Answer
The current in resistance 1 is $~~50.5~mA$
Work Step by Step
We can find the equivalent resistance of $R_2, R_3,$ and $R_4$ which are in parallel:
$\frac{1}{R_{234}} = \frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}$
$\frac{1}{R_{234}} = \frac{1}{50.0~\Omega}+\frac{1}{50.0~\Omega}+\frac{1}{75.0~\Omega}$
$\frac{1}{R_{234}} = \frac{3}{150~\Omega}+\frac{3}{150~\Omega}+\frac{2}{150~\Omega}$
$R_{234} = \frac{75.0~\Omega}{4}$
We can find the equivalent resistance in the circuit:
$R_{eq} = \frac{75.0~\Omega}{4}+100~\Omega$
$R_{eq} = \frac{75.0~\Omega}{4}+\frac{400~\Omega}{4}$
$R_{eq} = \frac{475~\Omega}{4}$
We can find the total current in the circuit:
$i = \frac{\mathscr{E}}{R_{eq}}$
$i = \frac{6.00~V}{475~\Omega/4}$
$i = \frac{24.0~V}{475~\Omega}$
$i = 50.5~mA$
Since all the current passes through resistance 1, the current in resistance 1 is $~~50.5~mA$
