Circuit reduction/Thevenin's equvalent resistance question.

[braceman]

Homework Statement

I have a question to find the thevenin equivalent of a circuit (attached). I have found the Vout but am struggling when trying to reduce the circuit to just the Zth value. All the examples I have don't show an impedance between the terminals and I can't find any examples to try and follow through on how to reduce it. it doesn't help that I've got one book saying look at the circuit from the terminals side, and another saying work FROM the other side (where the Vin was and is now short-circuited) towards the terminals. This configuration is proving problematic for me.

Homework Equations

Resistance series and parallel equations.

Re = R1+R2 etc...Re = R1R2/R1+R2

The Attempt at a Solution

I want to work this one out myself and don't want the answer (no point in blagging it if I can't do it again in the future!) but need some pointers.
Can anyone talk me through how to reduce a circuit like this or point me to a webby with some good examples. I've seen a suggestion to use the left hand side in a 'star' config and convert it to a 'delta', but my books haven't covered these so I'm guessing there's another normal way to do it. I'm betting it's just staring me in the face, but I haven't done this type of thing for a good few years. Any pointers would be greatly appreciated!

 

[gneill]

Can you post your circuit diagram as a gif or jpg? My (ancient, creaking) software can't sort out the format of your .docx.doc attachment, and I'm always leery of opening an unknown file with an application that has enough smarts to read/write my hard disk and run macros...

 

[braceman]

is this pic any good for you?

 

[gneill]

is this pic any good for you?

Much better, thanks. I'll take a look...

...Looks like the top rail and bottom rail are connected by the wire on the far left. That makes the top and bottom a single node, and you can move the connections all to one rail (the equivalent of "folding" the circuit along the central line). Does that help you out?

 

[braceman]

hopefully this is close...If it is, then next step would be to reduce the 2 parallel resistor pairs.

Then add the left and top middle resistors to make a single resistor, which would then sit in parallel with the final right resistor,,,and this would give us the final resistance (impedance) for Zth??

 

[gneill]

hopefully this is close...


If it is, then next step would be to reduce the 2 parallel resistor pairs.

Then add the left and top middle resistors to make a single resistor, which would then sit in parallel with the final right resistor,,,and this would give us the final resistance (impedance) for Zth??

Sounds like an excellent plan

 

[braceman]

R1R2/R1+R2 = (2 x 10/7)/2+10/7 = 5/6

 

[braceman]

R1+r2 = 5/6 + 5/2 = 10/3

 

[braceman]

R1R2/R1+R2 = (10Z/3 x 5Z/6) / (10Z/3 + 5Z/6) = 2Z/3

Therefore Zth - 2Z/3 (hopefully!)

 

[gneill]

R1R2/R1+R2 = (10Z/3 x 5Z/6) / (10Z/3 + 5Z/6) = 2Z/3

Therefore Zth - 2Z/3 (hopefully!)

Yup. Looks good.