Finding current and resistance for an immersion heater using V, I, R and power relationships.
This worked example treats an immersion element as a simple resistive load. Use Ohm’s law with the power relationship to distinguish current in amps from resistance in ohms. The same shortcuts do not describe every AC motor or pump.
The relationships. For this resistive element, V = I × R and P = V × I, using RMS voltage and current on AC. Therefore I = P / V and R = V / I = V² / P.
The situation. A 3 kW (3000 W) immersion heater runs from a 230 V supply.
Step 1 — find the current. I = P / V = 3000 / 230 = about 13.0 amps (A).
Step 2 — find the resistance. R = V / I = 230 / 13.04 = about 17.6 ohms. The same figure comes straight from the power rating, R = V² / P = 52,900 / 3000 = 17.6 ohms, which avoids rounding the current first.
Why it matters. The element draws about 13 A, but its resistance is about 17.6 Ω: these answer different questions. Load current helps select the circuit; cable installation conditions, protective-device requirements and the appliance instructions also matter.
Takeaway. Keep the resistance calculation tied to this resistive element at its stated operating condition. Electrical work needs suitable competence and verification under the rules applying to the work. In England, Approved Document P gives guidance for electrical safety in dwellings. Securely isolate and prove dead before work.
Common calculation traps
Keep units consistent. In this working, enter power in watts, voltage in volts, current in amps and resistance in ohms. Convert prefixes before substituting: using 3 instead of 3,000 W changes the result by a factor of 1,000. Use the voltage specified for the exercise or the actual equipment data for the task, and keep sufficient precision until the final answer.
The 17.6 Ω result describes the hot heater element, not its supply cable. It is also not a way to find an AC motor winding’s resistance from the motor’s rated watts and supply voltage. AC motors involve impedance, power factor and conversion to mechanical output. Use the manufacturer’s electrical data and specified tests for a motor or pump.
Finally, a calculation describes what a healthy circuit should do once it is energised; it says nothing about whether this circuit is energised right now. A conductor can still be live from a second supply, a borrowed neutral or stored charge, none of which shows up in the sum. Working out the current a load will draw is useful before you connect it — it never replaces safely isolating the circuit and proving it dead before you touch it.