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(C) For a $d.c.$ source,the inductor acts as a simple resistor because the frequency is zero. Thus,the resistance $R$ is given by $R = \frac{V}{I} = \

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$\frac{4}{\pi} 	imes 10^{-2} \ H$

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(C) For a $d.c.$ source,the inductor acts as a simple resistor because the frequency is zero. Thus,the resistance $R$ is given by $R = \frac{V}{I} = \frac{12 \ V}{4 \ A} = 3 \ \Omega$.For an $a.c.$ source,the impedance $Z$ is given by $Z = \frac{V}{I} = \frac{12 \ V}{2.4 \ A} = 5 \ \Omega$.The impedance of an $RL$ circuit is $Z = \sqrt{R^2 + X_L^2}$,where $X_L$ is the inductive reactance.Squaring both sides,$Z^2 = R^2 + X_L^2$,so $X_L^2 = Z^2 - R^2 = 5^2 - 3^2 = 25 - 9 = 16$.Thus,$X_L = 4 \ \Omega$.Since $X_L = 2 \pi f L$,we have $L = \frac{X_L}{2 \pi f} = \frac{4}{2 \pi 	imes 50} = \frac{4}{100 \pi} = \frac{4}{\pi} 	imes 10^{-2} \ H$.

When a coil is connected to a $d.c.$ source of $e.m.f.$ $12 \ V$,the current of $4 \ A$ flows in it. If the same coil is connected to a $12 \ V, 50 \ Hz$ $a.c.$ source,the current flowing in it is $2.4 \ A$. The self-inductance of the coil will be:

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