# How to find the equivalent resistance?

1. Sep 19, 2015

### haha1234

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
(a)I think there are two pairs of resistances are in parallel,one pair is the 40ohm and 50ohm resistances,and the other pair is the 30ohm and 60ohm resistances.But I'm not to sure
So the Req I've calculate is 152.2ohm...
but I have no idea in how to calculate the Req in (b)...
Thanks

2. Sep 19, 2015

### Hesch

a) is wrong because the 10Ω is not in series ( and not in parallel ).

Regard the middle 5 resistors, supply them by a voltage V ( e.g. 100V ), calculate the the current, I, that is passing them. Then Req = V / I + 20Ω + 80Ω.

b) can be solved, using the same method.

3. Sep 19, 2015

### Hesch

(a): By using Kirchhoffs 1. law ( KCL ), you must find ( by 100V ):

I60Ω = 0.811456A
I30Ω = 1.551312A

Itotal = 2.36277A → R5 = 100V / Itotal = 42.3232Ω → Req = 142.3232Ω

This can also be calculated using Thevenin equivalents.

Last edited: Sep 19, 2015
4. Sep 20, 2015

### haha1234

Um...I still cannot understand.
Could you explain why the total current is equal to I60Ω + I30Ω?
Why the value of the total current does not depend on other resistances?
And, how to find it by using Thevenin equivalents?
Thanks

Last edited: Sep 20, 2015
5. Sep 20, 2015

### Hesch

Look at the diagram in #1: There are these two, and no other paths, for the current to pass the circuit. You could also say:

Itotal = I40Ω + I50Ω
In reality there is no current at all, because no voltage is supplied. But to find the resistance of the circuit, you pick out 5 resistors and supply them by
V = 100V and find the resulting current, I. Then R = V / I.

You are just doing an experiment.
Remove the 10Ω resistor.
Again supply 100V across (30Ω+40Ω) and across (50Ω+60Ω).

Calculate the node voltage between 30Ω and 40Ω. This is the Thevenin voltage as for these two resistors.
The Thevenin resistance is the parallel value of these two resistors

Calculate the node voltage between 50Ω and 60Ω. This is the Thevenin voltage as for these two resistors.
The Thevenin resistance is the parallel value of these two resistors

Now reinsert the 10Ω between this and this equvalent, face to face, and calculate the voltages at each end of the resistor.
I40Ω and I50Ω can now be calculated by means of ohm's law. R = V / (I40Ω + I50Ω).

Last edited: Sep 20, 2015
6. Sep 21, 2015

### haha1234

um…
So the 10ohm resistance is set to be a load resistance ?
But I still cannot understand …
There will be an open circuit after we remove the resistance(
Oh…Thanks
Btw I read some solutions from the internet
The solution shows that V1 is equal to V2
But if we supply 100V to them,R1and R2 will get 100V,while R3,R4 and R5 need to share 100V together so I think V1 and V2 may not be the same…
I know I'm too stupid

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7. Sep 22, 2015

### epenguin

Problem (b) I can do in my head.

I get it wrong of course. But I do that with pen and paper anyway.

It's a fairly involved problem for a general circuit of that morphology. But here there are strong simplifying features - all the resistances are equal and it has a symmetry - so a general method is perhaps over-powered.

I take the overall potential difference A-C to be 1. To make it simple I'll let all those 9 individual resistances be 1 too. Because of the symmetry, the potential at B has got to be ½. (It must be ½ at D too, but I won't use that.) By the symmetry the current in EB must equal that in BF. (The previous statement also couldn't be true if this one were not also). There is, so to speak, no redistribution of current at node B. So the node B could equally well not be there, you can detach the upper E-F pathway from it and it makes no difference to any current or potential. So the problem reduces to one of two parallel pathways; the lower pathhway has a section with three parallel pathways within it. Not difficult. I get the circuit resistance to be 10/9. If all the individual resistances are R, that would be the circuit resistance is (10/9)R.

If and when this is right there are interesting extensions and generalisations I believe.

Last edited: Sep 22, 2015