How to Calculate the Equivalent Capacitance of an Infinite Ladder

An “infinite ladder of capacitors” is not completely defined without a circuit diagram. The standard interpretation is a repeating network of identical capacitors: one capacitor in series along the top branch, followed at each node by one capacitor connected to ground. The input terminals are the leftmost top node and ground.

For that topology, the infinite ladder has a finite equivalent capacitance:

$$C_{\mathrm{eq}}=\frac{\sqrt{5}-1}{2}C\approx 0.618C.$$

The result is smaller than one capacitor $C$ because the first capacitor is in series with the rest of the network.

The standard ladder topology

Label the capacitance of every capacitor by $C$. Starting at the input, the circuit consists of:

1. A series capacitor $C$.
2. A node connected to ground through a capacitor $C$.
3. The same pattern repeated indefinitely to the right.

Let $X$ denote the equivalent capacitance seen at the input terminals. Because the ladder is infinite and every section is identical, removing the first section leaves a network whose equivalent capacitance is also $X$.

This self-similarity is the key idea.

Set up the equivalent-capacitance equation

After the first series capacitor, the next node sees two branches to ground:

• the vertical capacitor, with capacitance $C$;
• the remaining infinite ladder, with equivalent capacitance $X$.

These two branches are in parallel, so their combined capacitance is

$$C_{\parallel}=C+X.$$

That parallel combination is in series with the first capacitor $C$. The equivalent capacitance of two capacitors in series is

$$C_{\mathrm{series}}=\frac{C_1C_2}{C_1+C_2}.$$

Therefore,

$$X=\frac{C(C+X)}{C+(C+X)} =\frac{C(C+X)}{2C+X}.$$

Solve the equation

Multiply both sides by $2C+X$:

$$X(2C+X)=C(C+X).$$

Expand both sides:

$$2CX+X^2=C^2+CX.$$

Rearrange into a quadratic equation:

$$X^2+CX-C^2=0.$$

Using the quadratic formula,

$$X=\frac{-C\pm\sqrt{C^2+4C^2}}{2} =\frac{-C\pm\sqrt{5}C}{2}.$$

Capacitance must be nonnegative, so discard the negative root. The physical solution is

$$\boxed{C_{\mathrm{eq}}=\frac{\sqrt{5}-1}{2}C}.$$

Since

$$\frac{\sqrt{5}-1}{2}\approx 0.618034,$$

the infinite ladder has an equivalent capacitance of about $61.8\%$ of the capacitance of one individual capacitor.

Numerical example

Suppose every capacitor has capacitance

$$C=10\ \mathrm{nF}.$$

Then

$$C_{\mathrm{eq}}=\frac{\sqrt{5}-1}{2}(10\ \mathrm{nF}) \approx 6.18\ \mathrm{nF}.$$

The first series capacitor alone would give an upper bound of $10\ \mathrm{nF}$, because the rest of the ladder cannot make the series combination larger than either series component. The result $6.18\ \mathrm{nF}$ is consistent with that bound.

Why an infinite circuit can have a finite answer

“Infinite” refers to the number of components, not to an infinite capacitance. Each additional section is connected through more series capacitors, so its influence on the input becomes progressively limited.

One way to verify convergence is to calculate finite ladders. If $X_n$ is the equivalent capacitance of a ladder with $n$ repeated sections, then

$$X_{n+1}=\frac{C(C+X_n)}{2C+X_n},$$

with $X_1=C/2$ for one series capacitor followed by one shunt capacitor. The sequence approaches $0.618034C$ as $n$ increases.

For example, in units where $C=1$:

Number of sections	Equivalent capacitance
1	$0.500$
2	$0.600$
3	$0.615$
4	$0.6176$
Infinite limit	$0.618034$

The infinite result is the limit of increasingly long finite ladders.

Common mistakes

Treating every capacitor as parallel

Capacitors are parallel only when they connect between the same two nodes. The vertical capacitors do not all share the same upper node, so they cannot simply be added together.

Treating every capacitor as series

The ladder contains branching nodes. Once a vertical capacitor connects to ground, the circuit is no longer one single series chain.

Choosing the negative quadratic root

The other algebraic solution is

$$X=-\frac{1+\sqrt{5}}{2}C,$$

which is negative for positive $C$. It does not represent the equivalent capacitance of ordinary passive capacitors.

Assuming every “infinite ladder” has this answer

The value depends on the topology and on the capacitances. If the series capacitors and shunt capacitors have different values, say $C_s$ and $C_p$, then the same method gives

$$X=\frac{C_s(C_p+X)}{C_s+C_p+X}.$$

After rearranging,

$$X^2+C_pX-C_sC_p=0,$$

so the positive solution is

$$X=\frac{-C_p+\sqrt{C_p^2+4C_sC_p}}{2}.$$

Other arrangements—such as a ladder with alternating series and parallel blocks, or capacitors connected between adjacent nodes without a ground rail—require a different equation. The circuit’s node connections must be specified before an equivalent capacitance can be calculated.