Worked exercise — infinite ladder (R_A=2, R_B=4)

Problem

An infinite ladder network has on each cell a resistor $R_A$ in series on the top wire and a resistor $R_B$ shunted between the two wires. With $R_A = 2\;\Omega$ and $R_B = 4\;\Omega$, find the equivalent resistance $Z_\infty$ seen at the input terminals.

$R_A$ in series on the top wire, $R_B$ shunted; the cell repeats to infinity.

Solution

Self-similarity trick. Adding one more cell leaves the ladder unchanged, so $Z_\infty$ satisfies $Z_\infty = R_A + \frac{R_B\,Z_\infty}{R_B + Z_\infty} \;\Longrightarrow\; Z_\infty^2 - R_A\,Z_\infty - R_A R_B = 0$

Substituting the values $R_A = 2\;\Omega$, $R_B = 4\;\Omega$: $Z_\infty^2 - 2\,Z_\infty - 2\cdot 4 = 0 \;\Rightarrow\; Z_\infty^2 - 2\,Z_\infty - 8 = 0$

Discriminant and positive solution: $\Delta = 4 + 32 = 36$ $Z_\infty = \frac{2 + \sqrt{36}}{2} = \frac{2 + 6}{2} = 4\;\Omega$

Immediate check with the self-similarity equation: $Z_\infty = R_A + \frac{R_B\,Z_\infty}{R_B + Z_\infty} = 2 + \frac{4\cdot 4}{4 + 4} = 2 + 2 = 4\;\Omega$ Correct.

$\ev{Z_\infty = 4\;\Omega}$

Collegamenti

Argomenti: Circuiti elettrici Concetti: Resistenze in serie e parallelo Competenze: Impostazione simbolica