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:
Answer & explanation
Correct answer: option 4
$ 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 $
$\Rightarrow L = \frac{70.7}{2000\pi} = 1.1 \times 10^{-2} henry$