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 3 A (rms) from a 230 V (rms) supply at a power factor of 0.75 lagging. Find the real power P it takes.

V = 230 VI = 3 Apf = cos φ = 0.75
P = W
2.Easy
120 VloadP = 1680 W20 A

A voltmeter, an ammeter and a wattmeter connected to a single-phase inductive load read 120 V, 20 A and 1680 W. Find the power factor cos φ of the load.

V = 120 VI = 20 AP = 1680 W
cos φ =
3.Medium
240 V116 kW, 0.65226 kW, 1I

Two loads are connected in parallel to a 240 V single-phase supply: a 16 kW load at a power factor of 0.65 lagging and a 26 kW heater (unity power factor). Find the overall power factor cos φ seen by the supply.

V = 240 VP1 = 16 kW, pf1 = 0.65 laggingP2 = 26 kW, pf2 = 1
cos φ =
4.Medium
240 V16 kW, 0.6217 kW, 0.7I

Two loads are connected in parallel to a 240 V single-phase supply: a 6 kW load at a power factor of 0.6 lagging and a 17 kW load at a power factor of 0.7 lagging. Find the overall power factor cos φ seen by the supply.

V = 240 VP1 = 6 kW, pf1 = 0.6 laggingP2 = 17 kW, pf2 = 0.7 lagging
cos φ =
5.Hard
100 V50 HzRs​ = 2 ΩXs​ = j15 ΩZL​ = ?

A 100 V (rms), 50 Hz source has an internal impedance Zs = 2 + j15 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.

Vs = 100 V (rms)f = 50 HzZs = 2 + j15 Ω
Pmax = W
6.Hard
P = 15 kWQ1​S1​φ1​QC​

A single-phase load takes 15 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 0.9. Find the new line current Inew.

P = 15 kWpf1 = 0.65pf2 = 0.9V = 240 Vf = 50 Hz
Inew = A

Answer key

  1. 1. P = 517.5 W
  2. 2. cos φ = 0.7
  3. 3. cos φ = 0.9135
  4. 4. cos φ = 0.672
  5. 5. Pmax = 1250 W
  6. 6. Inew = 69.44 A

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