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Electrician calculations with worked answers

Updated 16 September 2026 · 2 min read

Electrical calculations are where precision matters most, in an electrician's test and on site. A wrongly sized cable overheats; a miscalculated load trips breakers every evening. The examples below cover the calculations electricians use most, each with the formula, the substitution and the result. Every figure was checked 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. Find an unknown resistance

    A component drops 37.5 V while carrying 0.25 A. What is its resistance?

    Show the working
    1. Rearrange Ohm’s law for resistance: R = V ÷ I.
    2. R = 37.5 V ÷ 0.25 A
    3. R = 150 Ω

    Answer: 150 Ω

  2. Heat lost in a cable

    A cable with a resistance of 1.2 Ω carries 2 A. How much power is lost as heat in the cable?

    Show the working
    1. With current and resistance known, use P = I² × R.
    2. P = 2² × 1.2 = 4 × 1.2
    3. P = 4.8 W

    Answer: 4.8 W

  3. Voltage drop on a single-phase copper run

    A single-phase 230 V circuit uses a 25 m run of 6 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 × 25 × 20 × 0.0172) ÷ 6
    3. Vd = 2.87 V
    4. As a share of supply: 2.87 ÷ 230 × 100 = 1.25 %

    Answer: 2.87 V (1.25 % of 230 V)

  4. Daily and monthly energy use

    A household runs 5 lamps of 9 W for 6 hours a day and a 1500 W iron for 1 hours a day. How many kWh does it use per day, and over 30 days?

    Show the working
    1. Energy = power × time. Work in watt-hours, then divide by 1,000 for kWh.
    2. Lamps: 5 × 9 W × 6 h = 270 Wh
    3. Iron: 1500 W × 1 h = 1500 Wh
    4. Daily total: 1770 Wh = 1.77 kWh
    5. Over 30 days: 1.77 × 30 = 53.1 kWh

    Answer: 1.77 kWh a day, 53.1 kWh over 30 days

  5. Total resistance in series

    Resistors of 22 Ω, 82 Ω and 15 Ω are connected in series. What is the total resistance?

    Show the working
    1. In series, resistances add: R = R1 + R2 + R3.
    2. R = 22 + 82 + 15
    3. R = 119 Ω

    Answer: 119 Ω

  6. Find the current through a resistor

    There are 705 V across a 470 Ω resistor. What current flows through it?

    Show the working
    1. Rearrange Ohm’s law for current: I = V ÷ R.
    2. I = 705 V ÷ 470 Ω
    3. I = 1.5 A

    Answer: 1.5 A

  7. Heat lost in a cable

    A cable with a resistance of 0.4 Ω carries 4 A. How much power is lost as heat in the cable?

    Show the working
    1. With current and resistance known, use P = I² × R.
    2. P = 4² × 0.4 = 16 × 0.4
    3. P = 6.4 W

    Answer: 6.4 W

  8. Voltage drop on a single-phase copper run

    A single-phase 230 V circuit uses a 60 m run of 6 mm² copper cable carrying 13 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 × 60 × 13 × 0.0172) ÷ 6
    3. Vd = 4.47 V
    4. As a share of supply: 4.47 ÷ 230 × 100 = 1.94 %

    Answer: 4.47 V (1.94 % of 230 V)

The calculations electricians use most

Ohm's law links voltage, current and resistance, and almost every other calculation rests on it. Power calculations tell you how much current an appliance draws, which decides the cable and the protective device. Series and parallel resistance helps with fault-finding and with understanding circuits. Voltage drop shows whether a long cable run still delivers enough voltage at the far end. Energy calculations explain electricity use to customers in the kilowatt-hours their meter records.

Habits that prevent mistakes

Write a unit beside every number and convert before calculating: milliamps to amps, watts to kilowatts, minutes to hours. Estimate the answer roughly first, then compare. If a calculation says a small lamp draws more current than a water heater, something went wrong. For voltage drop, remember that current travels out and back, so the conductor length is doubled on a single-phase circuit.

Common questions

How do I find the current an appliance draws?

Divide its power in watts by the supply voltage. A 2,300 W heater on 230 V draws about 10 A. For motors and other inductive loads, the power factor also matters.

Why does voltage drop matter if the circuit still works?

Equipment can malfunction or wear out faster at low voltage, and motors may overheat. Cables must be sized so the drop stays within the limit the wiring standard allows, especially on long runs.

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