Voltage and effective resistance of an infinite lattice of resistors

[zenterix]

Homework Statement

The following problem appears in Chapter 4 of Purcell's Electricity and Magnetism:

Consider a two-dimensional infinite square lattice of 1 $\Omega$ resistors. That is, every lattice point in the plane has four 1 $\Omega$ resistors connected to it. What is the equivalent resistance between two adjacent nodes? This problem is a startling example of the power of symmetry and superposition. Hint: If you can determine the voltage drop between two adjacent nodes when a current of, say, 1 A goes in one node and comes out the other, then you are done. Consider this setup as the superposition of two other setups.

Relevant Equations

My issue is that I don't really even understand the setup.

I had imagined that there would only be four possibilities for current flow in an infinite lattice, because of symmetry:

The reasoning I had was that in the horizontal direction, if the current flows into a node from the left, then it must flow into the node on the right so that that node has current flowing in from the left. Thus, the flow on the horizontal is always either from left to right or from right to left.

Analogous reasoning leads to current always flowing from south to north or from north to south. Thus we get the four possibilities above.

However, it seems that it is necessary to consider where the current flows into the infinite lattice and where it leaves.

Here is the solution that appears in the book

Let the two adjacent nodes be labeled $N_1$ and $N_2$. Consider a setup where a current of 1 A flows into the lattice at $N_1$ and heads out to infinity in the two-dimensional plane. If you want, you can imagine a return lead connected around a rim very far away. By symmetry, the current in each of the four resistors connected to $N_1$ is 1/4 A (all flowing away from $N_1$).In particular, there is a current of 1/4 A flowing from $N_1$ to $N_2$.

Consider a second setup where a current of 1 A comes in from infinity in the two-dimensional plane and flows out of the lattice at $N_2$. Again,you can imagine a return lead connected around a rim very far away. By symmetry, the current in each of the four resistors connected to $N_2$ is1/4 A (all flowing toward $N_2$). In particular, there is a current of 1/4 Aflowing from $N_1$ to $N_2$.

If we superpose the above two setups, we now have 1 A entering the lattice at $N_1$, 1 A leaving the lattice at $N_1$, and no current at infinity. This is exactly the setup we wanted to create. The total current flowing from $N_1$ to $N21$ is 1/4+1/4 = 1/2 A. This current flows across a 1 $\Omega$ resistor, so the voltage drop from $N_1$ to $N_2$ is 1/2 V. The effective resistance is defined by $V=IR_{eff}$, where I is the current that enters and leaves the circuit (which is 1 A here). So we have $1/2 V = (1 A)Re_{eff}$, which gives $R_{eff} = 1/2\Omega$. Unfortunately, this quick method works only if the nodes are adjacent.

My first question is: what is a "return lead connected around a rim very far away"?
The solution above is assuming first that all the current comes in at one node and spreads from there. Then it assumes a second case where the current comes in somewhere in infinity at flows completely out through $N_2$.

Now, using Kirchoff's rules on a normal circuit, we get a linear system of equations, and so we can superimpose solutions to get another solution.

If the two scenarios described by the book are solutions to whatever the linear system of equations is for this infinite lattice, then their superposition is another solution.

After the superposition, we have a known current flowing between two points (one of which is the points at which current flows into the circuit, and the other of which is the points through which current leaves the circuit) through a known resistor and so we can calculate the voltage between the two points.

In the second scenario, where current comes in from infinity and leaves through $N_2$, why can we ignore where the current is flowing in?

 

[vela]

The reasoning I had was that in the horizontal direction, if the current flows into a node from the left, then it must flow into the node on the right so that that node has current flowing in from the left. Thus, the flow on the horizontal is always either from left to right or from right to left.

The problem with this reasoning is that it breaks the symmetry. Why would the current flow left to right vs. right to left? Because of the symmetry, there's no physical difference in the two directions so there'd be no way for the current to decide to pick one direction over the other.

My first question is: what is a "return lead connected around a rim very far away"?

If you were to build a large circuit, so big that "infinite" is a good approximation, you'd still need somewhere for the circuit injected at the node to eventually end up. That would be the return lead connected to the outer edge of the circuit.

In the second scenario, where current comes in from infinity and leaves through $N_2$, why can we ignore where the current is flowing in?

You're not ignoring where the current coming from. As you stated, the current is flowing in from infinity.

 

[Lord Jestocost]

Homework Statement: The following problem appears in Chapter 4 of Purcell's Electricity and Magnetism:

Consider a two-dimensional infinite square lattice of 1 $\Omega$ resistors. That is, every lattice point in the plane has four 1 $\Omega$ resistors connected to it. What is the equivalent resistance between two adjacent nodes? This problem is a startling example of the power of symmetry and superposition. Hint: If you can determine the voltage drop between two adjacent nodes when a current of, say, 1 A goes in one node and comes out the other, then you are done.

Maybe, the following article on "Infinite Grid of Resistors" might be of help for you:
https://www.mathpages.com/home/kmath668/kmath668.htm