An inductance coil has a reactance of $100 \Omega$. When an $\mathrm{AC}$ signal of frequency $1000 \mathrm{~Hz}$ is applied to the coil, the applied voltage leads the current by $45^{\circ}$. The self-inductance of the coil is:
Correct Option: 1
(1) Given,
Reactance of inductance coil, $Z=100 \Omega$
Frequency of $\mathrm{AC}$ signal, $v=1000 \mathrm{~Hz}$
Phase angle, $\phi=45^{\circ}$
$\tan \phi=\frac{X_{L}}{R}=\tan 45^{\circ}=1$
$\Rightarrow X_{L}=R$
Reactance, $Z=100=\sqrt{X_{L}^{2}+R^{2}}$
$\Rightarrow 100=\sqrt{R^{2}+R^{2}}$
$\Rightarrow \sqrt{2} R=100 \Rightarrow R=50 \sqrt{2}$
$\therefore X_{L}=50 \sqrt{2}$
$\Rightarrow L \omega=50 \sqrt{2}$ $\left(\because X_{L}=\omega L\right)$
$\Rightarrow L=\frac{50 \sqrt{2}}{2 \pi \times 1000}$ $(\because \omega=2 \pi v)$
$=\frac{25 \sqrt{2}}{\pi} \mathrm{mH}=1.1 \times 10^{-2} \mathrm{H}$