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 = 5 + j12 Ωper phase

A balanced Y-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 400 V. Find the magnitude of the line current IL.

VL = 400 VZ = 5 + j12 Ω per phase
IL = A
2.Easy
ABCVL​ = 690 VZ = 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 690 V. Find the magnitude of the phase current Iph flowing through each impedance.

VL = 690 VZ = 16 + j12 Ω per phase
Iph = A
3.Medium
ABCVL​ = 480 VIL​ = ?M18.5 kW, η = 95%cos φ = 0.82

A three-phase induction motor delivers 18.5 kW of mechanical output power at full load. At this load its efficiency is 95% and its power factor is 0.82 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.

Pout = 18.5 kWη = 95%cos φ = 0.82VL = 480 V
IL = A
4.Medium
ABCVL​ = 400 VZ = 6 + j8 Ωper phase

A balanced Δ-connected load with an impedance of 6 + j8 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 = 6 + j8 Ω per phase
P = kW
5.Hard
P = 75 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 75 kW at a power factor of 0.7 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 0.9 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 = 0.7pf2 = 0.9VL = 415 Vf = 60 Hz
Cph = µF
6.Hard
P = 75 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 75 kW at a power factor of 0.78 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 0.98 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 = 0.78pf2 = 0.98VL = 400 Vf = 60 Hz
Cph = µF

Answer key

  1. 1. IL = 17.76 A
  2. 2. Iph = 34.5 A
  3. 3. IL = 28.56 A
  4. 4. P = 28.8 kW
  5. 5. Cph = 619 µF
  6. 6. Cph = 745.1 µF

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