# electrons and batteries

Discussion in 'General Electronics Chat' started by Nathan Hale, Jul 3, 2015.

1. ### Nathan Hale Thread Starter Active Member

Oct 28, 2011
129
2
Hi folks!

i have a lamp that lights up when you give it 3v.
If i put together a circuit like the one in the picture below the lamp will not light up. ( i know the batteries are not connected electrically )

why doesn't the light turn on?
why dont the electrons flow from the negative terminal of battery "A" and end up at the positive terminal of battery "B" and be happy about it?
i mean after all ...these electrons just care about going from a higher potential to a lower potential right????
then why doesnt the lamp light up???

2. ### nsaspook AAC Fanatic!

Aug 27, 2009
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3,751
Lets rewrite the circuit a bit by making the open connection a switch from a 6 volt source of two 3 volt batteries in series to a lamp.

Why doesn't the lamp light up???

3. ### #12 Expert

Nov 30, 2010
17,895
9,314
Hint: What is the total resistance from + of battery A to - of battery B in post #1?

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4. ### Nathan Hale Thread Starter Active Member

Oct 28, 2011
129
2

well in your case the electrons got absolutely NO WHERE to go! at least in mine they have the option of going to the positive terminal of battery "B" and still do their job!! ( while continuously satisfying the requirement that all they have to do is to go from a higher potential to a lower potential to accomplish the task of lighting the bulb )

5. ### Nathan Hale Thread Starter Active Member

Oct 28, 2011
129
2
infinite resistance

6. ### crutschow Expert

Mar 14, 2008
16,534
4,455
Then I'm sure you can answer the question of about how much current flows through such a resistance, keeping in mind that no current can flow without a complete circuit.

7. ### Nathan Hale Thread Starter Active Member

Oct 28, 2011
129
2
but thats my question. why do the electrons care if the circuit is complete or not? as long as they got SOMEWHERE with lower potential to go? it seems to me that these electrons want to go back to their source battery only and absolutely no where else in order to accomplish the task!

8. ### nsaspook AAC Fanatic!

Aug 27, 2009
3,619
3,751
Exactly!
To the electrons at the gap the two diagrams are identical after a very short time. Infinite resistance gives no current and that makes the entire open circuit charge net neutral (equal potential as a system) after conductor electrons move to balance the electric field from the batteries. So the instant the batteries are connected to the wires, lamp and open switch there is movement for an instant as (charge) electrons move to balance the external field to make the conductors neutral then stop at the open. When the switch is closed the electrons now have a path to flow in response to the electric field from the batteries as current that lights the lamp when energy moves from the battery to power the lamp as the product of voltage and current.

Last edited: Jul 3, 2015
9. ### GopherT AAC Fanatic!

Nov 23, 2012
7,264
5,978
Lets say the electrons do flow as you suggest, then, after a minute of current flowing at 1.5 amps, you will have built up a charge of 90 coulombs in one battery and -90 coulombs in the other battery. As you open up the cap on your flashlight, the two batteries will have a very high electrostatic attractive force and you will never be able to separate them. Well, never is a big statement so you could go and calculate the force required to pull them apart.

http://www.physicsclassroom.com/class/estatics/Lesson-3/Coulomb-s-Law

Then you can realize that nature has made the requirement of a complete circuit, otherwise we will only be able to handle dead batteries in pairs because they will never come apart.

10. ### crutschow Expert

Mar 14, 2008
16,534
4,455
Repeat this mantra after me:
"Electrons cannot move from one place to another along a conductor unless there is a complete return conductive path."

As soon as an electron tries to leave a terminal the change in net charge will pull it right back. If an electron leaves a battery terminal there must be an electron entering the opposite terminal to keep the net charge value balanced.

Makes no difference what the battery voltage is. There can be no change in that voltage due to a change in the charge balance. Net charge must remain unchanged.

11. ### AnalogKid AAC Fanatic!

Aug 1, 2013
5,652
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What you're missing is that terms like higher and lower by definition need a reference. Higher than what? Lower than what? In your first schematic the two batteries do not share a common reference potential to anchor the terms "higher" and "lower". Since the left sides of both batteries are floating, the right sides, even though they are marked + and -, are in fact at the same potential. That is why you will read 6 V if you connect a meter to the two floating left side terminals. At the very low loop current of a DVM input, the light bulb is essentially a short circuit connecting the two right side terminals.

Another approach - leaving things exactly as they are in the first schematic, connect the DVM to the right side + and - terminals, and notice that the reading is 0 V. That explains the zero current through the bulb, and the zero illumination.

An individual battery's negative terminal always is more negative than its own positive terminal (hence the name). But that doesn't make it negative with respect to anything else in the universe unless unless there is a connection, a common reference potential.

ak

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12. ### wayneh Expert

Sep 9, 2010
13,624
4,416
They care because the circuit - the "pipe" - is their only path for movement for one potential to another. No path, no movement. I suppose the water analogy would be trying to pump water through a hose full of ice. Infinite resistance and no flow despite the pressure drop.

13. ### ian field AAC Fanatic!

Oct 27, 2012
5,548
980
Last night we had a pretty big thunderstorm with some nasty bolts of lightning - the clouds may have been a few miles up, that's quite some air gap, but then the clouds do charge up to a few million volts.

14. ### MrAl Distinguished Member

Jun 17, 2014
3,744
791
Hello,

The problem with the first circuit and thoughts is that the word "potential" is used as a shorthand for "potential difference". There is really no such thing as a potential, but that is used as shorthand assuming you know that there has to be a potential difference.
If you replace the LED/bulb with a resistance you will find that there is no potential difference because there is no voltage drop in the resistance because there is no current flow. You can only get current flow in a complete circuit in most regular cases that dont involve changing fields.

15. ### WBahn Moderator

Mar 31, 2012
20,230
5,755
The problem is the notion that the negative terminal of battery "A" and the positive terminal of battery "B" are not at the same potential. They ARE!

What you have is an electric potential (due to electric fields in the space between them) between the positive terminal of battery "A" and the negative terminal of battery "B" that results in the former being 6V higher than the latter.

Although a bit peripheral to your question, the following might help: