An inductor coil takes a current of 8 , A whe | vedclass

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(D) For the inductor coil at $50 \, Hz$:$X_L = \frac{V}{I} = \frac{100}{8} = 12.5 \, \Omega$.For the resistor:$R = \frac{V}{I} = \frac{100}{10} = 10 \

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$5 \sqrt{2} \, A$

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(D) For the inductor coil at $50 \, Hz$:$X_L = \frac{V}{I} = \frac{100}{8} = 12.5 \, \Omega$.For the resistor:$R = \frac{V}{I} = \frac{100}{10} = 10 \, \Omega$.At the new frequency $f' = 40 \, Hz$, the new inductive reactance $X_L'$ is:$X_L' = X_L 	imes \frac{f'}{f} = 12.5 	imes \frac{40}{50} = 10 \, \Omega$.When connected in series, the impedance $Z$ is:$Z = \sqrt{R^2 + (X_L')^2} = \sqrt{10^2 + 10^2} = 10\sqrt{2} \, \Omega$.The current $I$ in the series circuit is:$I = \frac{V}{Z} = \frac{100}{10\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, A$.

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