Capacitive reactance of a capacitor

by Alexander FufaeV

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

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

A capacitor to which an alternating voltage is applied has a complex non-ohmic resistance which is called capacitive reactance $ X_{\text C}(t) $.

Du kannst den kapazitiven Blindwiderstand ganz einfach berechnen. Du brauchst dafür lediglich die Wechselspannungsfrequenz $f$ und die Kondensator-Kapazität $C$:

Formula: Capacitive reactance

$$ \begin{align} \class{purple}{X_{\text C}} ~=~ -\frac{1}{2\pi \, f \, \class{purple}{C}} \end{align} $$

$\pi$ is a mathematical constant with the value $ \pi = 3.14 $. The minus sign in the formula states, that the alternating voltage at a capacitor lags behind the alternating current.

The unit of capacitive reactance is Ohm:

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

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

Formula: Capacitive reactance using angular frequency

$$ \begin{align} X_{\text C} ~=~ -\frac{1}{\omega \, C} \end{align} $$

If you use a very large AC frequency, then the capacitive reactance becomes very small and the capacitor easily lets the current through.
If, on the other hand, the AC voltage frequency is very low or even zero, i.e. if a DC voltage is applied, then the capacitive reactance becomes infinitely large. The capacitor does not allow any current to pass.

As you can see from Eq. 1 or 2, you can also use the capacitance $C$ to adjust the reactance $ X_{\text C} $ of the capacitor.

When we work with rms values of voltage and current, we are usually only interested in the magnitude $ |X_{\text C}| $ of the capacitive reactance:

Formula: Magnitude of the capacitive reactance

$$ \begin{align} |X_{\text C}| ~=~ \frac{1}{2\,\pi \, f \, C} \end{align} $$

Example: Calculate reactance of the capacitor

You apply a voltage of $ 230 \, \mathrm{V} $ to a capacitor with a capacitance of $10 \, \mathrm{nF} $. The voltage has a frequency of $50 \, \mathrm{Hz} $. Insert the values into the formula 3:

$$ \begin{align} |X_{\text C}| &~=~ \frac{1}{2\,\pi ~\cdot~ 50 \, \mathrm{Hz} ~\cdot~ 10 \cdot 10^{-9} \, \mathrm{F}} \\\\
&~=~ 318309 \, \mathrm{\Omega} \\\\
&~=~ 318 \, \mathrm{k\Omega} \end{align} $$

To determine the rms current $I_{\text{eff}}$ flowing through the capacitor, use the URI formula. Instead of using the ohmic resistance R, use the capacitive reactance $X_{\text C}$. Rearrange the URI formula according for the current:

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

Insert the $230 \, \mathrm{V} $ rms voltage and $ 318 \, \mathrm{k\Omega} $, then you get the rms current:

$$ \begin{align} I_{\text{eff}} &~=~ \frac{ 230 \, \mathrm{V} }{ 318 \, \mathrm{k\Omega} } \\\\
&~=~ 0.7 \, \mathrm{mA} \end{align} $$