An AC voltage V = 20sin 200π t is applied to a series LCR circuit which drives a current I = 10sin (200π t + π/3). The average power dissipated is: (a) 21.6 W (b) 200 W (c) 173.2 W (d) 50 W

Answer: (d). P = (V₀I₀)/(2)cos φ with φ = π/3 P = (20×10)/(2)×0.5 P = 50 W So the answer is (d) 50 W.

An AC voltage V = 20sin 200π t is applied to a series LCR circuit… | JEE Main 2024 (31 Jan, Shift 2)

Q32JEE Main 2024 · 31 JAN · SHIFT 2Easy

Power, Power Factor & Wattless Current

An AC voltage $V = 20\sin 200\pi t$ is applied to a series LCR circuit which drives a current $I = 10\sin\left(200\pi t + \frac{\pi}{3}\right)$ . The average power dissipated is:

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Answer: (d)

$P = \frac{V_0I_0}{2}\cos\phi$ with $\phi = \frac{\pi}{3}$
$P = \frac{20\times10}{2}\times0.5$
$P = 50\text{ W}$
So the answer is (d) 50 W.

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