An electric bulb rated 50 mathrm{~W}-200 math | vedclass

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Question

Rated power $\&$ voltage gives resistance

$\mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(200)^2}{50}=\frac{40000}{50}$

$\mathrm{R}=800$

$\m

Vedclass · Accepted Answer

$12.5 \mathrm{~W}$

Vedclass · Answer

Rated power $\&$ voltage gives resistance

$\mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(200)^2}{50}=\frac{40000}{50}$

$\mathrm{R}=800$

$\mathrm{P}=\frac{\left(\mathrm{V}_{	ext {applied }}ight)^2}{\mathrm{R}}=\frac{(100)^2}{800}$

$\mathrm{P}=12.5 	ext { watt }$

Hence option $1$ is correct.

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 :