What is the concept of equivalent resistance in electrical circuits?

[Ezio3.1415]

1.What is the equivalent resistance between A and B? (The diagonals do not intersect)

2.equations: I don't think I can provide any equation u don't know... Then again my last 2 posts were deleted... so no mistake this time... :D
Rs=R1+R2+...
1/Rp=1/R1+1/R2+...
concept of voltage,current etc...

3.I first thought the answer would be 2R+4 as from the first junction current would flow through the diagonal as it has no resistance... But I am not sure of that as the path after the diagonal has a resistance... That could effect as a resistance for the diagonal path... Then how to do it?

PS: Trust me its not my homework...

 

[CWatters]

The right hand side looks like a square. Fold it diagonally.

 

[gneill]

You can move a component's connection point anywhere along a continuous wire. So consider the components that connect to the diagonal ending at the bottom right corner. What happens if you slide those two resistor's connections up along the diagonal to its other end (essentially shortening the length of the diagonal wire until it becomes zero length)? How does this "new" version of the circuit look? See any opportunities for simplification?

 

[Ezio3.1415]

Your answers are not the same... And they give different values... 2R+4 & 2R+10

 

[Ezio3.1415]

How to be sure of its answer then?

 

[gneill]

Your answers are not the same... And they give different values... 2R+4 & 2R+10

No, I don't think they give either of those answers.

Can you show your work? How are you arriving at those results?

 

[Ezio3.1415]

"Fold it diagonally." 4 on the left and lower side becomes 8... they are in parallel... Re=4
now R,4,R are in series
2R+4

And according to ur method,bottom right corner goes to upper left... so 4^-1+4^-1+... ...=1 Oh now its 2R+1 ... I guess its right...

What's your answer?

 

[gneill]

"Fold it diagonally." 4 on the left and lower side becomes 8... they are in parallel... Re=4
now R,4,R are in series
2R+4

And according to ur method,bottom right corner goes to upper left... so 4^-1+4^-1+... ...=1 Oh now its 2R+1 ... I guess its right...

What's your answer?

"Folding diagonally" should achieve the same result as "sliding the connections" up the diagonal; The "folding" places the 4Ω resistors in parallel in the same manner.

So the final result that you've arrived at, 2R + 1 looks good.

 

[Ezio3.1415]

Oh now I understand... Previously I thought folding means superposing the resistor... Hahaha... However,your process makes more sense... And folding is saying what u said in a different way...

Thank you very much guys...