I am reading a basic electronics book: "There are no Electrons: Electronics for Earthlings" and I came across a clever passage about the fact that you need a closed circuit in order for current to flow. Here is the passage I am curious about:

"This has always bothered me: If the negative terminals of batteries have excess electrons (a negative charge) and the positive terminals of batteries have too few electrons (a positive charge) and opposites attract, why can't I hook a wire between the negative side of one battery and the positive side of a different battery and get any current? This truth is it won't work. No current will flow. Had someone been able to explain that to me, I probably would never have written this book."

Does anyone have a straight-forward answer to this question?

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    \$\begingroup\$ This is probably the best forum to ask this question, to my knowledge. Welcome to CH! \$\endgroup\$
    – J. Polfer
    Apr 29 '10 at 15:56
  • \$\begingroup\$ @mike: Could you edit the title of this question to be more descriptive? Maybe something like "Current flow in batteries?" \$\endgroup\$
    – pingswept
    Apr 29 '10 at 22:54
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    \$\begingroup\$ Actually a current will flow if you connect a conductor to any voltage, through simple electrostatics. Not noticable at most voltages, but see what happens when you touch a peice of metal to a 100,000kV line, even in a vaccumm with no earth, a sizeable current will flow to bring the metal to the same electrostatic charge. The problem with electrical education these days is that electrostatics is ignored when it comes to circuits, and therefore people can't understand things like batteries - or worse - they are taught that electricity has to involve the flow of electrons. \$\endgroup\$
    – Myforwik
    Sep 9 '12 at 2:02
  • \$\begingroup\$ Related to @Myforwik's point: electronics.stackexchange.com/questions/75743/… \$\endgroup\$
    – Phil Frost
    Jul 16 '13 at 15:46

The confusion here is from the initial poor description of how a battery works.

A battery consists of three things: a positive electrode, a negative electrode, and an electrolyte in between. The electrodes are made of materials that strongly want to react with each other; they are kept apart by the electrolyte.

The electrolyte acts like a filter that blocks the flow of electrons, but allows ions (positively charged atoms from the electrodes) to pass through. If the battery is not connected to anything, the chemical force is pulling on the ions, trying to draw them across the electrolyte to complete the reaction, but this is balanced by the electrostatic force-- the voltage between the electrodes. Remember-- a voltage between two points means there is an electric field between those points which pushes charged particles in one direction.

When you add a wire between the ends of the batteries, electrons can pass through the wire, driven by the voltage. This reduces the electrostatic force, so ions can pass through the electrolyte. As the battery is discharged, ions move from one electrode to the other, and the chemical reaction proceeds until one of the electrodes is used up.

Thinking about two batteries next to each other, linked by one wire-- there is no voltage between the two batteries, so there is no force to drive electrons. In each battery, the electrostatic force balances the chemical force, and the battery stays at steady state.

(I kind of glossed over what it means for two materials to "want" to react with each other. Google "Gibbs free energy" for more details on that. You might also google "Nernst equation.")


Forget the batteries for a second, thats just one of a thousand analogies you could use to describe voltage/current and the reason that no current flows has nothing to do with the electro-chemical properties of batteries, its far simpler.

The easiest way to think of it is this: Current will only ever flow in a loop, even in very complex circuits you can always break it down into loops of current, if there is no path for current to return to its source, there will be no current flow.

In your battery example, there is no return current path so no current will flow. There is obviously a more deep physics reason for why this works but as the question asked for a simple answer I'll skip the math, google Maxwell's Equations and how they are used in the derivation of Kirchhoff's voltage law.

Batteries do make a good example for this simply because they are current sources with completely isolated grounds. This example would be equally true of any other power source with a completely isolated "ground".

However, this is not an easy thing to find, for instance doing this with 2 bench supplies would likely make one of the bench supplies very unhappy, but thats not because the effect is different, the difference is that the bench supplies are likely both grounded to the electrical wiring in the building and as such there is a return path for current to flow through.

The water analogy for this also effective. Think of your battery example this way:

You have a water pump (battery A) connected to a pipe (the wire), and you have another water pump (battery B) connected to the same pipe (the wire) . Now in your example the there is no return path in the system so imagine that the pipe is full of water but capped off on both ends.

You hit the power switch on the pumps, what happens?

The answer is nothing, there is no where to move the water to, the pumps don't even spin. (ignore water turbulence like effects for this analogy).

Now if you were to connect the pipe in a loop and hit the switch the pumps would spin up (voltage) and water would flow (current).

If you used 2 difference speed pumps (different voltage batteries) and faced them toward each other one will over power and cause the other to spin in the wrong direction (burn out just like connecting a 9V and 6V battery in parallel).

If you connected both pumps pointing in the same direction you would get more water pressure (voltage) because the pumps are helping each other out (2 batteries in series).

  • \$\begingroup\$ Pumps and loops are no way of describing an electrical circuit unless you are trying to describe the process to a 5 year old. It's all about potential difference. Not about pumping or looping. Two points with > 0 potential difference between them will cause current to flow through a conductor. There does not need to be a "loop". \$\endgroup\$ Nov 2 '16 at 11:23

Let's say you have AA batteries, with 1.5 V each. Further, let's label them battery A and battery B. If you hook A+ to B-, what you actually get is a 3 V difference across A- to B+.

B+  -------------------
|                     |
B- _ A+  --           | 3V
     |    | 1.5 V     |
     A-  --------------

When you hook B- to A+, they are both at the same potential (they're hooked up with a wire, after all). B+ is 1.5V higher than this potential, and A- is 1.5V lower.

