A $220 \; V , 50 \; Hz$ AC source is connected to a $25 \; V$, $5 \; W$ lamp and an additional resistance $R$ in series (as shown in figure) to run the lamp at its peak brightness, then the value of $R$ (in ohm) will be
A$975$
B$875$
C$775$
D$675$
JEE MAIN 2022, Medium
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A$975$
a $P = Vi$
$5=25 i$
$i=\frac{1}{5}$
$V_{R}=i R$
$(220-25)=\frac{1}{5} R$
$R=195 \times 5=975 \Omega$
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