In an A.C circuit, V & I are given by V = 250sin (100t) volt and I = 10sin (100t + π/3) A respectively The power dissipated in the circuit is: (a) 2500 W (b) 625 W (c) 1250 W (d) 625√(2) W

Answer: (b). P = (V₀I₀)/(2)cos φ with phase difference φ = π/3. P = (250 × 10)/(2) × cos 60° = 1250 × 0.5. P = 625 W. So the answer is (b) 625 W.

In an A.C circuit, V & I are given by V = 250sin (100t) volt and I =… | JEE Main 2022 (30 Jul, Shift 2)

Q52JEE Main 2022 · 30 JUL · SHIFT 2Medium

Power, Power Factor & Wattless Current

In an A.C circuit, V & I are given by $V = 250\sin(100t)$ volt and $I = 10\sin\left(100t + \frac{\pi}{3}\right)\text{ A}$ respectively The power dissipated in the circuit is:

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Answer: (b)

$P = \frac{V_0I_0}{2}\cos\phi$ with phase difference $\phi = \frac{\pi}{3}$ .
$P = \frac{250 \times 10}{2} \times \cos 60^\circ = 1250 \times 0.5$ .
$P = 625\text{ W}$ .
So the answer is (b) 625 W.

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