Irodov 3.152: The Chain That Contains Itself – Amit & Deepali Sinha Foundation

Problem Irodov 3.152

An infinite ladder network starts at the terminals $A$ (upper wire) and $B$ (lower wire). It is built by repeating one link without end: a resistor $R_1 = 4.0$ $\Omega$ in the upper wire, followed by a resistor $R_2 = 3.0$ $\Omega$ connecting the upper wire to the lower one. The first link begins at $A$ with its $R_1$ . Find the resistance of the network between $A$ and $B$ .

The infinite ladder: each link is a series R_1 followed by a shunt R_2, starting with R_1 at A; the dashed rails continue without end. — The infinite ladder: each link is a series $R_1$ followed by a shunt $R_2$ , starting with $R_1$ at $A$ ; the dashed rails continue without end.

Given

$R_1 = 4.0$ $\Omega$
$R_2 = 3.0$ $\Omega$
infinitely many links

Find

the resistance

R

between

A

and

B

Solution

Step 1The ladder beyond the first link is the ladder itself

Remove the first link. What is left, seen from the two wires at the first shunt, is again an infinite chain of the same links starting with an $R_1$ , so its resistance is the sought $R$ itself. The whole network is therefore $R_1$ in series with $R_2$ and $R$ in parallel:

The first link, with the rest of the infinite chain replaced by its own resistance R. — The first link, with the rest of the infinite chain replaced by its own resistance $R$ .

R = R_1 + \frac{R_2\,R}{R_2 + R} \tag{1}

Step 2Solving the self-consistency condition

Multiplying (1) by $R_2 + R$ and collecting terms gives a quadratic in $R$ :

\begin{aligned} R\,(R_2 + R) &= R_1(R_2 + R) + R_2 R\\ R^2 - R_1 R - R_1 R_2 &= 0 \end{aligned}

The quadratic formula gives two roots:

R = \frac{R_1 \pm \sqrt{R_1^2 + 4R_1R_2}}{2}

The square root exceeds $R_1$ , so the lower sign gives a negative value; the resistance is the positive root. Numerically, with $R_1 = 4.0$ $\Omega$ and $R_2 = 3.0$ $\Omega$ , the root taken with $R_1$ outside it is

\sqrt{1 + \frac{4R_2}{R_1}} = \sqrt{1 + 3} = 2

R = \frac{R_1}{2}\left(1 + \sqrt{1 + \frac{4R_2}{R_1}}\right) = 6\ \Omega

(2)

Check

Substituting $R = 6$ $\Omega$ back into the right side of (1):

4\ \Omega + \frac{3 \cdot 6}{3 + 6}\ \Omega = 4\ \Omega + 2\ \Omega = 6\ \Omega

as required. Limits of (2): with $R_2 \to 0$ the first shunt shorts everything beyond it and $R \to R_1$ ; with $R_2 \to \infty$ no current can return to $B$ and $R \to \infty$ . The value $6$ $\Omega$ also lies between $R_1 = 4$ $\Omega$ (the shunt shorted) and $R_1 + R_2 = 7$ $\Omega$ (the chain cut after the first link).

Remark

The same trick (an infinite network is unchanged by removing one period) gives the input resistance of any periodic ladder, and it is the discrete form of the characteristic impedance of a transmission line. A finite ladder closed by a $6$ $\Omega$ resistor behaves exactly like the infinite one, and a long ladder closed by anything else approaches $6$ $\Omega$ quickly: each link the reader looks through weakens the effect of the far end.