Two $220\; V , 100 \;W$ bulbs are connected first in series and then in parallel. Each time the combination is connected to a $220 \;V \;AC$ supply line. The power drawn by the combination in each case respectively will be
AIPMT 2003, Medium
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$R=\frac{V^2}{P}=\frac{220 \times 220}{100}=484 \Omega$
In series, $R_{e q}=484+484+968 \Omega$
$\therefore P_{e q}=\frac{V^2}{968}=\frac{220 \times 220}{968}=50 \;watt$
In parallel, $R_{\text {eq }}=242 \Omega$
$\therefore P_{e q}=\frac{V^2}{242}=\frac{220 \times 220}{242}=200\; watt$.
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