A) zero
B) 10 W
C) 5 W
D) 2.5 W
Correct Answer: A
Solution :
Key Idea: Average power dissipated depends on phase difference between voltage and current. Average power dissipated in an AC circuit is given by. \[P=VI\,\cos \text{o }\!\!|\!\!\text{ }\] where V is voltage, \[I\] the current, \[\text{o }\!\!|\!\!\text{ }\] the phase difference Given, \[V=5\cos 2\pi ft\] Using \[\sin ({{90}^{o}}+\theta )=cos\theta \] we have \[V=5\sin \left( 2\pi ft+\frac{\pi }{2} \right)\] and \[I=2\sin 2\pi ft\] Hence, phase difference between V and \[I\]is \[\text{o }\!\!|\!\!\text{ =}\frac{\pi }{2}.\] \[p=VI\cos \text{o }\!\!|\!\!\text{ =}\,\text{VI}\,\text{cos}\frac{\pi }{2}=0\] Note: The given circuit is a wattless circuit, because average power dissipated is zero.
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