Irodov 3.153: Half the Current Takes the Shortcut – Amit & Deepali Sinha Foundation

Problem Irodov 3.153

An infinite plane grid of wires has square cells, the wires being joined at every crossing. Each piece of wire between two neighbouring joints has resistance $R_0$ . The points $A$ and $B$ are two neighbouring joints, the ends of one such piece. Find the resistance $R$ of the whole grid between $A$ and $B$ . Hint: use symmetry and the superposition principle.

A patch of the infinite square grid, wires joined at every crossing; A and B are neighbouring joints, the ends of one wire of resistance R_0. — A patch of the infinite square grid, wires joined at every crossing; $A$ and $B$ are neighbouring joints, the ends of one wire of resistance $R_0$ .

Let a current $I$ enter the grid at $A$ and leave it at $B$ . Then $R = (\varphi_A - \varphi_B)/I$ , and the voltage across the single wire $AB$ is its own current $I_0$ times $R_0$ :

I R = I_0 R_0 \tag{1}

So the whole problem is to find what share of $I$ runs through the wire $AB$ . The currents in a network of resistors obey linear equations, so the flow from $A$ to $B$ may be built as the sum of two simpler flows.

Given

square grid, infinite
$R_0$ between neighbouring joints
$A$ , $B$ neighbouring joints

Find

the resistance

R

between

A

and

B

Solution

Step 1A current fed in at

A

alone

Let $I$ enter at $A$ and spread out to infinity. The grid looks the same from $A$ in all four directions along the wires: a turn by $90^\circ$ about $A$ carries the grid into itself. So the four wires meeting at $A$ carry equal currents, and together they carry $I$ :

The current I fed in at A and spreading to infinity: I/4 leaves along each of the four wires at A, including AB. — The current $I$ fed in at $A$ and spreading to infinity: $I/4$ leaves along each of the four wires at $A$ , including $AB$ .

I_{AB}^{(1)} = \frac{I}{4}\quad\text{from } A \text{ toward } B \tag{2}

Step 2A current taken out at

B

alone

Now let $I$ come in from infinity and leave the grid at $B$ . The shift of the whole grid by one cell along $AB$ carries the grid into itself and $A$ onto $B$ , so this flow is the first one moved to $B$ with every current reversed. By the $90^\circ$ rotation about $B$ , the four wires at $B$ each bring $I/4$ into $B$ . The wire $AB$ is one of them, so its current again points from $A$ toward $B$ :

The current I drawn in from infinity and taken out at B: I/4 arrives at B along each of its four wires, again from A toward B in the wire AB. — The current $I$ drawn in from infinity and taken out at $B$ : $I/4$ arrives at $B$ along each of its four wires, again from $A$ toward $B$ in the wire $AB$ .

I_{AB}^{(2)} = \frac{I}{4}\quad\text{from } A \text{ toward } B \tag{3}

Step 3Superposition

Adding the two flows, the current $I$ entering at infinity in the second cancels the current $I$ leaving to infinity in the first, and what remains is exactly $I$ in at $A$ and $I$ out at $B$ . The current in $AB$ is the sum of (2) and (3):

I_0 = \frac{I}{4} + \frac{I}{4} = \frac{I}{2}

Putting $I_0 = I/2$ into (1) and dividing by $I$ :

R = \frac{R_0}{2}

(4)

Check

Foster’s theorem for a network of $N$ joints states that the effective resistances across the wires, each divided by that wire’s own resistance, sum to $N - 1$ . A large square grid has two wires per joint, about $2N$ in all, and far from the edges they are all equivalent, so each must give $(N - 1)/2N \to 1/2$ , which is $R/R_0$ exactly. The value also lies between the limits it must respect: below $R_0$ (the wire $AB$ alone) and above $R_0/4$ (the four wires at $A$ alone in parallel, with the rest of the grid shorted).

Remark

The argument never used the shape of the cells, only that every joint has the same number $z$ of wires and that the lattice looks the same from every joint along every wire. Each partial flow then puts $I/z$ through the wire $AB$ , so $R = 2R_0/z$ : $R_0/3$ for a triangular grid ( $z = 6$ ), $2R_0/3$ for a honeycomb ( $z = 3$ ), and $R_0/3$ again for neighbouring joints of an infinite cubic lattice of wires in space ( $z = 6$ ).