It's important to remember that a voltage is not an absolute value. It's a relative value. The B- _ A+ wire will be at one potential, and B+ and A- are relative to that potential.

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    \$\begingroup\$ ajs410, i have no idea why someone down voted you. Your answer is by far my favorite, electricity can be simplified, but a solid understanding of what you are explaining is required for a professional to grow. \$\endgroup\$
    – Kortuk
    Apr 30 '10 at 20:57
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    \$\begingroup\$ They probably voted me down because it looked like I was saying 1.5 + 1.5 = 3. What I wanted to stress, though, was how each terminal is relative to the others along the Y axis. B- and A+ are the same potential because they're hooked up by a wire. With no potential difference, there is no current flow. \$\endgroup\$
    – ajs410
    May 3 '10 at 18:27
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    \$\begingroup\$ ajs410, I still don't get it with your example. You said that B- and A+ are the same potential because they are hooked up by a wire. If you think about it B+ and A- are hooked up by a wire which should make them the same potential but why do you have current flow in that case and not the case when B- is hooked to A+? I have always wondered about this issue but noone has been able to explain it to me. I guess I am slow. LOL. \$\endgroup\$
    – ddtrn
    Jan 10 '11 at 7:07

I too have always found that the traditional layman's description of a battery to be misleading. Most people describe a battery as a storage container for electricity, but that doesn't explain why you can't dump the electricity from a battery to the ground, or why you can't have one battery feed another, like in your question above.

This may not be an accurate description of what is actually happening, but I find that a more understandable analogy is to describe a battery as a pump instead. The "energy" contained in the battery is used to drive the pump; it is not sent out over the wire. With this analogy, it is plainly obvious why both the positive and negative ends of a battery must be connected in a circuit. If, say, you connect only the negative electrode to ground, there is no current because there is no electricity coming in on the positive electrode that can be pumped out.


Technically, current may or may not flow when a wire is connected that way. It all depends on whether or not there is a potential difference in charges between those two terminals. If the difference is small, little/no current will flow. This holds true for any wire connected between any two terminals, anywhere.

However, current more than likely won't (depending upon the age/use of the battery). The reason why is because the voltage potential difference - the "excess holes on the positive end" and the "excess electrons on the negative end" - is relative to a given battery. There are excess electrons/holes on the ends of a given battery with respect to each other. That relationship may or may not hold true between one battery's negative terminal and another's positive terminal.

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    \$\begingroup\$ If there's a voltage on one battery relative to the other, then they have different static electric charges, which will be equalized after you connect them together. \$\endgroup\$
    – endolith
    Apr 29 '10 at 17:51
  • \$\begingroup\$ @endolith - True... so that means my answer is incorrect, isn't it? What could be said to correct my answer? \$\endgroup\$
    – J. Polfer
    Apr 29 '10 at 18:28
  • \$\begingroup\$ In his example it all stops at the setup since there is no return path for the current so no current will ever flow which is actually stated in the first line of the article you linked "As was previously mentioned, we need more than just a continuous path (circuit) before a continuous flow of electrons will occur" and covered in more depth in the previous section. \$\endgroup\$
    – Mark
    Apr 29 '10 at 22:13

The current that flows into a junction is always equal to the current that flows out of the junction. Therefore, current must always flow in a loop.

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    \$\begingroup\$ current is dq/dt. You can have current without a loop \$\endgroup\$
    – user5419
    Jan 20 '18 at 18:58

First. A very small current WILL flow for a very short time ... but it needs only a small amount of electric charge (static) to pass to build up sufficient backward voltage to nullify the forward voltage. Recall that any noticable static electricity like rubbing balloons on hair usually involves BIG voltages like (tens of) thousands of volts which is necessary to make a spark. Typical simple 1.5 volt cell cant make a spark (without some special help). Actually if you carefully suspended a pair of simple cell (batteries) in space or hanging on a string they would have a small dipole interaction and twist like a pair of magnets and then attract each other. Unfortunately this just turns out to be very small though calculable and finite forces. Perhaps this expt has been done sensitively to show this. Further, a the original problem is a bit like charging a capacitor if you think about it. With the unconected terminals being just the capacitor plates. The smallness of the terminals and the seperation is very different numerically from a good capacitor which has big area and small seperation which makes all the difference and thats why current is miniscule (but finite) in original problem.

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    \$\begingroup\$ When answering please answer the question fully, and explain the differences between the two scenarios (battery connected vs not connected). \$\endgroup\$
    – Voltage Spike
    Apr 19 '16 at 18:17

I think another thing which might be confusing you is that we say the voltage in a conductor is always constant. While this is mostly true it is kind of a lie. All conductors still have a finite resistance. Because of this if you hook up a wire between the two terminals of a battery one end of the wire will actually be at a different potential then the other and the current in the wire will follow ohms law. I don't know if you've ever tried shorting the two terminals of a battery like this but because the resistance of the wire is so low ohms law gives a very large current which will heat up the wire and burn you.

Now the reason that we usually say that the potential in a wire is constant is because there are usually other components in our circuit whose resistance is much larger then the wires. Because of this most of the voltage will end up being dropped across the other components in the circuit and there will only be a very small voltage drop from one end of a wire to the other. So we can simplify our problems by saying that the voltage is constant in the wires.

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    \$\begingroup\$ Except the OP's question involves two batteries with a single wire between them and no complete circuit. So no current flows, the resistance of the wire is thus irrelevant, and there's no voltage drop along the wire. In short, your answer is about cases that are different than the OP's question. \$\endgroup\$
    – gwideman
    Sep 23 '17 at 7:03