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 = 5 + j12 Ωper phase

A balanced Δ-connected load has an impedance of 5 + j12 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 = 5 + j12 Ω per phase
Iph = A
2.Easy
ABCVL​ = 415 VIL​ = ?NZ = 12 + j9 Ωper phase

A balanced Y-connected load has an impedance of 12 + j9 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 line current IL.

VL = 415 VZ = 12 + j9 Ω per phase
IL = A
3.Medium
ABCVL​ = 415 VZ = 24 + j7 Ωper phase

A balanced Δ-connected load with an impedance of 24 + j7 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 = 24 + j7 Ω per phase
P = kW
4.Medium
ABCVL​ = 415 VNZ = 12 + j9 Ωper phase

A balanced Y-connected load with an impedance of 12 + j9 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 = 12 + j9 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 0.75 lagging from a 400 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 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 = 0.75pf2 = 0.98VL = 400 Vf = 60 Hz
Cph = µF
6.Hard
P = 75 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 75 kW at a power factor of 0.75 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 0.98 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 75 kWpf1 = 0.75pf2 = 0.98VL = 690 Vf = 60 Hz
QC = kvar

Answer key

  1. 1. Iph = 31.92 A
  2. 2. IL = 15.97 A
  3. 3. P = 19.84 kW
  4. 4. P = 9.185 kW
  5. 5. Cph = 1125 µF
  6. 6. QC = 50.91 kvar

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