When current flowing in wire #1 creates a magnetic field which then generates a current in wire #2, that's inductive coupling. It's the principle utilized by every transformer.

Capacitive coupling is concerned with electric fields rather than magnetic ones. In practice, when wires are close enough to provide inductive coupling they're also going to be close enough for some capacitive coupling, so there will always be some capacitance between the primary and secondary windings of a transformer, for example.

Perhaps a couple of worked examples might help. Let's say you have a length of feeder with a measured capacitance between hot & neutral of 0.5uF (microfarads). Such a value is not hard to achieve with a moderately long cable.

Assuming a U.S. supply of 120V at 60Hz, the reactance will be

X = 1 / (2 x 3.142 x 60 x 0.0000005)
= 5300 ohms approx.

That reactance is directly across the supply, so (ignoring the series resistance of the wiring, which is negligible by comparison) there will be a capacitive current of

I = 120V / 5300 ohms = 0.0226 A

True enough, 22mA is hardly going to be a problem compared to the probable load on that feeder, and because it is reactive current it isn't actually consuming any power anyway.

BUT -- Consider that you could just as easily have 0.5uF between hot & ground.
See why trying to feed this with a sensitive GFI would give you a problem?

Now an example of series reactance. Assume power is going to a light fixture with a 2-wire loop from there to the switch. With the switch off and bulb removed, the hot wire from switch to light is effectively diconnected from the system. If you took a voltmeter and measured from that wire to ground (or neutral), you would expect a zero reading, right? This is where using a modern high-impedance digital meter can cause confusion.

Let's say the capacitance between the two wires of the switch loop is 0.001uF - Again, not too difficult to achieve.

The capacitive reactance is

X = 1 / (2 x 3.142 x 60 x 0.000000001)
= 2.66 megohms

Hey, that's pretty high isn't it? Surely that won't have much effect on anything?

Well, the input of a typical DMM is 10 megohms. So in effect we have 2.66 meg of capacitive reactance in series with a 10 meg resistance (the meter will actually have a small amount of capacitance of its own across that 10M,
but it's small so we'll ignore it for our present purpose - I'm trying to keep it as simple as possible).

OK, now although reactance and resistance are both measured in ohms, we can't just add them together because the voltages across them are not in phase. So we have to use Pythagoras, with impedance (Z) as the hypotenuse of a right-triangle:

Z = SQRT ( R^2 + X^2 )

So

Z = SQRT ( 10^2 + 2.66^2 ) = 10.35 Meg

Stick wih me, it gets eesier now [Linked Image]
Here's the simple bit:

I = 120V / 10.35 megohms = 11.59 uA.

Only 11.59 microamps going through the meter. Now work out the voltage across the meter:

V = 11.59uA x 10 megohms = 115.9 volts.

Say what??!! Nearly 116V to ground on a conductor that isn't connected to a power source? Yep, that's right, and it's all because of the capacitive coupling.

In practice, the capacitance of the meter would pull that figure down a little, and there would be other things to consider; e.g. If that switch cable had a ground wire then the capacitance between the open conductor & ground would have a considerable effect. But you get the idea.

As you observed, Cindy, at higher frequencies the reactance is lower for any given value of capacitance. When we get to talking about UHF TV or SHF satellite frequencies, moving a wire by a quarter-inch can somtimes mean the difference between perfect and nothing.

OK, you can all wake up at the back there now... [Linked Image]