PhysicsISC Class 12 · Alternating Current

Series LCR Circuit: Phasor Diagram, Impedance and Phase Angle

Updated 15 October 2026 · Fact-checked

In a series LCR circuit one current flows through R, L and C. Adding their voltages as phasors gives $V = I\sqrt{R^2 + (X_L - X_C)^2}$, so the impedance is $Z = \sqrt{R^2 + (X_L - X_C)^2}$ with $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$. The voltage leads the current by φ, where $\tan\phi = \frac{X_L - X_C}{R}$.

Understand Series LCR Circuit: Impedance and Phase

Building blocks. For $V = V_{0} sin ω t$ :

Resistor: $I = I_{0} sin ω t$ , in phase with V; $V_{0} = I_{0} R$ .
Inductor: $I = I_{0} sin (ω t - \frac{π}{2})$ : current lags voltage by 90°; $V_{0} = I_{0} X_{L}$ with inductive reactance $X_{L} = ω L = 2 π f L$ (rises with frequency).
Capacitor: $I = I_{0} sin (ω t + \frac{π}{2})$ : current leads voltage by 90°; $V_{0} = I_{0} X_{C}$ with capacitive reactance $X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C}$ (falls with frequency; infinite for d.c.).

Reactances are measured in ohms, like resistance, but they store and return energy instead of dissipating it.

Series LCR: phasor method. The current $I = I_{0} sin ω t$ is common, so draw it as the reference phasor along the x-axis. Then:

$V_{R} = I_{0} R$ along I;
$V_{L} = I_{0} X_{L}$ ahead of I by 90° (up);
$V_{C} = I_{0} X_{C}$ behind I by 90° (down).

$V_{L}$ and $V_{C}$ are in opposite directions, so combine them first: $V_{L} - V_{C}$ (taking $V_{L} > V_{C}$ ). The supply voltage is the vector sum:
$V_{0}^{2} = V_{R}^{2} + (V_{L} - V_{C})^{2} = I_{0}^{2} [R^{2} + (X_{L} - X_{C})^{2}] .$
So
$Z = \frac{V _{0}}{I _{0}} = R^{2} + (X_{L} - X_{C})^{2} = R^{2} + (ω L - \frac{1}{ω C})^{2},$
and the phase angle φ between V and I is given by
$tan ϕ = \frac{V _{L} - V _{C}}{V _{R}} = \frac{X _{L} - X _{C}}{R}, cos ϕ = \frac{R}{Z} .$
Then $V = V_{0} sin (ω t + ϕ)$ , or equivalently $I = I_{0} sin (ω t - ϕ)$ .

Character of the circuit.

$X_{L} > X_{C}$ : φ > 0, current lags: the circuit is inductive.
$X_{L} < X_{C}$ : φ < 0, current leads: capacitive.
$X_{L} = X_{C}$ : φ = 0, Z = R: resonance (see the next topic).

Special cases. RL circuit (no C): $Z = R^{2} + X_{L}^{2}$ , current lags by $tan^{- 1} \frac{X _{L}}{R}$ . RC circuit: $Z = R^{2} + X_{C}^{2}$ , current leads. Never add R and X as plain numbers: they are at right angles. The ISC 2024 examiners penalised $Z = R + X_{L}$ and the use of V = IR instead of V = IZ.

Voltages across components. The r.m.s. readings $V_{R}$ , $V_{L}$ , $V_{C}$ do not add up to the supply voltage, and $V_{L}$ or $V_{C}$ can each be larger than the supply voltage, because $V_{L}$ and $V_{C}$ partly cancel.

Identifying a component from equations. If $V = V_{0} sin ω t$ and $I = I_{0} sin (ω t - ϕ)$ with φ > 0, the current lags: the unknown is an inductor (with R in series). If the current leads, it is a capacitor. Then $Z = \frac{V _{0}}{I _{0}}$ , $R = Z cos ϕ$ , $X = Z sin ϕ$ , $f = \frac{ω}{2 π}$ , $I_{r m s} = \frac{I _{0}}{2}$ and power factor $= cos ϕ$ (2025 ISC 5-mark question).

Series LCR Circuit: Impedance and Phase: formulas and key facts

Inductive reactance: $X_{L} = ω L = 2 π f L$; Current lags by 90°.
Capacitive reactance: $X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C}$; Current leads by 90°.
Impedance: $Z = R^{2} + (X_{L} - X_{C})^{2}$; Ohms.
Phase angle: $tan ϕ = \frac{X _{L} - X _{C}}{R}, cos ϕ = \frac{R}{Z}$; φ > 0: inductive (lag).
Current: $I_{r m s} = \frac{V _{r m s}}{Z}, I_{0} = \frac{V _{0}}{Z}$; Use Z, not R.
Component voltages: $V^{2} = V_{R}^{2} + (V_{L} - V_{C})^{2}$; Phasor sum, not arithmetic.

