Answer
$2.3A$
Work Step by Step
We know that
$V=IR$
$\implies V=(0.795A)(13.8\Omega)=10.97V$
The current in $17.2\Omega$ resistor is given as
$I^{\prime}=\frac{10.97V}{17.2\Omega}=0.637A$
As the equivalent resistance of $4.11\Omega$, $8.45\Omega$ is parallel to $17.2\Omega$, thus the current in this combination is
$I^{\prime \prime}=\frac{V}{R_{4.11}+R_{8.45\Omega}}=0.873A$
Now the required current through the entire circuit can be calculated as
$I=0.795A+0.637A+0.873A$
$I=2.3A$
