# Effective resistance between two points problem

No.
Hint: Look for balanced wheatstone bridge.

There are two Wheatstone bridges . Can we ignore the bottom one while analysing the top one or vica versa ?

cnh1995
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Gold Member
There are two Wheatstone bridges . Can we ignore the bottom one while analysing the top one or vica versa ?
You can consider them simultaneously. The two bridge-resistors R and R/3 will be redundant.

Jahnavi
The circuit is symmetric - the left side is the mirror image of the right side. So if the voltage at A is 0 and the voltage at B is ##V_B## and you draw a line down the middle of the circuit, what does the voltage have to be along that line?

Thank you very much @tnich .

I think I understand why I was wrong in removing resistors between LM and QM .

Even though I have got the right answer , I am still doubtful about why we could remove resistor between T and M .

The thing which is troubling me is that - The middle line we have drawn is coinciding with the resistor between T and M .

tnich
Homework Helper
Thank you very much @tnich .

I think I understand why I was wrong in removing resistors between LM and QM .

Even though I have got the right answer , I am still doubtful about why we could remove resistor between T and M .

The thing which is troubling me is that - The middle line we have drawn is coinciding with the resistor between T and M .
Since you have solved the problem (and I assume that you used the delta star transformation to do that), I will stop hinting around and explain what I meant. Substitute two R/2 resistors in series for the resistor between L and Q. Call the point between those two resistors K. If we draw a line through T, M and K, the part of the circuit on the right is the mirror image of the part on the left. That means that the voltage drop from B to T, M or K must be the same as the voltage drop from T, M or K to A. So if the voltage at A is 0 and voltage at B is ##V_B##, then the voltage at T, M and K must be ##V_B/2##. Since T, M and K are all at the same voltage, you can combine them into one point without changing the current flow through any of the resistors. When you do that you have a circuit that you can solve using just the series and parallel rules. Also, now the resistor between T and M has both ends connected to the same point (because T and M are the same point), so no current flows through it and it does not contribute anything to the effective resistance between A and B.

Jahnavi
This is exactly how I got the answer first time in post#24

and I assume that you used the delta star transformation to do that)

No . I had actually employed the strategy as suggested by you when I wrote the answer in post#24 . After getting the right answer , then I tried applying Star Delta approach .

Please help me understand one thing i.e how could the middle line drawn ( to exploit symmetry ) coincide with the resistor between T and M .

I am still unsure how we could remove resistor between T and M .

Please reflect a little more on this symmetry aspect .

tnich
Homework Helper
This is exactly how I got the answer first time in post#24

No . I had actually employed the strategy as suggested by you when I wrote the answer in post#24 . After getting the right answer , then I tried applying Star Delta approach .