Question

In the given circuit, $V_{C}=50 \,V$ and $R=50\, \Omega$. The values of $C$ and $V_{R}$ are

• A. $3.3\,\mu F , \,60\, V$
• B. $3.3\, \mu F , \,98 \,V$
• C. $1.6 \,\mu F ,\,30 \,V$
• D. $2 \,\mu F , \,60 \,V$

Answer 1

Answer: (B)

$V_{C}^{2}+V_{R}^{2}=V^{2}$
$50^{2}+V_{R}^{2}=V^{2}$
$V_{R}^{2}=110^{2}-50^{2}$
$V_{R}=\sqrt{110^{2}-50^{2}}$

$V_{R}=\sqrt{160 \times 60} \simeq 98 \,V$
Then, $I_{v}=\frac{98}{50} (\therefore R=50\, \Omega)$
Also, $I_{v}=\frac{110}{\sqrt{R^{2}+X_{C}^{2}}} $
$\Rightarrow \frac{98}{50}=\frac{110}{\sqrt{50^{2}+X_{C}^{2}}}$
Find $X_{C}$ and then use $X_{c}=\frac{1}{\omega C}$ to find $C$.
$C=3.3\, \mu F$