resonance: reactances shake hands and cancel
X_L = X_C, so Z = R and the circuit acts like a plain resistor.
Use it when: Properties at resonance.
Only R eats power
P = I²R; L and C only borrow and return energy.
Use it when: Power calculations.

Series LCR Circuit and Resonance: solved examples

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

Example 1: Impedance, current and power · 3 marks · Short answerBoard level

A series circuit has R = 40 Ω, X_L = 90 Ω and X_C = 60 Ω, connected to a 220 V supply. Find (a) the impedance, (b) the current, (c) the phase angle and (d) the average power.

Try it first, then show the solutionSolution

(a) $Z = 4 0^{2} + (90 - 60)^{2} = 50$ Ω; (b) $I = \frac{220}{50} = 4.4$ A. (1)
(c) $tan ϕ = \frac{30}{40}$ , φ ≈ 36.9°; the circuit is inductive, so the current lags. (1)
(d) $P = V I cos ϕ = 220 \times 4.4 \times 0.8 = 774.4$ W (check: $I^{2} R = 19.36 \times 40 = 774.4$ W). (1)

Answer: (a) 50 Ω; (b) 4.4 A; (c) about 36.9°, current lagging; (d) about 774 W.

Example 2: Derivation: impedance by phasors and resonance · 5 marks · Long answerBoard level

(a) A series LCR circuit is connected to $v = v_{m} sin ω t$ . Draw the phasor diagram and derive the impedance and the phase angle. (3)
(b) Obtain the condition for resonance and the resonant frequency. State two features of the circuit at resonance. (2)

Try it first, then show the solutionSolution

(a) Take the current phasor as reference. $V_{R} = i_{m} R$ along it, $V_{L} = i_{m} X_{L}$ 90° ahead, $V_{C} = i_{m} X_{C}$ 90° behind. (1)
$V_{L}$ and $V_{C}$ are opposite, giving a net $i_{m} (X_{L} - X_{C})$ perpendicular to $V_{R}$ . By Pythagoras, $v_{m}^{2} = i_{m}^{2} [R^{2} + (X_{L} - X_{C})^{2}]$ , so $Z = \frac{v _{m}}{i _{m}} = R^{2} + (X_{L} - X_{C})^{2}$ . (1)
The angle between V and I: $tan ϕ = \frac{X _{L} - X _{C}}{R}$ . (1)
(b) The current $\frac{v _{m}}{Z}$ is largest when Z is least, i.e. when $X_{L} = X_{C}$ : $ω_{0} L = \frac{1}{ω _{0} C}$ , $ω_{0} = \frac{1}{L C}$ . (1)
At resonance Z = R (current largest), φ = 0 (power factor 1), and $V_{L} = V_{C}$ , each possibly larger than the source voltage. (1)

Answer: (a) Z = √(R² + (X_L − X_C)²), tan φ = (X_L − X_C)/R; (b) X_L = X_C, ω₀ = 1/√(LC); Z = R and φ = 0 at resonance.

Example 3: Voltages at resonance · 3 marks · Short answerHarder board level

A 0.5 H inductor, an 8 μF capacitor and a 50 Ω resistor are in series across a 200 V variable-frequency supply. Find (a) the resonant angular frequency and frequency, (b) the current at resonance, and (c) the rms voltage across the inductor and across the capacitor at resonance. Comment.

Try it first, then show the solutionSolution

(a) $ω_{0} = \frac{1}{0.5 \times 8 \times 1 0 ^{- 6}} = 500$ rad/s; $f_{0} = \frac{500}{2 π} \approx 79.6$ Hz. (1)
(b) Z = R: $I = \frac{200}{50} = 4$ A. (1)
(c) $V_{L} = I ω_{0} L = 4 \times 250 = 1000$ V; $V_{C} = \frac{I}{ω _{0} C} = 1000$ V. Each is five times the supply voltage, but they are opposite in phase and cancel. (1)

Answer: (a) 500 rad/s, about 79.6 Hz; (b) 4 A; (c) 1000 V each, cancelling.

Example 4: Power factor and wattless current · 2 marks · Very short answerHarder board level

A circuit draws 2 A from a 230 V supply with the current lagging the voltage by 30°. Find (a) the power factor, (b) the average power and (c) the wattless current.

