Wheatstone Bridge on breadboard

IonDim said:

So I take it that my calculations for solving the circuit are wrong as well? I took:

R4 and R5 in series: R4+R5
In parallel with R3: R3//(R4+R5)
and in series with R1 + R2: R1+R2+R3//(R4+R5)

Click to expand...

Again, what two points are you trying to find the resistance between? This matters!

The resistance between A and D is not the same as the resistance between B and C or the resistance between A and B.

If you want the resistance between A and D, then R4 and R5 are NOT in series. In order to be in series, any path from A to D that goes through R4 must also go through R5. This isn't the case.

How to use symmetry to get the equivalent resistance when the ratio of R1/R4 is equal to the ratio R2/R5 has been discussed already. In this case, symmetry demands that there be no voltage across R3, and hence no current through it, and therefore it can be removed from the circuit. If these ratios are not the same, then you need to use more powerful techniques because you simply do not have a circuit that can be reduced via series/parallel combinations of resistors. The two common ways of doing this are to apply a test voltage from A to D and analyze the resulting circuit to determine the current flowing in the test source. The equivalent resistance is then the test voltage divided by the test source. Another way is to use a delta-wye transform on one side of the circuit to transform it into an equivalent circuit that IS a series/parallel combination of resistors.