Infinite Resistor Grid

This is part one of a two part series in solving two classic resistor network problems.

In this part 1, the infinite Resistor Grid with equal value resistors throughout the grid is solved by invoking symmetry and superposition.

Part two finds the equivalent resistance between two diametrically opposed nodes in a resistor cube.

Two updates where I have further discussed the infinite resistor grid:

MORE HINTS in UPDATE 2:

[georgehnatiuk]

The original grid problem was uploaded on Feb 5, 2018 six months ago. There have been 2800 views as of today and no correct solution has been submitted.
NOTE: *the answer is NOT 2R/3, R/2, 5R/6 etc.* If you got that value then you are NOT applying symmetry correctly and you need to watch BOTH update videos. The actual equivalent resistance value across a diagonal is NOT a rational number. The resistance value between diagonals is NOT of the form P/Q where P and Q are integers.
VIEW THE UPDATES.
GH dated: August 9, 2018

I have further discussed the infinite resistor grid:

MORE HINTS in UPDATE 2:

George Hnatiuk

[georgehnatiuk]

With the use of the "substitution theorem" you do not replace voltage and current sources when using the substitution theorem for those are fixed and do not change. In the substitution theorem PASSIVE elements are replaced with a voltage source that is equal to the voltage across that element. Once a substitution is made, you lose information about the original network and can not go back unless you also calculate the current in the voltage source that is substituted.
I never use the theorem in practice for I have found it to have limited value. It is just an obvious statement of a property of linear circuits.

[rinzhler]

Why are we adding ia and ib for the current passing between nodes a and b ? Can we not just consider the i/4 passing through the node a is the same current which is passing through node b ?

[ayushanand]

Simple and easy to understand. Thank you.
From India

[smitnema]

Why we cnt say that current equally divide at node B When it connected to diagonally with A

[sureshpatil]

If the resistance R between A and B is removed then find the current shown by the ammeter if it is connected between A and B

[_dark_trader]

After super imposing at point A and B KCL is not followed... Why? Please clear this doubt.. Thank you

[harshevardhan]

what will be the voltage required to produce this current? Will it be infinity since the resistance across infinity and A is infinite?

[aiqaamin]

For which class it is? Can anyone tell

[sirichandanreddyanumandla]

sir could you make a video on methods of equipotential reduction

[jony]

Well in my PC the "hiss" (vinyl noise) is not that bad. I'm using headphones.

[jony]

Why don't this "trick" with the two current source not working in the second case?

[rakheeburagohain]

When using superposition, why don't we consider the other 3 branches?

Like if in AB, I/2 flows, and I/4 from the other 3 branches(into B), then shouldn't net current coming out be 5I/4?

[shivakumararm]

It is not divide equally since path resistance is not same...

[ezxd]

Why is R_AB = V_AB/I and not V_AB/(I/2)

[kylefafnis]

This is far more complicated than he would have you believe in the beginning of the video

[rafaelferreira]

Thank You very very much Professor!!!!

[akashverma]

Since the two current source is connected, It should be 2I.

[TheRumpusView]

The hiss is driving me And I want to watch this, having solved this problems myself many years ago.

[gopalajith]

This problem is easily solved by Er.DC Gupta. Text _ Disha publication. _jee main and advanced.