# Capacitance Impedes Circuit Current But THe More There Is The Less Impedance

#### Jsw123

Joined Jun 20, 2009
46
How could it be explained that capacitance creates capacitive reactance, which impedes current flow and would restrict current in a circuit beyond that which the resistance would do, yet the more capacitance there is in a circuit the lower the capacitive reactance?

#### ZCochran98

Joined Jul 24, 2018
221
There's the mathematical way to go about this, and then a semi-"intuitive" explanation. The semi-"intuitive" explanation (that's mostly accurate) relies on remembering how resistance (regular, boring resistance) works. The larger your conduction area (i.e.: the more surface area per unit length of conductor), the more current can be passed, which corresponds to a lower resistance. For instance, a 1m wire with 1 square mm of cross-sectional area has higher resistance than a 1m wire with 10 square mm, as there is less space to carry charge, so it takes more voltage to get the same current. Similarly, you can think of a capacitor: a larger capacitor is equivalent (effectively) to larger parallel plates. Thus, they can carry more charge for a given voltage than a smaller capacitor. If they can hold more charge, then they can carry more current [keep in mind: this is the very "hand-wavy" explanation]. So larger plates correspond to more current, which corresponds to lower impedance (because decreasing impedance increases current per voltage).

If you don't want the mathematical explanation of the same thing, stop reading here.

.
.
.

The mathematical explanation repeats the "hand-wavy" explanation, but with more rigor. I'll give it for completeness' sake.
The equation for a capacitor is defined as follows:
$C = \frac{q}{v}\rightarrow q = Cv$
Here, $$v$$ is charge. Lowercase letters are used to indicate time-dependent variables (capital letters will be for frequency-dependent or just constant). This equation is equivalent to my statement that the larger plates mean more charge can exist for a given voltage.
Taking the derivative (and assuming constant capacitance) w.r.t time, we get:
$\frac{dq}{dt} = C\frac{dv}{dt}$
The derivative was don because the term $$dq/dt$$ is the definition of current, so we have:
$i = C\frac{dv}{dt}$
This kind of leads into the idea of impedance (or, rather, admittance, because it looks vaguely similar to the conduction version of Ohm's law, $$I = GV$$, where $$G = R^{-1}$$). However, it's got a derivative in the time domain, which suggests that the frequency domain might be more "pleasant" to work with. So, to convert from time-dependence (with a derivative) to frequency-dependence (without a derivative), we take the Laplace transform (and ignoring initial conditions):
$\mathcal{L}\{i\} = C\mathcal{L}\left\{\frac{dv}{dt}\right\}$
$I = CVs$
The value $$s$$ is the frequency element, $$s = j\omega\ = j2\pi f$$. The Laplace transform basically asks, similarly to the Fourier transform, "how can we write something in the time domain as a bunch of sinusoids?"
If we rearrange the equation, we get:
$\frac{V}{I} = \frac{1}{Cs}$
The left term looks like Ohm's law, so we'll call it "impedance," specifically, capacitive reactance, $$X_c$$. Thus, we can write it:
$X_c = \frac{1}{Cs} = \frac{1}{j2\pi fC} = \frac{-j}{2\pi fC}$
Frequently, the $$-j$$ is replaced with $$1$$ because we often just care about the magnitude of the capacitive reactance.
This is the mathematical explanation of the semi-"intuitive" explanation I gave before.

#### Jsw123

Joined Jun 20, 2009
46
I understand well the theory of bigger surface area meaning more capacitance. It just doesn't make sense why it would be that if there were no capacitance in an AC series circuit, the resistance alone woud impede current. If capacitance were added, it would also impede current. However, the more of it there was the less it would add to the overall circuit impedance.

#### Papabravo

Joined Feb 24, 2006
19,575
I understand well the theory of bigger surface area meaning more capacitance. It just doesn't make sense why it would be that if there were no capacitance in an AC series circuit, the resistance alone woud impede current. If capacitance were added, it would also impede current. However, the more of it there was the less it would add to the overall circuit impedance.
Capacitance is not the only factor at play here. A resistor is not affected by the frequency of the current or voltage. But the reactance certainly is. The reactance is infinite at zero frequency regardless of the capacitance, effectively acting like an open circuit. That is whay in some applications it is referred to as a "blocking capacitor".
Sorry it doesn't make sense. You can tak it on faith or you can take a deep dive into the weeds. Your choice.

#### ZCochran98

Joined Jul 24, 2018
221
The thing to remember is that capacitance is a DC block, so it's more willing to impede low frequencies than high. This is also kind of (sort of) related to why higher capacitance makes the overall capacitive reactance contribution to impedance smaller. It really goes back to the amount of charge/current at play in the circuit (as I tried to explain in my first post), as that's how impedance is defined.

I apologize, but there's really not much other or better explanation I can offer without, as @Papabravo put it, "taking a deep dive into the weeds." Maybe someone else may have some better insight.

#### Papabravo

Joined Feb 24, 2006
19,575
The thing to remember is that capacitance is a DC block, so it's more willing to impede low frequencies than high. This is also kind of (sort of) related to why higher capacitance makes the overall capacitive reactance contribution to impedance smaller. It really goes back to the amount of charge/current at play in the circuit (as I tried to explain in my first post), as that's how impedance is defined.

I apologize, but there's really not much other or better explanation I can offer without, as @Papabravo put it, "taking a deep dive into the weeds." Maybe someone else may have some better insight.
A masterful job of expository writing.

#### BobTPH

Joined Jun 5, 2013
6,076
Your mistake is in thinking that capacitance is analogous to resistance. It is not, it is analogous to conductivity, which is the inverse of resistance. More conductivity or more capacitance means more current flows.

Bob

#### nsaspook

Joined Aug 27, 2009
10,674