Power dissipated on internal resistance of short-circuited voltage source

Suppose we have a voltage source with an EMF of $\mathcal{E}$ and an internal resistance $R$. If we connect to it a perfect wire with zero resistance, we get a short circuit. The value of the current is determined by: $$I=\frac{\mathcal{E}}{R+0}=\frac{\mathcal{E}}{R}$$

Then the voltage across the source would be $V=\mathcal{E}-IR=\mathcal{E}-\mathcal{E}=0$ volts. Now I want to find how much power is dissipated do tue the current inside the source (e.g. Joule heating inside a battery due to collisions of electrons with ions). The formula is pretty simple:

$$P=VI$$

So for this example $P=0 \cdot \frac{\mathcal{E}}{R}=0$

But we can rewrite the formula using Ohm's law:

$$P=VI=I^2 R$$

And then it would be $P=\frac{\mathcal{E}^2}{R} \neq 0$

We get here a contradiction. Where is my mistake here? Maybe I have a wrong understanding of voltage? Is it because we have an EMF i.e. some non-electrostatic force which does work despite the fact we have short-circuited the source? The voltage itself as I understand describes an electric field which is not present. So maybe we should use instead of $V$ in Joule's heating formula an emf? But then, it still confuses me a little, because the formula itself is derived from Ohm's law and the electrostatic fields.

• When you calculate the power dissipated due to heating of the internal resistance when shorting the battery, you have to use the EMF $$\mathcal{E}$$ of the voltage source in your third equation, not the nominal battery voltage external to the internal resistance, $$V$$, as you have done. It then agrees with your final equation, and there is no contradiction. – Bitrex Dec 2 '13 at 21:34

Draw the circuit using ideal circuit elements: Now, the series current is:

$$I = \dfrac{\mathcal{E}}{R_{internal}+ R_{load}}$$

The voltage across the internal resistance is:

$$V = \mathcal{E} \dfrac{R_{internal}}{R_{internal}+ R_{load}}$$

The power dissipated by the internal resistance can by found three equivalent ways:

$$P = VI = \dfrac{V^2}{R_{internal}} = I^2 R_{internal} = \mathcal{E}^2 \dfrac{R_{internal}}{(R_{internal}+ R_{load})^2}$$

Clearly, setting $R_{load} = 0$ yields:

$$I = \dfrac{\mathcal{E}}{R_{internal} + 0} = \dfrac{\mathcal{E}}{R_{internal}}$$

$$V = \mathcal{E} \dfrac{R_{internal}}{R_{internal + 0}} = \mathcal{E}$$

$$P = \dfrac{\mathcal{E}^2}{R_{internal}}$$

Simply put, the entire emf appears across the internal resistance. Zero volts appears across the source plus internal resistance due to the short circuit.

• Many thanks! So basically, my mistake was that I've calculated the total power including both the voltage rise across the source and the voltage drop across the resistor. The one delivers power and the other consumes it. Thanks again. – user35126 Dec 3 '13 at 12:50