$3$ identical bulbs are connected in series and these together dissipate a power $P$. If now the bulbs are connected in parallel, then the power dissipated will be
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(c) When bulbs are connected in series, $P = \frac{{{V^2}}}{{R'}} = \frac{{{V^2}}}{{3R}}$
When bulbs are connected in parallel,
$P' = \frac{{{V^2}}}{{R''}} = \frac{{{V^2} \times 3}}{R} = 3 \times 3P = 9P.$
When bulbs are connected in parallel,
$P' = \frac{{{V^2}}}{{R''}} = \frac{{{V^2} \times 3}}{R} = 3 \times 3P = 9P.$
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