question_answer

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

A) \[zero\]                                               

B) \[20\,\,V\]

C)  \[30\,\,V\]                        

D)  \[50\,\,V\]

Correct Answer: D

Solution :

For an \[LCR\] circuit the impedance \[(Z)\] is given by                    \[Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}\] where   \[{{X}_{L}}=\omega L=2\pi fl\] and        \[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2\pi fC}\] Given,   \[f=\frac{50}{\pi }Hz,\,\,R=300\Omega ,\,\,L=1H,\]                 \[C=20\mu C=20\times {{10}^{-6}}C\]. \[\therefore \]\[Z=\sqrt{\begin{align}   & {{(300)}^{2}}+ \\  & \left( 2\pi \times \frac{50}{\pi }\times 1-\frac{1}{2\pi \times \frac{50}{\pi }\times 20\times {{10}^{-6}}} \right) \\ \end{align}}\]                 \[Z=\sqrt{90,000+{{(100-500)}^{2}}}\]                 \[Z=\sqrt{90,000+16,000}=500\Omega \] Hence, current in circuit is given by                  \[i=\frac{V}{Z}=\frac{50}{500}\]                    \[=0.1\,\,A\] Voltage across capacitor is,  \[{{V}_{C}}=i{{X}_{C}}=\frac{i}{2\pi fC}=\frac{0.1}{2\pi \times \frac{50}{\pi }\times 20\times {{10}^{-6}}}\]                      \[=\frac{0.1\times {{10}^{6}}}{100\times 20}\] \[\Rightarrow \]               \[{{V}_{C}}=50\,\,V\].