# Battery, AC, DC and unconnected terminals....

Discussion in 'General Electronics Chat' started by antennaboy, Mar 10, 2013.

1. ### antennaboy Thread Starter Active Member

Jan 31, 2008
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Hello Forum,

If we take a 12V battery and connect only one terminal to ground (either the + or the -) and the other to a light bulb, nothing will happen. Why?

1) The circuit is open and DC current needs a closer metallic path to flow.

2) The terminal that gets connected to ground assumes the same potential as ground.

3) To make current flow we need a potential difference.

4) Once we connect the battery terminal to ground, a fast current will take plance that will transfer electric charge around and make the battery terminal+ wire+ ground at the same potential. So there is a transient situation that leads to an electrostatic situation eventually.
It is true that initially there is a potential difference between the battery terminal and ground but it cannot be sustained.

5) There is a potential difference between the battery terminals:12V. But we don't really know what the potential of each terminal is. Same goes for the potential of ground before electrostatic equilibrium is reached.
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AC case: AC current does not need a close path to flow. Circuits involving antennas are an example: they are open circuits where AC current flows back and forth.

Now, if we connect one terminal of an AC battery to a light bulb and we don't connect the other terminal to anything (just there in the air), will current flow through the light bulb or not? I don't think so. There will be a fast current that disappears super fast once we connect the light bulb until electrostatic equilibrium is reached. The terminal+wire+light bulb will becomes an equipotential conductor that changes with time..

At the same time, I feel like when an antenna is connected, AC current flows up and down the antenna arms....That would lead me to believe that the light bulb connected to a single terminal of a AC voltage source would get illuminated....

Thanks,
Antennaboy

2. ### crutschow Expert

Mar 14, 2008
12,565
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All AC voltages will radiate some energy but at the low frequency of a power line it is negligible. For the radiated power to be significant, the antenna equivalent length needs to be at least a quarter wavelength of the frequency (which is over 700 miles at power line frequencies)

There will also be a tiny current due to the capacitance between the AC terminals (or one terminal and ground if one side is grounded), but again that's a very tiny current at low frequencies, certainly not anywhere near enough to light a bulb.

3. ### MrChips Moderator

Oct 2, 2009
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Having potential difference alone is not sufficient to make a light bulb turn on. You also need current. To have sufficient current, the total resistance around the current loop must low enough to produce the desired current. Connecting two rods into the ground does not create a low enough resistance for appreciable current to flow. A light bulb requires a low resistance path around the loop for low frequency AC voltage to be able to turn on the light bulb. Radio frequency is a different story.

4. ### antennaboy Thread Starter Active Member

Jan 31, 2008
45
0
thanks!

Crutschow:

There will also be a tiny current due to the capacitance between the AC terminals (or one terminal and ground if one side is grounded), but again that's a very tiny current at low frequencies, certainly not anywhere near enough to light a bulb.

Is the small current you are referring the displacement current? That is not a real conduction current (electrons moving). It is simply a time rate of change of electric field. It is improperly called a current....

Mr Chips:
If there is current going back and forth on one of the wires attached to one of the AC source terminals and a light bulb is connected too, wouldn't that current flow through the bulb as well and brighten it up?

This is the set up: AC source, straight wire attached to + terminal, light bulb terminal attached to the straight wire. One of the light bulb terminals is left unconnected...

thanks,
antennaboy

5. ### crutschow Expert

Mar 14, 2008
12,565
3,079
No, I'm taking about an actual current due to the small capacitance between any terminal or wire and ground.

The current is much too small to light any practical lamp.

6. ### WBahn Moderator

Mar 31, 2012
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There are a few issues at play here that are not being properly distinguished.

There are a few mechanisms by which electromagnetic energy can traverse a non-conducting gap, for lack of a better catch-all term.

A circuit, usually by way of an antenna, can radiate coupled self-sustaining alternating electric and magnetic fields that carry energy away from the source never to return. The currents involved remain local to the circuit, but the energy is transmitted away.

A circuit can have an alternativing current that travels through a capacitive gap in which case the current on one side of the gap is the result of the changing electric fields in the gap.

A circuit can have an alternativing current that is inductively coupled across a gap in which case the current on one side of the gap is the result of the changing magnetic fields in the gap.

You most definitely can get a light to light up without it being physically connected to anything. The classic example is a fluorescent tube held near a tesla coil.

7. ### antennaboy Thread Starter Active Member

Jan 31, 2008
45
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Ok,

As you are saying, there is no connection between the AC source terminal and ground so no electrical, conduction current will flow.
This must be a displacement current, like the one between the plates of a capacitors....

I agree that the current would be too small....

Again, neither of the AC source terminals not grounded. This is a thought experiment. Imaging the AC source sitting on an insulated table. One terminal (say the negative) is connected to a wire. The other terminal (+) is connected to a straight wire which is connected to one of the terminals of an extremely low resistance light bulb....

