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​ = 415 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 415 V. Find the magnitude of the phase current Iph flowing through each impedance.

VL = 415 VZ = 12 + j5 Ω per phase
Iph = A
2.Easy
ABCVL​ = 208 VIL​ = ?NZ = 24 + j18 Ωper phase

A balanced Y-connected load has an impedance of 24 + j18 Ω per phase. It is supplied from a balanced three-phase source with a line-to-line voltage of 208 V. Find the magnitude of the line current IL.

VL = 208 VZ = 24 + j18 Ω per phase
IL = A
3.Medium
ABCVL​ = 690 VIL​ = 40 Acos φ = 0.7lagging

A balanced three-phase load draws a line current of 40 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 = 40 Acos φ = 0.7
P = kW
4.Medium
ABCVL​ = 400 VNZ = 6 + j8 Ωper phase

A balanced Y-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 = 120 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 120 kW at a power factor of 0.78 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 0.92 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 120 kWpf1 = 0.78pf2 = 0.92VL = 415 Vf = 50 Hz
Cph = µF
6.Hard
P = 75 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 75 kW at a power factor of 0.78 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 0.95 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 75 kWpf1 = 0.78pf2 = 0.95VL = 415 Vf = 60 Hz
Cph = µF

Answer key

  1. 1. Iph = 31.92 A
  2. 2. IL = 4.003 A
  3. 3. P = 33.46 kW
  4. 4. P = 9.6 kW
  5. 5. Cph = 834.5 µF
  6. 6. Cph = 182.4 µF

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