Answer
(a) $9.0V$
(b) decrease
(c) $8.6V$
Work Step by Step
(a) We know that when the switch is open then no current flows in the circuit and hence the internal resistance will be zero. As a result, the potential difference across the terminals of the battery will be $9.0V$.
(b) When the switch is closed then the current starts flowing and the potential difference decreases due to the internal resistance (that is, $V=\mathcal{E}-Ir$).
(c) We know that
$V=\mathcal{E}(1-\frac{r}{R_{eq}})$
We plug in the known values to obtain:
$V=(9.0V)(1-\frac{0.73\Omega}{(11+5.6+0.73)\Omega})$
$V=8.6V$