Capacitance develops whenever there are two metal conductors, charged, within a finite distance from each other and separated by a dielectric.

thanks,
antennaboy

8. ### praondevou AAC Fanatic!

Jul 9, 2011
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If you take a small light bulb and put it in series with a whip antenna connected to a transmitter it will light up. You can even use this method to adjust the antenna length or transmitter frequency on small DIY HF oscillators.

It depends mainly on your frequency. A 1m wire is not an appropriate antenna for a 60Hz AC voltage, i.e. a lamp connected to only one side will NOT light up.

EDIT: You could also see it like this: If no energy is transmitted from an antenna because it has the wrong impedance then there will be no current through the bulb.

And also you seem to believe there is AC current through a wire that is only connected to one side of an AC voltage source. There is not. The current will only be there if you allow energy to be transmitted via an antenna or close the loop via a conductor. Open the circuit breaker in your home , put a current clamp on the incoming wire and see if you can measure a current. You can not.

Last edited: Mar 10, 2013
9. ### WBahn Moderator

Mar 31, 2012
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4,701
The point you are overlooking is the magnitude of the currents involved.

A typical bulb has somewhere around 250mA to 500mA of current (whether we're talking about a 60W bulb running from 120VAC or a 3V bulb in a flashlight). How much current would it take to see the bulb "brighten up"? Let's be WAY overoptimistic and say that we could see a difference if there were just 1mA of current flowing.

Now let's say that your unconnected terminal was actually connected to a plate of metal that had an area of one square meter (a REALLY big plate) and the negative terminal of the battery was also connected to a similar plate and the plates are positioned parallel to each other a distance 10cm apart. How much AC voltage do we need across the gap at 60Hz to get 1mA of AC current flowing?

Our capacitor At 60Hz, the capacitance would be about

C = εA/d = (8.85x10^-12F/m)*(1m^2)/(0.1m) ~= 100pF

At 60Hz, the capacitive reactance is

Zc = 1/(2∏fC) ~= 27MΩ

So to get 1mA of current, you would need 27kV across the gap.

Now consider that the capacitance of the setup described is orders of magnitude higher than the unconnected terminal situation you are describing and you can appreciate the different scales you are talking about when talking about the tiny, minute transient charge movement needed to establish electrostatic equilibrium and the current needed to light a light bulb.

10. ### WBahn Moderator

Mar 31, 2012
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Capacitance exists whether the conductors are charged or not. Also, there is no requirment that the electrodes be conductors. Capacitance is simple the ratio of the voltage that develops when a charge separation occurs to the amount of charge that is separated. With insulating electrodes things get messy because the voltage isn't necessarily uniform across the gap and making it part of a circuit can be tricky, but it is still a capacitor.

11. ### antennaboy Thread Starter Active Member

Jan 31, 2008
45
0
Sure,

I think I am with you about the magnitudes of these current being super low, almost negligible....It all strongly depends on the frequency of the AC source (and other parameters).

But in principles, AC current does not need a closed conductive circuit to flow and in the case of an idealized light bulb, if we used the right frequency and voltage AC source,
it seems we could lighten that bulb up?

Why does it light up? Because there is a current through it? Why is there a current? Because there is a potential difference along that single terminal+light bulb.
This does not occur in the case of a DC source....

Clearly, circuits involving antennas have decent currents along the terminals (which represent the antenna arms).

The other point is that, if the frequency is right, it seems that engaging a single terminal of the AC source would be enough. The other terminal could be isolated, encapsulated....
I know it is not true but I can't come up with a clean physics explanation in terms of charge behavior. Capacitance and inductance are averaged, large scale concepts.

For those interested, I found this explanation online, invoking capacitance and inductance to explain things:

http://www.hottconsultants.com/pdf_files/dipoles-1.pdf
Thanks,
Antennaboy

12. ### antennaboy Thread Starter Active Member

Jan 31, 2008
45
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Given a simple dipole antenna, the current on each arm has a magnitude that is varying with position along the arms.

Current is the time derivative of charge.

So it is easy to see and imagine how current and charge is distributed along the antenna arms. How about voltage?
Does anyone know how voltage changes along the arms? Does it change from point to point too?

thanks
antennaboy

Jan 31, 2008
45
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14. ### WBahn Moderator

Mar 31, 2012
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That's what we've been saying! Did you not see where I pointed out the example of a fluorescent tube and a tesla coil?

Which is what we've been saying!

15. ### antennaboy Thread Starter Active Member

Jan 31, 2008
45
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Ok, sorry, I am kinda thick today.

How about the other terminal, not connected to the light bulb? Is it necessary? I think so.
what would happen if we completely covered it with dielectric material?