Internal resistance problem

A 12V battery has an internal resistance of $2.0Omega$. A load of variable resistance is connected across the battery and adjusted to have resistance equal to that of the internal resistance of the battery. Find the power dissipated.

The equation for internal resistance is
$$mathcal{E}=I(R+r)$$
where $R$ is the resistance in the circuit, $r$ is the internal resistance of the battery and $I$ is the current. We know that $mathcal{E}=12V$. The sum of internal and external resistance would be $4 Omega$, seeing as the variable resistor is set to $2 Omega$, which adds to the already exisiting $2 Omega$ from the battery.

The equation for power dissipated is

$$P=I^2R=frac{V^2}{R}$$

So I substitued the value for $V$ and the value for $R$ to get

$$P=frac{12^2}{4}=36$$

This is, however, wrong. Any ideas as to why?