A long solenoid connected to a 12,V, DC sourc | vedclass

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(C) For a $DC$ source,the inductor acts as a simple resistor because the frequency is $0$. The resistance $R$ is given by $R = V / I = 12\,V / 2\,A =

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$33$

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(C) For a $DC$ source,the inductor acts as a simple resistor because the frequency is $0$. The resistance $R$ is given by $R = V / I = 12\,V / 2\,A = 6\,\Omega$.When connected to an $AC$ source,the impedance $Z$ is $Z = V_{rms} / I_{rms} = 12\,V / 1\,A = 12\,\Omega$.The impedance of an $RL$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$,where $X_L = \omega L = 2\pi f L$.Squaring both sides: $Z^2 = R^2 + (2\pi f L)^2$.Substituting the values: $12^2 = 6^2 + (2 \pi 	imes 50 	imes L)^2$.$144 = 36 + (100\pi L)^2$.$(100\pi L)^2 = 108$.$100\pi L = \sqrt{108} \approx 10.39$.$L = 10.39 / (100 	imes 3.14159) \approx 10.39 / 314.16 \approx 0.033\,H$.Converting to millihenry: $L = 0.033 	imes 1000 = 33\,mH$.

$A$ long solenoid connected to a $12\,V\, DC$ source passes a steady current of $2\,A.$ When the solenoid is connected to a source of $12\,V\, RMS$ at $50\,Hz,$ the current flowing is $1\,A\, RMS.$ Then the inductance of the solenoid is ....$mH$.

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