# Finding equivalent resistance of a resistor network

In the circuit below, all of the resistors are 15 Ohms. I am trying to find the equivalent resistance between A and B. Can we argue that since RAD/RDC = RAO/ROC that AODC is a wheatstone bridge and thus say no current will flow through DO and remove DO from the circuit? Using the same argument, can we remove BO too? Can we then just simplify the circuit like below and then find the equivalent resistance using the series and parallel rules? • For what it’s worth, your analysis is wrong. You cannot remove the resistor between nodes O and D. It is wrong because of the current that flows through the resistor between nodes O and B. – user2233709 Nov 19 '17 at 15:58

I’d suggest you consider the schematic below: simulate this circuit – Schematic created using CircuitLab

What is the voltage between O1 and O2? What current would flow through a resistor you would add between those nodes? What if you replace that resistor with a wire?

• Is it zero? How do I know? – Tiash Nov 19 '17 at 12:30
• Given that only A and B are externally connected, you may calculate the voltages at nodes O1 and O2 as functions of the voltages at nodes A and B. – user2233709 Nov 19 '17 at 13:04
• Sorry, I am a little confused. How do I get the voltages at nodes A and B? – Tiash Nov 19 '17 at 13:09
• You don’t get them, you use them as variables. – user2233709 Nov 19 '17 at 14:01
• @Tiash Your question looks like homework. Isn’t it? I tried to give you hints, because I don’t think this site should be a homework service. But it’s hard to help you without knowing what you were taught… You wrote about a wheatstone bridge. If you realy understand what a wheatstone bridge is and how it works, I think you should be able to understand my hint. – user2233709 Nov 19 '17 at 15:56

Equivalent Resistor is 8 Ohm.

Equivalent resistor is independent of voltage applied at Node A and Node B, so, for simplicity assume that Node A is at 1V and Node B is at 0V. Split current and use kirchoff rule at 3 nodes. This problem can be solved fairly easily using Delta to Star Transformations. Recognizing the left and right delta and converting each resistor yields an equivalent resistance of 5R in each leg of the star. After that it is a simple matter of combining series and parallel resistors. simulate this circuit – Schematic created using CircuitLab

By looking at the circuit due to lack of symmetry, it can be said that D,O,B are not at same potential, so what you have done is incorrect if equivalent resistance is to be found across AB.

But the second inage is correct if you are finding resistance across AC. The first answer gives some insight why it should be correct for this case.