An inductance coil has a reactance of \(100\, \Omega\). When an \(AC\) signal of frequency \(1000\, Hz\) is applied to the coil, the applied voltage leads the current by \(45^{\circ}\). The self- inductance of the coil is

\(=\sqrt{{ R ^{2}+ x _{ L }^{2}}}=100\) \(...(i)\)

\(f =1000 Hz\) of applied \(AC\) signal

Voltage leads current bly \(45^{\circ}\)

\(\tan 45^{\circ}=\frac{ i X _{ L }}{ iR }=\frac{\omega L }{ R }\)

ie \(R = X _{ L }=\omega L\)

Putting in eqn \((i):\) \(\sqrt{ X _{ L }^{2}+ X _{ L }^{2}}=100\)

\(\sqrt{2} X _{ L }=100 \Rightarrow X _{ L }=50 \sqrt{2}\)

ie \(\omega L =50 \sqrt{2}\)

\(L =\frac{50 \sqrt{2}}{\omega}=\frac{50 \sqrt{2}}{2 \pi f }=\frac{25 \sqrt{2}}{\pi \times 1000} H\)

\(=1.125 \times 10^{-2} H\)