How to solve Series LCR Circuit: Impedance and Phase questions

Reactances first, then impedance, then current, then everything else.

1Compute $X_{L} = 2 π f L$ and $X_{C} = \frac{1}{2 π f C}$ (or read ω from the equation, $ω = 2 π f$ ).
2Find $Z = R^{2} + (X_{L} - X_{C})^{2}$ .
3Current: $I_{r m s} = \frac{V _{r m s}}{Z}$ (or peak with peak).
4Phase: $tan ϕ = \frac{X _{L} - X _{C}}{R}$ ; state lag (inductive) or lead (capacitive).
5Component voltages: $V_{R} = I R$ , $V_{L} = I X_{L}$ , $V_{C} = I X_{C}$ ; check $V_{R}^{2} + (V_{L} - V_{C})^{2} = V$ .
6Draw the phasor diagram if asked to explain or derive.

Series LCR Circuit: Impedance and Phase 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.

An inductive load on the mains

Suppose a ceiling-fan motor behaves like a resistance of 80 Ω in series with an inductive reactance of 60 Ω on the 230 V (r.m.s.), 50 Hz mains.

Find the impedance, the current and the phase angle.

Try it first, then show the solutionSolution

$Z = 8 0^{2} + 6 0^{2} = 100$ Ω.
$I = \frac{230}{100} = 2.3$ A.
$tan ϕ = \frac{60}{80}$ , φ ≈ 36.9°, current lagging.

Answer: 100 Ω; 2.3 A; current lags by about 37°.

Memory hook: Impedance triangle: R across, X up, Z the slope.

Voltages that do not add

Imagine a student who measures r.m.s. voltages of 40 V across R, 90 V across L and 60 V across C in a series circuit, and is puzzled because they add to 190 V.

What is the supply voltage?

Try it first, then show the solutionSolution

$V_{L}$ and $V_{C}$ are in antiphase: net reactive voltage $90 - 60 = 30$ V, at 90° to $V_{R}$ .
$V = 4 0^{2} + 3 0^{2} = 50$ V.

Answer: 50 V. Phasors, not plain numbers, add.

Memory hook: L and C fight; R stands at right angles.

Tricks to remember Series LCR Circuit: Impedance and Phase

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) voltage E leads I; in a capacitor (C) I leads E.
Use it when: Phase relations.
Subtract, square, add, root
$Z = R^{2} + (X_{L} - X_{C})^{2}$ .
Use it when: Impedance.
L likes high f, C likes low f
$X_{L}$ grows with f; $X_{C}$ shrinks with f.
Use it when: Frequency questions.
Lag means L
Current lagging → inductive; leading → capacitive.
Use it when: Identifying components.

Series LCR Circuit: Impedance and Phase: solved examples

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

Example 1: Lag or lead? · 1 mark · MCQBoard level

In a series LCR circuit, $X_{L} > X_{C}$ . The current:
(a) is in phase with the voltage
(b) leads the voltage
(c) lags the voltage
(d) is zero

Try it first, then show the solutionSolution

$tan ϕ = \frac{X _{L} - X _{C}}{R} > 0$ : the circuit is inductive.

Answer: (c).

Example 2: An RL circuit · 2 marks · Short answerBoard level

A coil of resistance 30 Ω and inductive reactance 40 Ω is connected to a 100 V (r.m.s.) a.c. supply. Find the impedance, the current and the phase angle.

Try it first, then show the solutionSolution

$Z = 3 0^{2} + 4 0^{2} = 50$ Ω; $I = \frac{100}{50} = 2$ A. (1)
$tan ϕ = \frac{40}{30}$ , φ ≈ 53.1°, current lagging. (1)

Answer: 50 Ω; 2 A; 53.1° lag.

Example 3: Derivation: impedance by phasors · 4 marks · Long answerHarder board level

Using a phasor diagram, derive the impedance of a series LCR circuit and the phase angle between the supply voltage and the current.

Try it first, then show the solutionSolution

Same current $I_{0}$ in all elements: take it as the reference phasor. $V_{R} = I_{0} R$ in phase; $V_{L} = I_{0} X_{L}$ leading by 90°; $V_{C} = I_{0} X_{C}$ lagging by 90°. (1)
$V_{L}$ and $V_{C}$ are opposite: net $V_{L} - V_{C}$ perpendicular to $V_{R}$ . (1)
$V_{0}^{2} = V_{R}^{2} + (V_{L} - V_{C})^{2} = I_{0}^{2} [R^{2} + (X_{L} - X_{C})^{2}]$ , so $Z = \frac{V _{0}}{I _{0}} = R^{2} + (X_{L} - X_{C})^{2}$ . (1)
$tan ϕ = \frac{V _{L} - V _{C}}{V _{R}} = \frac{X _{L} - X _{C}}{R}$ ; voltage leads current by φ when $X_{L} > X_{C}$ . (1)

Answer: $Z = R^{2} + (X_{L} - X_{C})^{2}$ ; $tan ϕ = \frac{X _{L} - X _{C}}{R}$ .

