The equivalent resistance between points A and B of an infinite network of resistance each of  1Ω connected as shown is

# The equivalent resistance between points A and B of an infinite network of resistance each of  $1\Omega$ connected as shown is

1. A

infinite

2. B

$2\Omega$

3. C

zero

4. D

$\left(\frac{1+\sqrt{5}}{2}\right)$

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### Solution:

${R}_{AB}=R+\frac{R×{R}_{AB}}{R+{R}_{AB}}$

${R}_{AB}=1+\frac{1×{R}_{AB}}{1+{R}_{AB}}$

${R}_{AB}\left(1+{R}_{AB}\right)=1+2{R}_{AB}$

${R}_{AB}+{R}_{AB}^{2}=1+2{R}_{AB}$

${R}_{AB}^{2}-{R}_{AB}-1=0$

$\frac{+1±\sqrt{1+4}}{2}=\frac{1+\sqrt{5}}{2}$

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