# An Introduction to Theoretical Physics (Part 2 of 5)

Posted on 2011/10/18 by

## Introduction to Part 2

In Part 1, the importance of abstraction was justified as a way to satisfy the human goal of knowing and predicting the most, while remembering the least.  It enables physicsists to describe the workings of the universe with a small collection of simple patterns.

Satisifed that the resistor addition equations can be mathematically derived from some of our present-day basic rules, it is now possible to advance to solve the problem at hand. To remind you, it was:

What is the total resistance between two opposite vertices of a cube of identical resistors?

The first thing to consider is what tools we have available and compare this with what we are attempting to solve.  Namely, we have equations for the addition of resistors in two dimension and yet have a diagram that is three dimensional.

With this insight, we could either try to make 3 dimensional equations, or try to find a way of drawing the resistor cube in two dimensions in a way that we can use the equations we already have.  We’ll take the latter of these options.

## The Resistor Square

Before we try to figure out a way to collapse a 3 dimensional diagram into a 2 dimensional diagram, let’s see if we can first look at a 2 dimensional analogy.

We’ll study the square, which is a two dimensional equivalent of a cube.  One way to think about this is to notice that a cube has 3 edges joining to each vertex and they are all at right angles to one another.  The vertices of a square have two edges joining together and they are at right angles to each other.  So they are the same type of shape, just with a different number of dimensions.

Maybe investigating the total resistance of a resistor square will aid with the resistor cube problem.  So, let’s ask the simpler question:

What is the total resistance between two opposite vertices of a squre of identical resistors?

Taking the diagram of a square of resistors, it is quite easy to see that it a parallel circuit with two resistors in each branch.  So we can redraw the square so it looks more familiar:

Now we can calculate the total resistance using our existing equations for series and parallel circuits.

$R_T=\frac{1}{\frac{1}{R+R}+\frac{1}{R+R}}$

Combing the Rs gives:

$R_T=\frac{1}{\frac{1}{2R}+\frac{1}{2R}}$

$R_T=\frac{1}{\frac{2}{2R}}$

This simplifies to:

$R_T=R$

So a square of identical resistors has the same resistance as just one resistor.

There is also another way to get the same answer, using a physical consideration.

## Equipotentials

Equipotentials are parts of an electrical circuit where there is no difference in electrical potential.  Just as no book can fall between two sides of a straight shelf, no charge can flow between equipotentials.  So, electrically connecting equipotentials has no effect on the circuit.  For this reason, the process is often called ‘dead-shorting’.

Dead shorting serves no practical purpose, but can be a great aid to the mind when looking at a complicated circuit.  Knowing that there is no electrical change change to the circuit means that doing it is just a way of redrawing the same diagram.  It’s look at the same thing from a different point of view.

It is often found by physicists and mathematicians that a slight change in notation can make a tremendous difference to the perceived difficulty of a problem.

With our square, we can reach the same result as above, but by transforming our diagram as follows:

The dotted green line represents the two vertices that are equipotentials.

Now applying the resistor addition rules, we get:

$R_T=\frac{1}{\frac{1}{R}+\frac{1}{R}}+\frac{1}{\frac{1}{R}+\frac{1}{R}}$

$R_T=\frac{1}{\frac{2}{R}}+\frac{1}{\frac{2}{R}}$

$R_T=\frac{2}{\frac{2}{R}}$

Resulting in:

$R_T=R$

Which, comfortingly, is the same as before.

Now, this may seems very silly to go to all that bother with the square, but this method has a tremendous advantage with the cube.  Admittedly, I have presented the material in a disingenuous way so far.  When I discovered this method, I figured it out when looking for the cube solution.  However, as the concept of an equipotential needs explaining, I thought it better to do this with a familiar 2-dimensional example.

Now to explore its real use.

Equipotentials of the Resistor Cube

For the cube, we know that the first vertex has 3 resistors joined to it.  As the resistors are identical, this means that the any electrons moving through these resistors will lose the same amount of potential energy.  Using a gravitational analogy, we could say that they all fall the same electrical height.  And this is true no matter what the rest of the circuit looks like.  Thus, the vertices on the other sides of these resistors are equipotentials.

The above diagram is a 2-dimensional projection of a the cube.  It’s the best we can do to represent it using a 2-dimensional screen.

In the diagram, the biggest potential (where we connect one side of our power supply) is the vertex at the bottom left.  As electrons move through the red edges, which represent resistors, they all loose the same amount of energy.  They end up at the green vertices, which all have the same potential - those vertices are equipotentials.

The same idea can be applied to the final vertex (the one opposite to the first vertex).  It will have 3 resistors joining to it.  As each resistor is identical, it will contribute the final potential difference from the start to the finish, so to equal the power supply.  This all follows from Kirchoff’s potential difference law.  So, the vertices on the other end of these resistors are equipotentials as well.

Now, we can think of our resistor cube circuit diagram with these additional connections:

So, instead of 8 vertices, we can now think of the circuit as having just 4, since two sets of three vertices are equipotential (represented by the dotted lines).

What about the resistors in the middle?  Well, they connect the two sets of verticies that we have now joined up, so we can think of all 6 of these as in parallel.  Lets redraw the diagram once again with those considerations (and with the resistors drawn back in):

I’ve draw the diagram so to keep count of how many vertices have been joined together (in blue) and how many resistors there are in between (in red).

This resistor cube has now been redrawn in a way that we can apply our two resistor addition rules.  As you can see, the diagram is a series of three parallel circuits.

Before we go onto solve the problem, we might ask if there is an easier way to add identical resistors in parallel.  Sure, we have the parallel resistors rule, as shown in Part 1, but this is designed for complicated parallel circuits where all the component resistors are different.  Is there an easier equation for this easier problem?

## N-resistors in Parallel

$\frac{1}{R_T} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\dotsb+\frac{1}{R_N}$

But our problem deals with identical resistors:

$R=R_1=R_2=R_3=\dotsb=R_N$

And so:

$\frac{1}{R_T}=\frac{N}{R}$

Finally:

$R_T = \frac{R}{N}$

The above equation is a useful tool to aid calculation.  It most certainly doesn’t tell allow us to find out anything new that the parallel resistor equation didn’t already tell us.  In fact, if the derivation is valid, we should get identical results from both equations.  It’s like figuring out that multiplying by 10 can be done by adding zeros to the end of the number, instead of doing long multiplication (in base 10, at least).  This tells us that for identical parallel resistors, we simply need to divide the resistance of one of them by how many there are.  Same sort of trick.

Yet, a physicist might also look for the physical significance of what they have found.  Even though this is a calculation shortcut, we are still talking about resistors and circuits.  Asking ourselves for the physical significance of this might reveal a different way of looking at something we already knew. In this example, the equation should not be too great a surprise.  A parallel circuit has the same potential difference across it as just one resistor.  This is because the potential difference is determined by the power supply, not the resistors.

Each resistor in a parallel circuit has the same potential difference across it as just one.  Thus, using Ohm’s Law, each resistor in the parallel circuit has the same current through it as the sole resistor circuit.  N resistors let N times as much current through.

Using these ideas with equations we have:

$R = \frac{V}{I}$

which is Ohm’s Law rearranged.

Going from a single resistor to a parallel circuit with current N times larger can be written as:

$I \rightarrow NI$

Remember that the potential difference is the same for both circuits, we can combine the equations to give:

$R = \frac{V}{(I)} \rightarrow \frac{V}{(NI)} = \left(\frac{V}{I}\right)\frac{1}{N} = \frac{(R)}{N}$

which shows that the resistance decrease by a factor of N as well, agreeing with the above result. Thinking of it this way, it might seem more ‘obvious’ than deriving it from the resistor rules.

Furthermore, this way, we still have a grasp of what is actually going on inside the wires of the circuit.  The equation is no longer just a consequence of the assumption that we know must be true (if our premises are true).  It becomes an inevitable consequences of the physical model we are using to describe electrical circuits.

## Solution for the Identical Resistor Cube

Now we have a quick way of reaching our solution.  The resistance of each parallel circuit can be easily calculated from the above equation.  Then, as these circuits are in series, all three values are added together.  In symbols:

$R_{T}=\frac{R}{3}+\frac{R}{6}+\frac{R}{3}$

$R_{T}=\frac{5}{6}R$

## Solution for the Olympiad Question

That was a very special case where the resistors were all identical, but what about the original problem from the Olympiad paper in Part 1?  There, the resistors are not all identical but have 4 different values.  This can be achieved by relabeling our two dimensional diagram for the purposes of the question.

So this problem, although slightly more mathematically demanding, can be solved with our previous work.

$R_{T}=\frac{3.0}{3}+\frac{1}{\frac{3}{8.0}+\frac{3}{12.0}}+\frac{1.0}{3}$

$R_{T}= \frac{54}{15}\Omega$

This is the solution to the original Olympiad question.

Any other set of values for the resistors can be solved by relabeling our collapsed 2D circuit diagram for the cube.  As with the middle section of the Olympiad solution, the shortcut for calculating parallel components cannot be used if the resistors are not all identical.

## Summary

You might wonder, with the problem already solved, why there are still 3 parts to read.  Now we have found a solution for a resistor cube, its an interesting question to ask (for a physicist, at least), if we can make further abstraction from our two dimensional circuit diagrams.  Check back for tomorrow’s Part 3 to learn more.