# Equivalent resistance of unbalanced Wheatstone Bridge [duplicate]

Equivalent resistance of a balanced Wheatstone Bridge is pretty simple: if the corresponding ratios are equal the just cancel the wire connecting the centre of two wires. But what if they are not equal? What if the cental wire remains in its position? What will be it's equivalent resistance then? I have tried to use the star delta transformation but it is becoming very complex. Can any one of you give a brief formula for calculating the resistance of unbalanced Wheatstone bridge, and of course show it from where it comes? [ • Your question is unclear and would benefit from a circuit diagram. Do you want the equivalent resistance as seen from the excitation of the Wheatstone bridge or as seen by the detector? Which central wire? Draw a clear diagram. Then, saying that using a star-delta transformation makes things complex is quite an overstatement. – Massimo Ortolano Aug 30 '16 at 17:39
• If you want help with the derivation, then show your work. If you just want a formula, try googling. – sammy gerbil Aug 30 '16 at 20:34

Sad as it is, there exists no brief formula for the equivalent resistance of that network. I'll give you some hints, but not the complete derivation because that's your duty.

You can obtain the equivalent resistance in two ways:

1. As you already guessed, by means of a $\Delta$-$Y$ transformation. The derivation is straightforward and the calculations are not complex, just a bit tedious.
2. By means of Middlebrook's extra element theorem (EET) , choosing $A$ as extra element. This is a powerful theorem of network theory, unfortunately rarely taught in undergraduate classes, which allows to find solutions in so-called "low-entropy" form.

The analysis of that network with the EET is carried out in full in . To show you that there's no brief solution I'll report the final result derived in , adapted to your (awful) notation ($\mathbin{||}$ denotes the parallel between two resistances). You can't get briefer than this:

$$R_\mathrm{eq} = (P+Q)\mathbin{||}(R+S)\frac{1+\dfrac{P\mathbin{||}Q+R\mathbin{||}S}{A}}{1+\dfrac{(Q+S)\mathbin{||}(P+R)}{A}}$$

Notice that the first factor is the equivalent resistance when there is no resistance on the diagonal: the EET decomposes the equivalent resistance in an ideal term multiplied by a discrepancy factor.

You can try to derive the result with a $\Delta$-$Y$ transformation, and check your result against the above formula for correctness.

 R. D. Middlebrook, "Null Double Injection and the Extra Element Theorem", IEEE Trans. Edu., 32, 167-180, 1989 online copy

 V. Vorperian, Fast Analytical Techniques for Electrical and Electronic Circuits, Cambridge University Press, 2011.