In an AC circuit, V and I are given by $V = 150\sin \left( {150{\rm{t}}} \right){\rm{\;volt\;and\;}}I = 150\sin \left( {150{\rm{t}} + \frac{{\rm{\pi }}}{3}} \right){\rm{amp}}$ . The power dissipated in the circuit is

NEET UG PHYSICS MCQ: ALTERNATING CURRENT

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Application +4 | -1

In an AC circuit, V and I are given by

V = 150\sin \left( {150\mathrm{t}} \right)\mathrm{\;volt\;and\;}I = 150\sin \left( {150\mathrm{t} + \frac{\mathrm{\pi }}{3}} \right)\mathrm{amp}

. The power dissipated in the circuit is

View answer and explanation

Correct option: B

5625 W

Option B (5625 W) is correct because the power dissipated is given by

P = V_{rms}I_{rms}\cos\phi

, where

\phi = \pi/3

, so

P = (150/\sqrt{2})(150/\sqrt{2})\cos(\pi/3) = 5625

Practice All Chapter MCQs →

\mathrm{AC}

V_C=I X_C

V=I Z

X_C

Z

V = 240\sin 120\;t

{X_L}

R