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If you wanted to calculate the energy supplied by a battery in time $t$ you would use $E=VIt$ where $I$ is the current through the battery. If the internal resistance is $r$ we could also use $E=\frac{V^2}{r}t$. So it must be that $\frac{V^2}{r}=VI$ or $V=Ir$. But this is false in general (as an example, take emf $12V$, current $0.5A$ and load resistance $15\Omega$) and I am wondering why it doesn't work.

Thanks!

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Your analysis doesn't apply to what you are applying it to. $P=V^2/r$ is the power dissipated in the internal resistance of the battery. It is the power dissipated by the internal resistance if the battery is shorted.

What you want is "the power supplied by the battery", which is the power dissipated in the external resistor: $P = V^2/R$, where care must be taken to insure that the $V$ in that equation is the actual voltage across the resistor. You may have to take into account that some of the battery's EMF will appear across the internal resistance, so the voltage across the external resistance will be somewhat less than the EMF of the battery. Details left to the reader.

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$E=\frac{V^2}{r}t$ only holds if $I=\frac{V}{r}$, which only holds if the internal resistance $r$ is the only resistance in the circuit. Taking into account the load resistance $R$ we would get $I=\frac{V}{r+R}$, and hence $E=\frac{V^2}{r+R}t$ (assuming the battery's voltage is constant; otherwise it would become an integral).

If you want to take the internal resistance $r$ into account, you can calculate the voltage $V_R$ at the external resistance as $V_R=R·I$ and the energy at the external resistance as $V_R·I·t = V·\frac{R}{r+R}·t$.

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protected by Qmechanic Apr 11 '17 at 10:25

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