Is there a way of solving Wheatstone Bridge using a combination of resistors (parallel or series) by changing the structure of the circuit, or some other way, instead of making the potential across the wire the same so that there is no flow of current through the diagonal wire? Can't we directly have a solution using a combination of resistors?
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$\begingroup$ See here: en.wikipedia.org/wiki/Y-%CE%94_transform. $\endgroup$– Ed VCommented Nov 17, 2022 at 12:50
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1$\begingroup$ @Ed V. You could indeed use the $\Delta$ to Y transform on the top three resistors of the bridge and hence, using series and parallel formulae, find the total equivalent resistance, $R$, between the top and bottom nodes. You could then find the condition for $R$ to be independent of the bridging resistance – if your maths is up to it. Mine isn't, except with hard labour. If the procedure could be carried out it would fulfil the OP's wish to obtain the bridge balance condition without referring to voltages or currents! $\endgroup$– Philip WoodCommented Nov 17, 2022 at 17:55
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$\begingroup$ @PhilipWood I tried it once on the Wheatstone bridge and it was tedious, to say the least. Still, once the equations have been wrestled into submission, as it were, it might have some utility for sensitivity analysis or the like. $\endgroup$– Ed VCommented Nov 17, 2022 at 18:33
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*. . . . instead of making the potential across the wire same so that the there is no flow of current through the diagonal wire?
Since that is the balance condition it would be rather difficult not to include it in any derivation?
There are two potential dividers.
$V_{\rm A} = \dfrac {R_3}{R_1+R_3}\cdot V$ and $V_{\rm B} = \dfrac {R_4}{R_2+R_4}\cdot V$
If $V_{\rm A} = V_{\rm B}$ then $\dfrac {R_3}{R_1+R_3}\cdot V = \dfrac {R_4}{R_2+R_4}\cdot V \Rightarrow R_2\,R_3 = R_1\,R_4$
