An inductance coil has a reactance of 100 Ω. When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45°. The self- inductance of the coil is:

$1.1 × 10^{-2} H$

$ tan\phi = \frac{X_L}{R} = 1 $

$\Rightarrow R = X_L $

$ Z = 100\Omega = \sqrt {R^2 + X_L^2}=  X_L \sqrt 2$

$ X_L = \frac{100}{\sqrt 2} = 70.7 \Omega $

$ 2\pi fL = 70.7 \Omega $

$\Rightarrow L = \frac{70.7}{2000\pi} = 1.1 \times 10^{-2} henry$