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(C) Given,resistance $R = 30 \Omega$ and inductive reactance $X_L = 20 \Omega$ at $f_1 = 50 	ext{ Hz}$.Since $X_L = 2 \pi f L$,we have $20 = 2 \pi (5

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$4 	ext{ A}$

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(C) Given,resistance $R = 30 \Omega$ and inductive reactance $X_L = 20 \Omega$ at $f_1 = 50 	ext{ Hz}$.Since $X_L = 2 \pi f L$,we have $20 = 2 \pi (50) L \implies 2 \pi L = \frac{20}{50} = 0.4 \Omega/	ext{Hz}$.When the frequency is changed to $f_2 = 100 	ext{ Hz}$,the new inductive reactance $X_L'$ is:$X_L' = 2 \pi f_2 L = (2 \pi L) 	imes 100 = 0.4 	imes 100 = 40 \Omega$.The impedance $Z$ of the coil is given by $Z = \sqrt{R^2 + (X_L')^2}$.$Z = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \Omega$.The current $I$ in the coil is $I = \frac{V}{Z} = \frac{200}{50} = 4 	ext{ A}$.

$A$ coil has a resistance of $30 \Omega$ and an inductive reactance of $20 \Omega$ at $50 \text{ Hz}$ frequency. If an $AC$ source of $200 \text{ V}$,$100 \text{ Hz}$ is connected across the coil,the current in the coil is

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