Part (a)
For the correct answer:
$38\ \Omega$ (or $37.5\ \Omega$)
When $S_1$ is closed and $S_2$ is open, current flows through $R_1$ and $R_2$ in series. Since resistors are identical, the 6.0 V is shared equally: $V_1 = 3.0\text{ V}$. Using $R = \frac{V}{I} = \frac{3.0\text{ V}}{0.080\text{ A}} = 37.5\ \Omega$.
Part (b)(i)
For the correct answer:
$0.24\text{ A}$
With both switches closed, $R_3$ is in parallel with the top branch. $I_3 = \frac{V}{R} = \frac{6.0\text{ V}}{37.5\ \Omega} = 0.16\text{ A}$. Total current = $0.080\text{ A} (branch 1) + 0.16\text{ A} (branch 2) = 0.24\text{ A}$.
Part (b)(ii)
For the correct answer:
The potential difference across $R_3$ is larger than across $R_1$, so more work is done passing charge through $R_3$.
Potential difference is work done per unit charge ($V = \frac{W}{Q}$). $R_3$ has the full 6.0 V, while $R_1$ has 3.0 V. More work done results in a larger energy transfer as heat.
