You need to use the DsFT to allow you to calculate the current and voltages at nodes A and B since m and n (the grid dimensions) are infinite; ref. equations (28) and (30). In your example above the current and voltage calculations at A and B would be different and hence the resistance will be different.

Dave

DsFT? What is it? And where are equations 28 and 30?

Pardon me, but I am a hobbyist, and a bad theorist as well.

 

Actually, it is a simple problem that is very misleading. The answer is 0.6 ohms.

 

being an infinite array of resistors, couldnt you also associate this problem with that of a cable baving a characteristic impedance?

Not sure how it would work for an ideal resistor since L = 0 and C = 0. Characteristic impedance simplifies to R/G.

DsFT? What is it? And where are equations 28 and 30?

Pardon me, but I am a hobbyist, and a bad theorist as well.

DsFT is also known as the Discrete-time Fourier Transform. The examples of it in use are in equations 28 and 30 scroll down.

Actually, it is a simple problem that is very misleading. The answer is 0.6 ohms.

Indeed it is. Why do you say it is a simple problem?

Dave

 

could you explain to me how do you arrive at the answer of .6 ohms?

 

could you explain to me how do you arrive at the answer of .6 ohms?

The complete answer is here. I am curious to see if nomurphy can show an easier method.

Dave

 

Sorry, but I had forgotten about this post.

The reason it's simple is because the network stretches to infinity (a very critical point, and a piece of misdirection that makes it seem so MUCH more difficult). But, this means that you have equivalent infinite series/parallel resistors from all sides that counteract each other and neither add to, nor subtract from, the value presented between the nodes (red dots).

Current will take the direct path of least resistance, which is the seven resistors between the nodes.

So, just calculate the series/parallel resistance of the seven resistors exisiting between the nodes.