Three-phase power: practice problems with step-by-step solutions

Free Three-phase power practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: balanced Y and Δ voltages and currents, three-phase power, power factor correction.

Free practice problemEasy
ABCVL​ = 415 VIL​ = ?NZ = 24 + j7 Ωper phase

A balanced Y-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 415 V. Find the magnitude of the line current IL.

VL = 415 VZ = 24 + j7 Ω per phase
A
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What you practise

  1. Easybalanced Y and Δ voltages and currents
  2. Mediumthree-phase power
  3. Hardpower factor correction

Key formulas

Y connection
VL = √3·Vph, IL = Iph
Δ connection
VL = Vph, IL = √3·Iph
Total real power
P = √3·VL·IL·cos φ = 3·Iph²·R
Apparent and reactive power
S = √3·VL·IL, Q = P·tan φ
Correction capacitors
QC = P(tan φ1 − tan φ2), C = QC/(3ωV²)
Voltage across each capacitor
Δ: VL Y: VL/√3
Motor input
Pin = Pout/η

Learn Three-phase power from scratch

Before this: AC power & power factor

1What three-phase power is

Almost all electricity is generated and distributed as three-phase AC. Three wires, called lines or phases (A, B and C), each carry a sine wave.

The three waves have the same size and frequency, but each one is shifted by one third of a cycle, 120°, from the next.

tABC120°

Think of three people pedalling a bicycle in turn: one pushes while the others recover. The total push stays smooth. In the same way, a balanced three-phase system delivers a constant total power, which makes motors run smoothly.

It also needs less copper than three separate single-phase circuits for the same power.

In this topic every load is balanced: all three phases have the same impedance. Then you can work out one phase and multiply by three.

Self-check By how much is each phase voltage shifted from the next one?
  1. 120°
  2. 180°
  3. 60°
  4. 90°

Answer: 120°. Three phases share one full cycle of 360°, so they are 360/3 = 120° apart.

2Y and Δ connections

A three-phase load has three impedances. They can be connected in two ways.

NABCYABCΔ
  • Y (star): each impedance goes from a line to a common star point N.
  • Δ (delta): each impedance sits between two lines, forming a triangle.

The voltage between two lines is the line voltage VL, the value normally given (400 V, 415 V, 208 V…). The voltage across one impedance is the phase voltage Vph. The same idea applies to the line current IL and the phase current Iph.

Y: VL = √3·Vph, IL = Iph
Δ: VL = Vph, IL = √3·Iph

In both cases, one phase follows Ohm’s law: Iph = Vph / |Z|, with |Z| = √(R² + X²).

Self-check A Y-connected load is fed from a 400 V (line-to-line) supply. What voltage is across each impedance?
  1. 133 V
  2. 231 V
  3. 400 V
  4. 693 V

Answer: 231 V. In Y, Vph = VL/√3 = 400/√3 = 231 V. 400 V is the Δ case, 693 V multiplies by √3, and 133 V divides by √3 twice.

3Three-phase power

The total power is three times the power of one phase. Written with line values, the same total becomes:

P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ
S = √3·VL·IL Q = P·tan φ

This formula is the same for Y and Δ. The power factor cos φ is the power factor of one phase impedance: cos φ = R/|Z|.

If you know the phase current and the resistance, you can also use P = 3·Iph²·R. Only the resistance takes real power.

Self-check A balanced load draws 10 A per line from a 400 V supply at a power factor of 0.9. What total real power does it take?
  1. 10800 W
  2. 3600 W
  3. 6235 W
  4. 6928 W

Answer: 6235 W. P = √3·400·10·0.9 = 6235 W. Leaving out √3 gives 3600 W, using 3 instead of √3 with line values gives 10800 W, and 6928 W forgets the power factor.

4The method step by step

  1. Note the connection (Y or Δ) and the line voltage VL.
  2. Find the phase voltage: Vph = VL/√3 for Y, Vph = VL for Δ.
  3. Find |Z| = √(R² + X²) and the phase current Iph = Vph/|Z|.
  4. Find the line current: IL = Iph for Y, IL = √3·Iph for Δ.
  5. Find the power: P = 3·Iph²·R, or P = √3·VL·IL·cos φ with cos φ = R/|Z|.

Work in one phase as long as you can. Switch to line values only at the end.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A balanced Y-connected load has an impedance of 6 + j8 Ω per phase and is fed from a 400 V (line-to-line) supply. Find the line current IL and the total real power P.

1

Phase voltage

In a Y connection each impedance sees the phase voltage.

Vph = 400 / √3 = 230.9 V
2

Impedance magnitude

|Z| = √(6² + 8²) = 10 Ω

The power factor is cos φ = R/|Z| = 6/10 = 0.6.

3

Phase and line current

Iph = 230.9 / 10 = 23.1 A

In Y the line current is the phase current, so IL = 23.1 A.

4

Total real power

P = 3·Iph²·R = 3·23.1²·6 = 9600 W
5

Check the answer

With line values: P = √3·400·23.1·0.6 = 9600 W. Both formulas agree.

Answer

IL = 23.1 A and P = 9.6 kW.

