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In an $\mathrm{AC}$ circuit, $V$ and $I$ are given by $V=150 \sin (150 t) \quad$ volt $\quad$ and $I=150 \sin \left(150 t+\frac{\pi}{3}\right)$ amp. The power dissipated in the circuit is
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The correct answer is:
$5625 \mathrm{~W}$
Given $V=150 \sin (150 t)$ volt
$$
I=150 \sin (150 t+\pi / 3) \mathrm{amp}
$$
$$
I_{0}=150 \mathrm{amp}
$$
$V_{0}=150$ volt
$P=\frac{1}{2} V_{0} I_{0} \cos \phi$
$P=0.5 \times 150 \times 150 \times \cos 60^{\circ}$
$P=5625 W$
$$
I=150 \sin (150 t+\pi / 3) \mathrm{amp}
$$
$$
I_{0}=150 \mathrm{amp}
$$
$V_{0}=150$ volt
$P=\frac{1}{2} V_{0} I_{0} \cos \phi$
$P=0.5 \times 150 \times 150 \times \cos 60^{\circ}$
$P=5625 W$
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