My name is Alexander FufaeV and here I write about:

Inductive Reactance of a Coil

Important Formula

Formula: Inductive Reactance of a Coil

$$\class{brown}{X_{\text L}} ~=~ 2 \pi \, f \, \class{brown}{L}$$ $$\class{brown}{X_{\text L}} ~=~ 2 \pi \, f \, \class{brown}{L}$$ $$f ~=~ \frac{ \class{brown}{X_{\text L}} }{ 2\pi \, \class{brown}{L} }$$ $$\class{brown}{L} ~=~ \frac{ \class{brown}{X_{\text L}} }{ 2\pi \, f }$$

Rearrange formula

What do the formula symbols mean?

Inductive reactance

$$ \class{brown}{X_{\text L}} $$ Unit $$ \mathrm{\Omega} $$

Inductive reactance is the complex component of the impedance (complex resistance). This reactance of the coil on the one hand inhibits the alternating current through the coil and on the other hand creates a phase shift between voltage $U_{\text{L}}(t)$ and current $I_{\text{L}}(t)$.

In the case of a DC circuit, the voltage frequency is $ f = 0 $. In this case, the induction coil has no reactance. Without an internal resistance, the coil would produce a short circuit when a DC voltage is applied to it.

Frequency

$$ f $$ Unit $$ \mathrm{Hz} = \frac{ 1 }{ \mathrm{s} } $$

Frequency at which the AC voltage changes its polarity: $$ U_{\text{L}}(t) ~=~ U_0 \, \cos(2\pi\, f \, t) $$

The alternating current flowing through the coil also changes its direction at this frequency.

With the angular frequency $\omega ~=~ 2\pi \, f$ the inductive reactance can be written as: $$ X_{\text L} ~=~ \omega \, L $$

Inductance

$$ \class{brown}{L} $$ Unit $$ \mathrm{H} = \frac{ \mathrm{Vs} }{ \mathrm{A} } $$

Inductance is a characteristic quantity of a coil and describes how much magnetic flux it can enclose inside when a certain current flows through the coil. The higher the inductance $L$, the greater the inductive reactance $ X_{\text L} $.

A time-dependent alternating voltage is applied to a coil.

If we apply an AC voltage $ U_{\text L}(t) $ to a coil of inductance $ L $ , then an AC current $ I_{\text L}(t) $ flows through the coil. The alternating voltage changes polarity with the frequency $f$. With this frequency, the alternating current also changes its direction.

A coil to which an alternating voltage is applied has a complex non-ohmic resistance which is called inductive reactance. This resistance is usually abbreviated as $ X_{\text L} $.

You can easily calculate the inductive reactance. You need the AC frequency $ f $ and the inductance $ L $ of the coil:

$$ \begin{align} \class{brown}{X_{\text L}} ~=~ 2 \pi \, f \, \class{brown}{L} \end{align} $$

$\pi$ is here a mathematical constant with the value $ \pi = 3.14 $. The unit of the inductive reactance is Ohm:

$$ \begin{align} \left[ X_{\text L} \right] ~=~ \mathrm{\Omega} \end{align} $$

By the way, the factor $ 2 \, \pi \, f $ in Eq. 1 is often combined to angular frequency $ \omega $:

$$ \begin{align} X_{\text L} ~=~ \omega \, L \end{align} $$

If you use a very large AC frequency, then the inductive reactance will also be very large and the coil will barely let the current through.
If, on the other hand, the AC voltage frequency is very small or even zero, that is if a DC voltage is applied, then the inductive reactance also becomes zero. The coil allows an arbitrarily high current to pass, which corresponds to a short circuit.

As you can see from Eq. 1 or 2, you can also use the inductance $ L $ to adjust the reactance of the coil.

Example: Calculate reactance of coil

You apply $ 230 \, \mathrm{V} $ to a coil with an inductance of $ 500 \, \mathrm{mH} $ (Millihenry). The applied rms voltage has a frequency of $ 50 \, \mathrm{Hz} $. Insert inductance and frequency into Eq. 1:

$$ \begin{align} X_{\text L} &~=~ 2 \, \pi ~\cdot~ 50 \, \mathrm{Hz} ~\cdot~ 500 \cdot 10^{-3} \, \mathrm{H} \\\\
&~=~ 157 \, \mathrm{\Omega} \end{align} $$

To determine the RMS current $I_{\text{eff}}$ flowing through the coil, use the URI formula. Instead of using the ohmic resistance $R$, use the inductive reactance $X_{\text L}$:

$$ \begin{align} I_{\text{eff}} ~=~ \frac{ U_{\text{eff}} }{ X_{\text L} } \end{align} $$

Insert the $230 \, \mathrm{V} $ RMS voltage and $ 157 \, \mathrm{\Omega} $, then you get the RMS current:

$$ \begin{align} I_{\text{eff}} &~=~ \frac{ 230 \, \mathrm{V} }{ 157 \, \mathrm{\Omega} } \\\\
&~=~ 1.5 \, \mathrm{A} \end{align} $$