How to find the equivalent resistance of this infinite circuit?

[zenterix]

TL;DR Summary: I am following the course 6.002 "Circuit and Electronics" on MIT OCW. There are no solutions to the problem sets. I would like to check my solution to one particular problem.

We are asked to find the equivalent resistance of the network

as viewed from its ports.

I simplified the network as

Where $R_{eq}$ is not only the equivalent resistance of the branch shown above, it is also the equivalent resistance of the entire network shown above (because the network shown above repeats itself infinitely).

Then

$$R_{eq}=R+\frac{R_{eq}R}{R+R_{eq}}$$

$$R_{eq}(R+R_{eq})=R(R+R_{eq})+RR_{eq}$$

$$R_{eq}^2=R^2+RR_{eq}$$

$$R_{eq}^2-RR_{eq}-R^2=0$$

$$\Delta = R^2+4R^2=5R^2$$

$$R_{eq}=\frac{R\pm R\sqrt{5}}{2}$$

$$=R\frac{1\pm\sqrt{5}}{2}$$

Is this result correct?

Here is the problem set in full if it is useful.

 

[erobz]

TL;DR Summary: I am following the course 6.002 "Circuit and Electronics" on MIT OCW. There are no solutions to the problem sets. I would like to check my solution to one particular problem.

We are asked to find the equivalent resistance of the network

View attachment 332529
as viewed from its ports.

I simplified the network as

View attachment 332530
Where $R_{eq}$ is not only the equivalent resistance of the branch shown above, it is also the equivalent resistance of the entire network shown above (because the network shown above repeats itself infinitely).Here is the problem set in full if it is useful.

Seems like a clear approach. If your algebra is correct I'd say you have it correct.

 

[DaveE]

$R=\frac{1-\sqrt{5}}{2}$ certainly isn't correct.

Have you checked your own work? How would you use that (positive) value to check?

 

[kuruman]

Your approach is fine. I think the circuit you want is the one shown below in which the resistance between points A and B is to be found is clearer about what you are replacing with what in view of the drawing provided by the problem.

 

[Baluncore]

Note that: (1±√5)/2 ; is the golden ratio, or its reciprocal.

An infinite ladder can be written as an infinite recurring continued fraction, and solved in the same equivalent way.

 

[Babadag]

According to (Endreny 1967)

Endrenyi, J., “Analysis of transmission tower potentials during ground faults" for a transmission line R1∞=R/2+sqrt(R^2+R^2/4)
See also:
https://ace.ucv.ro/sintes11/Volume2/5 ELECTRONICS/E17 - VINTAN Maria.pdf

 

[Steve4Physics]

According to (Endreny 1967)

Endrenyi, J., “Analysis of transmission tower potentials during ground faults" for a transmission line R1∞=R/2+sqrt(R^2+R^2/4)

If not already clear, note that "R/2+sqrt(R^2+R^2/4)" simplifies to $R \frac {1+\sqrt 5}2$.

 

[Gavran]

The second addend in the expression $R _ { eq } = R \frac { 1 } { 2 } + ( \pm R \frac { \sqrt 5 } { 2 } )$ is resistance and must be positive. This is why $R _ { eq } = R \frac { 1 + \sqrt 5 } { 2 }$ is the only acceptable solution.

 

[Sat-P]

TL;DR Summary: I am following the course 6.002 "Circuit and Electronics" on MIT OCW. There are no solutions to the problem sets. I would like to check my solution to one particular problem.

We are asked to find the equivalent resistance of the network

View attachment 332529
as viewed from its ports.

I simplified the network as

View attachment 332530
Where $R_{eq}$ is not only the equivalent resistance of the branch shown above, it is also the equivalent resistance of the entire network shown above (because the network shown above repeats itself infinitely).

Then

$$R_{eq}=R+\frac{R_{eq}R}{R+R_{eq}}$$

$$R_{eq}(R+R_{eq})=R(R+R_{eq})+RR_{eq}$$

$$R_{eq}^2=R^2+RR_{eq}$$

$$R_{eq}^2-RR_{eq}-R^2=0$$

$$\Delta = R^2+4R^2=5R^2$$

$$R_{eq}=\frac{R\pm R\sqrt{5}}{2}$$

$$=R\frac{1\pm\sqrt{5}}{2}$$

Is this result correct?

Here is the problem set in full if it is useful.

It is indeed but the answer can't be in negative so + one is right

 

[Sat-P]

Try puting R adjacent to 2R and 4R then 8R and so on , of resitance increases by a factor of two parallely till infinity. It is a must try

 

[Baluncore]

An infinite ladder is insensitive to the far end resistance, because that approximation is attenuated by the intermediate ladder.
Guess a far end value, say 1.5 ohms.
Then repeatedly take the reciprocal, and add one, until you get a stable value.
Solve; x = 1 + 1/x ; and you get the golden ratio = 1.618033