A) \[2\,A\]
B) \[1\,A\]
C) \[\frac{\sqrt{3}}{2}A\]
D) \[\frac{2}{\sqrt{3}}A\]
Correct Answer: A
Solution :
If the capacitance is removed, it is an L-R circuit \[\tan \phi =\frac{{{X}_{L}}}{R}=\tan {{60}^{o}}=\sqrt{3}\] If inductance is removed, it is a capacitance circuit or R-C circuit \[\left| \phi \right|\] is the same. \[\therefore \] \[\omega L=\frac{1}{\omega C}\]. This is a resonance circuit. \[\therefore \] \[Z=R\,;\,{{I}_{rms}}=\frac{{{E}_{rms}}}{R}\,;\,{{E}_{rms}}=200\,V\] \[\therefore \] \[{{I}_{rms}}=\frac{200\,\,V}{100\,\,\Omega }=2\,\,A\]You need to login to perform this action.
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