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 \[{{V}_{C}}=i\,{{X}_{C}}=\frac{i}{2\pi fC}=\frac{0.1}{2\pi \times \frac{50}{\pi }\times 20\times {{10}^{-4}}}\] \[=\frac{0.1\times {{10}^{6}}}{100\times 20}\,\,\,\Rightarrow \,\,\,{{V}_{C}}=50\,V\]You need to login to perform this action.
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