Answer
$R_1=27\Omega; R_2=13\Omega$
Work Step by Step
We can find $R_1$ and $R_2$ as follows:
When $R_1$ and $R_2$ are connected in series then applying Ohm's law, we obtain:
$6.0V-4.0V=IR_2$
$\implies I=\frac{2.0V}{R_2}$.......eq(1)
Similarly,when $R_1$ and $R_2$ are connected in series then applying Ohm's law, we obtain:
$6.0V=I_2R_2$
$\implies R_2=\frac{6.0V}{0.45A}$
$\implies R_2=13\Omega$......eq(2)
We plug in the value of $R_2$ in eq(1) to obtain:
$I=\frac{2.0V}{13\Omega}$
$I=0.15A$
As given that the potential difference for $R_1$ is $4.0V$
$\implies IR_1=4.0V$
$\implies R_1=\frac{4.0V}{I}$
$R_1=\frac{4.0V}{0.15A}$
$R_1=27\Omega$
