# Direction of current through a circuit with multiple batteries

So I've got a bit of a problem trying to understand this circuit in my physics book.

If current flows in the direction of the three arrows (I1, I2 and I3, couldn't find a way to name them) how can there be a complete loop? As far as I'm aware current can't travel both ways through the same cable. Are the batteries charging each other up or is there something else going on? Do the currents cancel each other out when they meet after leaving the positive pole?

The reason for me asking this is because I need to know the direction of the three currents "leaving" the positive sides of the battery in order to do a proper loop equation. Now, I realize that this might be a bit more theoretical than most other problems you answer on this site, but I thought I might as well give it a go.

Thank you in advance to anyone who might want to help me out.

simulate this circuit – Schematic created using CircuitLab

• On one hand, you are saying that this is a theoretical problem. On the other hand, you area wondering about one battery getting charged from another. Is this a textbook-type circuit analysis problem? In such king of problem, a battery is usually an ideal voltage source. It doesn't get charged or discharged, it simply provides a constant voltage. Or is this a practical problem with real batteries? Welcome to EE.SE, by the way. – Nick Alexeev Oct 21 '14 at 19:06

One of those currents can be regarded as "negative" to the other two. An arrow is just an arrow at the end of the day. If you did the calculations you would find one current is the opposite direction to the other two.

When solving a circuit like this, we choose reference directions for the current and then let the sign of the answer tell us the actual direction.

This is not unlike placing an ammeter in series with, say, the top branch with the red lead on the left and the black lead on the right.

If you read a positive value, you know the current is from left to right. If you read a negative value, you know the current is from right to left.

That's really all there is to it. Choosing a reference direction for your current variable is no different from choosing a reference direction for your ammeter leads.

You can't get the wrong answer by choosing the 'wrong' direction. If you choose left to right and I choose right to left, your answer will be the negative of my answer but both answers will be correct and give the same information.

As an aside, since all of the positive terminals are connected together, this circuit is especially easy to solve.

Consider the point to the right of battery 2. By Kirchoff's Current Law we know that the sum of currents into and out of any given point must equal 0, so just from that alone, it's certain that at least one of those currents is flowing against the arrow.

Stated another way, if we consider the arrows as absolute, then at least one of the currents is negative.

The sign of voltage is just showing the potential relative to some other node. (Just for Analogy: you define height relative to the sea level. But you can be lower than that)

The sign of the current is showing the direction of the current relative to the arrow, you painted on the schematics. If the flow of the current (btw: Electrons always flow against the direction of current) is in the opposite direction to your arrows, you simply get a negative sign to the current.

To point this out, I made some modifications to your circuit. First of all: you can combine the Resistors R1 and R5, as well as R4 and R3 as the same current flows through both of them.

simulate this circuit – Schematic created using CircuitLab

If you create a equation for the current, you get: $$i_1+i_2+i_3=0$$ Which means, that with the arrows pointing that way, you will always have at least one current being negative. $$i_1=-i_2 - i_3$$

You could still create a loop, just remember: When the arrow is pointing against the direction of the loop, you have to subtract that voltage. e.g. R1 -> BAT1 -> BAT2 -> R2: $$R_1*i_1 + 6V - 3V - R_2 * i_2 = 0V$$