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An electric bulb marked as 50W-200V is connected across a $100 \mathrm{V}$ supply. The present power of the bulb is
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2126 Upvotes
Verified Answer
The correct answer is:
$12.5 \mathrm{W}$
We know that
$$
R=\frac{V^{2}}{P}
$$
Given $\quad V=200 \mathrm{V}, P=40 \mathrm{W}$
and $\quad V=100 \mathrm{V}$
Hence, $\quad R=\frac{V^{2}}{P}=\frac{200 \times 200}{50}$
$$
\begin{array}{l}
P=\frac{V^{2}}{R}=\frac{100 \times 100 \times 50}{200 \times 200} \\
P^{\prime}=12.5 \mathrm{W}
\end{array}
$$
$$
R=\frac{V^{2}}{P}
$$
Given $\quad V=200 \mathrm{V}, P=40 \mathrm{W}$
and $\quad V=100 \mathrm{V}$
Hence, $\quad R=\frac{V^{2}}{P}=\frac{200 \times 200}{50}$
$$
\begin{array}{l}
P=\frac{V^{2}}{R}=\frac{100 \times 100 \times 50}{200 \times 200} \\
P^{\prime}=12.5 \mathrm{W}
\end{array}
$$
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