How it works
Electrical power is voltage times current, P = V·I. Rearranged, the current a device draws is its power divided by the supply voltage: I = P / V. That single step turns any appliance's wattage into the amps flowing through its cord and the circuit behind it.
The voltage matters as much as the wattage. The same 2,300 W kettle draws 10 A on a 230 V European supply but about 19 A on a 120 V North American one — which is why high-power appliances on lower-voltage grids need noticeably thicker cords and higher-rated circuits.
Worked example — watts to amps: a 2,300 W device on 230 V draws I = 2,300 / 230 = 10 A, and behaves like a resistance of V/I = 230 / 10 = 23 Ω. A 60 W bulb on the same supply draws only 60 / 230 ≈ 0.26 A. This assumes a resistive (unity power factor) DC or simple AC load; motors and electronics with a lower power factor draw somewhat more current for the same real power.
Use it in real life
Circuit loading: add up the amps of everything on a ring — a 16 A breaker trips once the total current, not the total wattage, crosses its rating. Converting each appliance's watts to amps tells you how close you are.
Cable and fuse sizing: wire gauge and fuse ratings are chosen in amps, so a device's current draw (watts ÷ volts) is what decides the cable, not its power label alone.
Solar and battery systems: low-voltage 12 V or 24 V systems turn modest wattages into large currents (a 600 W inverter draws 50 A at 12 V), which is why battery cabling is so thick.
Frequently asked questions
How do you convert watts to amps?
Divide the power in watts by the voltage: amps = watts ÷ volts. A 1,380 W appliance on a 230 V supply draws 1,380 / 230 = 6 A. You always need the voltage — watts alone don't fix the current.
Why does the same wattage draw more amps at a lower voltage?
Because power is voltage times current. To deliver the same watts at half the voltage, the current must double. That's why 120 V countries see higher appliance currents than 230 V ones for identical devices.
Does power factor change the answer?
For resistive loads (heaters, filament bulbs, DC) no — I = P/V is exact. For motors and switching electronics the current is higher than P/V because their power factor is below 1; divide by (voltage × power factor) for those.