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

<p>To find the power dissipation of the bulb when it's connected to a $100 \mathrm{V}$ supply instead of its rated $200 \mathrm{V}$ supply, we can use the relation between power (P), voltage (V), and resistance (R), which is given by $P = \frac{V^2}{R}$. The resistance of the bulb can be considered constant in this case, allowing us to calculate the change in power dissipation due to the change in voltage.</p> <p>First, let's find the resistance of the bulb based on its rated conditions:</p> <p>$P = \frac{V^2}{R} \Rightarrow R = \frac{V^2}{P}$</p> <p>Substituting the rated values, we get:</p> <p>$R = \frac{(200)^2}{50} = \frac{40000}{50} = 800 \, \Omega$</p> <p>Now, using this resistance, we can find the power dissipation when the bulb is connected to a $100 \mathrm{V}$ supply:</p> <p>$P = \frac{V^2}{R} = \frac{(100)^2}{800} = \frac{10000}{800} = 12.5 \, W$</p> <p>This means the power dissipation of the bulb when connected to a $100 \mathrm{V}$ supply is $12.5 \, W$.</p> <p>Therefore, the correct answer is Option C: 12.5 W.</p>