circuit analysiscircuit-designresistorsshort-circuitwheatstone-bridge
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?
You can use T to Pi transformation which would make things a little more practical. Reference to : Pi and T networks .
Hope it helped.
It's quite easy if you exploit the symmetry:
Because of the radial symmetry the 5 marked nodes have the same potential. Therefore there will be no current going throught the 5 marked resistors, i.e. you can treat them as being absent.
So the final resistance will be (R + R / 2) / 5
Best Answer
I’d suggest you consider the schematic below:
simulate this circuit – Schematic created using CircuitLab
What is the voltage between
O1andO2? What current would flow through a resistor you would add between those nodes? What if you replace that resistor with a wire?