An electric bulb rated $$50$$ W $$- 200$$ V is connected across a $$100$$ V supply. The power dissipation of the bulb is:

We need to find the power dissipated by a 50 W - 200 V bulb when connected to a 100 V supply.

The filament resistance can be found from the rated power and voltage using $$P = \frac{V^2}{R}$$. From this, $$R = \frac{V_{\text{rated}}^2}{P_{\text{rated}}} = \frac{(200)^2}{50} = \frac{40000}{50} = 800 \; \Omega$$.

When the bulb is connected to 100 V, the resistance remains $$800 \; \Omega$$, so the power dissipated is $$P = \frac{V^2}{R} = \frac{(100)^2}{800} = \frac{10000}{800} = 12.5 \; \text{W}$$.

Alternatively, since $$P \propto V^2$$ at constant $$R$$, one can write $$\frac{P}{P_{\text{rated}}} = \left(\frac{V}{V_{\text{rated}}}\right)^2 = \left(\frac{100}{200}\right)^2 = \frac{1}{4} \implies P = \frac{50}{4} = 12.5 \text{ W}$$.

The correct answer is Option (2): 12.5 W.