Question

An inductance coil has a reactance of $100\Omega$. When an AC signal of frequency$1000Hz$ is applied to the coil, the applied voltage leads the current by ${45}^0$. The self-inductance of the coil is:

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

Answer 1

The correct option is D

$1.1\times {10}^{-2}H$

Step 1: Given data and assumption.

Impudence of the coil, $Z=100\Omega$

Frequency, $f=1000Hz$

Current leads the voltage by an angle, $\varphi ={45}^0$

Step 2: Finding the self-inductance $L$ of the coil.

Since, It is clear from the given data that the inductance is not pure inductance. The given coil behaves as a series combination of L and R.

Therefore,

The applied voltage leads the current by ${45}^0$

$\tan \left(\varphi \right)=\frac{X_L}{R}$

Where, $X_L$is inductive reactance.

$\tan \left({45}^0\right)=\frac{X_L}{R}$

$\Rightarrow$ $1=\frac{X_L}{R}$

$\Rightarrow$ $X_L=R$ ……$\left(i\right)$

Now,

We know that

Impedance of the coil,

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

$\Rightarrow$ $Z=\sqrt{X_L^2+X_L^2}$ $\left[\because Fromequation\left(i\right)\right]$

$\Rightarrow$ $100=\sqrt{2X_L^2}$

$\Rightarrow$ $X_L=\frac{100}{\sqrt{2}}$ …..$\left(ii\right)$

As we know that,

$X_L=L\omega$

where $\omega$ is angular frequency

$L\omega =\frac{100}{\sqrt{2}}$

$\Rightarrow$$L=\frac{100}{2\times \pi \times f\times \sqrt{2}}$ $\left[\because \omega =2\pi f\right]$

$\Rightarrow$$L=\frac{100}{2\times 3.14\times 1000\times \sqrt{2}}$

$\Rightarrow$$L=1.1\times {10}^{-2}H$

Hence, option D is correct.