On this page
An infinite ladder of resistors looks impossible: you cannot add up infinitely many series and parallel steps by hand. Yet the question has a two-line answer once you see that the ladder repeats itself. This post explains why the trick works, solves a ladder with unequal resistors, shows how fast a finite ladder approaches the infinite answer, and tracks how the voltage dies away along the rungs.
Is this topic in the syllabus?
"Series and parallel combinations of resistors" is listed under Current Electricity in the JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus, and the JEE (Advanced) 2026 brochure lists "series and parallel arrangements of resistances". Note one change: the current NCERT Class 12 Current Electricity chapter (2026-27 reprint) no longer has a section on combining resistors and refers back to "the formulae we have derived earlier". Those rules now sit in the Class 10 Electricity chapter, section 11.6. Infinite ladders are not named in any syllabus; they are an exam application of the series and parallel rules. Check the current year's documents.
Why does the ladder equal itself plus one rung?
Each rung has a series resistor R₁ followed by a shunt resistor R₂ across the two rails. Call the input resistance x. If you cut off the first rung, what remains is still an infinite ladder of identical rungs, so it also has resistance x. Putting the first rung back, the shunt R₂ sits in parallel with x, and R₁ is in series with that pair:
x = R₁ + R₂x/(R₂ + x)
Multiply out and you get x² − R₁x − R₁R₂ = 0, so x = [R₁ + √(R₁² + 4R₁R₂)]/2. Take the positive root; a resistance cannot be negative.
Worked example 1. R₁ = 2 Ω and R₂ = 12 Ω.
x² − 2x − 24 = 0, which factorises as (x − 6)(x + 4) = 0.
So x = 6 Ω.
Check: the shunt 12 Ω in parallel with 6 Ω gives 12 × 6/18 = 4 Ω. Add 2 Ω and you are back at 6 Ω.
How does the voltage fall along the ladder?
The same self-similarity tells you about currents. Connect 12 V across the input of the 2 Ω / 12 Ω ladder.
Input current: 12/6 = 2 A.
The 4 Ω parallel block takes 4/6 of the voltage, so the first node sits at 8 V and the first shunt carries 8/12 ≈ 0.67 A.
The ladder after that node is again a 6 Ω ladder, so the next node gets 2/3 of 8 V, which is about 5.33 V, then 3.56 V, then 2.37 V.
The voltage falls by the same factor, 2/3, at every rung. That is a geometric sequence, and it is why very long ladders behave like the infinite one: the far rungs carry almost no current.
How quickly does a finite ladder reach the infinite answer?
Take the simplest case, every resistor equal to R, and build the ladder one rung at a time from the far end. One rung (series plus shunt) gives 2R. Adding rungs gives 5R/3 ≈ 1.667R, then 13R/8 = 1.625R, then 34R/21 ≈ 1.619R. These fractions are ratios of Fibonacci numbers. Srinivasan (1992) showed that the infinite network's resistance divided by R is the golden ratio, (1 + √5)/2 ≈ 1.618. By the fourth rung you are already within 0.1% of it.
There is a neat way to make a finite ladder exact. Terminate it with a resistor equal to x. For our 2 Ω / 12 Ω ladder, put 6 Ω at the far end: one rung then gives 2 + (12 parallel 6) = 6 Ω, and so does any number of rungs.
What if the rungs are not identical?
The trick needs the remainder to be a known multiple of the whole. Worked example 2. Every rung's resistors are double the previous rung's, starting with R and R. The remainder is the same ladder scaled by 2, so its resistance is 2x:
x = R + R(2x)/(R + 2x)
2x² − 3Rx − R² = 0
x = (3 + √17)R/4 ≈ 1.78R
If the rungs change in an irregular way, there is no shortcut and the network has to be reduced or solved with Kirchhoff's laws.
Which mistakes cost marks?
Writing x = R₁ + R₂ + x. The remainder is in parallel with the shunt, not in series with everything.
Keeping the negative root. It solves the quadratic but not the circuit.
Starting the repeat at the wrong place. Check that cutting off your "first rung" really leaves an identical ladder, including which resistor comes first.
Using the trick on a finite ladder without checking how many rungs there are or what terminates it.
For bridges, cubes and other symmetry shortcuts, read our guide to combination of resistors problems. If a network will not reduce, the Kirchhoff's law mistakes guide sets out a sign convention you can trust.
A practice routine
Day 1: Solve three ladders with different R₁ and R₂, and check each answer by substitution.
Day 2: For one of them, find the voltage at the first four nodes and the current in each shunt.
Day 3: Build a finite all-R ladder rung by rung to four rungs and compare with 1.618R.
Day 4: Try a scaled ladder like Worked example 2, then one with capacitors (series and parallel rules swap).
If you set up the self-similar equation and still get a strange answer, compare your steps with the Fermi Problem Library, which has a step-by-step solver that explains the reasoning rather than printing an answer.
Tonight, solve a ladder with R₁ = 3 Ω and R₂ = 6 Ω, then check your x by substituting it back.
