Explanation

(c) The simplified circuit is shown below
From figure,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 15\ I_{1} = 5I_{2}\ or\ I_{2} = \frac{15I_{2}}{5} = 3I_{1}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I = I_{1} + \ I_{2} = \frac{I_{2}}{3} + I_{2} = \frac{4I_{2}}{3}\ \ \ \ \ \ \ldots(i)\ $
But $\ \ \ \ I_{2}^{2}\ \times 5 = 42$
Or $I_{2}^{2} = \frac{42}{5} = 8.4$
Putting value of $I_{2}$ in Eq. (i), we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I = \frac{4}{3} \times \sqrt{8.4}$
Therefore, heat dissipated across $2\ \Omega$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = I^{2} \times 2 = \frac{16}{9} \times 8.4 \times 2$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \approx 30\ {Js}^{- 1}$