- #1
The above looks okay. Note that I completed the second equation (red bit).Jd303 said:Ok so i get:
Node 1: (Vs-V1)/R - (V1-2Vx)/R - (V1-Vo)/R = 0
Node 2: (V1-Vo)/R + Is - Vo/R = 0
Which leaves me with V1 - Vo = Vs
and -V1 + 2Vo = Is*R
Have I done anything wrong?
Jd303 said:Will Io not just equal Is if the output is shorted?
The above looks good for the open-circuit case (finding Vo). You should point out at this point that the above expression for Vo is also Vth (the Thevenin voltage).Jd303 said:Ok so I have tried attempted the theory used from gneill.
My final equations are:
V1 + Vo = Vs
-V1 + 2Vo = IsR
Therefore:
Vo = (Vs + IsR)/3
V1 = (2Vs - IsR)/3
... and since Vo is shorted at this point ...And short circuit current equates to:
Io = Is + (V1-Vo)/R
Egads! You're right. Very careless of me, and good catch on your part.uart said:I didn't get those equations. I think the problem is the node marked "2Vx" in reply #8 should have been labeled as "-2Vx".
I went through this quickly just now and I got the same answer for R_Thv as I previously got using the "test source" method outlined in reply #12.
Yeah no problems. It's always the simple mistakes that catch me out too.gneill said:Egads! You're right. Very careless of me, and good catch on your part.
Jd303 said:uart, using replacing the centre node with -2Vx I get the following equations:
5(V1) -3(Vo) = Vs
-V1 + 2Vo + IsR <----- not an equation
Io = Is + (V1 - Vo)/R <------ Solving for short circuit current, Vo should be zero
Rth = Vo/Io
Using the initial values set in comment #5 do you get Rth = 2.9kohms? I instead simply end up with the value of R
Jd303 said:Of coarse! Vo is not the same when the output is shorted, what an oversight! I didn't even pick that up from the hint that you tried to give me.
Thanks, I don't think i could possibly get this wrong now (But you never know)
Thevenin resistance is the equivalent resistance of a complex circuit with dependent sources, as seen from two specified terminals. It is a simplified representation of the original circuit, and is used to calculate the current and voltage at those terminals.
Thevenin resistance is calculated by replacing all independent sources with their internal resistances, and then finding the resistance between the two specified terminals. This can be done through various methods, such as nodal analysis or mesh analysis.
Thevenin resistance is significant because it allows us to simplify complex circuits and make them easier to analyze. It also helps us to determine the behavior of a circuit at a specific point, without having to analyze the entire circuit.
Yes, Thevenin resistance can be negative in a complex circuit with dependent sources. This typically occurs when the dependent sources are inverting, causing the equivalent resistance to have a negative value.
The presence of dependent sources can make the calculation of Thevenin resistance more complex, as it requires the use of additional equations and variables. However, the overall process is still the same, and the presence of dependent sources does not change the fundamental principles of Thevenin resistance calculation.