6Common mistakes

  • Putting the line voltage across a Y impedance. In Y each impedance sees VL/√3.
  • Forgetting that in Δ the line current is √3 times the phase current.
  • Dividing by R instead of |Z| when finding the current.
  • Using 3 with line values or √3 with phase values in the power formula.
Self-check The same 6 + j8 Ω impedances are now connected in Δ to the 400 V supply. What is the line current?
  1. 115 A
  2. 23.1 A
  3. 40 A
  4. 69.3 A

Answer: 69.3 A. In Δ, Vph = 400 V, so Iph = 400/10 = 40 A and IL = √3·40 = 69.3 A. 40 A is the phase current only, and 115 A divides by R instead of |Z|.

7How exam questions look

  • Easy: a Y or Δ load with an impedance such as 3 + j4 Ω. You find the line current or the phase current.
  • Medium: total real power, from the impedance or from VL, IL and the power factor. A motor question gives the output power and efficiency: first find Pin = Pout/η, then IL = Pin/(√3·VL·cos φ).
  • Hard: power factor correction with a three-phase capacitor bank. You find the reactive power QC or the capacitance per phase.

For correction, QC = P·(tan φ1 − tan φ2) as in single phase. Each of the three capacitors supplies one third of it:

C = QC / (3·ω·V²)

Here V is the voltage across each capacitor: VL for Δ, and VL/√3 for Y. So a Y bank needs three times the capacitance of a Δ bank.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
ABCVL​ = 400 VIL​ = ?Z = 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 line current IL.

VL = 400 VZ = 24 + j7 Ω per phase

Find IL (A)

Show hints
  1. In a Δ connection each impedance is connected between two lines.
  2. So each impedance sees the full line-to-line voltage, and IL = √3·Iph.
  3. Iph = 400/25 = 16 A.
Show the answer and the step-by-step solution

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.

Impedance magnitude

|Z| = √(24² + 7²) = 25 Ω

Phase current

Each impedance is connected between two lines and sees VL.

Iph = VL/|Z| = 400/25 = 16 A

Line current

IL = √3·Iph = √3·16 = 27.71 A

Answer

IL = 27.71 A
Example 2Medium
ABCVL​ = 400 VZ = 24 + j18 Ωper phase

A balanced Δ-connected load with an impedance of 24 + j18 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 = 24 + j18 Ω per phase

Find P (kW)

Show hints
  1. In a Δ connection each impedance sees the full line-to-line voltage.
  2. Find the phase current, then P = 3·Iph²·R.
  3. Iph = 13.33 A.
Show the answer and the step-by-step solution

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.

Phase voltage

Vph = VL = 400 V

Phase current

|Z| = √(24² + 18²) = 30 Ω
Iph = 400/30 = 13.33 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·13.33²·24 = 12.8 kW

Answer

P = 12.8 kW

Check: cos φ = R/|Z| = 0.8, and √3·VL·IL·cos φ gives the same power.

Example 3Hard
P = 60 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 60 kW at a power factor of 0.7 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 0.9 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 60 kWpf1 = 0.7pf2 = 0.9VL = 480 Vf = 60 Hz

Find Cph (µF)

Show hints
  1. First find QC = P·(tan φ1 − tan φ2).
  2. Each capacitor supplies QC/3.
  3. In Y each capacitor sees VL/√3: QC/3 = (VL/√3)²·ω·C.
Show the answer and the step-by-step solution

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ. Power factor correction: capacitors supply reactive power locally, QC = P·(tan φ1 − tan φ2). Each of the three capacitors supplies QC/3 = V²·ω·C, where V is the voltage across that capacitor: VL in a Δ bank, VL/√3 in a Y bank.

Reactive power before and after

The real power is unchanged, so each power factor fixes the reactive power: Q = P·tan φ.

Q1 = 60·tan(acos 0.7) = 60·1.02 = 61.21 kvar
Q2 = 60·tan(acos 0.9) = 60·0.4843 = 29.06 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 32.15 kvar

Capacitance per phase

In a Y bank each capacitor is connected between a line and the star point.

V = 480/√3 = 277.1 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 10718/(377·277.1²) = 370.2 µF

Answer

Cph = 370.2 µF

Questions and answers

What is Three-phase power?

Almost all electricity is generated and distributed as three-phase AC. Three wires, called lines or phases (A, B and C), each carry a sine wave.

The three waves have the same size and frequency, but each one is shifted by one third of a cycle, 120°, from the next.

How do I solve Three-phase power problems step by step?

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.

What are common mistakes in Three-phase power?
  • That is the phase current through one impedance. In a Δ connection the line current is √3 times larger.
  • That is the result for a Y connection. In a Δ connection each impedance sees the full line-to-line voltage.
  • Divide by the magnitude of the impedance |Z| = √(R² + X²), not by R alone.
What do the Three-phase power exercises cover?
  • Easy: balanced Y and Δ voltages and currents
  • Medium: three-phase power
  • Hard: power factor correction

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

AC power & power factor

Calculators

Three-phase power calculatorVoltage drop calculator

Courses with this topic

Circuit Analysis 2: AC circuitsElectrical Engineering FundamentalsPower Systems and Electrical MachinesPower Electronics and DrivesFE Electrical and Computer exam practice

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Printable worksheet

Three-phase power worksheet with answers
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