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(D) In an $LR$ circuit,the phase angle $\phi$ between voltage and current is given by $	an \phi = \frac{X_L}{R}$.Given $\phi = 45^{\circ}$,we have $\

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$\frac{1}{2 \pi}$

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(D) In an $LR$ circuit,the phase angle $\phi$ between voltage and current is given by $	an \phi = \frac{X_L}{R}$.Given $\phi = 45^{\circ}$,we have $	an 45^{\circ} = 1$.Therefore,$\frac{X_L}{R} = 1$,which implies $X_L = R$.Given $R = 100 \ \Omega$,so $X_L = 100 \ \Omega$.The inductive reactance is $X_L = 2 \pi f L$.Substituting the values,$100 = 2 \pi 	imes 100 	imes L$.Solving for $L$,we get $L = \frac{100}{2 \pi 	imes 100} = \frac{1}{2 \pi} \ H$.

An inductance coil has a resistance of $100 \Omega$. When an $A.C.$ signal of frequency $100 \ Hz$ is applied to the coil,the voltage leads the current by $45^{\circ}$. The inductance of the coil in henry is $\left[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]$

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