Ohm's Law is the single most useful piece of physics you will ever apply as an electrician. It explains why an undersized fuse blows, what happens when cable resistance rises in a fault, why a long cable run loses voltage, and how to size a resistor in a control circuit. If you can use V = IR fluently, every other electrical calculation follows from it.
The four quantities
Ohm's Law involves four related quantities. Understanding each one is essential before you can apply them in practice.
Voltage (V)— also called electromotive force or potential difference, measured in volts (V). It is the pressure that pushes current around a circuit. On a 230 V ring main, 230 V is what the supply pushes across the circuit. On a 120 V NEC circuit, the supply is 120 V.
Current (I)— the rate of flow of charge, measured in amperes (A). A 3 kW kettle on a 230 V supply draws 13 A. A 1,500 W hair dryer on a 120 V supply draws 12.5 A.
Resistance (R) — the opposition to current flow, measured in ohms (Ω). All conductors have resistance. A long, thin cable has more resistance than a short, fat one. Resistance is what causes voltage drop and heat generation.
Power (P) — the rate of energy transfer, measured in watts (W). Power is what does useful work at the load — it is also what heats a cable when current flows through its resistance.
The formulas
The three Ohm's Law formulas are rearrangements of the same relationship:
V = I × R— voltage equals current times resistanceI = V ÷ R— current equals voltage divided by resistanceR = V ÷ I— resistance equals voltage divided by current
The power formulas extend this:
P = V × I— power equals voltage times currentP = I² × R— power equals current squared times resistanceP = V² ÷ R— power equals voltage squared divided by resistance
The P = I²R formula is the one that explains cable heating. Double the current through a cable and you quadruple the heat generated. This is why overcurrent protection is so critical.
Application 1: finding load current
You need to wire a 3.5 kW instantaneous shower on a 230 V supply. What current does it draw?
- I = P ÷ V = 3,500 ÷ 230 = 15.2 A
You now know the design current Ib for the circuit. From there, you can size the cable and protective device. A 16 A MCB and 2.5 mm² cable will handle it in most installations (subject to derating and route length).
On a 120 V NEC circuit, the same 3.5 kW load would draw 3,500 ÷ 120 = 29.2 A — requiring a 30 A breaker and heavier wire. This is why 240 V is preferred for high-power loads in North American installations.
Application 2: voltage drop
A 25 A circuit runs 35 metres of 4 mm² copper cable (resistance approximately 4.61 mΩ/m). What is the voltage drop?
- Total resistance of the loop (out and back) = 2 × 35 × 4.61 ÷ 1000 = 0.323 Ω
- Voltage drop = I × R = 25 × 0.323 = 8.1 V
- As a percentage of 230 V: 8.1 ÷ 230 × 100 = 3.5%
BS 7671 allows 5% for power circuits. At 3.5% this is within limits, though only just — you may want to step up to 6 mm² if there is any chance of future load growth. The same logic applies under NEC and AS/NZS.
Application 3: fault current
Understanding fault current is critical for protective device selection. A line-to-earth fault at the end of a circuit creates a low-resistance path back to the source. The fault current is:
If = Uo ÷ Zs
Where Uo is the nominal line-to-earth voltage (230 V in the UK) and Zs is the total earth fault loop impedance in ohms. For a BS 7671 type B MCB to trip in 0.4 seconds (the disconnection time required for a 230 V socket circuit), it needs to see a fault current of at least 5× its rating. A 32 A type B MCB needs at least 160 A. That means Zs must not exceed 230 ÷ 160 = 1.44 Ω.
This is why earth fault loop impedance testing matters — and why running a long circuit in thin cable can put you outside the disconnection time limits even with a perfectly good protective device.
Application 4: checking a heating element
A 230 V fan heater element measures 26.5 Ω cold. What is its rated power?
- P = V² ÷ R = 230² ÷ 26.5 = 52,900 ÷ 26.5 = 1,996 W ≈ 2 kW
This is a useful diagnostic check. If the measured resistance gives a power figure well below the nameplate rating, the element has partially failed. If it gives a much higher figure, something else is wrong with your measurement.
Power factor and AC circuits
The formulas above assume a purely resistive load. In AC circuits with motors, transformers, or fluorescent lighting, the load is partly reactive — it stores and releases energy without consuming it. The ratio of real power to apparent power is the power factor (pf), always between 0 and 1.
For single-phase AC: P = V × I × pf
A motor drawing 20 A at 230 V with a power factor of 0.85 consumes: 230 × 20 × 0.85 = 3,910 W of real power, but demands 230 × 20 = 4,600 VA of apparent power from the supply. The cable and protective device must be sized for the apparent power (the actual current drawn), not just the real power.
Run the numbers
The free Voltix Ohm's Law calculator solves for any of the four quantities instantly. For voltage drop on a specific circuit, the voltage drop calculator uses the correct conductor resistance tables for NEC, BS 7671 and AS/NZS rather than requiring you to look up resistance per metre manually.