Quick Thevenin Equiv Question

Always works? I've yet to see a situation where the general methods dont apply.
1)Remove the load.
2)Find the open circuit voltage across terminals AB (Vth)
3)Find the resistance looking in from the terminals (Rth) by shorting the voltage source or replace a current source with an open circuit.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/thevenin.html

 

3. Deactivate all independent sources. Insert a voltage source Vs between AB and measure the resultant current I; then RTh = Vs/I. Alternatively, insert a current source Is between AB and the resultant voltage V is measured; RTh = V/Is. You may choose any nonzero values for Vs and Is, e.g. Vs = 1V and Is = 1A.

 

Ah, yes. I see your point. Well it turns out that what you for the independent source circuit cannot work for with only dependent sources.

http://en.wikibooks.org/wiki/Electronics/Thevenin/Norton_Equivalents#Thevenin_Equivalents

 

Yes, the method is applicable on all type of circuits.

 

Hi U can visit www.powerelectrical.blogspot.com to see about Thevenin's theorem and more topics in theorems. Also to get clear idea when dependent sources present visit ths link
www.eas.asu.edu/~holbert/ece201/ECE201Lect-13.ppt[/URL]