AC impedance & phasors: practice problems with step-by-step solutions

Also called: phasor analysis · complex impedance

Free AC impedance & phasors practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: series RL, series RLC, capacitor voltage and resonance.

Free practice problemEasy
120 V50 HzR = 40 ΩL = 150 mH

A series RL circuit is connected to a 120 V (rms), 50 Hz supply. Find the rms voltage VR across the resistor.

V = 120 V (rms)f = 50 HzR = 40 ΩL = 150 mH
V
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What you practise

  1. Easyseries RL
  2. Mediumseries RLC
  3. Hardcapacitor voltage and resonance

How to solve AC impedance & phasors problems

  1. Convert the frequency: ω = 2πf.
  2. Find each reactance: XL = ωL and XC = 1/(ωC).
  3. Add impedances like resistors, with Z = R + jX.
  4. Find the magnitude |Z| = √(R² + X²) and the angle φ = atan(X/R), then use Ohm’s law with phasors.

Common mistakes

  • Using f instead of ω.
  • Adding magnitudes of impedances instead of complex values.
  • Forgetting that capacitive reactance is negative in the complex impedance.

Key formulas

Impedances
ZR = R, ZL = jωL, ZC = 1/(jωC)
Angular frequency
ω = 2πf
Magnitude and angle
|Z| = √(R² + X²), φ = atan(X/R)
Phasor Ohm’s law
V = I·Z

Learn AC impedance & phasors from scratch

Before this: Series & parallel resistors

1What AC and impedance are

The mains supply is alternating current (AC): the voltage swings up and down like a sine wave, many times per second.

The frequency f is the number of full swings per second, in hertz (Hz). Mains is 50 Hz in Europe and 60 Hz in North America.

AC calculations often use the angular frequency:

ω = 2π·f (rad/s)

AC voltages and currents are normally given as rms values. An rms value does the same heating work as that DC value. “230 V” mains means 230 V rms.

In AC circuits, inductors and capacitors also limit the current, not only resistors. Their combined effect is called the impedance Z, measured in ohms.

Self-check A supply runs at 50 Hz. What is its angular frequency ω?
  1. 157 rad/s
  2. 314 rad/s
  3. 50 rad/s
  4. 7.96 rad/s

Answer: 314 rad/s. ω = 2π·50 = 314 rad/s. Using f itself (50) or dividing by 2π (7.96) are the usual slips; 157 forgets the factor 2.

2Reactance of an inductor and a capacitor

An inductor and a capacitor resist AC in a way that depends on frequency. This frequency-dependent opposition is called reactance X, in ohms.

XL = ω·L XC = 1 / (ω·C)
  • An inductor opposes fast changes. Its reactance XL grows with frequency. At DC (f = 0) it is a plain wire.
  • A capacitor blocks DC but lets fast changes through. Its reactance XC falls as frequency rises.

Always use L in henries and C in farads: 100 mH = 0.1 H and 10 µF = 10·10−6 F.

Self-check What is the reactance of a 100 mH inductor at 50 Hz?
  1. 0.0318 Ω
  2. 31.4 Ω
  3. 5 Ω
  4. 5000 Ω

Answer: 31.4 Ω. XL = ωL = 2π·50·0.1 = 31.4 Ω. Using f instead of ω gives 5 Ω, and forgetting to convert mH to H gives 5000 Ω.

3Impedance and the phase angle

In a resistor, voltage and current rise and fall together. In an inductor the current lags a quarter cycle behind the voltage; in a capacitor it leads by a quarter cycle.

Because of this shift you cannot add R and X as plain numbers. You add them at right angles, written as a complex number with j:

Z = R + j·(XL − XC)
RXL​|Z|φ

The size of the impedance and its angle come from the right triangle:

|Z| = √(R² + X²) φ = arctan(X / R)

Ohm’s law works with the magnitude: I = V / |Z|. The angle φ tells how far the current lags the voltage. A positive φ means inductive (current lags); a negative φ means capacitive.

Self-check A series circuit has R = 30 Ω and XL = 40 Ω. What is |Z|?
  1. 10 Ω
  2. 2500 Ω
  3. 50 Ω
  4. 70 Ω

Answer: 50 Ω. |Z| = √(30² + 40²) = √2500 = 50 Ω. Adding the two (70 Ω) ignores the right angle, and 2500 forgets the square root.

4The method step by step

  1. Convert the frequency: ω = 2π·f.
  2. Find each reactance: XL = ωL and XC = 1/(ωC). Use H and F.
  3. Find the net reactance X = XL − XC (series circuit).
  4. Find |Z| = √(R² + X²) and, if needed, φ = arctan(X/R).
  5. Use Ohm’s law: I = V/|Z|. Then the voltage across one part is I times its R or X.

In a series circuit the same current flows through every part. That is why you can find I first and use it everywhere.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A series RL circuit with R = 50 Ω and L = 100 mH is connected to a 120 V (rms), 60 Hz supply. Find |Z|, the rms current I and the phase angle φ.

1

Look at the circuit

120 V60 HzR = 50 ΩL = 100 mHI

There is no capacitor, so XC = 0.

2

Angular frequency and reactance

ω = 2π·60 = 377 rad/s
XL = ωL = 377·0.1 = 37.7 Ω
3

Impedance

50 Ω37.7 Ω62.6 Ωφ
|Z| = √(50² + 37.7²) = 62.6 Ω
4

Current and phase angle

I = V / |Z| = 120 / 62.6 = 1.92 A
φ = arctan(37.7 / 50) = 37.0°

The angle is positive, so the current lags the voltage, as expected for an inductive circuit.

