Explanation

(d) $R_{total} = 2 + \frac{6 \times 1.5}{6 + 1.5} = 3.2k\Omega$
$(a)I = \frac{24V}{3.2k\Omega} = 7.5mA = I_{R_{1}}$
$I_{R_{2}} = \left( \frac{R_{L}}{R_{L} + R_{2}} \right)I\ $
$I = \frac{1.5}{7.5} \times 7.5 = 1.5mA$
$I_{R_{L}} = 6mA\ $
$(b)V_{R_{L}} = \left( I_{R_{L}} \right)\left( R_{L} \right) = 9V\ $
$(c)\ \frac{P_{R_{1}}}{P_{R_{2}}} = \frac{\left( I_{R_{1}}^{2} \right)R_{1}}{\left( I_{R_{2}}^{2} \right)R_{2}} = \frac{(7.5)^{2}(2)}{(1.5)^{2}(6)} = \frac{25}{3}$
(d) When $R_{1}$and $R_{2}$are interchanged, then
$\frac{R_{2}R_{L}}{R_{2} + R_{L}} = \frac{2 \times 1.5}{3.5} = \frac{6}{7}k\Omega$
Now potential difference across $R_{L}$ will be
$V_{L} = 24\left\lbrack \frac{\frac{6}{7}}{6 + \frac{6}{7}} \right\rbrack 3V$
Earlier it was 9V
Since, $P = \frac{V^{2}}{R}or\ P \propto V^{2}$
In new situation potential difference has been decreased three times. Therefore, power dissipated will decrease by a factor of 9.