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 = 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 400 V. Find the magnitude of the line current IL.

VL = 400 VZ = 12 + j16 Ω per phase
IL = A
2.Easy
ABCVL​ = 400 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 400 V. Find the magnitude of the line current IL.

VL = 400 VZ = 12 + j16 Ω per phase
IL = A
3.Medium
ABCVL​ = 480 VZ = 12 + j16 Ωper phase

A balanced Δ-connected load with an impedance of 12 + j16 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 480 V. Find the total real power P absorbed by the load in kW.

VL = 480 VZ = 12 + j16 Ω per phase
P = kW
4.Medium
ABCVL​ = 480 VZ = 18 + j24 Ωper phase

A balanced Δ-connected load with an impedance of 18 + j24 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 480 V. Find the total real power P absorbed by the load in kW.

VL = 480 VZ = 18 + j24 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 0.72 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 0.98 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 120 kWpf1 = 0.72pf2 = 0.98VL = 415 Vf = 50 Hz
Cph = µF
6.Hard
P = 200 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 200 kW at a power factor of 0.75 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 0.98 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 200 kWpf1 = 0.75pf2 = 0.98VL = 480 Vf = 60 Hz
QC = kvar

Answer key

  1. 1. IL = 34.64 A
  2. 2. IL = 11.55 A
  3. 3. P = 20.74 kW
  4. 4. P = 13.82 kW
  5. 5. Cph = 562.4 µF
  6. 6. QC = 135.8 kvar

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