Irodov 3.116: The Ladder That Contains Itself – Amit & Deepali Sinha Foundation

Problem Irodov 3.116

An infinite ladder is built by repeating one link without end. Each link has two identical capacitors of capacitance $C$ : one in series along the top wire, then one across from the top wire to the bottom wire. The two input terminals are the left ends of the top and bottom wires, so the first element met from the input is a series capacitor. Find the capacitance of the ladder between its input terminals.

The infinite ladder: each link is a series C on the top wire followed by a shunt C to the bottom wire; the dashes continue both wires without end. — The infinite ladder: each link is a series $C$ on the top wire followed by a shunt $C$ to the bottom wire; the dashes continue both wires without end.

Given

each link: two capacitors $C$ , series then shunt
infinitely many links

Find

the input capacitance

C_x

Solution

Step 1Removing one link leaves the same ladder

Call $C_x$ the capacitance sought. Detach the first link: what stays to its right is again an infinite ladder of the same links, starting with a series $C$ , so between the rails at that point it presents the same capacitance $C_x$ .

The first link kept and the rest of the ladder replaced by its own capacitance C_x, which sits in parallel with the first shunt C. — The first link kept and the rest of the ladder replaced by its own capacitance $C_x$ , which sits in parallel with the first shunt $C$ .

Step 2One equation for

C_x

In the figure, the rest of the ladder $C_x$ stands in parallel with the first shunt, giving $C + C_x$ , and that block is in series with the first series capacitor. The result must be $C_x$ again:

\frac{1}{C_x} = \frac{1}{C} + \frac{1}{C + C_x}

Multiplying through by $C\,C_x(C + C_x)$ :

\begin{align*} C(C + C_x) &= C_x(C + C_x) + C\,C_x \\ C^2 + C\,C_x &= C_x^2 + 2C\,C_x \\ C_x^2 + C\,C_x - C^2 &= 0 \tag{1}\end{align*}

Step 3The physical root

The quadratic (1) has the roots

C_x = \frac{-C \pm \sqrt{C^2 + 4C^2}}{2} = \frac{-1 \pm \sqrt{5}}{2}\,C

A capacitance is positive, so only the upper sign survives:

C_x = \frac{\sqrt{5} - 1}{2}\,C \approx 0.62\,C

(2)

Check

Put the answer back: with $C_x = 0.618\,C$ , the block $C + C_x = 1.618\,C$ in series with $C$ gives $1.618/2.618\,C = 0.618\,C$ , which is $C_x$ again. Bounds: the ladder can be no larger than its first series capacitor alone (a series element always lowers the total), so $C_x < C$ ; and cutting the ladder after the first link leaves $C$ in series with $C$ , which is $C/2$ , and adding links only adds capacitance in parallel, so $C_x > C/2$ . The value $0.62\,C$ sits between the two.

Remark

The number $(\sqrt{5} - 1)/2 = 0.618$ is the reciprocal of the golden ratio. Cutting the ladder after $n$ links gives $C/2$ , $3C/5$ , $8C/13$ , $21C/34$ , …: ratios of neighbouring Fibonacci numbers, taking every other pair, which converge on $0.618$ quickly, so in practice a handful of links already behave like the infinite ladder.