Question

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

• A. $1.1\times {10}^{-2}\text{H}$
• B. $1.1\times {10}^{-1}\text{H}$
• C. $5.5\times {10}^{-5}\text{H}$
• D. $6.7\times {10}^{-7}\text{H}$

Answer 1

The correct option is A $1.1\times {10}^{-2}\text{H}$

Given,
Reactance of inductance coil, $Z=100\Omega$
Frequency of AC signal, $v=1000\text{Hz}$
Phase angle $\varphi ={45}^{\circ }$
$\tan \varphi =\frac{X_L}{R}=\tan {45}^{\circ }=1$
$\Rightarrow X_L=R$

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

$\Rightarrow 100=\sqrt{R^2+R^2}$

$\Rightarrow \sqrt{2}R=100\Rightarrow R=50\sqrt{2}$

$\therefore X_L=50\sqrt{2}$

$\Rightarrow L\omega =50\sqrt{2}$

$(\because X_L=\omega L)$

$\Rightarrow L=\frac{50\sqrt{2}}{2\pi \times 1000}(\because \omega =2\pi v)$

$\Rightarrow L=\frac{25\sqrt{2}}{\pi }=1.1\times {10}^{-2}\text{H}$