Answer
a)
$ V_1: 12.0V $
$ V_2: 4.8V $
$ V_3: 7.2V $
b)
$ R_1: 0.12A $
$ R_2: 0.02A $
$ R_3: 0.02A $
Work Step by Step
(a) We know that $ R_1$ is connected across the battery; the potential difference across it will be same as the emf, which is $12.0V $. The resistances $ R_2$ and $ R_3$ are in series and their combined potential difference will be the emf. Since $ R_3$ is $\frac{3}{2}$ times $ R_2$, the potential difference across $ R_3$ will also be $\frac{3}{2}$ times the potential difference across $ R_2$. Let $ x $ be the potential difference across $ R_2$:
$ x+\frac{3}{2}x=12$
$\implies x=4.8V $
Now the potential difference across $ R_3$ will be $12-4.8=7.2V $. Thus, the potential difference across each resistor is as follows:
$ V_1: 12.0V $
$ V_2:4.8V $
$ V_3:7.2V $
(b) We can find the required currents as follows:
$ I_1=\frac{V_1}{R_1}$
$ I_1=\frac{12.0V}{100\Omega}$
$ I_1=0.12A $
The current through $ R_2$ is
$ I_2=\frac{V_2}{R_2}$
We plug in the known values to obtain:
$ I_2=\frac{4.8V}{200\Omega}$
$ I_2=0.02A $
As $ R_2$ and $ R_3$ are in series, hence the current though them will be the same. Thus, the current through each resistor is as follows:
$ R_1: 0.12A $
$ R_2: 0.02A $
$ R_3: 0.02A $
