Thevenin equivalent, (superposition?)

[Color_of_Cyan]

Homework Statement

http://imageshack.us/a/img14/2387/homeworktest2prob2.jpg

a. Find the Thevenin equivalent circuit with respect to terminals a, B for the circuit.

b. Find also the Norton equivalent circuit with respect to the terminals A, B, for the circuit

Homework Equations

V = IR

current division, voltage division,

KCL, KVLThevenin procedure, Norton procedure.

The Attempt at a Solution

With the load cut out, 4Ω disappears and

25V/5Ω + 3A = Vb/20Ω

100V + 60V = Vb

Vb = 160V, VTh = 160VI need help mainly with getting RThI don't know if I can just simplify the resistors from A to B with that current source (do I cut it out so it becomes an open circuit? I know you put a wire to short the 25V while doing it though).

Trying RTh = VTh/Ino instead, then a wire goes from a to b, then once I do that I try to use superposition (so might need help here with that to

With the 3A cut out the equivalent R is 8.3Ω with 25V so I' = 3A

then with the 25V silenced the current is just I'' = 3A, right? Because that's the total current in?

So 3A + 3A = 6A

Ino = 6ASo for RTh I get 160V / 26.6A

= 6ΩBut I don't think it's correct.

 

[gneill]

Your node equation is not correct. The potential difference across the 5Ω resistor is not 25V. What potentials are at either end of this resistor?

I find that the simplest way to write a node equation is to assume that either all currents flow into the node or all currents flow out. Then all the terms can be written on one side of the equation and their sum must equal zero.

Since there are no dependent supplies, you can suppress all the supplies and calculate the net resistance of the network as seen from terminals a-b.

 

[Color_of_Cyan]

Ahh thanks, my bad, the nodal equation should be:

(25V - Vb)/5 + 3A = Vb/20Ω

So Vb = VTh = 32V then.For the second part do you mean that by suppressing everything that I can do that with the current source too (but only to find RTh)? Would it then just be 4Ω?

 

[gneill]

Ahh thanks, my bad, the nodal equation should be:

(25V - Vb)/5 + 3A = Vb/20Ω

So Vb = VTh = 32V then.

Yes

For the second part do you mean that by suppressing everything that I can do that with the current source too (but only to find RTh)? Would it then just be 4Ω?

Yes. Suppress all sources. Remember that current sources aren't suppressed in the same way that voltage sources are...

 

[Color_of_Cyan]

Ah ok, so it's 8Ω then. Thanks again.