This post will deviate from computer-related posts and document a life-hack calculation instead. This question came up when my girlfriend and I were packing for a road trip in our F-150 truck. We were thinking of bringing an electric kettle with us, but we wanted to know if the truck’s battery would be up to the task of boiling the kettle. This post documents the calculations for posterity.

F-150 Battery

Voltage : 12V
Capacity: 72Ah
Cranking Amps: 900A (we can safely draw 900A for 30s, with voltage staying above 7.2V)

The Kettle

Average Power Usage: 2200W
Time to boil: 200s

Battery Usage

Amount of Energy needed: 2200W * (1h * 200s / 3600s) = 122.22Wh
Plus inefficiency of the inverter (85%): 122.22Wh / 0.85 = 143.78Wh
Amp-hours of battery needed: 143.78Wh / 12V = 11.98Ah
So, boiling a kettle drains: 11.98Ah / 72Ah = 16.7% of battery


Kettle will continuously draw:

11.98Ah / (1h * 200s / 3600s) = 217.82 Amps

If we assume a linear degradation of voltage, then time to 7.2V (based on battery’s Cold Cranking Amps (CCA)) at 217.82A draw is:

30s / (217.82A/900A) = 123.95 seconds

So the battery can supply the necessary current for 123.95 seconds, but we need 200 seconds to boil the kettle. So, even though we only need 16.7% of battery, the rate of consumption takes the battery voltage below 7.2V for

(200s - 123.95s) = 76.05 seconds


Boiling a kettle uses 16.7% of battery capacity, but for the last 76 seconds the battery is pushed into operating outside of the designed parameters due to rate of discharge. Don’t know how serious or not serious that is. Please comment if you know more! We used the kettle in the truck, and it worked, so this calculation is supported by empirical evidence :).