Imagine a battery charging a capacitor to deliver a set voltage for a given time.

If we decide to double the voltage delivered by the capacitor, then we must double the amount of charge (current) delivered to it which doubles the current drain on the battery.

I would have thought that this would double the rate of battery drain; but the power equation P = V^2/R suggests that doubling the voltage squares the amount of work delivered by the capacitor (and implies that the amount of work needed to charge it was a squared increase), even though the current delivered is only double the original amount.

So a 10A/hr battery, load 5A/hr will last 2 hours --- double the voltage delivery, should last 1 hr. (i.e. 10A drawn in the hour).

But: 10W/hr doing 5W of work per hour will last 2 hours -- double the voltage delivery, you square the work done, i.e. 25W (therefore will last 15 minutes)

Therefore I cannot understand how the two measures can be converted between each other. It seems that the current delivery capacity (A-hr) is dependent on factors other than 'stored current'.

I know I've gone wrong somewhere, just can't see where. Thank you.