Alexander Fufaev
My name is Alexander FufaeV and here I write about:

Capacitive Reactance of a Capacitor Simply Explained

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 \(C\):

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Formula: Capacitive reactance
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\(\pi\) 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:

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Unit of capacitive reactance
\left[ X_{\text C} \right] ~=~ \mathrm{\Omega}
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By the way, the factor \( 2 \, \pi \, f \) in Eq. 1 is often combined to angular frequency \( \omega \):

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Formula: Capacitive reactance using angular frequency
X_{\text C} ~=~ -\frac{1}{\omega \, C}
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  • 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:

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Formula: Magnitude of the capacitive reactance
|X_{\text C}| ~=~ \frac{1}{2\,\pi \, f \, C}
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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:

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Example calculation: Capacitive reactance
|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}

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:

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Formula: RMS current using capacitive resistance
I_{\text{eff}} ~=~ \frac{ U_{\text{eff}} }{ X_{\text C} }

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

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Example calculation: Determine rms current using capacitive reactance
I_{\text{eff}} &~=~ \frac{ 230 \, \mathrm{V} }{ 318 \, \mathrm{k\Omega} } \\\\
&~=~ 0.7 \, \mathrm{mA}