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Q.
A $50\, Hz$ AC source of $20$ volts is connected across $R$ and $C$ as shown in figure. The voltage across $R$ is $12$ volt. The voltage across $C$ is
Voltage across capacitor lags behind current by $90^{\circ}$.
The given circuit is a $C-R$ series circuit, $V_{R}$ is in phase with $i$, while $V_{C}$ lags behind $i$ by $90^{\circ}$. Hence, resultant potential is
$V =\sqrt{V_{R}^{2}+V_{C}^{2}}$
Given, $V_{R} =12\, V,\, V=20\, V$
$\therefore (20)^{2} =(12)^{2}+V_{C}^{2}$
$\Rightarrow V_{C}^{2} =(20)^{2}-(12)^{2}$
$=400-144 =256$
$\Rightarrow V_{C} =\sqrt{256}=16\, V$