Example 4: Voltages across each element · 3 marks · Short answerHarder board level

A series circuit has R = 24 Ω, $X_{L}$ = 50 Ω and $X_{C}$ = 18 Ω across 200 V (r.m.s.). Find the current, the voltage across each element and the phase angle. Comment on $V_{L}$ .

Try it first, then show the solutionSolution

$Z = 2 4^{2} + 3 2^{2} = 40$ Ω; $I = 5$ A. (1)
$V_{R} = 120$ V, $V_{L} = 250$ V, $V_{C} = 90$ V; check $12 0^{2} + 16 0^{2} = 200$ V. (1)
$tan ϕ = \frac{32}{24}$ , φ ≈ 53.1°, current lagging; $V_{L}$ exceeds the supply because $V_{C}$ partly cancels it. (1)

Answer: 5 A; 120 V, 250 V, 90 V; 53.1° lag.

Example 5: Identifying the hidden component · 3 marks · Short answerHOTS / competency

An emf $E = 10 sin (100 π t)$ V applied to a resistor in series with an unknown component X gives $I = 2 sin (100 π t + \frac{π}{6})$ A. Identify X and find R, its reactance and its value.

Try it first, then show the solutionSolution

The current leads the emf, so X is a capacitor. $Z = \frac{10}{2} = 5$ Ω. (1)
$R = Z cos 30° \approx 4.33$ Ω; $X_{C} = Z sin 30° = 2.5$ Ω. (1)
$C = \frac{1}{ω X _{C}} = \frac{1}{100 π \times 2.5} \approx 1.27 \times 1 0^{- 3}$ F. (1)

Answer: Capacitor; R ≈ 4.33 Ω, $X_{C}$ = 2.5 Ω, C ≈ 1.27 mF.

Example 6: A coil on d.c. and a.c. · 4 marks · Case-basedHOTS / competency

Suppose a coil draws 3 A from a 12 V d.c. supply but only 2.4 A from a 12 V (r.m.s.), 50 Hz a.c. supply.
(i) Find its resistance. (1 mark)
(ii) Find its impedance and reactance on a.c. (1 mark)
(iii) Find its inductance. (1 mark)
(iv) Find the power factor and the average power on a.c. (1 mark)

Try it first, then show the solutionSolution

(i) On d.c. only resistance acts: R = $\frac{12}{3} = 4$ Ω.
(ii) $Z = \frac{12}{2.4} = 5$ Ω; $X_{L} = 25 - 16 = 3$ Ω.
(iii) $L = \frac{X _{L}}{2 π f} = \frac{3}{100 π} \approx 9.5$ mH.
(iv) $cos ϕ = \frac{4}{5} = 0.8$ ; $P = V I cos ϕ = 12 \times 2.4 \times 0.8 \approx 23$ W.

Answer: (i) 4 Ω; (ii) 5 Ω, 3 Ω; (iii) ≈ 9.5 mH; (iv) 0.8, ≈ 23 W.

Where students lose marks in Series LCR Circuit: Impedance and Phase

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

Mistake: Adding R and X directly
The ISC 2024 examiners saw $Z = X_{L} + R$ instead of the square-root formula.
Wrong working
$Z = 314 + 400 = 714$ Ω.
Right working
$Z = 31 4^{2} + 40 0^{2} \approx 509$ Ω.
Do this: Draw the impedance triangle and use Pythagoras.
Usually costs: 2 marks.
Mistake: Using V = IR
Also in 2024: current found with R instead of Z.
Do this: $I = \frac{V}{Z}$ ; only at resonance is Z = R.
Usually costs: 1 mark.
Mistake: Not identifying the component
In 2025 candidates could not tell from the equations that a lagging current means an inductor.
Do this: Compare phases: current lagging → L; leading → C.
Usually costs: 1 mark.
Mistake: Confusing ω with f and I₀ with I_rms
2025: angular frequency was given as frequency, and peak used as r.m.s.
Do this: $f = \frac{ω}{2 π}$ ; $I_{r m s} = \frac{I _{0}}{2}$ .
Usually costs: 1 mark each.

