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Q.
In the circuit given in figure, $V_{ c }=50\, V$ and $R=50\, \Omega$. The values of $C$ and $V_{ R }$ are
Alternating Current
Solution:
$V_{ C }^{2}+V_{ R }^{2}=V^{2}$
$50^{2}+V_{ R }^{2}=110^{2} $
$\Rightarrow V_{ R }=\sqrt{160 \times 60}=98 \,V$
Then, $ I_{ v }=\frac{\sqrt{160 \times 60}}{50} (\therefore R=50\, \Omega)$
Also, $I_{ v }=\frac{110}{\sqrt{R^{2}+X_{ C }^{2}}} $
$\therefore \frac{98}{50}=\frac{110}{\sqrt{50^{2}+X_{ C }^{2}}}$
Flux $X_{ c }$ and now using $X_{ C }=\frac{1}{\omega C}$, we get $C=104\, \mu F$
