My name is Alexander FufaeV and here I write about:

Capacitive Reactance of a Capacitor Simply Explained

Important Formula

Formula: Capacitive Reactance of a Capacitor

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

Rearrange formula

What do the formula symbols mean?

Capacitive reactance

$$ \class{purple}{X_{\text C}} $$ Unit $$ \mathrm{\Omega} $$

Capcitive reactance is the complex part of the capacitor impedance (complex resistance). This capacitance reactance, on the one hand, allows the alternating current to flow through the capacitor and, on the other hand, it creates a phase shift between voltage $U_{\text{C}}(t)$ and current $I_{\text{C}}(t)$.

In the case of a DC circuit, the voltage frequency is $ f = 0 $. In this case, the capacitor has an infinite reactance and thus the capacitor does not conduct current.

Frequency

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

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

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

With the angular frequency $\omega ~=~ 2\pi \, f$ the capacitive reactance can be written as: $$ X_{\text C} ~=~ -\frac{1}{\omega \, C} $$

Capacitance

$$ C $$ Unit $$ \mathrm{F} = \frac{ \mathrm{C} }{ \mathrm{V} } $$

It is a characteristic quantity of the capacitor and tells how many charges must be brought onto the capacitor to charge the capacitor to the voltage $ 1 \, \text{V} $.

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 $C$:

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

$\pi$ 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 minus sign in the capacitive reactance causes a phase shift between current and voltage at the capacitor.

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 $:

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

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:

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

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:

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