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I've been learning about the concept of Wheatstone bridge in current electricity chapter. I attempted to solve this problem by that concept:

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I don't know if my method is right. I tried to take the resistances inside the dotted lines as one unit( represented by a box) :

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Since all the resistances have the same value and by Balanced WB , the 'box' is neglected. However I feel like this is a wrong method. Is it so ? If not please state an alternate formulae ( the question is to calculate equivalent resistance between A and B where all resistance are same

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  • $\begingroup$ What is the problem you are trying to solve? $\endgroup$
    – Bernhard
    Commented Jan 1, 2021 at 10:01

3 Answers 3

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The network is symmetric about the line joining points A and B. This implies no current would flow through the resistors which are perpendicular to the line joining A and B.

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  • $\begingroup$ Since Parvathy J tells us that all the resistances are equal, this answer is essentially right, though only the two resistors perpendicular to AB and in the dotted box can be ignored. We can show this either by invoking symmetry or by considering potentials. $\endgroup$
    – Philip Wood
    Commented Jan 1, 2021 at 23:07
  • $\begingroup$ Yes the method of using the perpendicular axis of symmetry may be appropriate here. I didn't think of that earlier.Thank you aneet kumar. $\endgroup$
    – Parvathy
    Commented Jan 2, 2021 at 5:53
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You cannot replace the dashed bounding box with one equivalent resistance in your second picture. The box in the first picture has four wires connected to it, while your replacement only has two. Essentially this means that you have assumed the resistances at the top and bottom of your box to be zero. This contradicts with your statement that they are all equal.

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You need to apply the delta-wye transformation back and forth (combining resistors when you can) until you have simplified the circuit to a single resistance between A and B.

The Wheatstone Bridge method is used to determine the value of an unknown resistor. It is not a method for practical use in solving circuits like this.

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