# How does symmetry allow a rapid determination of the current between $A$ and $B$?

The following was originally given to me as a homework question at my physics 2 course:

Consider the following circuit The difference of potentials between the point $V_{1}$and the point $V_{2}$ is $4.4$ volts, the resistance of all the resistors is the same $R=1\Omega$.

Find the current between point $A$ and point $B$.

The answer given is simply $0$ and the argument was just the pair of words using symmetry''.

I don't really understand the answer:

First, it is not completely symmetric: There is a difference of potentials so the potential at the point $V_{1}$ is not the same as the potential at the point $V_{2}$.

Secondly: How can I see that the symmetric structure will give me that the current between $A$ and $B$ is $0$ ?

Also, I would appreciate to see a calculation of this current to get a better feel for whats going on, I know the rule $V=IR$ (which seems the most useful here, but I also know other rules that can be used), but I don't understand how to use this rule to find the current. The current that comes from the source, when reaches the point that must choose it's way, sees no difference between the two paths (symmetry) , so half of it flows through one way and the other part flows in the second way. It means that, $I_1=I_2$ , So the potential difference across yellow resistors is the same. It means that the potential of point $\mathbf{A}$ is equal to potential of point $\mathbf{B}$ :

$$I_1=I_2\to V_A=4.4-I_1R \text{ , }V_B=4.4-I_2R\to V_A=V_B$$

So there isn't any potential difference across the blue resistor, and the current through it is 0, and it can be omitted from the circuit without any change in the behavior of the circuit.

Symmetry is a powerful way to obtain a result by inspection. Here's how it works in this case.

Remove the resistor between the A & B nodes. Now, it should be easy to see by inspection that the voltage between the A & B nodes must be zero for any $V_1$ and $V_2$.

The reason is symmetry. The left hand path is identical to the right hand path so it must be the case that $V_A = V_B$.

Since the voltage across A & B is zero, replacing the resistor between those nodes will not change the circuit since, with zero volts across the resistor, there will be zero current through it.

• Another way of seeing it without removing the resistor first: if there is a voltage over the resistor, then either $V_A < V_B$ or $V_A > V_B.$ But since the rest of the circuit is symmetric (under flipping $A \leftrightarrow B$), this cannot happen. Jun 6 '13 at 21:04
• @Vibert, you should post that as an answer. Jun 6 '13 at 21:10