A) 106 W
B) 150 W
C) 5625 W
D) zero
Correct Answer: C
Solution :
(c) 5625 W Compare \[V=150\,\sin \,(150\,t)\]with \[V={{V}_{0}}\sin \,\omega t\]. we get, \[{{V}_{0}}=150\,V\] Compare \[l=150\,\sin \left( 150t+\frac{\pi }{3} \right)\] with \[l={{l}_{0}}\sin \,(\omega t+\phi )\]. we get \[{{l}_{0}}=150\,A,\,\phi =\frac{\pi }{3}=60{}^\circ \] The power dissipated in AC circuit is \[P=\frac{1}{2}{{V}_{0}}{{l}_{0}}\cos \,\phi =\frac{1}{2}\times 150\times 150\times \cos \,60{}^\circ \] \[=\frac{1}{2}\times 150\times 150\times \frac{1}{2}=5625\,W\]
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