Question

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

• A. $\text{5}0\text{ V}$
• B. $\text{2}0\text{ V}$
• C. zero
• D. $\text{3}0\text{V}$

Answer 1

Answer: (A)

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