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 VIL​ = ?Z = 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 line current IL.

VL = 690 VZ = 4 + j3 Ω per phase
IL = A
2.Easy
ABCVL​ = 480 VZ = 12 + j5 Ωper phase

A balanced Δ-connected load has an impedance of 12 + j5 Ω per phase. It is supplied from a balanced three-phase source with a line-to-line voltage of 480 V. Find the magnitude of the phase current Iph flowing through each impedance.

VL = 480 VZ = 12 + j5 Ω per phase
Iph = A
3.Medium
ABCVL​ = 690 VIL​ = ?M11 kW, η = 91%cos φ = 0.82

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 0.82 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.

Pout = 11 kWη = 91%cos φ = 0.82VL = 690 V
IL = A
4.Medium
ABCVL​ = 400 VZ = 6 + j8 Ωper phase

A balanced Δ-connected load with an impedance of 6 + j8 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 400 V. Find the total real power P absorbed by the load in kW.

VL = 400 VZ = 6 + j8 Ω per phase
P = kW
5.Hard
P = 300 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 300 kW at a power factor of 0.8 lagging from a 415 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 = 300 kWpf1 = 0.8pf2 = 0.95VL = 415 Vf = 50 Hz
Cph = µF
6.Hard
P = 250 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 250 kW at a power factor of 0.75 lagging from a 415 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. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 250 kWpf1 = 0.75pf2 = 0.9VL = 415 Vf = 60 Hz
Cph = µF

Answer key

  1. 1. IL = 239 A
  2. 2. Iph = 36.92 A
  3. 3. IL = 12.33 A
  4. 4. P = 28.8 kW
  5. 5. Cph = 778.7 µF
  6. 6. Cph = 1531 µF

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