Effective resistance between two points problem

  • Thread starter Jahnavi
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  • #26
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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 ?
 
  • #27
cnh1995
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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.
 
  • #28
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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 .
 
  • #29
tnich
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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.
 
  • #30
848
102
This is exactly how I got the answer first time in post#24 :smile:

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 .

Your approach is quite admirable !

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 .
 
  • #31
tnich
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This is exactly how I got the answer first time in post#24 :smile:



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 .

Your approach is quite admirable !

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 .
If the voltage is the same at T and M, then there is no current and no voltage drop across the resistor between them. So removing the resistor will not change any voltages or currents anywhere else in the circuit. Therefore, the effective resistance between A and B is the same with or without it.
 
  • #32
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102

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