An infinite combination of resistors, each having resistance R = 4 Ω, is given below. What is the net resistance between the points A and B? (Each resistance is of equal value, R = 4)
- (a)0 Ω
- (b)2 + 2√5 Ω
- (c)2 + √5 Ω
- (d)∞ Ω
Correct — B, 2 + 2√5 Ω. An infinite ladder looks identical to itself if you strip off one section, so let the whole thing be a single resistance X between A and B. Adding one more identical section in front must give the same X — one series R followed by R in parallel with the rest X — so X = R + (R x X)/(R + X). Clearing fractions gives X^2 - R x X - R^2 = 0, whose positive root is X = R(1 + √5)/2. Putting R = 4 Ω, X = 4(1 + √5)/2 = 2(1 + √5) = 2 + 2√5 Ω, about 6.47 Ω.
- (a)0 Ω — Zero resistance would mean A and B are joined by a perfect short. A network built from finite 4 Ω resistors, however connected, can never present zero resistance between two points unless a plain wire links them, which the ladder does not.
- (c)2 + √5 Ω — This keeps the correct '2 +' part but mis-simplifies 4(1 + √5)/2. Distributing the 4 correctly gives 4/2 + (4/2)√5 = 2 + 2√5, so the √5 term must also carry the factor 2 — the value is 2 + 2√5, not 2 + √5.
- (d)∞ Ω — An infinite number of resistors does not force infinite resistance. Each shunt arm offers current an extra path, so the series builds up to a finite limit, R(1 + √5)/2. Only a chain with no shunt paths (pure series to infinity) would diverge to ∞.
An infinite repeating (ladder) network is solved by self-similarity: because removing one section leaves an identical infinite network, the resistance of the whole equals the resistance of the whole with one section added in front. That single condition turns the infinite circuit into one algebraic equation you can solve.
Do not try to add resistors one by one — set the unknown total X and use the fact that one more section reproduces X. A section here is a series R feeding an R that sits in parallel with the remaining ladder X, giving X = R + (R x X)/(R + X). This becomes a quadratic X^2 - R x X - R^2 = 0; take the positive root because resistance cannot be negative.
- Because an infinite ladder is self-similar, adding one identical section in front leaves its resistance unchanged.
- The self-consistency condition is X = R + (R x X)/(R + X), which reduces to X^2 - R x X - R^2 = 0.
- The positive root is X = R(1 + √5)/2, so the network resistance is the golden ratio (1 + √5)/2 ≈ 1.618 times R.
- For R = 4 Ω, X = 2(1 + √5) = 2 + 2√5 Ω ≈ 6.47 Ω.
- An infinite chain of finite resistors converges to a finite value — it is neither 0 nor ∞.
- Trying to sum the resistors term by term instead of using self-similarity.
- Assuming infinitely many resistors must give infinite resistance.
- Taking the negative root of the quadratic — resistance must be positive.
NDA/UPSC pose an infinite (repeating) resistor ladder and ask for the resistance between two nodes — set X, write the self-consistency equation, solve the quadratic and keep the positive root.
No directly related past PYQ was found.
- practice — not a real PYQ
In an infinite resistor ladder, what happens to the total resistance when one more identical section is added at the front?
- (a)It increases by R
- (b)It decreases by R
- (c)It stays exactly the same
- (d)It doubles
Answer(c) It stays exactly the same — self-similarity of the infinite ladder is the property that lets you solve it.
- practice — not a real PYQ
For the same infinite ladder with each resistor R = 1 Ω, the net resistance between A and B is
- (a)1 Ω
- (b)(1 + √5)/2 Ω ≈ 1.618 Ω
- (c)2 Ω
- (d)∞ Ω
Answer(b) (1 + √5)/2 Ω ≈ 1.618 Ω — the resistance is R(1 + √5)/2, the golden ratio times R.