question_answer

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