# Does the 'mAh' rating of a battery have something to do with its power?

I'm curious about the 'mAh' of a battery: how can this impact the power of the battery?

I've done some research on the internet, and most of the articles I found explain about the 'amount of charge stored' inside the battery, or in other words, the 'capacity' of the battery. So the 'mAh' will tell how long the battery can last if a certain amount of current is drawn. But what about the power, does it have anything to do with the power as well? Can anyone explain it?

I'm asking because I have an RC (remote control) car, where usually it uses a Ni-Cd 700 mAh 6V battery. I tried using other batteries; for example I plugged in a standard 6V drycell battery (4 x 1.5V), and it will work, but the car runs very slow, as if it's running under low battery power. Is it because of the 'mAh'? They're the same voltage, 6 volts, but how can the power be different?

I remember a formula from back when I was in school: $V=IR$. So with the internal resistance inside the car, this means the current drawn will be the same. But why can the Ni-Cd battery boost the car?

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The mAh hour rating of a battery is how much energy it stores. It's simply the number of mA it can deliver for an hour (in theory)

The power of a battery is how much energy it can deliver in a certain time. Since Power is current * voltage, and the voltage of a battery is (almost) fixed, then the maximum power is related to the maximum current. Which depends on the internal resistance, which depends on mostly the on the type of battery chemistry and then on the specific design of the electrodes.

So a car battery is designed to produce very high currents (ie high power) to start a car so hasvery low internal resistance, a also NiCad has low internal resistance and so can run a toy car. A disposable battery has highest internal resistance so can produce lowest maximum current and the least power

ps. you can picture the internal resistance as a real resistor built into the battery. Suppose that your dry cell battery has an internal resistance of 2 Ohms and your car's motor has a resistance of 4 Ohms then there is only 4volts of the 6volts from the battery across the motor. Now put in the NiCad with a resistance of 0.1 Ohms and there is 5.9Volts across the motor

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I support the answer given by Martin Beckett.

The energy of the battery(that you used) is, 700mAh*6V. The voltage applied across the impedance is constant 6V. Depending on the impedance, the current flows through it. If the current requirement is high(more than 700mA, it runs for 1 hour at that current), you need to get a battery of higher current. Or, you can attach them in parallel. So that, you'll get 700*4mAh. But, the terminal voltage will be 1.5V. So, you need to get 4*4 set of batteries to increase the power delivered to your impedance(motor here).

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