In an AC circuit E = 200sin (50t) volt and I = 100sin (50t + π/3) mA. The power dissipated in the circuit is (matrixsin 30° = cos 60° = 0.5 sin 60° = cos 30° = √(3)/2matrix) (a) 20 watt (b) 15 watt (c) 10 watt (d) 5 watt

Answer: (d). P = (E₀I₀)/(2)cos φ with φ = 60° P = (200×0.1)/(2)×0.5 P = 5 W So the answer is (d) 5 watt.

In an AC circuit E = 200sin (50t) volt and I = 100sin (50t + π/3) mA… | MHT CET 2024 (3 May, Shift 1)

Q30MHT CET 2024 · 3 MAY · SHIFT 1Easy

Power, Power Factor & Wattless Current

In an $AC$ circuit $E = 200\sin(50t)$ volt and $I = 100\sin\left(50t + \frac{\pi}{3}\right)\text{ mA}$ . The power dissipated in the circuit is

$\left(\begin{matrix}\sin 30^\circ = \cos 60^\circ = 0.5 \\ \sin 60^\circ = \cos 30^\circ = \sqrt{3}/2\end{matrix}\right)$

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Answer: (d)

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