3
\$\begingroup\$

For capacitive reactance, I am presented this formula: $$X_C = -\frac{1}{2\pi fC} $$ Where clearly, the capacitive reactance is inversely proportional to the capacitance.

This confuses me.

We usually talk about capacitive reactance in relation to capacitors, which have a rather high capacitance. From this formula, I would expect their reactive capacitance to be small, and the reactive capacitance of elements with low capacitance to become very high. That is, a simple wire should always have a much higher reactive capacitance than a capacitor.

What am I getting wrong here? Does this formula only apply to the circuit-elements we define as capacitors? If so, what makes them so special that we can apply this formula to them, but not to other elements which also have a (very small) capacitance?

\$\endgroup\$
  • \$\begingroup\$ A wire isn't a capacitor. A capacitor has two conductors. Wire has one. \$\endgroup\$ – JRE Aug 5 at 7:15
  • 1
    \$\begingroup\$ It's right. The problem is that your brain is off on a tangent. Suppose there is no capacitance between two wires? This means there is "no connection at all." So that's the same thing as infinite impedance. Which is what you get from the formula if you plug in zero capacitance. That's a good thing. \$\endgroup\$ – jonk Aug 5 at 7:15
  • \$\begingroup\$ Are you expecting the opposite results from Xl ? \$\endgroup\$ – user105652 Aug 5 at 7:21
  • \$\begingroup\$ @jonk that clarified a confusion of me. I started to misinterpret the reactance as "a current flowing back" (i guess from the name, as well as reading about how reactance can come about). Therefore, i was assuming that for this formula to hold, an infinite current would have to flow back between the two wires you mentioned. Interpreting reactance analogous to resistance, giving infinite resistance between the wires of course makes more sense in this scenario than an infinite opposed current. \$\endgroup\$ – LeonTheProfessional Aug 5 at 7:21
  • 2
    \$\begingroup\$ @LeonTheProfessional Yeah. This "reactance" is to be treated similarly to "resistance." There is another component, phase, that rises up. But for now, you can just ignore it. The main idea is that a capacitor can be treated similarly to a resistor and that its reactance is similar to resistance. The difference being that a capacitor's reactance is frequency-dependent while a resistor is the same resistance no matter the frequency. \$\endgroup\$ – jonk Aug 5 at 7:36
6
\$\begingroup\$

The same formula applies. Wires have low capacitance to their surroundings so they have high Xc. Capacitors have high capacitance in comparison so they have low Xc. Thus the wires are a smaller AC load (less current flows) and capacitors are a larger AC load (more current flows).

It is no different from resistance, high resistance means small load and small current, low resistance means high load and high current.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ You are talking about two unconnected wires (for example in parallel) and the current flowing between them, aren't you? Because i would assume that a single wire should still have a low Xc in relation to itself, and nearly optimal current flow. \$\endgroup\$ – LeonTheProfessional Aug 5 at 7:29
  • \$\begingroup\$ Capacitance is always between two conductors. Yes I was talking about capacitance between two wires, but even if there is a single wire held in free air, it will have capacitive coupling to surroundings, like earth or humans, so it will have some femtofarads of capacitance. If you wind up a wire into a coil, there will be capacitance between the turns of the same wire too. For a straight piece of wire, the capacitance of the wire to itself is pretty small, so it will have large Xc. \$\endgroup\$ – Justme Aug 5 at 7:50
3
\$\begingroup\$

A straight wire will, as you say, have a very low capacitance and therefore a very high capacitive reactance (dependent on frequency). What you may be overlooking is that the capacitive reactance is only one part of the wire's impedance (R + jX), the other parts being inductive reactance and resistance. As the resistance in a wire will be significantly lower than the reactance components and also not frequency dependent it dominates the overall impedance as the capacitive reactance is effectively in parallel with the wire's resistance component not in series with it.

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

There's a trick for making a low capacitance, high reactance, capacitor: just twist two wires together. These "gimmick" capacitors were perhaps more common in the past, but may still be found in the wild. So, yes, wires have capacitance to other conductors.

| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

The formula features \$C\$ in the \$1/C\$ reciprocal configuration because in fact the impedance of a capacitor to the flow of AC current in fact decreases with increased capacitance.

For instance, a 1 \$nF\$ capacitor will not pass 60Hz AC very well at all; it appears nearly an open circuit to that frequency. A 100 \$\mu F\$ capacitor, much larger, passes 60HZ AC much better.

Now a stretch of copper wire conducts; it passes not only very low frequencies very well, but even DC. So does that mean that, paradoxically, a wire is a very large capacitor? No, it doesn't. A piece of wire has almost no capacitance at all; because it conducts, it cannot spontaneously maintain a separation of charges. Wire conducts very well because it has low resistance.

We can model a wire and a capacitor it like this:

schematic

simulate this circuit – Schematic created using CircuitLab

That is to say, a wire or a capacitor can both be be modeled as a resistance in parallel with a capacitance. (If we presently cared about inductance, we would add that also, and for completeness we would include an inductor model.)

A wire has a very low capacitance, which means that the C branch of the circuit has a very high \$X_C\$ and therefore very high impedance. Therefore all the current flows because of the low resistance; effectively, the tiny capacitance is shorted out by the low resistance.

A capacitor is basically an open circuit; it blocks the flow of DC, except for some small leakage. That leakage can be modeled as a large resistor shunting an ideal capacitance.

Parallel impedances sum using the \${1/Z_{\text total} = 1/Z_1 + 1/Z_2 + ... + 1/Z_n}\$ formula. Wire has a very low resistive Z, and so that vastly dominates over its huge capacitive Z. A capacitor has a relatively low capacitive Z, which dominates over its huge resistive Z.

So in the diagram on the left, the overall Z is that of the R branch, and in the diagram on the right, the overall Z is that of the C branch.

In other words, we can explain everything using the principle "electricity takes the path of least resistance impedance".

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

Examine the capacitance of Twisted Pairs (2 wires) versus capacitance of a Coaxial Cable (a wire inside a tube).

A wire does have very high capacitance reactance as it interacts with stored charges in its surroundings.

As you try to send a signal from A to B, a wire in isolation requires the lowest amount of charge.

On the other hand, a wire over a sheet of metal (a plane), requires MORE charge but will be faster.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

"Just a wire", and for that matter the "textbook electrical circuit", only exists in "electrically short" setups where the longest wire is shorter than the wavelength of the highest frequency component in use by an order of magnitude or more. Beyond that, you have to always think of pairs of wires as transmission lines - loose wires will become part of a transmission line with whatever other wire carries the return current. If the shape/layout of the resulting transmission line is wack, the whole setup will behave like a transmission line that is wack. Long overland AC lines are not just called "transmission lines" because they transmit electricity on a line, they actually behave with all the RF magic in play - a line of several hundred miles is no longer truly electrically short at 60Hz.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.