In an AC circuit, V and I are given by V = 150 sin ( 150 t ) volt and I = 150 sin ( 150 t + 3 π ) amp. The power dissipated in the circuit is :

Use peak values and the phase difference. P = 2 1 V 0 I 0 cos ϕ P = 2 1 × 150 × 150 × cos ( 3 π ) = 5625 W

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Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACAC current and voltageMedium2 minQB

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

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Answer:

Solution:

Use peak values and the phase difference.

$P = \frac{1}{2} V_{0} I_{0} cos ϕ$

$P = \frac{1}{2} \times 150 \times 150 \times cos (\frac{π}{3}) = 5625 W$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:AC current and voltage

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