question_answer

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

A)  50V               

B)  20V

C)  zero               

D)  30 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, \[{{X}_{L}}=\omega L=2\pi {{f}_{l}}\] and        \[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2\pi fC}\] Given,   \[f=\frac{50}{\pi }Hz,\,\,R=300\,\Omega \] and \[L=1\,H\] and        \[C=20\,\mu C=20\times {{10}^{-6}}C\]                 \[Z=\sqrt{{{(300)}^{2}}+\left( \begin{align}   & 2\pi \times \frac{50}{\pi }\times 1 \\  & -\frac{1}{2\pi \times \frac{50}{\pi }\times 20\times {{10}^{-6}}} \\ \end{align} \right)}\]                 \[Z=\sqrt{90000+{{(100-500)}^{2}}}\] \[\Rightarrow \] \[Z=\sqrt{90000+160000}=\sqrt{250000}\] \[\Rightarrow \] \[Z=500\,\Omega \]S Hence, the current in the circuit is given by                 \[i=\frac{V}{Z}=\frac{50}{500}=0.1\,A\] Voltage across capacitor is \[{{V}_{C}}=i\,{{X}_{C}}=\frac{i}{2\pi fC}=\frac{0.1}{2\pi \times \frac{50}{\pi }\times 20\times {{10}^{-4}}}\] \[=\frac{0.1\times {{10}^{6}}}{100\times 20}\,\,\,\Rightarrow \,\,\,{{V}_{C}}=50\,V\]