A) \[305\text{ }W\]
B) \[210\text{ }W\]
C) Zero W
D) \[242\text{ }W\]
Correct Answer: D
Solution :
| When capacitance is taken out, the circuit is LR. |
| \[\therefore \,\,\tan \phi =\frac{\omega L}{R}\] |
| \[\Rightarrow \,\,\,\omega L=R\,\tan \phi =200\times \frac{1}{\sqrt{3}}=\frac{200}{\sqrt{3}}\] |
| Again, when inductor is taken out, the circuit is CR. |
| \[\therefore \,\,\tan \phi =\frac{1}{\omega CR}\] |
| \[\Rightarrow \,\,\frac{1}{\omega c}=R\tan \phi =200\times \frac{1}{\sqrt{3}}=\frac{200}{\sqrt{3}}\] |
| Now, \[Z=\sqrt{{{R}^{2}}+{{\left( \frac{1}{\omega C}-\omega L \right)}^{2}}}\] |
| \[=\sqrt{{{(200)}^{2}}+{{\left( \frac{200}{\sqrt{3}}-\frac{200}{\sqrt{34}} \right)}^{2}}}=200\Omega \] |
| Power dissipated \[={{V}_{rms}}{{I}_{rms}}\cos \phi \] |
| \[={{V}_{rms}}.\frac{{{V}_{rms}}}{Z}.\frac{R}{Z}\] \[\left( \because \,\,\cos \phi =\frac{R}{Z} \right)\] |
| \[=\frac{{{V}^{2}}_{rms}R}{{{Z}^{2}}}=\frac{{{(200)}^{2}}\times 200}{{{(200)}^{2}}}=\frac{220\times 220}{200}=242W\] |
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