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​ = 415 VZ = 8 + j6 Ωper phase

A balanced Δ-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 415 V. Find the magnitude of the phase current Iph flowing through each impedance.

VL = 415 VZ = 8 + j6 Ω per phase
Iph = A
2.Easy
ABCVL​ = 600 VZ = 24 + j18 Ωper phase

A balanced Δ-connected load has an impedance of 24 + j18 Ω per phase. It is supplied from a balanced three-phase source with a line-to-line voltage of 600 V. Find the magnitude of the phase current Iph flowing through each impedance.

VL = 600 VZ = 24 + j18 Ω per phase
Iph = A
3.Medium
ABCVL​ = 415 VIL​ = 10 Acos φ = 0.75lagging

A balanced three-phase load draws a line current of 10 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.75 lagging. Find the total real power P drawn by the load in kW.

VL = 415 VIL = 10 Acos φ = 0.75
P = kW
4.Medium
ABCVL​ = 400 VIL​ = ?M18.5 kW, η = 91%cos φ = 0.86

A three-phase induction motor delivers 18.5 kW of mechanical output power at full load. At this load its efficiency is 91% and its power factor is 0.86 lagging. The supply has a line-to-line voltage of 400 V. Find the line current IL.

Pout = 18.5 kWη = 91%cos φ = 0.86VL = 400 V
IL = A
5.Hard
P = 75 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 75 kW at a power factor of 0.78 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 0.98 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 = 690 Vf = 60 Hz
Cph = µF
6.Hard
P = 100 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 100 kW at a power factor of 0.72 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 0.92 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 100 kWpf1 = 0.72pf2 = 0.92VL = 400 Vf = 60 Hz
QC = kvar

Answer key

  1. 1. Iph = 41.5 A
  2. 2. Iph = 20 A
  3. 3. P = 5.391 kW
  4. 4. IL = 34.12 A
  5. 5. Cph = 250.4 µF
  6. 6. QC = 53.79 kvar

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