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

### However

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

### Summary

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 :).