Physics - Current Electricity Question with Solution | TestHub

k open: two 9 Ω branches in parallel across 9 V, each carries 1 A: $P_{6\Omega \times 2}=2\times 1^2\times 6=12\mathrm{W}$

k closed: (3 Ω ∥ 6 Ω) + (6 Ω ∥ 3 Ω) = 2 + 2 = 4 Ω, I = 9/4 A; each parallel section has 4.5 V: $P_{3\Omega \times 2}=2\times \frac{4.5^2}{3}=\frac{27}{2}\mathrm{W}$

$\frac{P_{3\Omega }(\mathrm{closed})}{P_{6\Omega }(\mathrm{open})}=\frac{27/2}{12}=\frac{9}{8}$ Hence (B).