Thevinen Equivalent Analysis Prob

Thread Starter


Joined Mar 5, 2014
I came across a problem while searching for extra problems on the internet for practice.and came across one im having trouble with. I know to figure out a equivalent circuit, we need to find Voc(open circuit across load) and Isc(short circuit across load) to then figure out rL =(Voc/Isc).

Problem 1)
using nodal analysis,and current leaving is positive as my convention.

I decided to source transform the 4A independent source with the 4ohm resistor in parallel to a 16V source with a 4 ohm resistor in series.To try and reduce the nodes for nodal analysis.

But I kinda get stuck with the 2V source,I know itd be difficult to figure out a KCL since the 2V source doesn't have a resistance to divide by. I would like to apply nodal analysis at A and B's nodes. I just having trouble simplifying the circuit.Any help would be appreciated.

I have another problem after this,but ill post once I figure this puppy out:)



Joined Mar 31, 2012
Q1) Which node is your reference node (i.e., "ground")? There's one obvious choice, but unless you indicate it, we are left to guess and/or mind read.

Q2) Assuming you make the obvious choice, what is the voltage at node V2?

Q3) Can you write KCL for node V1?

Hint: While do a source transformation is perfectly legal (though why you didn't combine the 4Ω resistor and the 6Ω resistor before transforming eludes me), it is easier to leave it as is because nodal analysis is about KCL and KCL loves current sources.


Joined Mar 6, 2009
Whilst it's perfectly reasonable to use the short circuit current and open circuit voltage approach to find the Thevenin resistance, I'm curious that you haven't mentioned the idea (in the absence of any controlled sources) of replacing sources either with a short (as with voltage sources) or an open circuit (as with current sources). The Thevenin resistance is the resistance seen looking back into the terminals with the aforementioned replacements applied.

Thread Starter


Joined Mar 5, 2014
if I chose the 6 ohm bottom,right beside the 2V source's (-) as ground.

The KCL: for determining Voc.

at v1:

-4A+v1(1/4Ω+1/6Ω+1/3Ω)-2V/3 = 0

v1(1/4Ω+1/6Ω+1/3Ω)= 14/3A
(left out the 2.5 resistor since theres no current going through it for our Voc

at v2:

would the voltage at v2 just be the voltage source 2V. v2=2V