When alternating current is passed through L - R series circuit, the power factor is √(3)/2 and R = 50 Ω, then the value of L is [cos π/6 = √(3)/2, sin π/6 = 1/2, tan π/6 = 1/√(3)] (a) 1/2π (b) √(3)/2π (c) (1)/(2√(3)π) (d) (1)/(√(3)π)

Answer: (c). cos φ = √(3)/2 ⇒ φ = π/6, and tan φ = XL/R = 1/√(3). XL = (50)/(√(3)) Ω. Taking supply frequency f = 50 Hz, ω = 100π rad/s. L = (XL)/(ω) = (50)/(√(3) × 100π) = (1)/(2√(3)π) H. So the answer is (c) (1)/(2√(3)π) H.

When alternating current is passed through L - R series circuit, the… | MHT CET 2025 (22 Apr, Shift 1)

Q27MHT CET 2025 · 22 APR · SHIFT 1Medium

Power, Power Factor & Wattless Current

When alternating current is passed through $L - R$ series circuit, the power factor is $\frac{\sqrt{3}}{2}$ and $R = 50\ \Omega$ , then the value of L is

$\left[\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}, \sin\frac{\pi}{6} = \frac{1}{2}, \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}\right]$

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Answer: (c)

$\cos\phi = \frac{\sqrt{3}}{2} \Rightarrow \phi = \frac{\pi}{6}$ , and $\tan\phi = \frac{X_L}{R} = \frac{1}{\sqrt{3}}$ .
$X_L = \frac{50}{\sqrt{3}}\ \Omega$ .
Taking supply frequency $f = 50$ Hz, $\omega = 100\pi$ rad/s.
$L = \frac{X_L}{\omega} = \frac{50}{\sqrt{3} \times 100\pi} = \frac{1}{2\sqrt{3}\pi}\text{ H}$ .
So the answer is (c) $\frac{1}{2\sqrt{3}\pi}$ H.

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