A) \[{{10}^{4}}W\]
B) 100 W
C) 2.5 kW
D) 10 W
Correct Answer: C
Solution :
\[E=100\,\sin \,t\] \[I=100\,\sin \left( 100t+\frac{\pi }{3} \right)mA\] Comparing these equations with \[E={{E}_{0}}\,\sin \omega t\] and \[I={{I}_{0}}\,\,\sin \,(\omega t+\phi )\] \[{{E}_{0}}=100,\,{{I}_{0}}=100,\,\phi =\frac{\pi }{3}\] Energy loss \[P={{E}_{rms}}\times {{I}_{rms}}\,\cos \phi \] \[=\frac{{{E}_{0}}}{\sqrt{2}}\times \frac{{{I}_{0}}}{\sqrt{2}}\cos \phi \] \[\frac{100}{\sqrt{2}}\times \frac{100}{\sqrt{2}}\times \cos \frac{\pi }{3}\] \[=2.5kW\]
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