In an A.C. circuit, the potential difference ' V ' and current 'I' are given respectively by V = 100sin (100t) V, I = 100sin (100t + π/3) mA The power dissipated in the circuit will be [Given → cos π/3 = 1/2 ] (a) 10⁴ W (b) 10 W (c) 2.5 W (d) 5 W

Answer: (c). P = (V₀I₀)/(2)cos φ with φ = π/3 P = (100×0.1)/(2)×1/2 P = 2.5 W So the answer is (c) 2.5 W.

In an A.C. circuit, the potential difference ' V ' and current 'I'… | MHT CET 2024 (2 May, Shift 1)

Q31MHT CET 2024 · 2 MAY · SHIFT 1Easy

Power, Power Factor & Wattless Current

In an A.C. circuit, the potential difference ' $V$ ' and current 'I' are given respectively by $V = 100\sin(100t)\text{ V}, I = 100\sin\left(100t + \frac{\pi}{3}\right)\text{ mA}$ The power dissipated in the circuit will be [Given $\rightarrow \cos\frac{\pi}{3} = \frac{1}{2}$ ]

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Answer: (c)

$P = \frac{V_0I_0}{2}\cos\phi$ with $\phi = \frac{\pi}{3}$
$P = \frac{100\times0.1}{2}\times\frac{1}{2}$
$P = 2.5\text{ W}$
So the answer is (c) 2.5 W.

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Power, Power Factor & Wattless Current question from MHT CET 2024 (2 May, Shift 1)

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