A) \[\frac{{{V}^{2}}}{({{R}^{2}}+{{\omega }^{2}}{{L}^{2}})}R\]
B) zero
C) \[\frac{{{V}^{2}}}{{{L}^{2}}}R\]
D) \[\frac{{{V}^{2}}}{{{Z}^{2}}}R\]
Correct Answer: B
Solution :
\[{{P}_{av}}={{V}_{rms}}\times {{I}_{rms}}\cos \phi \] If circuit has only inductance, then \[\phi =\frac{\pi }{2}\] \[\therefore \] \[{{P}_{av}}=0\]
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