In an AC circuit, V and I are given by V = 150 sin ⁡ 150 t V and I = 150 sin ⁡ 150 t + π 3 A . The power dissipated in the circuit is

Power P = 1 2 V 0 I 0 cos ⁡ ϕ = 0.5 × 150 × 150 × cos ⁡ 60 ° = 22500 4 = 5625 W

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Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACQuality and Power factorEasy2 minPYQ_2020

PhysicsEasysingle choice

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

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Answer:

Solution:

$Power P = \frac{1}{2} V_{0} I_{0} \cos ϕ$

$= 0.5 \times 150 \times 150 \times \cos 60 ° = \frac{22500}{4}$

$= 5625 W$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:EMI/ACSubtopic:Quality and Power factor

⏱ 2mℹ️ Source: PYQ_2020

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