Cube
Imagine a cube, where each edge is a \(1 \Omega\) resistor.
What is the effective resistance between opposite corners of the cube?
The effective resistance is \(\frac{5}{6} \Omega\) .
Proof
Let’s call one of the corners point \(A\) , through which we will force 1 A of current into the circuit. Call the opposite corner point \(B\) , through which we will draw 1 A of current from the circuit. Thus overall, 1 A is flowing from \(A\) to \(B\) . We will set the voltage at \(B\) to \(V_B = 0\) .
Now look at the 3 corners adjacent to \(A\) , call them \(A_1\) , \(A_2\) and \(A_3\) . By symmetry, the current through an edge between \(A\) and any \(A_i\) is:
\[I_A = \frac{1}{3} \text{ A}\]Thus the voltage drop \(V_1\) , between \(A\) and any point \(A_i\) is given by:
\[V_1 = I_A R = \frac{1}{3} \text{ V}\]Now look at the 3 corners adjacent to \(B\) , call them \(B_1\) , \(B_2\) and \(B_3\) . The situation is the same as above with the current flowing in the opposite direction, hence the voltage drop \(V_2\) between any point \(B_i\) and \(B\) is given by:
\[V_2 = \frac{1}{3} \text{ V}\]Now there are 6 equivalent ways for the current to flow in the middle resistor, between the points \(A_i\) and \(B_i\) . Thus the current flowing through each path is \(\frac{1}{6}\) A. The voltage drop \(V_3\) between any connected \(A_i\) and \(B_i\) is given by:
\[V_3 = \frac{1}{6} \text{ V}\]The total voltage drop between \(A\) and \(B\) can be found by following any path from \(A\) to \(B\) , thus:
\[V_A = V_B + V_2 + V_3 + V_1 = \frac{5}{6} \text{ V}\]Finally, use Ohm’s law to find the effective resistance:
\[R_{\text{eff}} = \frac{V}{I} = \frac{\left(\frac{5}{6}\right)}{1} = \frac{5}{6} \text{ } \Omega\]Infinite lattice
Imagine an infinitely extending 2D square lattice. Between any two adjacent intersection points is a resistor of \(1 \Omega\) .
What is the effective resistance between two adjacent intersection points?
The effective resistance is \(\frac{1}{2} \Omega\) .
Proof
Label the two adjacent points \(A\) and \(B\) . We will superimpose two ways of applying current to this circuit:
- Force 1 A into the circuit at point \(A\) while holding the voltage 0 at infinity. Since each path out of \(A\) is equivalent, the current from \(A\) to \(B\) must be \(I_1 = \frac{1}{4}\) A.
- Draw 1 A into the circuit at point \(B\) while holding the voltage 0 at infinity. Since each path out of \(A\) is equivalent, the current from \(A\) to \(B\) must be \(I_2 = \frac{1}{4}\) A.
Now we superimpose the two cases to get the situation where we have 1 A of current flowing from \(A\) to \(B\) , giving a current \(I = I_1 + I_2\) through the resistor:
\[I = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \text{ A}\]The voltage difference between \(A\) and \(B\) is, by Ohm’s law:
\[V = IR = 1 \times \frac{1}{2} = \frac{1}{2} \text{ V}\]But overall, 1 A of current is going from \(A\) to \(B\) , thus the effective resistance, by Ohm’s law is:
\[R_{\text{eff}} = \frac{V}{I} = \frac{\left(\frac{1}{2}\right)}{1} = \frac{1}{2} \text{ } \Omega\]