Search any question & find its solution
Question:
Answered & Verified by Expert
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:
Options:
Solution:
1766 Upvotes
Verified Answer
The correct answer is:
$12.5 \mathrm{~W}$
Rated power \& voltage gives resistance
$\begin{aligned}
& \mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(200)^2}{50}=\frac{40000}{50} \\
& \mathrm{R}=800 \\
& \mathrm{P}=\frac{\left(\mathrm{V}_{\text {applied }}\right)^2}{\mathrm{R}}=\frac{(100)^2}{800} \\
& \mathrm{P}=12.5 \text { watt }
\end{aligned}$
$\begin{aligned}
& \mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(200)^2}{50}=\frac{40000}{50} \\
& \mathrm{R}=800 \\
& \mathrm{P}=\frac{\left(\mathrm{V}_{\text {applied }}\right)^2}{\mathrm{R}}=\frac{(100)^2}{800} \\
& \mathrm{P}=12.5 \text { watt }
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.