Beating a dead horse... the necessity of complete circuits

Discussion in 'General Electronics Chat' started by vpoko, Oct 9, 2013.

Jan 5, 2012
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I've had this explained to me before, and it just hasn't clicked. I'm hoping maybe another wording might do it.

Why is a complete circuit necessary, at the scale/from the point of view of charge carrier? Of course I know that if you connect a wire between the positive terminal of one battery to the negative terminal of another, no charge will flow (unless the batteries' other terminals are connected to each other as well).

If I'm an electron on the negative side of one battery, it's crowded; there are lots of electrons on that side of the battery (call it the source battery). Now I'm given an escape path to the other battery's positive terminal (call it the target battery). That (the positive) side of the target battery has a shortage of electrons. Why can't I take the path unless other electrons from the target battery are given a path to go to the source battery? Why does it have to be an exchange? I understand that Kirchhoff's laws exist, but why? What force acts on the electron to keep it from migrating through a conductor unless there's also a second conductor to the other terminals?

2. crutschow Expert

Mar 14, 2008
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A few of them may migrate to the other terminal due to the stray capacitance between the two batteries and ground. But after the charge is equalized on this capacitance as determined by the batteries' emf, there is no path for further travel of the electrons. Any more electrons that move from one battery to the other need an ohmic return path. For a simplified analogy think of water being pumped by the pump pressure (emf) through a pipe from one closed tank to another. If you don't have a return pipe, the tank pressure builds up until it reaches the pump pressure and no more water can flow.

Make sense?

RamaD, vpoko and Brownout like this.

Jan 5, 2012
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Sort of, but here's where my confusion arises: the battery isn't a like a tank with a single chamber and equal pressure throughout. The negative side of the battery has a high concentration of electrons, and the positive side has a low concentration. So if we think of the battery as a container with two chambers, one filled with water, and the other empty (or with only a little bit of water), then shouldn't I be able to pump more water into the empty (positive) half without worrying how full the negative half is?

4. Brownout Well-Known Member

Jan 10, 2012
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If charge was allowed to flow into the positive side of a battery, that would change the battery's voltage. But the voltage of a battery must remain constant (until charge is allowed to flow from the negative to the positive terminal, discharging the battery) In order for the voltage to remain constant, some charge must flow out of the negative side. So, the battery is like the full tank in the analogy. In order to put something in, you must take something out.

vpoko likes this.

Jan 5, 2012
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Wait a sec, or do the electrons from the negative half still influence how "full" the positive half is because they have fields surrounding them that repel other electrons?

Jan 5, 2012
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Thank you both, I think I can feel some gears turning in my brain.

7. nsaspook AAC Fanatic!

Aug 27, 2009
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It's not a force that keeps it from moving in a open circuit, it's a lack of a (net) force. In a perfect conductor the free electrons will move to neutralize the electrical force from one end of the conductor to the other resulting in a zero net electrical force within the wire over time creating current flow to keep the wire neutral. If there is no difference in electrical force from one end of the wire to the other (after than an initial electron movement at the instant of connection to equalize charge across the wire) there is no energy to cause continuous (non-random) motion.

A battery does not get more charge or lose charge, there is a fixed amount of charge (ions) that are separated by the chemical reaction into the anode and cathode plates creating an electrical field that causes the electrons in the conductors to move.

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8. crutschow Expert

Mar 14, 2008
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OK. Let me modify the analogy. That tank capacities are small. It's the pump energy that determines how much water can flow. If there is not a return pipe, the amount that can flow is small. With a return pipe the amount of water that can flow is limited only by energy of the pump.

Likewise, the amount of electrons that can flow to fill up the battery capacitance is very small compared to the number of electrons that flow in a steady current. Understand that the amount of electrons that can flow out of a battery in a compete circuit is much greater than the number of electrons that are in excess in a static condition. In a normal battery operation the current keeps flowing until the chemical energy that generates the emf is used up. That's many orders of magnitude greater than the small amount of excess electrons that are at the negative terminal of a static battery.

Last edited: Oct 9, 2013
vpoko likes this.

Jan 5, 2012
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Thank you, this was the key sentence for me! I think everyone else who has tried to explain it to me (in this thread and elsewhere) said something similar, but now it makes sense.

Jan 5, 2012
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Yes, that also did it! I'm so happy to finally understand this concept that's been bugging me for so long that I could jump out of my chair. Thanks!

11. ErnieM AAC Fanatic!

Apr 24, 2011
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Electrons don't bunch up!

Take a battery and some wire and a light bulb, but do not connect them. Count all the electrons and make a map.

Now connect the light and the battery with the wire. Watch the bulb glow and map the electrons again.

IT IS EXACTLY THE SAME MAP.

Electrons do not drive the current. Electrons ARE the current as they glacially move from one place to another, always replacing one electron while they in turn are replaced.

Electrons are induced to move by an electric field, which is what the voltage makes.

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12. WBahn Moderator

Mar 31, 2012
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nsaspook and vpoko like this.

Jan 5, 2012
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Thank you, your article is AWESOME! I'm glad it's a common misunderstanding, because it was driving me crazy for a while.

14. Efron Member

Oct 10, 2010
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May I suggest to also think about the antenna effect of a piece of wire connected to the output of an oscillator.

The output voltage of such oscillator will change, making the output more positive (less electrons) or more negative (more electrons). This change is somehow transferred to the piece of wire so that electrons will move from one side of the wire to the other and vice-versa with the same frequency that the oscillator.

This produces an electromagnetic radiation from that wire.

At the end you have electrons moving in a wire only connected to one side of the circuit, but the other side is not connected.

15. nsaspook AAC Fanatic!

Aug 27, 2009
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It's not more or less electrons that transfer to the wire (antenna), it's electromagnetic fields that surround the wire and move at close to light-speed from one end to the other. (Using simplified retarded potentials) If the wire is long (like λ/4)compared to the wavelength (speed of change) of the fields there is a potential difference from the feed point of the wire to the end causing the electrons (charge carriers with lines of force) to move (current) to neutralize the electric field inside the wire. As the electrons move in response their charge fields lines (that extend into free space) move (they can't move instantaneously so they kink) and propagate EM waves.

16. WBahn Moderator

Mar 31, 2012
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No, you may not suggest any such thing!

You still have a circuit, of course, but the details need to take into account Maxwell's displacement current.

It's basically the same idea as a capacitor except that the other plate is very, very large.