# How to find points with same potential while solving an equivalent resistance problem?

Lately, I've been reading about techniques to reduce networks and find their equivalent resistance/capacitance. While doing this, I came across the cube resistance problem and many other problems (eg. resistors on tetrahedron etc.), where the authors have argued that certain points on the figure have the same potential. But, none of them have explained a procedure which would allow one to use this technique for other problems. So I've two questions:

1. Does the figure need to be symmetrical in some manner if one has to use this technique?

2. How should one go about finding points with the same potential?

I've tried a couple of things: Suppose that we were required to find the equivalent resistance across the main diagonal of a cube. Then usually I would distribute the currents and look for branches carrying the same current. From this, I would try to deduce the points having the same potential. But of course, this technique hasn't worked so any hints or suggestions will be valuable.

The general idea is to find and exploit symmetries in the network. A symmetry means that if you change something about the problem, it remains the same. Generalized method for dealing with circuit involving symmetry? links to a basic introduction to a formal procedure for identifying the symmetries of the network, and the effect these symmetries have on the relation between the inputs and outputs of the circuit. However, in many cases the symmetries are most easily identified visually from a diagram of the circuit.

For example, the following infinite ladder of resistors looks exactly the same if you add another "unit" at the front. This suggests a method of finding the total resistance $R_\infty$ which is the same as $R$ in series with $R$ || $R_\infty$ :
$$R_\infty=R+\frac{R_\infty R}{R_\infty+R}$$ This is a quadratic equation which can be solved to find $R_\infty$.

In your cube problem, resistors $a$, $b$ and $c$ are in equivalent positions - ie if you rotate the cube about an axis through AB you can replace $a\to b\to c\to a$ without making any difference to the resistance between A and B. This symmetry means that the points marked $\alpha$ are all at the same potential, as are those marked $\beta$.

Without affecting the circuit we can connect wires between the points marked $\alpha$ - and likewise between those marked $\beta$ - because no current will flow through them. The cube is then equivalent to the following series of parallel resistors :

http://www.rfcafe.com/miscellany/factoids/kirts-cogitations-256.htm

Points which are connected by an ideal wire - which means anything that can carry a current with zero resistance - will be at the same potential. This is a direct consequence of Ohm's law, $\Delta V = IR$. A section of ideal wire is basically a resistor with zero resistance. If $R = 0$ for this resistor/wire, then $\Delta V = 0$, meaning that the change in potential across the resistor/wire is zero.

Other than that, there aren't really any shortcuts; you have to solve the equations. (You can sometimes recognize a symmetry in the circuit design or something which makes it obvious that the equations will tell you two points are at the same potential. But I'm not counting that, since it's not a general technique.)

• If the zero resistance wires are in parallel will not they make the equivalent resistance of the combination zero?In the diagram given on this page:physics.stackexchange.com/questions/298657/… the correct equivalent resistance is 2r/5 but i think that zero resistance wires make the equivalent resistance zero. What do you think? – MrAP Dec 18 '16 at 18:17
• @MrAP I suggest you read the answers to the question you linked. – David Z Dec 18 '16 at 18:48

In general one can solve all such circuits with a method called "modified nodal analysis" (https://en.wikipedia.org/wiki/Modified_nodal_analysis) which reduces the circuit to a set of linear equations represented by a matrix. This has been automated in circuit simulation software. If you want to try this for yourself, Linear Devices has a free software called LTSpice: http://www.linear.com/designtools/software/#LTspice that works very nicely.

LTSpice will, of course, only give you a numerical result, rather than solve the equations explicitly. This is enough for engineering purposes, but if you want to perform a real analysis of a circuit, you would have to solve the actual equations formally.

For purely linear circuits there are a few software tools that can actually simplify the formal equations for you, but you still end up with a set of linear equations that has no simple solutions other than in terms of matrices and their determinants, inverses and characteristic polynomials.

## protected by Qmechanic♦Aug 13 '16 at 22:45

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).