Three-phase power worksheet with answers

Free printable Three-phase power 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
ABCVL​ = 690 VZ = 4 + j3 Ωper phase

A balanced Δ-connected load has an impedance of 4 + j3 Ω per phase. It is supplied from a balanced three-phase source with a line-to-line voltage of 690 V. Find the magnitude of the phase current Iph flowing through each impedance.

VL = 690 VZ = 4 + j3 Ω per phase
Iph = A
2.Easy
ABCVL​ = 400 VIL​ = ?Z = 24 + j7 Ωper phase

A balanced Δ-connected load has an impedance of 24 + j7 Ω per phase. It is supplied from a balanced three-phase source with a line-to-line voltage of 400 V. Find the magnitude of the line current IL.

VL = 400 VZ = 24 + j7 Ω per phase
IL = A
3.Medium
ABCVL​ = 480 VIL​ = ?M11 kW, η = 91%cos φ = 0.9

A three-phase induction motor delivers 11 kW of mechanical output power at full load. At this load its efficiency is 91% and its power factor is 0.9 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.

Pout = 11 kWη = 91%cos φ = 0.9VL = 480 V
IL = A
4.Medium
ABCVL​ = 690 VIL​ = 60 Acos φ = 0.95lagging

A balanced three-phase load draws a line current of 60 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.95 lagging. Find the total real power P drawn by the load in kW.

VL = 690 VIL = 60 Acos φ = 0.95
P = kW
5.Hard
P = 80 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 80 kW at a power factor of 0.7 lagging from a 690 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.95 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 80 kWpf1 = 0.7pf2 = 0.95VL = 690 Vf = 60 Hz
Cph = µF
6.Hard
P = 150 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 150 kW at a power factor of 0.8 lagging from a 480 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 150 kWpf1 = 0.8pf2 = 0.9VL = 480 Vf = 60 Hz
QC = kvar

Answer key

  1. 1. Iph = 138 A
  2. 2. IL = 27.71 A
  3. 3. IL = 16.15 A
  4. 4. P = 68.12 kW
  5. 5. Cph = 308.2 µF
  6. 6. QC = 39.85 kvar

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