Hi there!

How would you go about finding the short circuit (s.c.) current \(I_{ab} \) and the open circuit (o.c.) voltage \(V_{ab} \) in order to calculate the equivalent resistance as seen from the terminal ab?

\( R_{ab} = \frac{V_{o.c.}} {I_{s.c.}} = \frac{V_{ab}} {I_{ab}} \)

I have been trying to use the node voltage method for the s.c. case with node c as the principal node and node d as reference. This does not give the correct result.

Is it correct to assume then that the voltage at point c is always 20 V?

Is it correct to assume that the voltage at point a is always 12 V?

Would it be better to assume that there are 2 principal nodes c and d and the reference node e when using the node voltage method?

 

You can just omit the voltage sources (short circuit) and calculate the equivalent resistance. Here, (5||2 ) +2 is the equivalent resistance Rab.

If you are using the node voltage method with node d as reference you cannot assume voltage at c as 20v as voltage at c is 20 more than the point 'e'and not 'd'. To know the voltage at c with d as reference you can use KCL at point c or KVL for the loop with voltage source 20v and 5 and 2 ohm resistors.

 

As the branch C-D is open circuited so no current flows.
Vcd= I*2ohm
Vab=Vcd-12V using KVL.
I=20/7A