# Can a battery charger be too powerful for a rechargeable battery?

I got the impression that a regular iPhone charger can charge the iPhone and the iPhone won't become too hot while charging, and the charging time is standard, but if using the 10W iPad charger to charge the iPhone, then it is 2A and it can make the iPhone hotter while charging and the charging time will be less?

But I = V / R, so V is the same at 5V, and R is the same, so it seems like I should be the same, and it shouldn't affect charging time or making the iPhone become hotter?

Unless if the standard iPhone charger outputting 5W is below the required power, and so it is charging with a lower than needed power, so when the iPad charger is used, then now more current will in fact go through, and so up to a point, when a 50W, 100W, or 300W charger is used, it will all be the same?

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Please consider accepting a few answers. –  drN Oct 25 '12 at 17:10

A battery has certain energy holding capacity. It can store energy supplied to it at a certain rate. This is the 'mAh' rating. i.e., when you recharge a 12V, 100mAh battery, it takes 1 hour to charge if you supply 100mA at 12V. If you supply more 500mAh, it takes 12 mins to fully recharge. If you use a charger of 500mAh, 12V, for recharging 100mAh, 12V, it takes less time. But, the amount of current flow is high. Let us consider the load resistance(ipod resistance) as constant(which is not as it is a semi-conductor device, used for explaining this problem). The power loss in a resistance is I^2*R. This is high if you use a 500mAh charger. The power loss is appeared as heat in a circuit with resistance.

BTW, some batteries just short(lose the capacity of holding energy) if you use a high capacity charger.

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Any phone charger has an internal resistance, $R_{int}$, so when you try and pull current from it there is a drop in the charger voltage of $IR_{int}$. The higher the current the larger the drop, and this ultimately limits the current the charger can produce. A higher current charger will have a lower internal resistance so more current can be drawn from it before it's voltage drops too far to be effective.