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If the power factor is 1/2 in a series RL circuit with $R = 100 Ω$ . If AC mains, 50 Hz is used, then L is

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$\frac{\sqrt{3}}{π}$ henry

$π$ henry

$\sqrt{3}$ henry

$\sqrt{3} π$ henry

answer is A.

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Detailed Solution

$z = \sqrt{R^{2} + ω^{2} L^{2}}$

$\cos ϕ = \frac{R}{z} \Rightarrow \frac{1}{2} = \frac{100}{z} = Z = 200 Ω$

$R^{2} + ω^{2} L^{2} = Z^{2} \Rightarrow ω^{2} L^{2} = Z^{2} - R^{2}$

$\Rightarrow ω^{2} L^{2} = 4 \times 10^{4} - {(100)}^{2} = 3 \times 10^{4} \Rightarrow ω L = 100 \sqrt{3}$

$\Rightarrow 2 \times π \times 50 \times L = 100 \sqrt{3} \Rightarrow L = \frac{100 \times \sqrt{3}}{2 π \times 50} = \frac{\sqrt{3}}{π}$

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