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​ = 208 VIL​ = ?NZ = 12 + j16 Ωper phase

A balanced Y-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 208 V. Find the magnitude of the line current IL.

VL = 208 VZ = 12 + j16 Ω per phase
IL = A
2.Easy
ABCVL​ = 690 VIL​ = ?Z = 20 + j15 Ωper phase

A balanced Δ-connected load has an impedance of 20 + j15 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 = 20 + j15 Ω per phase
IL = A
3.Medium
ABCVL​ = 415 VZ = 7 + j24 Ωper phase

A balanced Δ-connected load with an impedance of 7 + j24 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 = 7 + j24 Ω per phase
P = kW
4.Medium
ABCVL​ = 415 VIL​ = 30 Acos φ = 0.85lagging

A balanced three-phase load draws a line current of 30 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.85 lagging. Find the total real power P drawn by the load in kW.

VL = 415 VIL = 30 Acos φ = 0.85
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 400 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 250 kWpf1 = 0.72pf2 = 0.98VL = 400 Vf = 50 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 400 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 150 kWpf1 = 0.75pf2 = 0.98VL = 400 Vf = 50 Hz
QC = kvar

Answer key

  1. 1. IL = 6.004 A
  2. 2. IL = 47.8 A
  3. 3. P = 5.787 kW
  4. 4. P = 18.33 kW
  5. 5. Cph = 1261 µF
  6. 6. QC = 101.8 kvar

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