# Charging a battery

So we have a charger outputting V volts and $$I$$ Amperes and two rechargable batteries of the same capacity $$Q$$ coulombs but different emf's $$V_1\&V_2$$.

Which battery would charge faster?

Approach:

So since the charge stored in both the batteries is $$Q$$ C, and the battery provides a current of $$I$$, which is $$I \frac{C}{sec}$$ so the time taken should be the same in both the cases. ($$\frac{Q}{I}sec$$).

However the power provided by the adapter in both the cases is the same and the energy stored in the battery is different in both the cases so this answer is not possible.

Where did I go wrong?

• – probably_someone Oct 17 '18 at 14:07
• Related post by OP: physics.stackexchange.com/q/435110/2451 – Qmechanic Oct 17 '18 at 14:14
• It is not possible for the charger to control both the voltage and the current. If the charger controls the voltage, then the battery will control the current and vice versa. – Solomon Slow Oct 17 '18 at 16:58

## 2 Answers

Where did I go wrong?

The first 'wrong' is stipulating that the charge voltage across $$V_C$$ and current through $$I_C$$ are fixed by the charger. This is quite unrealistic. Why? Because you can't independently specify these when charging a battery.

To see this, stipulate that the uncharged battery emf is $$V_{B,uc}$$ and that the internal resistance of the battery is $$r_B$$.

If follows that a charger with voltage across $$V_C \gt V_{B,uc}$$ must be supplying a charging current $$I_C$$ equal to

$$I_C = \frac{V_C - V_{B,uc}}{r_B}$$

and so the charging current and voltage are not independent. Typically, a battery charger will limit the charging current to a safe value by controlling $$V_C$$ and so, in the example you give, the charge currents may be the same but the charging voltage will not be. Thus, you can't say that the power provided by the adapter to each battery during the charging process is the same.

The second 'wrong' is assuming that all of the power from the charger goes to charge the battery. Some of the power, $$I^2_C\cdot r_B$$, is dissipated by the internal resistance (the battery warms up during charging).

Finally, as others have pointed out, a battery (or cell) stores energy and not electric charge. If two batteries have the same (energy) capacity (typically given in watt-hours), then for the same charging current, the battery with the largest emf will finish charging first.

For example, and at the risk of simplifying too much, assume you have a 6V and a 12V battery each with the same capacity and 'small' internal resistance.

If both (fully discharged) batteries are charged with a 1A charging current, the 12V battery will become fully charged in essentially half the time of the 6V battery.

• I was talking about the time taken to charge the battery from 0 to 100% – harshit54 Oct 17 '18 at 17:49

Distilled from probably someone's comments in chat:

Rechargeable batteries don't store charge, they store chemical energy. You assume that the batteries both have a "stored charge" of $$Q$$, when in reality they don't do that. They store energy, not charge. They're not just capacitors. The energy capacity of the battery and its emf are mostly unrelated.

Battery capacity (the rating in ampere-hours $$\text{Ah}$$ you find on many batteries) is an integrated current (an amount of current multiplied by a time over which that current is produced), which happens to have the same units as charge, but is not the same quantity, and doesn't mean that any charge is stored in the battery. You can see the same thing happen in other areas of physics - torque and energy have the same units of force multiplied by distance, but are not equivalent.

A battery is basically a water pump, but one that pumps electrons rather than water. Suppose a water pump could make water flow at 1 kg/s for 500 hours before breaking down. Then the capacity of this water pump would be 500 (kg/s)*hours, or 1.8 million L. This doesn't mean that the pump contains 1.8 million L of water before it starts running; obviously, it's pushing water that's already there, because it's a pump. Batteries are the same way. To find the energy stored in a battery, multiply its capacity in coulombs by the voltage it produces (this is always smaller than the emf of the battery, due to the battery's internal resistance). The product of the capacity and the emf is the maximum amount of electrical energy that one could extract from the battery; whenever there is a nonzero current flowing through the battery, some of this energy will be wasted as heat due to the internal resistance. `

• Re, "...happens to have the same units as charge, but is not the same quantity." Using your water analogy, that's like saying that the "gallons" of water in a reservoir are not the same "gallons" that are moved by a pump. Imagine two scenarios: (1) We have a thousand gallons of water in a tank, and we use the pump to pump it all out, and (2) The same pump is in a loop that contains ten gallons, and it re-circulates the water around the loop a hundred times.Different gallons? You and I know that ten gallons does not equal a thousand, but the pump saw a thousand gallons go by in either case. – Solomon Slow Oct 17 '18 at 17:16