AC voltage source in an open circuit

[moderate]

Please see the attached illustration.

This is clearly an open circuit.

The voltage source is producing a voltage at some amplitude and frequency.

If one were to measure with an oscilloscope the potential difference at the point where the mouse pointer is with respect to the ground, what would be observed?

Or, to simplify the question (you don't really even need the resistor there) -- if one were to measure the potential difference between one pole of an AC voltage source and ground, would anything be observed, or would it be a flat line?

-Thanks.

 

[Dale]

You would measure the voltage.

 

[berkeman]

Please see the attached illustration.

This is clearly an open circuit.

The voltage source is producing a voltage at some amplitude and frequency.

If one were to measure with an oscilloscope the potential difference at the point where the mouse pointer is with respect to the ground, what would be observed?

Or, to simplify the question (you don't really even need the resistor there) -- if one were to measure the potential difference between one pole of an AC voltage source and ground, would anything be observed, or would it be a flat line?

-Thanks.

Measuring at the bottom side of the AC source would not show you anything with respect to ground, since there is no current flowing with the open circuit. If you measure at the top of the AC source with respect to ground, you would see the full AC voltage.

BTW, in the figure, you should try to make it a habit to come off of the ground symbol vertically with your wire, at least for a grid point or two. When it comes off horizontally like that, it's not real clear that you've connected up to the ground connection point with the wire. Just a schematic clarity tip.

 

[moderate]

BTW, in the figure, you should try to make it a habit to come off of the ground symbol vertically with your wire, at least for a grid point or two. When it comes off horizontally like that, it's not real clear that you've connected up to the ground connection point with the wire. Just a schematic clarity tip.

I felt that the schematic looked awkward, but decided not to do anything about it Sounds like adding the extra piece would have fixed it.

Measuring at the bottom side of the AC source would not show you anything with respect to ground, since there is no current flowing with the open circuit. If you measure at the top of the AC source with respect to ground, you would see the full AC voltage.

That is what I thought as well.

But, what about this: the AC source is creating a constantly changing voltage.

Consider the rising part of the sinusoidal waveform: the electric potential is becoming larger.
As the potential increases, wouldn't it be creating a flow of electrons towards the AC source (from the distant end of the wire)?

I know that what I wrote above is wrong (based on circuit laws), but can't figure out why.

 

[berkeman]

I felt that the schematic looked awkward, but decided not to do anything about it Sounds like adding the extra piece would have fixed it.



That is what I thought as well.

But, what about this: the AC source is creating a constantly changing voltage.

Consider the rising part of the sinusoidal waveform: the electric potential is becoming larger.
As the potential increases, wouldn't it be creating a flow of electrons towards the AC source (from the distant end of the wire)?

I know that what I wrote above is wrong (based on circuit laws), but can't figure out why.

When the top end of the AC source is open like that (an ideal open with no parasitic capacitance to ground), there is no current flow, so there is no current flow throught the resistor to ground, so there is no voltage at the mouse probe point.

But, add in a small parasitic capacitance of say 1pF from the top AC source point back around to the ground symbol, and things change. Now what do you measure at the mouse probe point?

 

[Toasty2k]

When the top end of the AC source is open like that (an ideal open with no parasitic capacitance to ground), there is no current flow, so there is no current flow throught the resistor to ground, so there is no voltage at the mouse probe point.

But, add in a small parasitic capacitance of say 1pF from the top AC source point back around to the ground symbol, and things change. Now what do you measure at the mouse probe point?

I may be over simplifying, but I believe you're making this problem too difficult. AC vs DC makes no difference in the basic analysis. The meter is attached to an open ckt, but there is a current path through a load resistor over which the full supply voltage will appear. However even if all resistors were removed, you would still read the supply voltage. Why? because your meter is a load, albeit a very high impedance one. Consider testing a dry cell, out of circuit, with a meter. In the original ckt the value of the meter, scope or whatever's impedance is very high, think megohm, relative to the ckt resistances, 100s of ohms) and swamps them. Clearly the reading will not be the full supply value, but very close to it.

 

[bjacoby]

But, what about this: the AC source is creating a constantly changing voltage.

Consider the rising part of the sinusoidal waveform: the electric potential is becoming larger.
As the potential increases, wouldn't it be creating a flow of electrons towards the AC source (from the distant end of the wire)?

I know that what I wrote above is wrong (based on circuit laws), but can't figure out why.

