Explanation

(b) Resistance of series combination of $3\ \Omega$ and $1\ \Omega$ is $R_{1} = 3 + 1 = 4\Omega,\ R_{2} = 8\Omega$
Let $i$ be the total current in the circuit
Current through$\ R_{1}$ is$\ i_{1} = \frac{i \times R_{2}}{R_{1} + R_{2}} = \frac{i \times 8}{12} = \frac{2i}{3}$
Current through$\ R_{2}$ is$\ i_{2} = \frac{i{\times R}_{1}}{R_{1} + R_{2}} = \frac{i \times 4}{12} = \frac{i}{3}$
Power dissipated in $3\Omega$ resistor is $P_{1} = i_{1}^{2} \times 3\ \ \ $…(i)
Power dissipated in $8\Omega$ resistor is $P_{2} = i_{2}^{2} \times 8\ \ \ \ \ $…(ii)
$\therefore\frac{P_{1}}{P_{2}} = \frac{i_{1}^{2} \times 3}{i_{2}^{2} \times 8}\ or,\frac{P_{1}}{P_{2}} = \frac{(2i/3)^{2} \times 3\ \ }{(i/3)^{2} \times 8} = \frac{12}{8} = \frac{3}{2}$
$P_{1} = \frac{3}{2} \times P_{2} = \frac{3}{2} \times 2 = 3watt$
$\therefore$ Power dissipated across $3\Omega$ resistor is 3 watt