A) \[~\text{5}0\text{ V}\]
B) \[\text{2}0\text{ V}\]
C) zero
D) \[\text{3}0\text{V}\]
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 \[{{X}_{L}}=\omega L=2\pi {{f}_{1}}\] and \[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2\pi fC}\] Given \[f=\frac{50}{\pi }Hz,\] \[R=300\Omega \] and \[L=1H\] and \[C=20\mu C=20\times {{10}^{-6}}C\] \[Z=\sqrt{{{(300)}^{2}}+\left( \frac{^{2\pi \times \frac{50}{\pi }\times 1}1}{2\pi \times \frac{50}{\pi }\times 20\times {{10}^{-6}}} \right)}\] \[Z=\sqrt{90000+{{(100-500)}^{2}}}\] \[\Rightarrow \] \[Z=\sqrt{90000+160000}=\sqrt{250000}\] \[\Rightarrow \] \[Z=500\Omega \] Hence, the current in the circuit is given by \[i=\frac{V}{Z}=\frac{50}{500}=0.1A\] Voltage across capacitor is \[{{V}_{C}}=i{{X}_{C}}=\frac{i}{2\pi fC}=\frac{0.1}{2\pi \times \frac{50}{\pi }\times 20\times {{10}^{-4}}}\] \[=\frac{0.1\times {{10}^{6}}}{100\times 20}\] \[\Rightarrow \] \[{{V}_{C}}=50V\]You need to login to perform this action.
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