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 :
JEE MAIN 2024, Diffcult
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Rated power $\&$ voltage gives resistance
$\mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(200)^2}{50}=\frac{40000}{50}$
$\mathrm{R}=800$
$\mathrm{P}=\frac{\left(\mathrm{V}_{\text {applied }}\right)^2}{\mathrm{R}}=\frac{(100)^2}{800}$
$\mathrm{P}=12.5 \text { watt }$
Hence option $1$ is correct.
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