# Effective Resistance in a Circuit (with many paths)

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1. Oct 29, 2014

### cavalieregi

1. The problem statement, all variables and given/known data
Hi I am working on a study task. This is the schematic of the circuit. (all resistors are 1 k ohms.)

I need to calculate the following values of resistance that will be seen between the pairs of nodes, indicated by the subscripts.

REF , RCD and RAB

(That is, REF refers to the total resistive effect between node E and node F. This is the resistance that would be 'felt' or 'seen' by a current injected at node E and extracted at node F. Multiple paths means multiple options, and a lowering of the effective resistance.

2. The attempt at a solution
I understand how to calculate effective resistance for series or parallel although I am not sure how I would approach this as I am confused by the alternative paths and hence how to lower effective resistance between the nodes.

2. Oct 30, 2014

### CWatters

Not sure what you mean by "Alternative paths"? Parallel paths ?

Just have a go and show your working so we can see where/if you have gone wrong.

3. Oct 30, 2014

### cavalieregi

Okay for example if current was injected into node E the resistance between E and F going up would be 1k ohms + 1k ohms + 1k ohms = 3k ohms but that current could also flow through the right path and resistance between E and F would be = 1k ohm. This would be the other path.

The question states 'REF refers to the total resistive effect between node E and node F. This is the resistance that would be 'felt' or 'seen' by a current injected at node E and extracted at node F. Multiple paths means multiple options, and a lowering of the effective resistance.'

I am not sure how you would get this effective resistance considering the many paths the current could take.

4. Oct 30, 2014

### Staff: Mentor

Many paths?? How many exactly, for E-F?

These are parallel paths here. You know how to caculate the resultant resistance for resistors in parallel?

5. Oct 30, 2014

### cavalieregi

RT = 1/(1/R1+1/R2+...+1/RN) right. I am just confused by the question as it says if you inject current at node E here will it travel via the right or up to get to node F. We need to find effective resistance between the nodes. The questions says you must take into account these extra paths to the node F.

6. Oct 30, 2014

### Staff: Mentor

Current will make its way taking all available paths. The easiest path will carry a heavier current, the path of highest resistance will carry the least current. These currents are always in accord with Ohm's Law.

If the paths are parallel, you can treat them as resistors in parallel and calculate a single equivalent resistance of the multiple paths.

7. Oct 30, 2014

### CWatters

There are actually three paths between E and F...

a) R6+R5+R2
b) R3
c) R4 (assuming the one on the right is R4)

These are all in parallel with each other so the equivalent resistance is

REF = (R6+R5+R2) // R3 // R4

where // means "in parallel".

8. Nov 1, 2014

### cavalieregi

Cheers this answers all my questions.