Answer
(a) 3.5 A
(b) 4.5 A
(c) 3.15 A
(d) 3.25 A
Work Step by Step
(a) The combination of $R_{2}$ and $R_{3}$is
$$R_{23} = R_{2} + R_{3} = 30\mathrm{~\Omega} $$
This combination is in parallel with $R_{1}$, so the equivalent resistance for it is
\begin{align*}
R_{\text{eq}} = \dfrac{R_{1}R_{23}}{R_{1} + R_{23} } = \dfrac{(15 \mathrm{~\Omega})(30 \mathrm{~\Omega})}{15\mathrm{~\Omega} + 30 \mathrm{~\Omega} } =10 \mathrm{~\Omega}
\end{align*}
Using this value of $R_{\text{eq}} $ in Ohm's law to get the current
\begin{align*}
I_{ab} = \dfrac{V}{R_{\text{eq}}} = \dfrac{35.0 \,\text{V}}{10 \mathrm{~\Omega}} = \boxed{3.5 \,\text{A}}
\end{align*}
(b) When the battery is connected to $bc$, $R_{1}$ and $R_{3}$ are in series and their equibalent is $R_{13} = R_{1} + R_{3} = 35 \mathrm{~\Omega} $. This combination is in parallel with $R_{2}$. So, their combination is
\begin{align*}
R_{\text{eq}} = \dfrac{R_{2}R_{13}}{R_{2} + R_{13} } = \dfrac{(10 \mathrm{~\Omega})(35 \mathrm{~\Omega})}{10 \mathrm{~\Omega} + 35 \mathrm{~\Omega} } =7.78 \mathrm{~\Omega}
\end{align*}
Again we apply Ohm's law to get $I_{bc}$
\begin{align*}
I_{bc} = \dfrac{V}{R_{\text{eq}}} = \dfrac{35.0 \,\text{V}}{7.78 \mathrm{~\Omega}} = \boxed{4.5\,\text{A}}
\end{align*}
(c) The same steps as on parts (a) and (b)
\begin{align*}
R_{\text{eq}} = \dfrac{R_{3}R_{12}}{R_{3} + R_{12} } = \dfrac{(20 \mathrm{~\Omega})(25 \mathrm{~\Omega})}{20 \mathrm{~\Omega} + 25 \mathrm{~\Omega} } =11.11 \mathrm{~\Omega}
\end{align*}
Use Ohm's law to get the current across (ac)
\begin{align*}
I_{ac} = \dfrac{V}{R_{\text{eq}}} = \dfrac{35.0 \,\text{V}}{11.11 \mathrm{~\Omega}} = \boxed{3.15 \,\text{A}}
\end{align*}
(d) To find the current $I_{bc}$
\begin{equation*}
I_{bc} = \dfrac{\varepsilon}{R_{eq} + r}
\end{equation*}
Where $r$ is the internal resistance of the battery $3\mathrm{~\Omega}$, $R_{eq} = 7.78 \mathrm{~\Omega}$.
\begin{align*}
I_{bc} = \dfrac{\varepsilon}{R_{eq} + r} = \dfrac{35.0 \,\text{V}}{7.78 \mathrm{~\Omega} + 3 \mathrm{~\Omega} } = \boxed{3.25 \,\text{A}}
\end{align*}
