| In the figure shown \[{{v}_{1}},\]\[{{v}_{2}},\]\[{{v}_{3}}\] are AC voltmeters and A is AC ammeter. The readings of \[{{v}_{1}},\]\[{{v}_{2}},\]\[{{v}_{3}}\] and A are 10 V, 20 V, 20 V, 2A respectively. Find the values of R, C, L and the source voltage \[{{v}_{s}}.\] If the inductor is short circuit then what will be the readings of \[{{v}_{1}},\]\[{{v}_{2}}.\] and A. |
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A) \[2\sqrt{5}\,\,\text{volts}\]
B) \[4\sqrt{5}\,\,\text{volts}\]
C) \[6\sqrt{5}\,\,\text{volts}\]
D) \[8\sqrt{5}\,\,\text{volts}\]
Correct Answer: B
Solution :
| Let \[{{I}_{f}}\] be the rms current through the circuit then |
| \[{{I}_{f}}=2A,\] \[\frac{{{I}_{r}}}{\omega \,C}=20\,V,\] \[{{I}_{r}}\,\omega L=20\,V\] |
| and \[{{I}_{r}}\,R=10\,V\] |
| Solving we get |
| \[R=5\,\Omega ,\] \[C=\frac{1}{\pi }\times {{10}^{-\,3}}F\] |
| and \[L=\frac{1}{10\pi }H\] |
| \[\therefore \] \[{{v}_{s}}=\text{source}\,\,\text{voltage}={{I}_{r}}\sqrt{{{R}^{2}}+{{\left( \omega \,L-\frac{1}{\omega \,C} \right)}^{2}}}\] |
| \[=\sqrt{{{({{I}_{r}}R)}^{2}}{{\left( {{I}_{r}}\,\omega \,L-\frac{{{I}_{r}}}{\omega \,C} \right)}^{2}}}\] |
| \[=\sqrt{{{10}^{2}}+{{(20-20)}^{2}}}=10\,\,\text{volts}\] |
| Now, after the inductor is shorted |
| \[{{I}_{r}}=\frac{{{\operatorname{v}}_{s}}}{\sqrt{{{R}^{2}}+\frac{1}{{{\omega }^{2}}{{C}^{2}}}}}=\frac{10}{\sqrt{25+100}}=\frac{2}{\sqrt{5}}\] ampere |
| \[{{\operatorname{v}}_{1}}={{I}_{r}}\] \[R=2\sqrt{5}\,\,\text{volts}\] |
| \[{{v}_{2}}=\frac{{{I}_{r}}}{\omega \,C}=4\sqrt{5}\,\,\text{volts}\] Ans. |
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