Physics - Current Electricity Question with Solution | TestHub

After the closing of$\mathrm{ }\mathrm{S},\mathrm{ }\mathrm{C}$bulb or lamp become short-circuited






the potential across$\mathrm{A}$and$\mathrm{B}$before the closing of$\mathrm{S}=\frac{V}{3}$

Potential across$\mathrm{A}$and$\mathrm{B}$after the closing of$\mathrm{S}=\frac{V}{2}$

As$P\propto V^2$

$\therefore \frac{P_{\mathrm{i}}}{P_{\mathrm{f}}}=\mathrm{ }\frac{4}{9}$

$P_{\mathrm{f}}=\frac{P_{\mathrm{i}}\times 9}{4}=\mathrm{ }2.25\mathrm{ }{P}_{\mathrm{i}}$