Question

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Answer 1

Consider the infinite ladder to right of AB.

The ladder is infinitely made up by a 1Ω resistor in parallel and then a 1Ω resistor in series to it (seen from right to left). So its equivalent resistance (say R₁) doesn't change if we additionally connect a 1Ω resistor in parallel and then a 1Ω resistor in series to it because the ladder remains the same.

Thus,

$\small\text{ \sf{\longrightarrow\dfrac{R_1}{R_1+1}+1=R_1} }$

$\small\text{ \sf{\longrightarrow\dfrac{R_1}{R_1+1}=R_1-1} }$

$\small\sf{\longrightarrow R_1=(R_1)^2-1}$

$\small\sf{\longrightarrow (R_1)^2-R_1-1=0}$

Since $\small\sf{R_1>0,}$

$\small\text{ \sf{\longrightarrow R_1=\dfrac{1+\sqrt{1+4}}{2}} }$

$\small\text{ \sf{\longrightarrow R_1=\dfrac{1+\sqrt5}{2}\ \Omega} }$

Similarly, the ladder to left of AB is infinitely made up by a 2Ω resistor in parallel and then a 1Ω resistor in series to it (seen from left to right), so the equivalent resistance (say R₂) doesn't change if we additionally connect a 2Ω resistor in parallel and then a 2Ω resistor in series to it because the ladder remains the same.

Thus,

$\small\text{ \sf{\longrightarrow\dfrac{2R_2}{R_2+2}+2=R_2} }$

$\small\text{ \sf{\longrightarrow\dfrac{2R_2}{R_2+2}=R_2-2} }$

$\small\sf{\longrightarrow 2R_2=(R_2)^2-4}$

$\small\sf{\longrightarrow (R_2)^2-2R_2-4=0}$

Since $\small\sf{R_2>0,}$

$\small\text{ \sf{\longrightarrow R_2=\dfrac{2+\sqrt{4+16}}{2}} }$

$\small\sf{\longrightarrow R_2=(1+\sqrt5)\ \Omega}$

Now $\small\sf{R_1}$ and $\small\sf{R_2}$ are in parallel connection so their equivalent resistance is,

$\small\text{ \sf{\longrightarrow R_{12}=\dfrac{1+\sqrt5}{2+1}} }$

$\small\text{ \sf{\longrightarrow R_{12}=\dfrac{1+\sqrt5}{3}\,\Omega} }$

[N.B.: If resistors having resistances $\small\sf{R}$ and $\small\sf{kR}$ are in parallel connection then their equivalent resistance is $\small\text{ \sf{\dfrac{kR}{k+1}.} }$]

This $\small\text{ \sf{R_{12}} }$ is in parallel with the 1Ω resistor connecting A and B.

Hence the equivalent resistance across AB is,

$\small\text{ \sf{\longrightarrow R_{AB}=\dfrac{\frac{1+\sqrt5}{3}}{\frac{1+\sqrt5}{3}+1}} }$

$\small\text{ \sf{\longrightarrow\underline{\underline{R_{AB}=\dfrac{\sqrt5+1}{\sqrt5+4}\,\Omega}}} }$

Answer 2

Answer:

The resistance across AB = (√5 + 1)/(√5 + 4) ohm

#Be Brainly

Hope it helps!!