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 VZ = 24 + j7 Ωper phase

A balanced Δ-connected load has an impedance of 24 + j7 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 phase current Iph flowing through each impedance.

VL = 400 VZ = 24 + j7 Ω per phase
Iph = A
2.Easy
ABCVL​ = 480 VIL​ = ?Z = 16 + j12 Ωper phase

A balanced Δ-connected load has an impedance of 16 + j12 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 = 16 + j12 Ω per phase
IL = A
3.Medium
ABCVL​ = 480 VIL​ = ?M45 kW, η = 94%cos φ = 0.85

A three-phase induction motor delivers 45 kW of mechanical output power at full load. At this load its efficiency is 94% and its power factor is 0.85 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.

Pout = 45 kWη = 94%cos φ = 0.85VL = 480 V
IL = A
4.Medium
ABCVL​ = 400 VNZ = 8 + j6 Ωper phase

A balanced Y-connected load with an impedance of 8 + j6 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 400 V. Find the total real power P absorbed by the load in kW.

VL = 400 VZ = 8 + j6 Ω per phase
P = kW
5.Hard
P = 150 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 150 kW at a power factor of 0.7 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 0.92 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 150 kWpf1 = 0.7pf2 = 0.92VL = 415 Vf = 50 Hz
QC = kvar
6.Hard
P = 300 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 300 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 0.95 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 300 kWpf1 = 0.75pf2 = 0.95VL = 400 Vf = 60 Hz
QC = kvar

Answer key

  1. 1. Iph = 16 A
  2. 2. IL = 41.57 A
  3. 3. IL = 67.74 A
  4. 4. P = 12.8 kW
  5. 5. QC = 89.13 kvar
  6. 6. QC = 166 kvar

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