A) 106 W
B) 150 W
C) 5625 W
D) zero
Correct Answer: C
Solution :
Given \[V=150\sin (150t)volt\] \[I=150\sin (150t+\pi /3)amp\] and \[{{I}_{0}}=150amp\] and \[{{V}_{0}}=150\,volt\] \[p=\frac{1}{2}{{V}_{0}}{{I}_{0}}\cos \phi \] \[p=0.5\times 150\times 150\times \cos {{60}^{o}}\] \[p=5625W\]You need to login to perform this action.
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