question_answer

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) \[\text{3}0\text{V}\]

Correct Answer: A

Solution :

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 \[{{V}_{C}}=i{{X}_{C}}=\frac{i}{2\pi fC}=\frac{0.1}{2\pi \times \frac{50}{\pi }\times 20\times {{10}^{-4}}}\] \[=\frac{0.1\times {{10}^{6}}}{100\times 20}\]    \[\Rightarrow \]               \[{{V}_{C}}=50V\]