ECN Forum Forums General Topics of the Trade General Discussion Area capacitive coupling

Capacitive Coupling in 14 words:

Current in wire 1 causes magnetic field. Magnetic field causes current in wire 2.

Actually the wires can also "transfer" current because of the charge on one wire affecting the charge on the other (the like charges repel and opposite charges attract thing). It's probably both the magnet field thing and electrical charge thing that makes it happen but you asked for brevity.

And the charge that "accumulates" on the bus bar isn't capacitive "coupling" although it does happen because the system acts like a big capacitor. The grounded set of wires acts like one capacitor plate and the hot wires running along side those act like the other capacitor plate.

I like your questions. What's your story? What do you do? Why all the questions?

[This message has been edited by maintenanceguy (edited 12-27-2001).]

I thought that was induction.
Don(resqcapt19)

i've been an electrician for several years now but realize i'll always be learning or relearning, just when you think you know something, poof, gone or relearned differently.

anyway what are we calling the force that "transforms" energy from the primary to the secondary? isnt it induction?

See what I mean ? There is very little PRACTICAL difference between capacitive and inductive coupling ! Hard to tell the difference for the field electrician, and they both produce nearly the same results, unless we get into an RF discussion, and that means communications (usually wireless) and scif rated computer rooms.

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
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...

Paul,

Excellent post!!! Very well structured.

Scott SET

Thank you (Blush!)

I'm just visiting old ground here. I did a couple of articles on inductive and capacitive reactance, resonance, etc. for Popular Electronics a few years ago.

Cindy,
I nearly forgot this: You were working on an assumption that a lower capacitive reactance might always be better.

In the words of an old song, "It ain't necessarily so."

In the case of leakage current from hot ground in a long feeder causing GFI trips, you would want to aim for a lower capacitance (i.e. a higher reactance) to reduce that unwanted current.

That's not to say that the higher capacitance might not have some beneficial effects, such as helping to filter out harmonics.

In some cases you're not aiming for reactance as high or as low as possible, but for a certain optimum value. The capacitor in a split-phase motor is a good example.


[This message has been edited by pauluk (edited 12-27-2001).]

seriously dont worry about answering this if its too much, i copied all of this to a word document and tried to sort out mostly scotts and pauls replies, but if you dont answer anything else, please tell me how the electricians in the UK refer to 1/2" EMT or 3/4" flex, for instance if i asked them for
a stick of 1/2" EMT would they say: 'eres yer 13-millimeter EMT matey also think i've heard 1/2" rigid referred to as 16 metric maybe, not 13?

You said this: Capacitive Coupling effect is present whenever there's current flowing, or available to flow on circuits, and there's any difference in Potential - either externally or internally to the conductors.

Questions: Do you have examples of the difference in potential "externally"? Vs "internally"?

You said this: Let's say you have a length of feeder with a measured capacitance between hot & neutral of 0.5uF (microfarads). 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.

Questions: What do you mean “that reactance is directly across the supply”? Potential difference? Electric field? Does the “line charging” ‘couple’ hot to hot, hot to neutral, hot to ground, and neutral to ground? Does capacitive coupling result in reactance and current flow only between those conductors coupled to each other? In other words, lines A and B are coupled, but lines B and C are not close enough to couple, so how does the capacitive coupling current flow in line A or line B relate to line C? If line B touches line C does it discharge? It seems like we’re talking about 2 effects, capacitive coupling results in current flow, and capacitive coupling results in reactance that inhibits current flow.

You said this: 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.
Also you said: 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.

Questions: What is the difference? Is reactive current like a different sine wave, different phase, or beyond my comprehension at this point?

Questions: In the example there is 2.66 megohms capacitive reactance in the light switch legs tested with a digital meter. Why is there 10 megohms of resistance in the meter? Why can you use 120v in the I=ExR formula since we started with an open circuit without voltage on it?

You said this: In some cases you're not aiming for reactance as high or as low as possible, but for a certain optimum value. The capacitor in a split-phase motor is a good example.

Question: Is there a short explanation for this, or is that another thread?
thanks -C-

Wow, that's some batch of questions!

