Question

In L-C-R circuit, $f=\frac{50}{\pi }Hz$, $V=50V$, $R=300\Omega$. If $L=1H$ and $C=20\mu C$, then the voltage across capacitor is :

• A. $50V$
• B. $20V$
• C. Zero
• D. $30V$

Answer 1

Answer: (A)

$50V$

For an L-C-R circuit, the impedance $\left(Z\right)$ is given by
$\because Z=\sqrt{R^2+{(X_L-X_C)}^2}$
where, $X_L=\omega L=2\pi f_1$
and $X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$
Given, $f=\frac{50}{\pi }Hz$, $R=300\Omega$ and $L=1H$
and $C=20\mu C=20\times {10}^{-6}C$
$Z=\sqrt{{\left(300\right)}^2+(2\pi \times \frac{50}{\pi }\times 1-\frac{1}{2\pi \times \frac{50}{\pi }\times 20\times {10}^{-6}})}$
$Z=\sqrt{90000+{(100-500)}^2}$
$\Rightarrow Z=\sqrt{90000+160000}=\sqrt{250000}$
$\Rightarrow Z=500\Omega$
Hence, the current in the circuit is given by
$i=\frac{V}{Z}=\frac{50}{500}=0.1A$
Voltage across capacitor is
$V_C=i{X}_C=\frac{i}{2\pi fC}=\frac{0.1}{2\pi \times \frac{50}{\pi }\times 20\times {10}^{-4}}$
$=\frac{0.1\times {10}^6}{100\times 20}\Rightarrow V_C=50V$