Q164 / 180 · KCET 2011 · Physics

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

  • A
    $106 \mathrm{~W}$
  • B
    $150 \mathrm{~W}$
  • Correct answer
    C
    $5625 \mathrm{~W}$
  • D
    zero

Solution

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 $P=\frac{1}{2} V_{0} I_{0} \cos \phi$ $P=0.5 \times 150 \times 150 \times \cos 60^{\circ}$ $P=5625 W$

Hence, the correct answer is option: 'C'.

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