Answer
$P_{20} = 45W$
$P_5 = 20W$
Work Step by Step
power dissipated by $10\Omega$
$ P= I^2R = \frac{V_{10}^2}{R}$
$ 40W = I^2 *10\Omega$
$ \implies I^2 = 4$
$\implies I=2A$
and $ 40W = \frac{V^2}{10\Omega}$
$ V_{10}^2 = 400W\Omega$
$V_{10} = 20V$
but current is same for $5\Omega $ band $10\Omega$
thus power dissipated by $5\Omega$
$P_5 = I^2R = 2^2 *5 W = 20W$
also $P_5 = \frac{V_5^2}{R}4$
$V_5^2 = 20* 5 V^2$
$V_5 = 10V$
now $V_{20} = V_5 + V_{10}$
$ V_{20} = 10V +20V =30V
P_{20} = \frac{V_{20}^2 }{20}$
$P_{20} = \frac{30^2}{20}W$
$P_{20} = 45W$