What you have discovered is that "circuit laws" aren't exact "laws" at all but merely an approximation to an actual situation by using some simplifying assumptions. So the "zero current" result is true within the approximate structures of circuit theory but not quite true in reality. In reality one would be better to use a more comprehensive model which would be Field Theory and Maxwell's equations. If you do that you discover that what happens is that when the generator changes voltage there are fields created in the wire to the terminal that "slosh" the electrons just enough so that an excess builds up on the end to counter the fields from the generator. And that creates the potential. So in that case yes indeed, that "slosh" represents a current. But all this is way outside the approximations of circuit theory.

 

[Toasty2k]

What you have discovered is that "circuit laws" aren't exact "laws" at all but merely an approximation to an actual situation by using some simplifying assumptions. So the "zero current" result is true within the approximate structures of circuit theory but not quite true in reality. In reality one would be better to use a more comprehensive model which would be Field Theory and Maxwell's equations. If you do that you discover that what happens is that when the generator changes voltage there are fields created in the wire to the terminal that "slosh" the electrons just enough so that an excess builds up on the end to counter the fields from the generator. And that creates the potential. So in that case yes indeed, that "slosh" represents a current. But all this is way outside the approximations of circuit theory.

True - If you do an AC analysis you can determine instantaneous values of current an voltage across the circuit elements, which will not necessarily be in phase with one another, and indeed if the parasitic inductances and capacitance and their impedances of the real life conductors and components are considered you'll find currents circulating ("sloshing") even in a driven "open" circuit. A high impedance meter (VTVM or FET front end)or O'scope's lead held in free space will indicate an AC voltage relative to ground and circuits don't get much more open than that.

 

[jimmyct]

Im not sure i really understand the question as it just seems so basic. If you measure the voltage, with an oscilloscope, on a controlled voltage source with respect to ground, you should see the same output voltage and frequency. In most circuit sim tools, such as pspice (or cadence), you should be able to do a transient analysis (over say 10 * 1/200khtz) to see around 10 periods, each will have amplitude of whatever was specified. Be careful though as some simulation programs have separate voltage sources for frequency and time domain plots, ie, in pspice/cadence, its vac (freq) vs. vsin (time) and there are even others like vpulse for creating transient. You want vsin.

 

[rcgldr]

You'd probably detect voltage and could draw significant current with a tranformer due to the field. Some low cost electricity grids use a single AC wire, relying on ground for the return path.

wiki_single_wire_return.html

The field is strong enough to liight up flouresent tube stuck in the ground near high voltage lines.

http://www.stopgeek.com/richard-boxs-light-field.html

It's also possible to draw current simply by tapping into a single wire a few feet apart, using a transformer like device. Some electrical companise use a similar device to move along wires to check for issues. If you draw a significant amount of current through a "passive" device, such as a transformer simply placed in the vicinity of the AC wire, the electrical companies will eventually detect this and where it's happening (some farmers have tried this).

 

[moderate]

What you have discovered is that "circuit laws" aren't exact "laws" at all but merely an approximation to an actual situation by using some simplifying assumptions. So the "zero current" result is true within the approximate structures of circuit theory but not quite true in reality. In reality one would be better to use a more comprehensive model which would be Field Theory and Maxwell's equations. If you do that you discover that what happens is that when the generator changes voltage there are fields created in the wire to the terminal that "slosh" the electrons just enough so that an excess builds up on the end to counter the fields from the generator. And that creates the potential. So in that case yes indeed, that "slosh" represents a current. But all this is way outside the approximations of circuit theory.

Thank you for your answers (I did construct the circuit physically, and measured an alternating voltage with an oscilloscope).

It seems that I have placed too much faith in circuit theory. It was good to confirm my suspicions.

My follow-up question then would be (and I don't even have an idea of how I would approach it): how would it be possible to determine how much of the voltage that I am measuring is due to capacitive coupling leading to the current returning to the other terminal and how much of the observed voltage is due simply to the electrons "sloshing" around?

Would it be possible to determine this?

I.e., maybe 90% of the current is the "sloshing" and 10% is capacitive coupling, or something like that.

Thank you again for your input.

 

[berkeman]

Thank you for your answers (I did construct the circuit physically, and measured an alternating voltage with an oscilloscope).

It seems that I have placed too much faith in circuit theory. It was good to confirm my suspicions.

My follow-up question then would be (and I don't even have an idea of how I would approach it): how would it be possible to determine how much of the voltage that I am measuring is due to capacitive coupling leading to the current returning to the other terminal and how much of the observed voltage is due simply to the electrons "sloshing" around?

Would it be possible to determine this?

I.e., maybe 90% of the current is the "sloshing" and 10% is capacitive coupling, or something like that.

Thank you again for your input.

Don't get too crazy with this "sloshing" concept. I believe he is still just referring to parasitic capacitance currents. If not, a more quantitative explanation would be needed.