AC current limiting

Thread Starter

Da iawn

Joined Nov 17, 2019
The context is a kettle supplied at 220v AC from an inverter. The kettle is rated at 850 watts, resulting in a current of about 75amps on the 12v side.
I would like a circuit I can plug in between the inverter and the kettle to limit the current on the 220v side, and consequently reduce the current on the 12v side down to around 40 amps.
This is to protect a lead acid auxiliary battery in s camper.

I know the simple answer is to get a lower wattage kettle, and it's not the most effective way of heating water in this context but due to a combination of real life factors it would be very convenient to be able to plug the kettle in to the inverter and run the engine for the time it takes to boil at times where naked flames are banned.
Any suggestions.
For background my electronics experience is mainly from the 70s although I'm confident with circuit con


Joined Dec 2, 2017
You could cut the power in half on the 220 side with a high current diode, assuming the kettle doesn't need the other half of the cycle. (for a timer or something)


Joined Dec 29, 2008
... Wondering if it would be feasible to reduce the high side 220 vac to 110 vac using a 2:1 transformer rated at 1 kva. The 12 volt battery load amps should be about half the original at 37 amps or so. The transformer might weigh 5 or 10 lbs. The heating wattage would only be about half at maybe 425 watts. There should not be any critical problem with running a resistive load at half voltage. The only difference would be an extended heating period necessary to achieve the required temperature.


Joined Jan 8, 2017
Re post #4
NO. The current at at 12 volts would be about 19 amps (Probably a little more due to losses in the transformer.) The power to the kettle would be 212 Watts. Remember P = V^2/R. To reduce the power to half would require a transformer with a secondary voltage of about 155 volts.