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When a resistance R is connected in series with an element A, the electric current is found to be lagging behind the voltage V by angle $30^{0}$ . When the same resistance is connected in series with element B, current leads by $60^{0}$ . When R, A, B are connected in series, the current now leads voltage by $θ$ which is equal to $\tan^{- 1} (K / \sqrt{3})$ , then the value of K is (assume same AC source is used in all cases)

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Detailed Solution

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$ϕ_{1} = - 30^{0} ϕ_{2} = + 60^{0} \tan ϕ = \frac{X_{L}}{R} \tan ϕ_{2} = \frac{X_{C}}{R} X_{L} = \frac{R}{\sqrt{3}} X_{C} = R \sqrt{3}$

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$\tan θ = \frac{(X_{C} - X_{L})}{R} = \frac{\frac{R \sqrt{3} - R}{\sqrt{3}}}{R} = \frac{2}{\sqrt{3}}$

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