You're wiring a 240V circuit and the load is a 24-ohm heating element. Before touching anything, you need the current draw confirmed so you can size the cable and select the correct circuit breaker.
Ohm's Law Calculator
Electrical
V = I × R · I = V ÷ R · R = V ÷ I · P = V × I
Example: Solve for V — Current 10A, Resistance 24Ω → V = 240V · Power = 2400W
Reference: AS/NZS 3000 Wiring Rules
Results are estimates for planning purposes. Verify with current standards and a qualified professional.
1 What this calculator does
Solves Ohm's Law for any unknown variable -- Voltage (V), Current (I), Resistance (R) or Power (W). Enter any two known values and calculate the third. Shows all four values and power in watts for the circuit.
2 Formula & professional reasoning
Ohm's Law: V = I x R
Solve for Voltage: V = I (A) x R (Ohms)
Solve for Current: I = V (V) / R (Ohms)
Solve for Resistance: R = V (V) / I (A)
Solve for Power: P = V x I = I^2 x R = V^2 / R
Ohm's Law describes the relationship between voltage (electrical pressure), current (flow of electrons) and resistance (opposition to flow). The power formula extends this to show the rate of energy consumption. For a circuit protection perspective: current determines cable sizing (higher current needs thicker cable), resistance determines heat generation (high resistance wastes energy as heat) and power determines circuit breaker rating.
3 Worked examples
⚠️ Illustrative example only — not clinical or professional instruction.
Basic
Size a circuit for a 24-ohm heating element
Given: Voltage: 240V | Resistance: 24 Ohm | Solve for: Current
Working:
I = V / R = 240 / 24 = 10A | Power = V x I = 240 x 10 = 2,400WAnswer: Current: 10A | Power: 2.4 kW
💡 10A current on 240V -- use 2.5mm² cable rated for 15-20A and a 16A circuit breaker. This is a 2.4kW load -- confirm the circuit breaker size with AS 3000 and the specific installation conditions.
Standard
Calculate the voltage drop across a resistor
Given: Current: 5A | Resistance: 48 Ohm | Solve for: Voltage
Working:
V = I x R = 5 x 48 = 240V | Power: 5 x 240 = 1,200WAnswer: Voltage: 240V | Power: 1.2 kW
💡 Both approaches confirm the same circuit -- V=IR and I=V/R are two sides of the same equation. Useful cross-check when confirming a circuit calculation.
Advanced
Find resistance of an unknown load
Given: Voltage: 240V | Current: 6.5A | Solve for: Resistance
Working:
R = V / I = 240 / 6.5 = 36.9 Ohm | Power = 240 x 6.5 = 1,560WAnswer: Resistance: 36.9 Ohm | Power: 1.56 kW
💡 Useful for identifying an unknown or suspected fault condition. If measured current differs significantly from the calculated value for the known resistance, a fault may be present.
4 Sanity check
Australian standard residential voltage
240V single phase | 415V three phase
US standard voltage
120V single phase (residential outlets) | 240V split phase (large appliances) | 480V three phase (commercial)
Cable current ratings (approximate, AS 3000)
1.5mm^2: 15A | 2.5mm^2: 20A | 4mm^2: 27A | 6mm^2: 35A | 10mm^2: 47A
Actual rating depends on installation method, ambient temperature and grouping factor.
This is a reference tool only
All electrical work in Australia must comply with AS/NZS 3000 and be performed by a licensed electrician
5 Common errors
| Error | Cause | Consequence | Fix |
|---|---|---|---|
| Using Ohm's Law for AC circuits without considering power factor | Applying the DC formula to inductive or capacitive AC loads | Current underestimated for motors and transformer loads -- circuit protection undersized | For resistive loads (heating elements, incandescent lights): Ohm's Law applies directly. For inductive loads (motors, compressors, fluorescent ballasts): the apparent power (VA) is higher than real power (W). Include power factor (typically 0.8-0.95) when sizing for these loads. |
| Calculating from nominal voltage without checking actual site voltage | Assuming 240V when site may be 230V or 245V | Calculated current slightly wrong -- minor for protection sizing but relevant for sensitive equipment | For precise engineering calculations, measure actual voltage at the supply point. For most field calculations, using 230V (the standard nominal voltage in Australia) is appropriate. |
| Treating circuit current and fuse rating as the same thing | Setting the fuse or circuit breaker to exactly the calculated current | Nuisance tripping under normal inrush current or legitimate transient loads | Circuit breaker rating is selected above the continuous current. For resistive loads, a circuit breaker rated at 125-150% of the continuous load current is typical. Always follow AS 3000 for cable and protection selection. |
| Not accounting for voltage drop in long cable runs | Using Ohm's Law for the load resistance only | Voltage at the load is lower than calculated -- load underperforms or fails to operate | For cable runs longer than 20-30m, calculate the cable resistance (using resistance per metre from cable data sheets) and add it to the load resistance. Voltage drop in the cable reduces the voltage available at the load. |
6 Reference & regulatory links
7 Professional workflow
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