Answer
The current at point $a$ is $~~3.94~A$
Work Step by Step
The equivalent resistance of $R_1$ and $R_2$ in series in the $S_3$ branch is $20.0~\Omega+10.0~\Omega$ which is $30.0~\Omega$
We can find the equivalent resistance of $R_1$ and $R_2$, along with $R_1$ which is in parallel in the $S_2$ branch:
$\frac{1}{R_{eq}} = \frac{1}{20.0~\Omega}+\frac{1}{30.0~\Omega}$
$\frac{1}{R_{eq}} = \frac{3}{60.0~\Omega}+\frac{2}{60.0~\Omega}$
$R_{eq} = 12.0~\Omega$
We can find the equivalent resistance of the $S_2$ and $S_3$ branch:
$R_{eq} = 10.0~\Omega+12.0~\Omega = 22.0~\Omega$
We can find the equivalent resistance of the $S_2$ and $S_3$ branch, along with $R_1$ which is in parallel:
$\frac{1}{R_{eq}} = \frac{1}{20.0~\Omega}+\frac{1}{22.0~\Omega}$
$\frac{1}{R_{eq}} = \frac{11}{220~\Omega}+\frac{10}{220~\Omega}$
$R_{eq} = 10.48~\Omega$
We can find the equivalent resistance in the circuit:
$R_{eq} = 20.0~\Omega+10.48~\Omega = 30.48~\Omega$
We can find the current in the circuit:
$i = \frac{120~V}{30.48~\Omega} = 3.94~A$
The current at point $a$ is $~~3.94~A$
