How Do You Calculate Total Resistance in a Complex Circuit?

[VisualStudio]

Homework Statement

In the above diagram, calculate the total resistance between A and C

R1 = R2 = R3 = R5 = 3.3 Ω

R4 = R7 = R8 = 6.6 Ω

R6 = R9 = R10 = 2.4 Ω

Homework Equations

Sum/Product

The Attempt at a Solution

Can anyone give me an insight on where to start for this?
So my current guess is that, R8+R10 = Parallel, meaning that then becomes in series with R9, I really don't know where to start with this

 

[berkeman]

Homework Statement

View attachment 75113
In the above diagram, calculate the total resistance between A and C

R1 = R2 = R3 = R5 = 3.3 Ω

R4 = R7 = R8 = 6.6 Ω

R6 = R9 = R10 = 2.4 Ω

Homework Equations

Sum/Product

The Attempt at a Solution

Can anyone give me an insight on where to start for this?
So my current guess is that, R8+R10 = Parallel, meaning that then becomes in series with R9, I really don't know where to start with this

Welcome to the PF.

I don't see a lot of simplifications in the circuit. Just write the KCL equations and solve. Put a 1V source between A and C, and see how much current you get. :-)

 

[VisualStudio]

Hi, the question is to calculate the total resistance, therefore we don't need to put in a correct source, correct?

 

[berkeman]

Hi, the question is to calculate the total resistance, therefore we don't need to put in a correct source, correct?

But that network is complicated enough that I don't see easy ways to use series & parallel combinations to collapse the network down to a resistor between A & C. Instead, just put an imaginary 1V voltage source between A & C, and calculate the node voltages and currents. That is one way to figure out the equivalent resistance.

 

[vela]

Can anyone give me an insight on where to start for this?
So my current guess is that, R8+R10 = Parallel, meaning that then becomes in series with R9, I really don't know where to start with this

R8 and R10 aren't in parallel. R9 and R10 are, but their combination isn't in series with anything else. There are two other pairs in parallel as well. Note also that R1 is shorted.

If you know the delta-wye transformation, you could use that, but Berkeman's suggestion is probably easiest once you simplify the circuit as much as you can.

 

[VisualStudio]

Ok thanks for your help, i'll be having another crack at it tommorrow. So what is exactly in series and parallel please?

 

[vela]

When two elements are In series, the current that goes through one must go through the other.

When two elements are in parallel, they're connected to the same two nodes, so the voltage across them is identical.

 

[VisualStudio]

So do you think it's wise that I add a voltage source to this circuit? And where would i place it?

 

[vela]

If you can't simplify it down to one resistor, then yes, you need to add a source. You're trying to find the resistance between A and C, right? Where do you think you should place the source?

 

[VisualStudio]

Sorry for the late reply, I think delta star transformation would be the way about it. Where would I start with this?

 

[VisualStudio]

Could anyone help me with this?

 

[gneill]

First make all the parallel resistance simplifications that are available. Pay attention to what R1 is connected to.
Then identify an obvious candidate for Y to Delta transformation to generate more parallel resistor opportunities.

 

[mfb]

You can use series and parallel resistors until you get a network with 5 resistors where this does not work any more (plus one resistor that can be treated separately).
That smaller network will need a Delta/Y transformation or Kirchhoff's laws, but start with the easier steps first.