Try it first, then show the solutionSolution

(a) cos 30° ≈ 0.866; (b) $P = 230 \times 2 \times 0.866 \approx 398$ W. (1)
(c) $I sin ϕ = 2 \times 0.5 = 1$ A. (1)

Answer: (a) about 0.866; (b) about 398 W; (c) 1 A.

Example 5: Retuning a circuit to resonance · 5 marks · Long answerHOTS / competency

A series circuit with R = 40 Ω, L = 0.4 H and C = 50 μF is connected to a 220 V, 50 Hz supply.
(a) Find X_L, X_C, Z and the current. (2)
(b) Find the power. Is the circuit inductive or capacitive? (1)
(c) What capacitor should replace C to make the circuit resonate at 50 Hz, and what are the new current and power? (2)

Try it first, then show the solutionSolution

(a) $X_{L} = 2 π \times 50 \times 0.4 \approx 125.7$ Ω; $X_{C} = \frac{1}{2 π \times 50 \times 50 \times 1 0 ^{- 6}} \approx 63.7$ Ω. (1)
$Z = 4 0^{2} + 62. 0^{2} \approx 73.8$ Ω; $I = \frac{220}{73.8} \approx 2.98$ A. (1)
(b) $P = I^{2} R \approx 2.9 8^{2} \times 40 \approx 356$ W; X_L > X_C, so the circuit is inductive. (1)
(c) $C = \frac{1}{ω ^{2} L} = \frac{1}{( 100 π ) ^{2} \times 0.4} \approx 25.3$ μF. (1)
At resonance Z = R = 40 Ω: I = 5.5 A and P = 5.5² × 40 = 1210 W. (1)

Answer: (a) about 125.7 Ω, 63.7 Ω, 73.8 Ω and 2.98 A; (b) about 356 W, inductive; (c) about 25.3 μF, giving 5.5 A and 1210 W.

Example 6: Voltmeter readings in an LCR circuit · 4 marks · Case-basedHarder board level

Read the passage and answer.

In a series LCR circuit carrying 2 A (rms), AC voltmeters read 60 V across R, 120 V across L and 40 V across C.
(a) Find the supply voltage. (1)
(b) Find the power factor and say whether the current leads or lags. (1)
(c) Find R, X_L and X_C. (1)
(d) Why is the supply voltage not 60 + 120 + 40 = 220 V? (1)

Try it first, then show the solutionSolution

(a) $V = 6 0^{2} + (120 - 40)^{2} = 100$ V. (1)
(b) $cos ϕ = \frac{V _{R}}{V} = 0.6$ ; V_L > V_C, so the circuit is inductive and the current lags. (1)
(c) R = 60/2 = 30 Ω, X_L = 120/2 = 60 Ω, X_C = 40/2 = 20 Ω. (1)
(d) The three voltages are not in phase; they add as phasors, not as numbers. (1)

Answer: (a) 100 V; (b) 0.6, current lags; (c) 30 Ω, 60 Ω, 20 Ω; (d) voltages add as phasors.

Where students lose marks in Series LCR Circuit and Resonance

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

Mistake: Adding V_R, V_L and V_C as numbers.
Forgetting the phase differences.
Wrong working
V = 60 + 120 + 40 = 220 V
Right working
V = √(60² + (120 − 40)²) = 100 V
Do this: Always add as phasors.
Usually costs: Most of the numerical.
Mistake: Writing Z = R + X_L + X_C.
Treating reactances like resistances in series.
Do this: Z = √(R² + (X_L − X_C)²).
Usually costs: The answer marks.
Mistake: Using P = VI for an LCR circuit.
Forgetting the power factor.
Do this: P = VI cos φ = I²R.
Usually costs: 1 mark.
Mistake: Getting lead or lag wrong from the phasor diagram.
Not checking which reactance is larger.
Do this: X_L > X_C: inductive, current lags. X_C > X_L: capacitive, current leads.
Usually costs: ½ to 1 mark.
Mistake: Writing the Q factor or analytical solution in a phasor question.
Older books include them.
Do this: For 2026-27, use phasors only; sharpness of resonance (Q factor) and LC oscillations are not assessed.
Usually costs: Time.

