Answer
$r = 0.3\Omega$
Work Step by Step
Based on the formula:
$V = \varepsilon - Ir$
We need $I$ to solve for $r$.
We know that there are two $4W$ lightbulbs, so $P= 8W$.
Thus,
$P = IV$
$I = \frac{P}{V} = \frac{8}{11.8} = 0.678A$
Now knowing $I$ we can solve for r in the above formula:
$V = \varepsilon - Ir$
$r = \frac{\varepsilon - V}{I} = \frac{12-11.8}{.678} = 0.295$ or $0.3A$
