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​ = ?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 690 V. Find the magnitude of the line current IL.

VL = 690 VZ = 12 + j9 Ω per phase
IL = A
2.Easy
ABCVL​ = 480 VIL​ = ?Z = 7 + j24 Ωper phase

A balanced Δ-connected load has an impedance of 7 + j24 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 line current IL.

VL = 480 VZ = 7 + j24 Ω per phase
IL = A
3.Medium
ABCVL​ = 415 VNZ = 24 + j18 Ωper phase

A balanced Y-connected load with an impedance of 24 + j18 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 + j18 Ω per phase
P = kW
4.Medium
ABCVL​ = 480 VIL​ = ?M45 kW, η = 91%cos φ = 0.86

A three-phase induction motor delivers 45 kW of mechanical output power at full load. At this load its efficiency is 91% and its power factor is 0.86 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.

Pout = 45 kWη = 91%cos φ = 0.86VL = 480 V
IL = A
5.Hard
P = 50 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 50 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. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 50 kWpf1 = 0.72pf2 = 0.98VL = 400 Vf = 50 Hz
QC = kvar
6.Hard
P = 100 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 100 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 0.92 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 100 kWpf1 = 0.75pf2 = 0.92VL = 400 Vf = 60 Hz
Cph = µF

Answer key

  1. 1. IL = 26.56 A
  2. 2. IL = 33.26 A
  3. 3. P = 4.593 kW
  4. 4. IL = 69.16 A
  5. 5. QC = 38.04 kvar
  6. 6. Cph = 252 µF

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