In an AC circuit E = 50sin (500t), I = 600sin (500t + π/3) mA. What is the power dissipated in the circuit? [ cos 60° = 0.5 ] (a) 10 W (b) 75 W (c) 7.5 W (d) 50 W

Answer: (c). P = (E₀I₀)/(2)cos φ with φ = 60° P = (50×0.6)/(2)×0.5 P = 7.5 W So the answer is (c) 7.5 W.

In an AC circuit E = 50sin (500t), I = 600sin (500t + π/3) mA. What… | MHT CET 2026 (11 Apr, Shift 1)

Q3MHT CET 2026 · 11 APR · SHIFT 1Easy

Power, Power Factor & Wattless Current

In an AC circuit $E = 50\sin(500t),\ I = 600\sin\left(500t + \frac{\pi}{3}\right)\text{ mA}$ . What is the power dissipated in the circuit? [ $\cos 60^\circ = 0.5$ ]

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Answer: (c)

$P = \frac{E_0I_0}{2}\cos\phi$ with $\phi = 60^\circ$
$P = \frac{50\times0.6}{2}\times0.5$
$P = 7.5\text{ W}$
So the answer is (c) 7.5 W.

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