5

Check the answer

The resistor takes 1.92·50 = 95.8 V and the inductor 1.92·37.7 = 72.2 V. Added as plain numbers that is 168 V, more than the supply.

Added at right angles it is right: √(95.8² + 72.2²) = 120 V, the supply voltage.

Answer

|Z| = 62.6 Ω, I = 1.92 A and φ = 37.0° (current lagging).

6Common mistakes

  • Using f instead of ω = 2πf in XL and XC.
  • Adding R, XL and XC as plain numbers. R is at right angles to X; XL and XC point in opposite directions.
  • Forgetting that the capacitive reactance subtracts: X = XL − XC.
  • Leaving L in mH or C in µF, or using a calculator set to radians when you want φ in degrees.
Self-check A series RLC circuit has R = 40 Ω, XL = 50 Ω and XC = 20 Ω. What is |Z|?
  1. 110 Ω
  2. 50 Ω
  3. 70 Ω
  4. 80.6 Ω

Answer: 50 Ω. X = 50 − 20 = 30 Ω, so |Z| = √(40² + 30²) = 50 Ω. Adding XC instead of subtracting it gives √(40² + 70²) = 80.6 Ω; 110 Ω adds everything as plain numbers.

7How exam questions look

  • Easy: a series RL circuit on a 50 Hz or 60 Hz supply. You find |Z| or the rms current I.
  • Medium: a series RLC circuit at various frequencies. You find the current I or the phase angle φ, which is negative when XC is larger than XL.
  • Hard: a series RLC circuit where you find the capacitor voltage, the current or the resonant frequency.

For the capacitor voltage, find I first and then VC = I·XC. Near resonance it can be larger than the supply voltage.

At resonance XL = XC, so they cancel and |Z| = R. The supply frequency does not matter for this:

f0 = 1 / (2π·√(L·C))

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
V100 HzC = 1 µF

Find the reactance XC of a 1 µF capacitor at 100 Hz.

C = 1 µFf = 100 Hz

Find XC (Ω)

Show hints
  1. Turn L and C into reactances: XL = 2πfL and XC = 1/(2πfC).
  2. Convert µF to F first: multiply by 10⁻⁶.
  3. ω = 628.3 rad/s.
Show the answer and the step-by-step solution

AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.

Reactance

ω = 2πf = 2π·100 = 628.3 rad/s
XC = 1/(ωC) = 1/(628.3·1·10⁻⁶) = 1592 Ω

Answer

XC = 1592 Ω
Example 2Medium
Vf0​ = 1000 HzRL = 200 mHC = ?

A series RLC circuit with L = 200 mH must resonate at 1000 Hz. Find the capacitance C needed.

f0 = 1000 HzL = 200 mH

Find C (µF)

Show hints
  1. At resonance XL = XC, which gives f₀ = 1/(2π√(LC)).
  2. Rearrange: C = 1/((2πf₀)²·L).
  3. (2πf₀)² = 3.948 × 10⁷ s⁻².
Show the answer and the step-by-step solution

AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.

Resonance condition

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC).

ω₀ = 2π·1000 = 6283 rad/s
C = 1/(ω₀²·L) = 1/(6283²·0.2) = 0.1267 µF

Answer

C = 0.1267 µF
Example 3Hard
VfR = 20 ΩL = 200 mHC = 1 µF

A series RLC circuit has R = 20 Ω, L = 200 mH and C = 1 µF. Find its quality factor Q at resonance.

R = 20 ΩL = 200 mHC = 1 µF

Find Q

Show hints
  1. First find the resonant frequency f₀ = 1/(2π√(LC)).
  2. Q = ω₀L/R = (1/R)·√(L/C).
  3. f₀ = 355.9 Hz, Q = 22.36.
Show the answer and the step-by-step solution

AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.

Resonant frequency

f0 = 1/(2π·√(0.2·1·10⁻⁶)) = 355.9 Hz

Quality factor

Q compares the reactance at resonance with the resistance: Q = ω₀L/R.

Q = (1/R)·√(L/C) = (1/20)·√(0.2/(1·10⁻⁶)) = 22.36

Answer

Q = 22.36

Questions and answers

What is AC impedance & phasors?

The mains supply is alternating current (AC): the voltage swings up and down like a sine wave, many times per second.

The frequency f is the number of full swings per second, in hertz (Hz). Mains is 50 Hz in Europe and 60 Hz in North America.

How do I solve AC impedance & phasors problems step by step?
  1. Convert the frequency: ω = 2πf.
  2. Find each reactance: XL = ωL and XC = 1/(ωC).
  3. Add impedances like resistors, with Z = R + jX.
  4. Find the magnitude |Z| = √(R² + X²) and the angle φ = atan(X/R), then use Ohm’s law with phasors.
What are common mistakes in AC impedance & phasors?
  • Using f instead of ω.
  • Adding magnitudes of impedances instead of complex values.
  • Forgetting that capacitive reactance is negative in the complex impedance.
What do the AC impedance & phasors exercises cover?
  • Easy: series RL
  • Medium: series RLC
  • Hard: capacitor voltage and resonance

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

Series & parallel resistors

Continue with

AC power & power factorBode plots & marginsLC resonance, Q factor & bandwidthTransformers

Calculators

RLC impedance calculator

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Circuit Analysis 2: AC circuitsElectrical Engineering FundamentalsPhysics 2: Electricity and MagnetismPower Systems and Electrical MachinesFE Electrical and Computer exam practice

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Printable worksheet

AC impedance & phasors worksheet with answers
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