Q. Reactance across supply, etc.

In the case of the long feeder, it just means that the 0.5uF capacitance is in parallel with the supply.

Have you studied the construction of a basic capacitor? You can make a very crude one by taking a couple of strips of aluminum kitchen foil, sandwiching a slightly wider strip of thin insulating material between them and on top, then rolling the whole lot into a cylinder with one wire connected to each foil. In earlier times capacitors were often just that, with waxed paper as the insulation.

The point is that it is nothing more than two strips of metal held a short distance apart by an insulating dielectric. If you took a 0.5uF capacitor and connected it between a hot breaker terminal and neutral busbar at a panel, you would get 22mA of current flowing through that capacitor.

A cable is also two long strips of metal with insulating material between them, so if you make it long enough you can get the same 0.5uF (or more). The capacitor just packs it into a much smaller space by deliberately minimizing the gap between the conductors and getting the most surface area in the smallest possible space.

So, the 0.5uF capacitor above has one foil connected to the hot and the other to the neutral. The two wires in the long feeder cable are also connected one to hot, one to neutral, so it's a parallel capacitance.

You'll get such a capacitance between any two such conductors which are close enough. If you took a length of, say, 3-wire armored cable, there would be a measurable capacitance line A to line B, B to C, and A to C. There would also be capacitance between each of those lines and the grounded armor.

The longer the cable, the higher the capacitance will be. The greater the spacing between conductors, the lower the capacitance.

Regarding your two effects: In the very broadest of terms, just placing capacitance in parallel with the supply will cause additional current to flow, while adding capacitance in series with some other device will increase the overall impedance and reduce the current flow, so you're on the right track.

HOWEVER, this is not necessarily the case where the rest of the circuit contains inductance (e.g. a motor, fluorescent lamp ballast etc.). The explanation for that had better wait until later.

Q. Difference between reactive/resistive ohms, why we can't add them directly.

Basically because there is a phase difference between the resistive and the capacitive portion of the circuit. In this case of the meter, it is a series circuit. The voltage across the resistive section (the meter) will be in phase with the current, but the voltage across the capacitive section will lag the current by 90 degrees.

Q. Why 10 meg input on a DMM.

That's the way the circuitry is designed. High-impedance meters are needed for some sensitive electronics to avoid loading effects, and that's the field in which the DMM first appeared. 10M has just become a common design value, although there are others.

Q. Why did we use 120V on the I=E/R (or more precisely, I=E/Z).

At first look you might think that the meter started on an open circuit, but we've already seen that that "open" conductor was really capacitively coupled to the energized hot wire.

The impedance (Z) calculated was that of the meter resistance (R) and the cable reactance (X) in series, so we need to use the voltage appearing across the whole series circuit. That's from the energized hot wire to neutral, i.e. 120V.

Q. Optimum value of reactance.

I'd better not get into too long an explanation, but the power-factor correction capacitor would be a good example.

Inductance (e.g. a fluoro. ballast) causes a phase shift between current and voltage on the supply which is undesirable (i.e. the power factor is less than 1).

Fitting a p.f. capacitor of the correct value can bring I and E back in phase (or very nearly so). Too small and the phase shift will not be enough to do that; too big and you'll "overshoot" and take it out of phase again in the other direction.

Notes on conduit later... Must eat!

Cindy,

If you went into a store here and asked for any size EMT, you'd be more likely to get a reply like "EMT guv? What's that then?"

Just another of those American things which hardly anybody here knows about, along with Edsels and Gilligan's Island!

We don't have an EMT equivalent. There was some thinwall conduit used way back pre-war, but it hasn't been made for decades.

Heavy-gauge steel conduit is available in galvanized and black enamel finishes. The smallest sizes are 20 and 25mm, which replaced the old 3/4 and 1" types.

I've never heard of a 16mm gauge, though it might exist in some other country.

We do have 15 mm copper water pipe though. It replaced the 1/2-inch type in the 1970s, but is almost exactly the same size, because the 1/2-inch was measured as the bore and the 15mm is outside diameter.