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

VL = 690 VZ = 5 + j12 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​ = 690 VIL​ = ?M30 kW, η = 93%cos φ = 0.8

A three-phase induction motor delivers 30 kW of mechanical output power at full load. At this load its efficiency is 93% and its power factor is 0.8 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.

Pout = 30 kWη = 93%cos φ = 0.8VL = 690 V
IL = A
4.Medium
ABCVL​ = 480 VIL​ = ?M30 kW, η = 88%cos φ = 0.86

A three-phase induction motor delivers 30 kW of mechanical output power at full load. At this load its efficiency is 88% 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 = 30 kWη = 88%cos φ = 0.86VL = 480 V
IL = A
5.Hard
P = 75 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 75 kW at a power factor of 0.8 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 0.98 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 75 kWpf1 = 0.8pf2 = 0.98VL = 415 Vf = 60 Hz
Cph = µF
6.Hard
P = 300 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 300 kW at a power factor of 0.8 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 0.92 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 = 0.8pf2 = 0.92VL = 480 Vf = 60 Hz
Cph = µF

Answer key

  1. 1. IL = 91.93 A
  2. 2. IL = 47.8 A
  3. 3. IL = 33.74 A
  4. 4. IL = 47.68 A
  5. 5. Cph = 631.8 µF
  6. 6. Cph = 373 µF

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