AC power & power factor worksheet with answers

Free printable AC power & power factor worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
P = ?QS = V·Iφ

A single-phase load draws 2 A (rms) from a 230 V (rms) supply at a power factor of 0.95 lagging. Find the real power P it takes.

V = 230 VI = 2 Apf = cos φ = 0.95
P = W
2.Easy
240 VloadP = 840 W5 A

A voltmeter, an ammeter and a wattmeter connected to a single-phase inductive load read 240 V, 5 A and 840 W. Find the power factor cos φ of the load.

V = 240 VI = 5 AP = 840 W
cos φ =
3.Medium
240 V127 kW, 0.75223 kW, 0.95I = ?

Two loads are connected in parallel to a 240 V single-phase supply: a 27 kW load at a power factor of 0.75 lagging and a 23 kW load at a power factor of 0.95 leading. Find the total current I drawn from the supply.

V = 240 VP1 = 27 kW, pf1 = 0.75 laggingP2 = 23 kW, pf2 = 0.95 leading
I = A
4.Medium
P = 32 kWQSφ

A single-phase motor takes 32 kW at a power factor of 0.7 lagging from a 400 V supply. Find the line current I.

P = 32 kWpf = 0.7V = 400 V
I = A
5.Hard
P = 50 kWQ1​S1​φ1​QC​

A single-phase load takes 50 kW at a power factor of 0.8 lagging from a 230 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.95. Find the new line current Inew.

P = 50 kWpf1 = 0.8pf2 = 0.95V = 230 Vf = 60 Hz
Inew = A
6.Hard
230 V1000 HzRs​ = 50 ΩXs​ = j20 ΩZL​ = ?

A 230 V (rms), 1000 Hz source has an internal impedance Zs = 50 + j20 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.

Vs = 230 V (rms)f = 1000 HzZs = 50 + j20 Ω
Pmax = W

Answer key

  1. 1. P = 437 W
  2. 2. cos φ = 0.7
  3. 3. I = 219.1 A
  4. 4. I = 114.3 A
  5. 5. Inew = 228.8 A
  6. 6. Pmax = 264.5 W

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