the current $I$ through the battery is
$$
7.5 \mathrm{~mA}
$$
the potential difference across $R_L$ is $18 \mathrm{~V}$
ratio of powers dissipated in $R_1$ and $R_2$ is 3
if $R_1$ and $R_2$ are interchanged, magnitude of the power dissipated in $R_L$ will decrease by a factor of 9.
Solution
$R_{\text {total }}=2+\frac{6 \times 1.5}{6+1.5}=3.2 \mathrm{k} \Omega$
(A) $I=\frac{24 \mathrm{~V}}{3.2 \mathrm{k} \Omega}=7.5 \mathrm{~mA}=I_{R_1}$ $I_{R_2}=\left(\frac{R_L}{R_L+R_2}\right) I$ $I=\frac{1.5}{7.5} \times 7.5=1.5 \mathrm{~mA}$
(B) $V_{R_L}=\left(I_{R_L}\right)\left(R_L\right)=9 \mathrm{~V}$
(C) $\frac{P_{R_1}}{P_{R_2}}=\frac{\left(I_{R_1}^2\right) R_1}{\left(I_{R_2}^2\right) R_2}=\frac{(7.5)^2(2)}{(1.5)^2(6)}=\frac{25}{3}$
(D) When $R_1$ and $R_2$ are interchanged, then
$
\frac{R_2 R_L}{R_2+R_L}=\frac{2 \times 1.5}{3.5}=\frac{6}{7} \mathrm{k} \Omega
$
Now potential difference across $R_L$ will be
$
V_L=24\left[\frac{6 / 7}{6+6 / 7}\right]
$
Earlier it was $9 \mathrm{~V}$
Since, $\quad P=\frac{V^2}{R}$ or $P \propto V^2$
In new situation potential difference has been decreased three times. Therefore, power dissipated will decrease by a factor of 9
