Question

In the circuit given in figure, $V_c=50V$ and $R=50\Omega$. The values of $C$ and $V_R$ are

• A. $3.3mF,60V$
• B. $104\mu F,98V$
• C. $52\mu F,98V$
• D. $2\mu F,60V$

Answer 1

Answer: (B)

$V_C^2+V_R^2=V^2$

$50^2+V_R^2=110^2$
$\Rightarrow V_R=\sqrt{160\times 60}=98V$
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 $C=104\mu F$