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The equivalent resistance of the infinite network given below is :

Asked in Re-NEET 2022 · Symmetric and complex networks

Figure: Symmetric and complex networks

(1) (1+√5) Ω
(2) 2 Ω
(3) (1+√2) Ω
(4) (1+√3) Ω

Answer: (4) (1+√3) Ω

Step-by-step solution

Removing the first unit leaves the same infinite network, x, across its rung.

x=1+1+(1· x)/(1+x), so x(1+x)=2(1+x)+x.

x²-2x-2=0, so x=(2+√12)/2=1+√3 Ω (the positive root).

Why the other options are wrong

(1) Solves x²-2x-4=0, taking the last term as 4; one unit adds 2 Ω in series and 1 Ω in parallel, giving x²-2x-2=0.
(2) Counts only the two series resistors of the first unit and ignores the rest of the ladder.
(3) This is the answer for a ladder with only one series resistor per unit, x=1+x/(x+1); here each unit has two.

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