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Electrical Engineer calculations with worked answers

Updated 17 September 2026 · 2 min read

Engineering calculations decide whether an installation is safe and whether it is affordable, and they are where written tests concentrate their marks. The examples below show the method for the calculations engineers use most, with each formula, substitution and result set out, and every figure recalculated by independent code before publication.

How this guide was made

Written for this site with AI assistance, then checked automatically before publishing. The topic table, sample questions, worked examples and study plan are generated from this site’s question bank: each worked example is recalculated by separate code, and each sample answer says where it came from.

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Worked examples

Try each one on paper first. Every figure is generated and then recalculated by separate code before it is published.

  1. Real power in a balanced three-phase load

    A balanced three-phase motor is supplied at 400 V line-to-line, draws 80 A line current, and runs at a power factor of 0.85. What real power does it take?

    Show the working
    1. For a balanced three-phase load: P = √3 × VL × IL × power factor.
    2. P = 1.732 × 400 × 80 × 0.85
    3. P = 47112 W = 47.11 kW

    Answer: 47.11 kW

  2. Voltage drop on a single-phase copper run

    A single-phase 230 V circuit uses a 15 m run of 10 mm² copper cable carrying 20 A. Taking the resistivity of copper as 0.0172 Ω·mm²/m, what is the voltage drop, and what percentage of 230 V is it?

    Show the working
    1. Current flows out and back, so the conductor length is doubled: Vd = (2 × L × I × ρ) ÷ A.
    2. Vd = (2 × 15 × 20 × 0.0172) ÷ 10
    3. Vd = 1.03 V
    4. As a share of supply: 1.03 ÷ 230 × 100 = 0.45 %

    Answer: 1.03 V (0.45 % of 230 V)

  3. Secondary voltage from the turns ratio

    An ideal transformer has 5500 primary turns and 200 secondary turns. If 11000 V is applied to the primary, what is the secondary voltage?

    Show the working
    1. For an ideal transformer, voltage follows turns: Vs ÷ Vp = Ns ÷ Np.
    2. Vs = 11000 × 200 ÷ 5500
    3. Vs = 400 V

    Answer: 400 V

  4. Power from voltage and current

    A load connected to 48 V draws 0.5 A. How much power does it use?

    Show the working
    1. Power is voltage times current: P = V × I.
    2. P = 48 V × 0.5 A
    3. P = 24 W

    Answer: 24 W

  5. Capacitive reactance

    What is the reactance of a 470 µF capacitor on a 50 Hz supply?

    Show the working
    1. Capacitive reactance: Xc = 1 ÷ (2πfC), with C in farads.
    2. Xc = 1 ÷ (2 × 3.1416 × 50 × 470 × 10⁻⁶)
    3. Xc = 6.77 Ω

    Answer: 6.77 Ω

  6. Real power in a balanced three-phase load

    A balanced three-phase motor is supplied at 400 V line-to-line, draws 10 A line current, and runs at a power factor of 0.9. What real power does it take?

    Show the working
    1. For a balanced three-phase load: P = √3 × VL × IL × power factor.
    2. P = 1.732 × 400 × 10 × 0.9
    3. P = 6235 W = 6.24 kW

    Answer: 6.24 kW

  7. Voltage drop on a single-phase copper run

    A single-phase 230 V circuit uses a 20 m run of 1.5 mm² copper cable carrying 25 A. Taking the resistivity of copper as 0.0172 Ω·mm²/m, what is the voltage drop, and what percentage of 230 V is it?

    Show the working
    1. Current flows out and back, so the conductor length is doubled: Vd = (2 × L × I × ρ) ÷ A.
    2. Vd = (2 × 20 × 25 × 0.0172) ÷ 1.5
    3. Vd = 11.47 V
    4. As a share of supply: 11.47 ÷ 230 × 100 = 4.99 %

    Answer: 11.47 V (4.99 % of 230 V)

  8. Secondary voltage from the turns ratio

    An ideal transformer has 2000 primary turns and 115 secondary turns. If 400 V is applied to the primary, what is the secondary voltage?

    Show the working
    1. For an ideal transformer, voltage follows turns: Vs ÷ Vp = Ns ÷ Np.
    2. Vs = 400 × 115 ÷ 2000
    3. Vs = 23 V

    Answer: 23 V

The calculations that carry design work

Three-phase power links a load in kilowatts to the current a cable and breaker must carry, and power factor sits in the middle of that relationship. Voltage drop decides whether a long run needs a larger conductor than its current rating alone suggests. Transformer ratios connect primary and secondary voltages and, with the rating, give the currents each side. Reactance explains why motors and capacitor banks behave as they do on an alternating supply.

Method that survives pressure

Set out every calculation the same way: formula, substitution with units, result with units, then a sanity check. Keep line and phase quantities visibly separate, and write kW, kVAr and kVA rather than a bare number, because the three are only interchangeable when the power factor is one. When a result looks surprising, check the factor you applied last before doubting the formula.

Common questions

How do I move between kW, kVA and power factor?

Apparent power in kVA multiplied by power factor gives real power in kW. Rearranging gives the other direction. Cables and transformers are rated in kVA because they carry current regardless of how much of it does useful work.

Why does voltage drop matter more on long runs?

Drop is proportional to length as well as current, so a long cable can satisfy its current rating and still deliver too little voltage at the far end. Size for both, then take the larger conductor.

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