Exam tips

A neat phasor diagram (I as reference, V_R along it, V_L up, V_C down, resultant V at angle φ) earns marks in every derivation.
Series LCR questions carried 3- and 5-mark parts in recent board papers, for example 2023 (55/1/1 Q27) and 2024 (55/1/1 Q25 and 55/5/1 Q32(a)); 2026 (55/1/1 Q6) set a 1-mark phasor item.
Always state whether the circuit is inductive or capacitive when asked for the phase.
At resonance, list: X_L = X_C, Z = R, I maximum, φ = 0, power factor 1.
Check power with both VI cos φ and I²R; they must agree.

Practice questions on Series LCR Circuit and Resonance

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
At resonance in a series LCR circuit, the phase difference between the current and the source voltage is
(a) 0 (b) π/4 (c) π/2 (d) π
Hint
X_L = X_C.
Answer
(a) 0.
Q2 · 2 marks · Very short answerBoard level
Define power factor. What is its value for (a) a pure resistor and (b) a pure inductor?
Hint
P = VI cos φ.
Answer
Power factor is cos φ, the ratio of average power to VI (also R/Z in a series circuit). (a) 1; (b) 0.
Q3 · 1 mark · Very short answerHarder board level
Find the resonant frequency of a series circuit with L = 0.1 H and C = 10 μF.
Hint
ω₀ = 1/√(LC).
Answer
$ω_{0} = \frac{1}{1 0 ^{- 6}} = 1000$ rad/s; $f_{0} \approx 159$ Hz.
Q4 · 3 marks · Short answerHarder board level
A series circuit has R = 6 Ω, X_L = 12 Ω and X_C = 4 Ω across 100 V. Find Z, I, the power factor and the power.
Hint
Net reactance 8 Ω.
Answer
Z = √(36 + 64) = 10 Ω; I = 10 A; cos φ = 6/10 = 0.6; P = 100 × 10 × 0.6 = 600 W (= I²R).
Q5 · 1 mark · Very short answerHarder board level
A current of 10 A flows in a circuit with power factor 0.8. Find the wattless component of the current.
Hint
sin φ = √(1 − cos²φ).
Answer
sin φ = 0.6, so the wattless current is 10 × 0.6 = 6 A.
Q6 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): At resonance, the voltage across the inductor in a series LCR circuit can be greater than the supply voltage.
Reason (R): At resonance the voltages across L and C are equal in size and opposite in phase, so they cancel and the supply voltage appears entirely across R.
(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
V_L = IX_L with I = V/R.
Answer
(a). Since $V_{L} = \frac{V}{R} X_{L}$ , V_L exceeds V whenever X_L > R; this is possible only because V_L is cancelled by V_C.
Q7 · 3 marks · Short answerHOTS / competency
A series LCR circuit is driven below its resonant frequency. Is it inductive or capacitive? Does the current lead or lag? What happens to the current as the frequency is slowly raised to resonance and beyond?
Hint
Compare X_L and X_C below ω₀.
Answer
Below ω₀, X_C > X_L, so the circuit is capacitive and the current leads. As the frequency rises, X_C falls and X_L rises, so Z falls and the current rises to its maximum V/R at resonance (in phase). Beyond resonance X_L > X_C, the circuit becomes inductive (current lags) and the current falls again.

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.Impedance of a series LCR circuit?Show answer
Z = √(R² + (X_L − X_C)²).
2.Resonant angular frequency?Show answer
ω₀ = 1/√(LC).
3.Power factor of a series LCR circuit?Show answer
cos φ = R/Z.
4.Impedance at resonance?Show answer
Z = R (minimum).
5.Average power in an AC circuit?Show answer
P = VI cos φ.
6.What is wattless current?Show answer
The component I sin φ, 90° out of phase with V, which uses no average power.

Series LCR Circuit and Resonance: frequently asked questions

Can an RL or RC circuit show resonance?

No. Resonance needs both an inductor and a capacitor, so that their reactances can cancel at one frequency.

Why is the current largest at resonance?

At resonance X_L = X_C, so they cancel and the impedance falls to its smallest value, R. With the smallest opposition, the current V/R is largest.

Why can the voltage across L exceed the supply voltage?

At resonance the current V/R can be large, so I X_L can be larger than V. The voltage across C is equal and opposite at every instant, so the two cancel and the supply only has to provide V across R.

What is the significance of power factor?

It tells what fraction of VI is actually used as power. A low power factor means a large current for little useful power, wasting energy as heat in the supply lines.

Is the Q factor in the 2026-27 syllabus?

No. The series LCR circuit is studied with phasors only; sharpness of resonance (Q factor), the analytical solution and LC oscillations are not assessed.