Does this mean that the battery only outputs current equal to the amount of current draw? I always thought that batteries outputted a constant output, but apparently after some research I found this to be false. Can someone help explain?

UPDATE What I mean is that if I have a load that requires 100ma and then I get a battery that releases 150mAh, what does that mean? Does the battery always give 150mA or does that mean that the load only draws 100mA from the 150mA battery


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  • \$\begingroup\$ Are you familiar with things like ohms law? \$\endgroup\$ – PlasmaHH Jan 16 '15 at 13:42
  • \$\begingroup\$ Yes. I understand that Current, Resistance, and Voltage are interrelated in electricity and electromagnetism. \$\endgroup\$ – Jdude2345 Jan 16 '15 at 13:52
  • \$\begingroup\$ The milliamp hour rating gives you an idea of how much total power a battery can provide - literally, current * time. Also, that in conjunction with the "C" rating gives you an idea of high-load performance, for example a "20C" 500mAh battery might be useful for briefly powering a 20*.5 = 10 amp load (for 3 minutes), while a "10C" battery of the same capacity may have too much internal impedance to provide more than 5 amps without the voltage severely sagging. \$\endgroup\$ – Chris Stratton Jan 17 '15 at 21:05

A fresh battery is mostly a voltage source with some resistance, called the battery's internal resistance, in series. The current produced by the battery is therefore dependent on the load. In rough terms, the battery sets the voltage, and the load decides how much current to draw at that voltage.


A battery is considered to be a constant-voltage source and, as such, will output whatever current the load requires in accordance with Ohm's law: \$ {E = IR} \$ , where E is the battery voltage in volts, I is the load current in amperes, and R is the load resistance in ohms.

In order to solve for the load current we can rearrange the formula like this: \$ I = \frac{E}{R} \$ and, assuming a 12 volt battery and a 12 ohm load, we'll have:

$$ I = \frac{E}{R} = \frac{12V}{12\Omega} = 1\ ampere $$

You can see, then, that if the battery voltage remains constant and the load resistance changes, the current into the load will change as demanded by the load.

Since some work has to be done to get charge out of the battery, into the load, and back to the battery through the wiring, that amounts to extra resistance and, since it's in series with the load and the battery, will add to the R term in the denominator of the formula, reducing the current through the load or, out of the battery, since current in a series circuit is everywhere the same.


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