Question 232 of 380
An electric bulb rated is connected across a supply. The power dissipation of the bulb is:
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Answer: (B)
Rated power \& voltage gives resistance
$\begin{aligned}
& \mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(200)^2}{50}=\frac{40000}{50} \\
& \mathrm{R}=800 \\
& \mathrm{P}=\frac{\left(\mathrm{V}_{\text {applied }}\right)^2}{\mathrm{R}}=\frac{(100)^2}{800} \\
& \mathrm{P}=12.5 \text { watt }
\end{aligned}$
$\begin{aligned}
& \mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(200)^2}{50}=\frac{40000}{50} \\
& \mathrm{R}=800 \\
& \mathrm{P}=\frac{\left(\mathrm{V}_{\text {applied }}\right)^2}{\mathrm{R}}=\frac{(100)^2}{800} \\
& \mathrm{P}=12.5 \text { watt }
\end{aligned}$