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:

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Aicharya · Accepted Answer

Rated power \& voltage gives resistance $\begin{aligned} & \mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(200)^2}{50}=\frac{40000}{50} \ & \mathrm{R}=800 \ & \mathrm{P}=\frac{\left(\mathrm{V}_{	ext {applied }}ight)^2}{\mathrm{R}}=\frac{(100)^2}{800} \ & \mathrm{P}=12.5 	ext { watt } \end{aligned}$

An electric bulb rated $50 \mathrm{~W}-200 \mathrm{~V}$ is connected across a $100 \mathrm{~V}$ supply. The…

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