Exam tips

Show the phasor diagram with the current as reference in every derivation.
State whether the circuit is inductive or capacitive with the phase angle.
Keep r.m.s. with r.m.s. and peak with peak.
Give Z, I and φ with units (Ω, A, degrees or radians).

Practice questions on Series LCR Circuit: Impedance and Phase

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 impedance of a series LCR circuit is:
(a) $R + X_{L} + X_{C}$
(b) $R^{2} + (X_{L} - X_{C})^{2}$
(c) $R^{2} + X_{L}^{2} + X_{C}^{2}$
(d) $R + X_{L} - X_{C}$
Hint
Phasors at right angles.
Answer
(b).
Q2 · 2 marks · Short answerBoard level
A 300 Ω resistor and a 1 H inductor are in series across 220 V (r.m.s.), 50 Hz. Find the reactance, impedance and current.
Hint
$X_{L} = 2 π f L$ .
Answer
$X_{L} \approx 314$ Ω; $Z \approx 434$ Ω; $I \approx 0.51$ A.
Q3 · 2 marks · Short answerHarder board level
A 60 Ω resistor and a capacitor of reactance 80 Ω are in series across 200 V (r.m.s.). Find the current, the phase angle and the voltage across each.
Hint
Z = 100 Ω.
Answer
2 A; current leads by 53.1°; $V_{R}$ = 120 V, $V_{C}$ = 160 V.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In a series LCR circuit, the r.m.s. voltage across the inductor can be larger than the supply voltage.
Reason (R): The voltages across L and C are in antiphase and partly cancel.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Phasor addition.
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
In a series circuit the r.m.s. voltages across R, L and C are 30 V, 70 V and 30 V. Find the supply voltage and the power factor.
Hint
$V_{R}^{2} + (V_{L} - V_{C})^{2}$ .
Answer
50 V; $cos ϕ = \frac{30}{50} = 0.6$ .
Q6 · 3 marks · Short answerHOTS / competency
For $E = 200 sin (314 t)$ V and $I = 4 sin (314 t - \frac{π}{4})$ A in a series RL circuit, find R, $X_{L}$ and L.
Hint
Z = 50 Ω, φ = 45°.
Answer
R = $X_{L}$ = $50 cos 45° \approx 35.4$ Ω; $L = \frac{35.4}{314} \approx 0.113$ H.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose R = 20 Ω, L = 0.2 H and C = 100 µF are connected in series to a 220 V (r.m.s.), 50 Hz supply.
(i) Find $X_{L}$ and $X_{C}$ . (1 mark)
(ii) Find Z and the current. (1 mark)
(iii) Find the phase angle and say whether the current leads or lags. (1 mark)
(iv) Find the voltage across the inductor and comment. (1 mark)
Hint
Net reactance ≈ 31 Ω.
Answer
(i) $X_{L} \approx 62.8$ Ω, $X_{C} \approx 31.8$ Ω. (ii) $Z \approx 36.9$ Ω, $I \approx 5.96$ A. (iii) $tan ϕ = \frac{31.0}{20}$ , φ ≈ 57°, lagging. (iv) $V_{L} \approx 375$ V, more than the 220 V supply, because $V_{C}$ partly cancels it.

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.Inductive reactance formula?Show answer
$X_{L} = ω L = 2 π f L$ .
2.Capacitive reactance formula?Show answer
$X_{C} = \frac{1}{ω C}$ .
3.R = 6 Ω, X_L = 8 Ω: Z?Show answer
10 Ω.
4.Phase angle of a series LCR circuit?Show answer
$tan ϕ = \frac{X _{L} - X _{C}}{R}$ .
5.In a pure capacitor, does current lead or lag?Show answer
It leads the voltage by 90°.

Series LCR Circuit: Impedance and Phase: frequently asked questions

Why can't I add the voltages across R, L and C?

They are not in phase. V_R is in phase with the current, V_L leads by 90° and V_C lags by 90°, so they must be added as vectors (phasors).

What is impedance?

The total opposition of an a.c. circuit to current: Z = \frac{V}{I} in ohms. It combines resistance and reactance as \sqrt{R^2 + (X_L - X_C)^2}.

Does a capacitor let d.c. through?

No. For d.c., f = 0 and X_C = \frac{1}{2\pi fC} is infinite, so no steady current flows once it is charged.

What does the ISC 2027 syllabus need here?

Pure R, L and C circuits with phasor diagrams, X_L and X_C and their graphs against f, and the series LCR circuit by the phasor method: Z, the phase angle and the current.

How was this tested recently?

In 2024 a 5-mark question found Z and I for an LR circuit and sketched Z against f; in 2025 a 5-mark question identified a component from E and I and found I_{rms}, f, the power factor and Z.