# is the voltage drop of a battery that is being short circuited the same as the battery's advertised voltige

So I know this is a pretty fundamental question, but I'll ask it any way. lets say you have a 12V battery, and the positive is connected to the negative directly by a wire with negligible resistance... is the voltage drop between the positive terminal and the negative terminal still 12V without a resistor? I know this is a short circuit by the way.

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## 2 Answers

No, quite opposite. For short-cut between both terminals, voltage drop between terminals is exactly 0 V! All of the advertised voltage drop is on the internal resistor of the battery.

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Ok that makes a lot of sense. thank you very much. –  AlexW.H.B. May 6 '12 at 10:27

As a slightly more elaborate model, the voltage drop between the terminals will not be exactly zero. You can model the situation as a voltage source $V_0$ in series with a big resistor (the battery's internal resistor) $R$ and a small resistor $r$ (the wire). The voltage drop across the wire is then $V_0r/R$, which is of course very close to zero.

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Indeed, but question was negligible resistance, so I've used that assumption :) –  Pygmalion May 6 '12 at 15:39