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, \[{{V}_{C}}=i{{X}_{C}}=\frac{i}{2\pi fC}=\frac{0.1}{2\pi \times \frac{50}{\pi }\times 20\times {{10}^{-6}}}\] \[=\frac{0.1\times {{10}^{6}}}{100\times 20}\] \[\Rightarrow \] \[{{V}_{C}}=50\,\,V\].You need to login to perform this action.
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