AC power & power factor: practice problems with step-by-step solutions

Also called: real and reactive power · power factor correction

Free AC power & power factor practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: P and Q of a load, line current, power factor correction.

Free practice problemEasy
PQ = ?S = V·Iφ

A single-phase load draws 11 A (rms) from a 240 V (rms) supply at a power factor of 0.7 lagging. Find the reactive power Q it takes.

V = 240 VI = 11 Apf = cos φ = 0.7
var
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What you practise

  1. EasyP and Q of a load
  2. Mediumline current
  3. Hardpower factor correction

Key formulas

Real power
P = V·I·cos φ
Reactive power
Q = V·I·sin φ
Apparent power
S = V·I = √(P² + Q²)
Power factor
pf = cos φ = P/S
Correction capacitor
QC = P(tan φ1 − tan φ2), C = QC/(ωV²)

Learn AC power & power factor from scratch

Before this: AC impedance & phasors

1What AC power is

In a DC circuit, power is simply P = V·I. In an AC circuit with an inductor or a capacitor, the current is shifted in time against the voltage by an angle φ.

Because of that shift, part of the energy only flows back and forth between the supply and the load. It does no useful work, but the wires still have to carry the current for it.

Think of a glass of beer. The beer is the useful part, the foam takes up space but you cannot drink it, and the glass has to hold both.

  • Real power P, in watts (W): the useful part that becomes heat, light or motion (the beer).
  • Reactive power Q, in volt-ampere reactive (var): the part that flows back and forth (the foam).
  • Apparent power S, in volt-amperes (VA): what the supply and the wires must carry (the glass).
Self-check Which kind of power turns into heat, light or mechanical work?
  1. All three equally
  2. Apparent power S, in VA
  3. Reactive power Q, in var
  4. Real power P, in W

Answer: Real power P, in W. Only the real power P does useful work. Q only moves back and forth, and S is the total the supply has to deliver.

2The power triangle and the power factor

The three powers form a right triangle. The angle φ is the same as the angle of the load impedance.

PQSφ
S = V·I P = S·cos φ Q = S·sin φ
S = √(P² + Q²) Q = P·tan φ

The power factor is cos φ = P/S. It tells what fraction of the apparent power does useful work. A power factor of 1 is ideal.

Motors and other coils have a lagging power factor: the current lags the voltage. Capacitors give a leading power factor.

Self-check A load draws 10 A from a 230 V supply at a power factor of 0.8. What real power does it take?
  1. 1380 W
  2. 1840 W
  3. 2300 W
  4. 2875 W

Answer: 1840 W. P = V·I·cos φ = 230·10·0.8 = 1840 W. 2300 W is the apparent power S, 1380 W uses sin φ = 0.6 instead of cos φ, and 2875 W divides by the power factor.

3The method step by step

  1. Write down what you know: V, I, P, the power factor cos φ, and whether it is lagging or leading.
  2. Find the angle: φ = arccos(power factor). Then you also have sin φ and tan φ.
  3. If V and I are given, start with S = V·I. Then P = S·cos φ and Q = S·sin φ.
  4. If P is given, use Q = P·tan φ and S = P / cos φ.
  5. For the line current, use I = P / (V·cos φ), with P in watts.

Draw the triangle with the numbers you know. One look shows whether your answer has the right size: S is always the longest side.

4A worked example

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

Worked example

A single-phase motor takes 4.6 kW at a power factor of 0.8 lagging from a 230 V supply. Find the line current I and the reactive power Q.

1

Find the angle

φ = arccos(0.8) = 36.9°

So sin φ = 0.6 and tan φ = 0.75.

2

Line current

Only the real part of the power does work, so divide by the power factor. Use watts, not kilowatts.

I = P / (V·cos φ) = 4600 / (230·0.8) = 25 A
3

Reactive power

Q = P·tan φ = 4.6·0.75 = 3.45 kvar
4

Check the answer

4.6 kW3.45 kvar5.75 kVAφ

S = V·I = 230·25 = 5750 VA = 5.75 kVA. Then Q = S·sin φ = 5.75·0.6 = 3.45 kvar, the same result.

Also √(4.6² + 3.45²) = 5.75, so the triangle closes.

Answer

I = 25 A and Q = 3.45 kvar.

5Power factor correction

A low power factor means a large current for the same useful power. That costs money in thicker cables and losses, so it is often corrected.

A capacitor connected in parallel with the load supplies reactive power locally. The real power P stays the same; only the reactive power taken from the supply drops.

QC = P·(tan φ1 − tan φ2)
12 kW16 kvarS1​φ1​7 kvar

Example: a 12 kW load at power factor 0.6 has tan φ1 = 1.33, so Q1 = 16 kvar. To reach 0.8 (tan φ2 = 0.75) it may take only Q2 = 9 kvar. The capacitor supplies 16 − 9 = 7 kvar.

On a 400 V supply the current falls from 12000/(400·0.6) = 50 A to 12000/(400·0.8) = 37.5 A.

