Irodov 3.151: A Termination That Hides the Length – Amit & Deepali Sinha Foundation

Problem Irodov 3.151

A ladder network is connected between the terminals $A$ (upper wire) and $B$ (lower wire). Directly across $A$ and $B$ sits a resistor $R$ . It is followed by a number of identical cells, each made of a resistor $2R$ in the upper wire followed by a resistor $R$ connecting the upper wire to the lower one. After the last cell the upper wire carries one more resistor $2R$ , and the ladder is closed by a resistor $R_x$ from the end of the upper wire to the end of the lower wire. For what value of $R_x$ is the resistance between $A$ and $B$ independent of the number of cells?

The ladder between A and B: a shunt R across the terminals, then cells of a series 2R and a shunt R; the dashed gap stands for any number of cells; the last 2R is closed by R_x. — The ladder between $A$ and $B$ : a shunt $R$ across the terminals, then cells of a series $2R$ and a shunt $R$ ; the dashed gap stands for any number of cells; the last $2R$ is closed by $R_x$ .

Given

shunts $R$
series resistors $2R$
terminating resistor $R_x$
any number of cells

Find

R_x

for which

R_{AB}

does not depend on the number of cells

Solution

Step 1The ladder as one operation repeated

Every shunt $R$ looks to the right into a series $2R$ followed by whatever the ladder holds beyond it. If that remainder has resistance $x$ , the shunt and the branch $2R + x$ are in parallel between the same two wires, so the resistance seen at that shunt is

f(x) = \frac{R\,(2R + x)}{R + 2R + x} = \frac{R\,(2R + x)}{3R + x} \tag{1}

The far end of the ladder seen from the last shunt: R across the points C, D, in parallel with 2R and R_x in series. — The far end of the ladder seen from the last shunt: $R$ across the points $C$ , $D$ , in parallel with $2R$ and $R_x$ in series.

Seen from the last shunt the remainder is $R_x$ alone. Each step toward $A$ applies $f$ once more, so with $n$ shunts in all

R_{AB} = \underbrace{f\bigl(f(\cdots f}_{n}(R_x)\cdots)\bigr)

Splitting off the whole part of (1) shows that $f$ increases steadily with $x$ , since the subtracted fraction shrinks as $x$ grows:

f(x) = \frac{R\,(3R + x) - R^2}{3R + x} = R - \frac{R^2}{3R + x}

Step 2The condition for independence

One more cell inserted at the far end gives the same chain of operations with $R_x$ replaced by $f(R_x)$ . Since $f$ never sends two inputs to one output, the two chains agree only if that replacement changes nothing, that is, if $R_x$ reproduces itself under (1):

R_x = \frac{R\,(2R + R_x)}{3R + R_x}

Clearing the denominator gives a quadratic in $R_x$ :

\begin{aligned} 3RR_x + R_x^2 &= 2R^2 + RR_x\\ R_x^2 + 2RR_x - 2R^2 &= 0 \end{aligned}

The quadratic formula gives its two roots:

R_x = R\left(-1 \pm \sqrt{3}\right)

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

R_x = R\left(\sqrt{3} - 1\right) \approx 0.73\,R

(2)

With this termination every shunt sees $R_x$ beyond it and passes $R_x$ on, so the whole ladder between $A$ and $B$ also has the resistance $R_x$ , whatever the number of cells.

Check

Substituting (2) into (1), then multiplying top and bottom by $2 - \sqrt{3}$ , whose product with $2 + \sqrt{3}$ is $1$ :

f(R_x) = \frac{R\,(1 + \sqrt{3})}{2 + \sqrt{3}} = R\,(1 + \sqrt{3})(2 - \sqrt{3}) = R\,(\sqrt{3} - 1)

which is $R_x$ again. Bounds: the value must lie below the bare shunt $R$ , since the branch beside it can only lower the resistance, and $0.73\,R < R$ .

Remark

$R(\sqrt{3} - 1)$ is also the resistance of an infinite ladder of these cells, because an infinite ladder is unchanged by adding one more cell in front. Terminating a finite ladder in that resistance makes it indistinguishable from the infinite one at its input: this is the idea of a matched load, the characteristic resistance that lets a finite transmission line or attenuator behave as if it went on forever.