A) \[2\sqrt{5}\] volts
B) \[4\sqrt{5}\] volts
C) \[6\sqrt{5}\] volts
D) \[8\sqrt{5}\] volts
Correct Answer: B
Solution :
[B]| Let \[{{I}_{r}}\]be the rms current through the in circuit then |
| \[{{I}_{r}}=2A,\] \[\frac{{{I}_{r}}}{\omega C}=20V,\] \[\omega L=20V\] and \[{{I}_{r}}R=10V\] |
| Solving we get \[R=5\Omega ,\] \[C=\frac{1}{\pi }\times {{10}^{-3}}F\] and \[L=\frac{1}{10\pi }\,\,H\] |
| \[\therefore {{V}_{S}}=\]Source Voltage |
| \[={{I}_{r}}\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}\] |
| \[=\sqrt{(I,{{R}^{2}})+{{\left( I{{,}_{r}}\omega L-\frac{1}{\omega C} \right)}^{2}}}\] |
| \[=\sqrt{{{10}^{2}}+{{(20-20)}^{2}}}=10\,volts\] |
| Now, after the inductor is shorted |
| \[{{I}_{r}}=\frac{{{v}_{s}}}{\sqrt{{{R}^{2}}+\frac{1}{{{\omega }^{2}}{{C}^{2}}}}}=\frac{10}{\sqrt{25+100}}=\frac{10}{\sqrt{5}}\] |
| ampere \[{{v}_{1}}={{I}_{r}}R=2\sqrt{5}\,\,volts\] |
| \[{{v}_{2}}=\frac{{{I}_{r}}}{\omega C}=4\sqrt{5}\,\,volts\,\,Ans.\] |
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