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(D) In the initial circuit,the resistors $R$ and $6\,\Omega$ are in series. The potential difference across the $6\,\Omega$ resistor is $V_{6\Omega} =

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$4$

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(D) In the initial circuit,the resistors $R$ and $6\,\Omega$ are in series. The potential difference across the $6\,\Omega$ resistor is $V_{6\Omega} = 3\,V$.Using Ohm's law,the current in the circuit is $I = \frac{V_{6\Omega}}{6\,\Omega} = \frac{3}{6} = 0.5\,A$.The total voltage is $V = 5\,V$. In a series circuit with $R$ and $6\,\Omega$,$V = I \sqrt{(R+6)^2} = I(R+6)$. Thus,$5 = 0.5(R+6)$,which gives $R+6 = 10$,so $R = 4\,\Omega$.When $R$ is replaced by an inductor $L$ with reactance $X_L$,the circuit becomes an $L-R$ series circuit. The current remains $I = 0.5\,A$.The impedance of the new circuit is $Z = \sqrt{X_L^2 + 6^2}$.Using $V = IZ$,we have $5 = 0.5 \sqrt{X_L^2 + 36}$.$10 = \sqrt{X_L^2 + 36} \implies 100 = X_L^2 + 36 \implies X_L^2 = 64 \implies X_L = 8\,\Omega$.The potential difference across the inductor is $V_L = I \cdot X_L = 0.5 	imes 8 = 4\,V$.

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