Inductive Reactance of a Coil

Level 2 requires school mathematics. Suitable for pupils.

Updated by Alexander Fufaev on 01/12/2022

Illustration : 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:

Formula: Inductive reactance

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

$\pi$ 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:

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

Video - Inductive reactance of a coil briefly explained