See figure attached for problem statement.

I keep confusing myself trying to reduce this resistor network into one single resistor.

Is there an efficient method for dealing with resistor networks in this configuration?

The resistors are sorta in a DELTA configuration so this is probably the key to solving the equivalent resistance of this network, however I'm not still not seeing it.

Does anyone have any ideas, or any suggestions then can make to me so I can make a decent attempt at it?

Thanks again!

 

Well I really don't like this kind of question.
To find Req you need to use Wye-delta several times.
http://en.wikipedia.org/wiki/Y-Δ_transform#.CE.94-load_to_Y-load_transformation_equations

And my answer is Req = 10/9*R = 90Ω

 

Well I really don't like this kind of question.
To find Req you need to use Wye-delta several times.
http://en.wikipedia.org/wiki/Y-%CE%94_transform#.CE.94-load_to_Y-load_transformation_equations

And my answer is Req = 10/9*R = 90Ω

This stuff is pretty confusing, visually atleast.

I had a question about what you did but I think I figured it out, hence why I edited this post

 

Using Delta to Wye conversion is tedius, I prefer to use simple series-parallel redrawal and then apply Node Voltage Method, here, you can see the process of redrawal, first redrawal, second redrawal. You will see that there is a little Bridge circuit in there, yeah, beautiful circuit, ain't it.

The first is normal redrawal,
The second is the Node Voltage method applied

I used a 7 volt battery to find out the current and....resistance of course!

RESISTANCE IS: 90 ohms, (just as Jony said)

 

I prefer to use simple series-parallel redrawal

This is what I tried first but I'm not confident enough in my abilities to redraw it correctly.

Is there any method to your madness? Aside from experience and intution of course!

 

This is another novel solution using circuit symmetry.

 

Yes, there's method, Jegues, gee, I wanna go step by step, ok:

You see the two top resistors in the first redrawal? one is 81 ohms and the other is twice as much, i.e. 162 ohm. So, the 7volt battery feeds 86.4mA to the 81 ohm, then it feeds half as much to the 162 (7 divided by 81...and so on)

Now you have the two currents that you see in the current source redrawal.
If you're wondering why there's 9 resistors in the original circuit and only 7 in the current source circuit, is because the last 3 resistors to the right can be made into one.

Now, what you see in the current source redrawal are somehow 7 resistors all 81 ohms.

And there is a little bridge circuit in there! you solve that little bridge circuit separately, using a battery to get the current, to get the one resistance.

Then you end up with 3 resistors in parallel, 2 current sources in parallel which can be added into one. The you solve it and you will get the voltage for the bottom resistor next to the 7 volt battery of the original circuit, which happens to be 3.5V.

Now you got it.!

 

Don't let my 'explanation' confuse you even dasdasdmore!

 

http://www.allaboutcircuits.com/vol_1/chpt_10/1.html

http://www.allaboutcircuits.com/vol_1/chpt_10/13.html