# Capacitive reactance and capacitance

Discussion in 'General Electronics Chat' started by Unicorn Notreal, Feb 1, 2016.

1. ### Unicorn Notreal Thread Starter New Member

Dec 24, 2015
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I don't understand why is the capacitance of a capacitor is inversely proportional to the quantity of charges !
And also why the capacitive reactance is inversely proportional to the capacitance ?

2. ### crutschow Expert

Mar 14, 2008
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3,496
You have it wrong. The value of the capacitance is proportional to the amount of charge stored: C = Q/V.

As the above equation shows, the larger the capacitance, the more charge that is moved for a given voltage (Q=CV), thus the "resistance" to moving that charge (the reactance) is reduced as the capacitance is increased.

Make sense?

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3. ### Unicorn Notreal Thread Starter New Member

Dec 24, 2015
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Yes, sorry I was probably asleep while reading the equation.
But what about the capacitive reactance (Xc=1/2πυC) why is it inversely proportional to the capacitance?

4. ### crutschow Expert

Mar 14, 2008
13,843
3,496
The larger the capacitance, the easier it is to move charge (Q=CV) through the capacitor, so it's reactance (resistance to charge movement) is inversely proportional to capacitance.

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5. ### MrChips Moderator

Oct 2, 2009
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$Reactance = \frac{1}{\omega C}$

$Susceptance =\omega C$

If we ignore the j bit, we can say

$Admittance =\omega C$

Does that make you happy?

As C increases we admit more AC signal.

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6. ### cmartinez AAC Fanatic!

Jan 17, 2007
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Sorry for the dumb question, but what is "the j bit"?

7. ### Papabravo Expert

Feb 24, 2006
10,409
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The "j bit" is a reference to the imaginary unit. Mathematicians use a lower case i, but Electrical Engineers might confuse that with current, so we use the letter j instead.
Reactance is is a real number, but impedance is a complex number.

Also:

$\frac{1}{j \omega C}=-j \frac{1}{\omega C}$

If the square root of -1 gives you a funny feeling, just think of j as a rotation operator.

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