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​ = 400 VIL​ = ?Z = 6 + j8 Ωper phase

A balanced Δ-connected load has an impedance of 6 + j8 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 = 6 + j8 Ω per phase
IL = A
2.Easy
ABCVL​ = 690 VZ = 12 + j16 Ωper phase

A balanced Δ-connected load has an impedance of 12 + j16 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 = 12 + j16 Ω per phase
Iph = A
3.Medium
ABCVL​ = 415 VNZ = 4 + j3 Ωper phase

A balanced Y-connected load with an impedance of 4 + j3 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 415 V. Find the total real power P absorbed by the load in kW.

VL = 415 VZ = 4 + j3 Ω per phase
P = kW
4.Medium
ABCVL​ = 690 VIL​ = 150 Acos φ = 0.7lagging

A balanced three-phase load draws a line current of 150 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.7 lagging. Find the total real power P drawn by the load in kW.

VL = 690 VIL = 150 Acos φ = 0.7
P = kW
5.Hard
P = 250 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 250 kW at a power factor of 0.72 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 0.95 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 250 kWpf1 = 0.72pf2 = 0.95VL = 480 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 0.75 lagging from a 690 V (line-to-line), 50 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 Δ. Find the capacitance Cph of each capacitor in µF.

P = 150 kWpf1 = 0.75pf2 = 0.95VL = 690 Vf = 50 Hz
Cph = µF

Answer key

  1. 1. IL = 69.28 A
  2. 2. Iph = 34.5 A
  3. 3. P = 27.56 kW
  4. 4. P = 125.5 kW
  5. 5. Cph = 1828 µF
  6. 6. Cph = 184.9 µF

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