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Q. If voltage across a bulb rated $220\, V , 100\, W$ drops by $2.5 \%$ of its rated value, the percentage of the rated value by which the power would decrease is

KCETKCET 2022Current Electricity

Solution:

$P =\frac{ V ^{2}}{ R }$
$P \propto V ^{2}$
$\frac{\Delta P }{ P } \times 100=2 \frac{\Delta V }{ V } \times 100$
$=2 \times 2.5=5 \%$