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​ = ?NZ = 6 + j8 Ωper phase

A balanced Y-connected load has an impedance of 6 + j8 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 = 6 + j8 Ω per phase
IL = A
2.Easy
ABCVL​ = 480 VIL​ = ?NZ = 18 + j24 Ωper phase

A balanced Y-connected load has an impedance of 18 + 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 = 18 + j24 Ω per phase
IL = A
3.Medium
ABCVL​ = 480 VIL​ = ?M7.5 kW, η = 94%cos φ = 0.84

A three-phase induction motor delivers 7.5 kW of mechanical output power at full load. At this load its efficiency is 94% and its power factor is 0.84 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.

Pout = 7.5 kWη = 94%cos φ = 0.84VL = 480 V
IL = A
4.Medium
ABCVL​ = 480 VIL​ = 10 Acos φ = 0.9lagging

A balanced three-phase load draws a line current of 10 A from a supply with a line-to-line voltage of 480 V. The power factor is 0.9 lagging. Find the total real power P drawn by the load in kW.

VL = 480 VIL = 10 Acos φ = 0.9
P = kW
5.Hard
P = 250 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 250 kW at a power factor of 0.78 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. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 250 kWpf1 = 0.78pf2 = 0.98VL = 480 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 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 0.95 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 300 kWpf1 = 0.8pf2 = 0.95VL = 690 Vf = 60 Hz
QC = kvar

Answer key

  1. 1. IL = 23.09 A
  2. 2. IL = 9.238 A
  3. 3. IL = 11.42 A
  4. 4. P = 7.482 kW
  5. 5. Cph = 1725 µF
  6. 6. QC = 126.4 kvar

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