Why does wire not have large capacitive reactance?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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:

^{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".

Answer 5

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.

Answer 6

"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.