# Potential problem

I have a question about the following circuit. I found it in a old textbook and I don't understand the solution.

We know that the potential difference between points A and C is 5V, and that point A is on higher potential. So logically current will flow form A to C. We also know that the resistance for every wire is 1 ohm/meter. If AB=BC=CD=DA=10 cm find the current trough every wire. Let's assume that G is the central point. I don't care about the numerical solution, I am interested in how the current will flow.

This is the circuit picture from the solution. They say that the potential in points D and B is equal and because of that current should not pass through BD at all.

My original thesis was that there will be a difference between B and G because the length of BC + BG > CG, the same analogy applies to D and G. My question is, why will there be no difference in potential between B and G(central point) and D and G(central point)?

Update: I think I understand now. The voltage across CG,AB and CD will be equal because it is a parallel circuit. This means that points B,G and D will have equal potential. I used resistance to determine the voltage which is a wrong approach.

• > "because it is a parallel circuit" -- I do not understand this. What is a parallel circuit? The important fact is that all points B,G,D are half-points for current path that goes through them, so at all three points the potential is half of potential of A. Commented Mar 7, 2021 at 12:01

To see this, let's cut the wires from $$B$$ to $$G$$ and from $$D$$ to $$G$$ for a moment. Notice that $$B$$ is exactly halfway between $$A$$ and $$C$$ (in terms of resistance along that path), $$D$$ is also halfway, and so is $$G$$. That means they are all at exactly half the potential difference between $$A$$ and $$C$$. Connecting these three points with a wire will not have any effect, because they already are all at the same potential.

• But the distance between A and B and C and A is a/2. The distance between G and A is sqrt(a)/2 which means that the resistance is not equal. Commented Mar 7, 2021 at 11:28
• But I think I figured it out, this comes down ta a parallel circuit and that means that the potential difference between those points will be equal. I have been looking at it form a resistance perspective. Commented Mar 7, 2021 at 11:30
• But the total distance between $A$ and $C$ along the diagonal is $\sqrt{2}a$, so $G$ at a distance of $a/\sqrt{2}$ is exactly halfway between along that path, so half the voltage between $A$ and $C$ must have dropped at $G$.
– noah
Commented Mar 7, 2021 at 11:36

There will be no difference in potential, all points B,G,D have the same potential.

The reason for this is symmetry of the circuit and the fact there can be only one solution to the potentials, so it is the symmetric one. If you try to prescribe potential difference to pair BG, Ohm's law will imply current flows through BG. Then currents in AG and GC can't be the same. The symmetry of currents is violated and you will find Kirchhoff's equations for currents do not hold.

• I am not sure I understand your approach, when I plug in the current and then split it(not equally) at every junction the current will be preserved. Commented Mar 8, 2021 at 10:50
• I meant it is not possible to satisfy Ohm's law for every line and Kirchhoff's current rules at all points. These laws, when written as equations for potentials and currents, determine unique solution in which potentials B,G,D are the same. If we add another condition such as $\varphi_G>\varphi_D$, the system has no solution. Commented Mar 8, 2021 at 13:24