# How can a circuit function with two negative battery terminals facing each other?

Here is a drawing of the circuit that is confusing me:

I don't quite understand how batteries work in this diagram. If a battery has a negative and positive terminal, there must be a barrier preventing them from neutralizing one another, so how can the potential from either negative terminal ever make it through the top half of the circuit without passing through a battery?

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I don't think Physics.SE is the right place to be asking this however the answer is that a battery can be thought of as a voltage difference and a resistor. If one battery is a greater voltage difference than the other you'll still have a current loop. –  Brandon Enright Apr 22 '13 at 23:58
@BrandonEnright Thank you for your comment. I posted it here because it's part of my physics homework. –  Jim Apr 23 '13 at 0:18
Personally I think it's a good question. Homework questions are acceptable here if they're about understanding the concept rather than just getting the answer - and that is clearly the case here. –  Nathaniel Apr 23 '13 at 1:24

To put Zetta's answer in more visual terms: A battery "pushes out" electrons on the negative pole while "sucking in" electrons on the positive side. In the diagram you draw the 9V battery pushes electrons way more strongly than the 3V battery, so they move from the 9V to the 3V.

An analogy with tilted boards may also help: Suppose you tilt a board to the left and put a ball on it, on the ground floor of a building. It will roll to the left. If you put the same board on the second floor, but tilt it to the right, the ball will roll to the right, even though both ends of the board are now located higher than before.

What matters is not the absolute value, but the difference!

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As Brandon said above, the reason is that since one battery is of a greater voltage than the other, they do not entirely neutralize one another to the point where there is no current. To demonstrate this a bit more rigorously, consider the current produced by each battery. This analysis is possible through the principle of superposition:

The current produced by the 9V battery can be calculated by simply removing the 3V battery altogether, and calculating the loop voltage with only the 9V battery in place. Applying some circuit analysis to the reduced circuit, the voltage produced by the 9V source (flowing clockwise) is $1 A$.

Doing the same for the 3V battery, and removing the 9V, the current produced by the 3V battery is calculated to produce $-1/3A$ in the clockwise direction (it is actually producing a $1/3A$ current in the opposite direction, because it is oriented in the reverse direction of the 9V battery).

So because these two currents are not equal and opposite, no neutralization of current occurs in the loop.

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