A) \[P=\sqrt{2}{{E}_{0}}{{I}_{0}}\]
B) \[P=\frac{{{E}_{0}}{{I}_{0}}}{\sqrt{2}}\]
C) \[\operatorname{P}=zero\]
D) \[\operatorname{P}=\frac{{{E}_{0}}{{I}_{0}}}{2}\]
Correct Answer: C
Solution :
We know that power consumed in a.c. circuit is given by, \[\operatorname{P}=\,\,{{E}_{rms}}.\,{{I}_{rms}}\cos \,\phi \] Here, \[E=\,\,{{E}_{0}}\,sin\omega t\] \[I\,\,=\,\,{{I}_{0}}\sin \left( \omega t-\frac{\pi }{2} \right)\] which implies that the phase difference, \[\phi =\frac{\pi }{2}\] \[\therefore \,\,\,\,P={{E}_{rms}}.\,{{I}_{rms}}.\cos \frac{\pi }{2}=0\] \[\left( \because \,\,\cos \frac{\pi }{2}=0 \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
