Capacitive reactance and capacitance

Thread Starter

Unicorn Notreal

Joined Dec 24, 2015
5
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 ?
 

crutschow

Joined Mar 14, 2008
23,349
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 ?
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?
 

Thread Starter

Unicorn Notreal

Joined Dec 24, 2015
5
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?
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?
 

crutschow

Joined Mar 14, 2008
23,349
Read my second paragraph again.
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.
 

MrChips

Joined Oct 2, 2009
19,277
\(Reactance = \frac{1}{\omega C}\)

\(Susceptance =\omega C\)

If we ignore the j bit, we can say

\(Admittance =\omega C\)

Hence admittance increases with C.
Does that make you happy?

As C increases we admit more AC signal.
 

cmartinez

Joined Jan 17, 2007
6,464
\(Reactance = \frac{1}{\omega C}\)

\(Susceptance =\omega C\)

If we ignore the j bit, we can say

\(Admittance =\omega C\)

Hence admittance increases with C.
Does that make you happy?

As C increases we admit more AC signal.
Sorry for the dumb question, but what is "the j bit"?
 

Papabravo

Joined Feb 24, 2006
12,405
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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