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 \[\text{an}\] \[AC\] circuit is given by. \[P=VI\cos \phi \] where \[V\] is voltage, \[I\] the current, \[\phi \] 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\[\phi =\frac{\pi }{2}\]. \[P=VI\cos \phi =VI\cos \frac{\pi }{2}=0\] Note: The given circuit is a watt less circuit, because average power dissipated is zero.
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