can you solve this interesting problem?

[wanily1983]

space paper 377 ohms

hi,guys
when interviewed, i meet this interesting problem. the figure undermentioned is the resistor network, the question is calculating the Resistance between node A and node B.

 

[flatulent]

The standard solution is to use superposition. First put 1 Amp into node a and follow it through the network. Find the voltage difference between nodes A and B. Then do the same for node B with -1 Amp. Sum the two voltage differences. Then the resistance is the total voltage drop divided by 1 Amp.

 

[mostafa0020]

paralling an infinite number of resistors with a resistance R will result mathematically into an equivalent resistance value equals to 2/3 R, here is the conclusion :-

for a 3 resistors Xi,R,R parallel with a resistor R , the equivalent value will be :
=((Xi+2R)*R)/(Xi+3R)
For a series with X tends to zero as i tends to infinity , this results in final equivalent resistance value of 2*R/3.

Mostafa

Added after 8 minutes:

sorry
the right side will result in 2R/3
the left the same
teh final will be 2R/6 // R = R/4.

My answer may needs other verifications

 

[wanily1983]

i am sorry for post the same topic twice.
by the way, Mostafa, "For a series with X tends to zero as i tends to infinity", what this mean? i am not clear about that. could you tell for detailed.

 

[mostafa0020]

ok wanily let's check this series i made to represent your case
from right meshes:-

R(i)=(((R(i-1)+2R)*R)/(R(i-1)+3R)

R(0) = R

Take R1 = 3/4 R

R2 = 11R/16
...
The values will decrese
getting R (∞) will be if we put X= R(i-1) tends to zero

resulting in R(∞) = 2/3 R

From left the same
both 2R/3 parallel results in 2R/6

Parallel with R

Resulting in R/4

It is fuzzy some, but , interesting, plz tell me who asked u this question and in which country , thx

Added after 15 minutes:

sorry
i gave these valuse neglecting upper and lower meshes, taking into consideration the same reduction will result in 3 2R/3 resistors right // with R and 3 2R/3 left

means an equivalent parallel of R

the total equivalent resistance between A & B will be R/2.

 

[Mr.Cool]

how is the answer not zero?

as each resistor gets paralleled with its neighbor, the equivalent resistance gets smaller. with an infinit # of resistors, that equivalent resistance is zero. on the left side of A/B and on the right side of A/B (and possibly above & below?)

Mr.Cool

 

[flatulent]

mostafa0020 got the correct answer. R/2

It is a lot easier to note that the array is infinite and that the currents injected as I described above go one quarter into each resistor connected to the node. They individually produce a (I/4)R voltage drop. Adding because of superposition gives a total of (I/2)R volts. Then the equivalent resistance is r=E/I or[( I/2)R]/I which sorts out to R/2

 

[Mr.Cool]

i'll have to try that in PSpice

 

[mostafa0020]

Thankx flatulent, the same answer achieved..

 

[wanily1983]

dear mostafa0020 and flatulent, thank you for your wonderful reply!
i am a chinese student. the problem is from a Doctor of Berkeley and he is working in a chinese company now.

 

[mostafa0020]

Indeed Wanli I thank u and flatulent for this problem,it is an impressive one, i may test engineering students using this problem!!

 

[wanily1983]

hi,flatulent
i am still puzzled with your idea. Assume this condition that there ia only one resistorbetween node A and node B, using ur method,we can get the resistance between A and B is 2*R,which is not right.

 

[flatulent]

The method I used assumes that the current return is at infinity. In real world circuits any current source will have to have two connections at specific points.

In any circuit you can put a voltage or current source between any two points and calculate the current or voltage. Then divide to get the resistance.

The stated problem is the discrete equivalent of sheet resistance. The four point probe calculations are done with the method I suggested.

A piece of history: Before the days of computers, many clever methods were used to obtain solutions to problems. There was a special paper coated with a conducting chemical with a sheet resistance of 377 ohms per square. This was called "space paper" and used for transmission line impedance predictions. Two conductors of the shape and orientation of the cross section of the transmission line in question were pressed onto the space paper. The resistance between them was measured with the usual DC multimeter. The reading was the impedance of the transmission line.

The field plots of two dimensional electric fields could be done with a conductive liquid in a tray with a glass bottom. The metal pieces with the proper shape were put in the liquid and an AC voltage was put between them. A probe was put in the water to measure the voltage at each point. There was graph paper under the glass bottom. The readings were taken down by hand and later made into graphs on paper.

 

[zorro]

Flatulent, it is a pleasure to read your comments.

Now we can propose a variation on this theme: which is the resistance between two points located on a diagonal of the mesh?
Regards

Z