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​ = 480 VZ = 6 + j8 Ωper phase

A balanced Δ-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 480 V. Find the magnitude of the phase current Iph flowing through each impedance.

VL = 480 VZ = 6 + j8 Ω per phase
Iph = A
2.Easy
ABCVL​ = 400 VIL​ = ?NZ = 8 + j6 Ωper phase

A balanced Y-connected load has an impedance of 8 + j6 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 = 8 + j6 Ω per phase
IL = A
3.Medium
ABCVL​ = 415 VNZ = 5 + j12 Ωper phase

A balanced Y-connected load with an impedance of 5 + j12 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 = 5 + j12 Ω per phase
P = kW
4.Medium
ABCVL​ = 480 VIL​ = ?M75 kW, η = 95%cos φ = 0.84

A three-phase induction motor delivers 75 kW of mechanical output power at full load. At this load its efficiency is 95% 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 = 75 kWη = 95%cos φ = 0.84VL = 480 V
IL = A
5.Hard
P = 300 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 300 kW at a power factor of 0.78 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 0.95 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 = 0.78pf2 = 0.95VL = 400 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 0.8 lagging from a 690 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 = 200 kWpf1 = 0.8pf2 = 0.98VL = 690 Vf = 50 Hz
Cph = µF

Answer key

  1. 1. Iph = 48 A
  2. 2. IL = 23.09 A
  3. 3. P = 5.095 kW
  4. 4. IL = 113 A
  5. 5. Cph = 942.2 µF
  6. 6. Cph = 243.8 µF

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