The capacitance follows from QC = V²·ωC:

C = QC / (2π·f·V²)
Self-check A 20 kW load has tan φ1 = 1 and must reach tan φ2 = 0.5. How much reactive power must the capacitor supply?
  1. 10 kvar
  2. 20 kvar
  3. 30 kvar
  4. 40 kvar

Answer: 10 kvar. QC = P·(tan φ1 − tan φ2) = 20·(1 − 0.5) = 10 kvar. 20 kvar is all of the old reactive power, which would correct to a power factor of 1.

6Common mistakes

  • Mixing up cos φ and sin φ. P goes with cos φ, Q with sin φ.
  • Using kW with V and getting the current 1000 times too small. Convert to watts first.
  • Multiplying by the power factor when finding the current. The current is larger than P/V, so you divide.
  • Thinking the capacitor changes the real power. It changes only Q and therefore S and I.
Self-check A load takes 2 kW at a power factor of 0.8 from a 250 V supply. What is the line current?
  1. 0.01 A
  2. 10 A
  3. 6.4 A
  4. 8 A

Answer: 10 A. I = P/(V·cos φ) = 2000/(250·0.8) = 10 A. Leaving P in kW gives 0.01 A, multiplying by the power factor gives 6.4 A, and 8 A ignores the power factor.

7How exam questions look

  • Easy: a load with given V, I and power factor. You find the real power P or the reactive power Q.
  • Medium: a motor with given P (in kW), power factor and V. You find the line current I or the reactive power Q.
  • Hard: power factor correction. You find the capacitor’s reactive power QC, its capacitance C or the new line current.

For the new line current after correction, use the new power factor: Inew = P / (V·cos φ2).

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

Worked examples

Example 1Easy
P = ?QS = V·Iφ

A single-phase load draws 9 A (rms) from a 400 V (rms) supply at a power factor of 0.9 lagging. Find the real power P it takes.

V = 400 VI = 9 Apf = cos φ = 0.9

Find P (W)

Show hints
  1. Real, reactive and apparent power form a right triangle: P = S·cos φ, Q = S·sin φ and S = V·I, where cos φ is the power factor.
  2. P = V·I·cos φ.
  3. S = 3600 VA, φ = 25.84°.
Show the answer and the step-by-step solution

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Apparent power and angle

S = V·I = 400·9 = 3600 VA
φ = arccos(0.9) = 25.84°

Real power

P = S·cos φ = 3600·0.9 = 3240 W

Answer

P = 3240 W
Example 2Medium
P = 35 kWQ = ?Sφ

A single-phase motor takes 35 kW at a power factor of 0.7 lagging from a 230 V supply. Find the reactive power Q it draws.

P = 35 kWpf = 0.7V = 230 V

Find Q (kvar)

Show hints
  1. Real, reactive and apparent power form a right triangle: P = S·cos φ, Q = S·sin φ and S = V·I, where cos φ is the power factor.
  2. Q = P·tan φ.
  3. φ = 45.57°.
Show the answer and the step-by-step solution

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Angle

φ = arccos(0.7) = 45.57°, tan φ = 1.02

Reactive power

Q = P·tan φ = 35·1.02 = 35.71 kvar

Answer

Q = 35.71 kvar
Example 3Hard
P = 45 kWQ1​S1​φ1​QC​

A single-phase load takes 45 kW at a power factor of 0.65 lagging from a 240 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.95. Find the new line current Inew.

P = 45 kWpf1 = 0.65pf2 = 0.95V = 240 Vf = 50 Hz

Find Inew (A)

Show hints
  1. Real, reactive and apparent power form a right triangle: P = S·cos φ, Q = S·sin φ and S = V·I, where cos φ is the power factor.
  2. The capacitor supplies QC = P·(tan φ₁ − tan φ₂) and does not change P.
  3. QC = 37.82 kvar.
Show the answer and the step-by-step solution

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Reactive power before and after

The real power P does not change; only the reactive power drawn from the supply does.

Q1 = P·tan φ1 = 45·tan(49.46°) = 52.61 kvar
Q2 = P·tan φ2 = 45·tan(18.19°) = 14.79 kvar
QC = Q1 − Q2 = 37.82 kvar

New line current

Inew = P/(V·cos φ2) = 45000/(240·0.95) = 197.4 A

Answer

Inew = 197.4 A

Questions and answers

What is AC power & power factor?

In a DC circuit, power is simply P = V·I. In an AC circuit with an inductor or a capacitor, the current is shifted in time against the voltage by an angle φ.

Because of that shift, part of the energy only flows back and forth between the supply and the load. It does no useful work, but the wires still have to carry the current for it.

How do I solve AC power & power factor problems step by step?

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

What do the AC power & power factor exercises cover?
  • Easy: P and Q of a load
  • Medium: line current
  • Hard: power factor correction

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

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AC impedance & phasors

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Three-phase power

Calculators

Power factor correction calculator

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Printable worksheet

AC power & power factor worksheet with answers
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