Current Electricity · Example 3.33

Equivalent Resistance of an Infinite Ladder

Find the equivalent resistance between A and B for the infinite ladder of series resistors R and shunt resistors 2R.

Series R along the top, shunt 2R at every node, repeating forever.

⬇ self-similar trick

Beyond C–D the network is identical to the whole ladder, so its resistance is again x.

Given

Series element (top rail): R per section.
Shunt element (each rung): 2R.
The ladder extends to ∞.

Asked

Equivalent resistance R_AB (call it x).

Key Idea + Formula

Infinity has no end to start from — but adding one more section to an infinite ladder gives back the same infinite ladder. So:

\[ x = R + \big(2R \parallel x\big) = R + \frac{2R\,x}{2R + x} \]

Solution

STEP 1 — Write the self-consistency equation

The first series R, then the first shunt 2R in parallel with "the rest" (= x again): \[ x = R + \frac{2R\,x}{2R + x} \]

STEP 2 — Clear the fraction

\[ (x - R)(2R + x) = 2R\,x \;\Longrightarrow\; x^2 - Rx - 2R^2 = 0 \]

STEP 3 — Solve the quadratic (keep the positive root)

\[ x = \frac{R \pm \sqrt{R^2 + 8R^2}}{2} = \frac{R \pm 3R}{2} \;\Longrightarrow\; x = 2R \]

Check

\( R + (2R \parallel 2R) = R + \dfrac{2R\cdot2R}{4R} = R + R = 2R \) ✓

Equivalent Resistance R_AB = 2R

Short Cut

One ready formula for any infinite ladder with series a and shunt b:

\[ x = \frac{a + \sqrt{a^2 + 4ab}}{2} \]

Here a = R, b = 2R: \( x = \dfrac{R + \sqrt{R^2 + 8R^2}}{2} = \dfrac{R + 3R}{2} = \) 2R. The "+1 section returns the same ladder" idea also works for infinite resistor grids and capacitor/spring chains — set the whole equal to a piece plus itself.

Given → Asked → Formula → Solution → Short Cut · self-similar fixed-point method