# Thevenin equivalence

1. Oct 13, 2016

### eehelp150

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

How do I do Thevenin voltage and impedance with a dependent source? Any hints would be greatly appreciated.

2. Oct 13, 2016

### The Electrician

You just did a network with a dependent source using nodal analysis. Solve for the voltage across the 300 ohm resistor in this network; that will be Vth.

3. Oct 13, 2016

### eehelp150

At the "right side" of node V, is it (V+2V)/300 or (V-2V)/300?
Would the nodal equation be:
(V-9)/600 + V/(-j300) + (V+2V)/300 = 0?

4. Oct 13, 2016

### The Electrician

The thing is, you have a supernode here if you're going to do nodal analysis.

See how they have two mesh currents, I1 and I2 shown? I would take that as an invitation to use mesh analysis.

5. Oct 13, 2016

### The Electrician

I see some of your posts changing, but there is no indication of editing. I think you may be deleting a post and then reposting. Don't do that; use the edit function.

When you delete and then repost it can be very confusing.

Do you know how to deal with a supernode? If you don't, perhaps you should do a mesh analysis first.

6. Oct 13, 2016

### The Electrician

That equation is correct, but you need the voltage on the right side of the dependent source--that will be Vth.

7. Oct 13, 2016

### Staff: Mentor

No worries. That would just be 3V. Solve for V and multiply by 3

8. Oct 13, 2016

### eehelp150

Solving for V in that equation, I get V = 9/5 Volts. Is that correct? What would V300ohm be?

9. Oct 13, 2016

### Staff: Mentor

No, the voltage should be complex thanks to the capacitor in the circuit.

10. Oct 13, 2016

### eehelp150

Interesting. I "re-pressed" the solve button on Wolfram alpha and got:
V = (63-18i)/53.

11. Oct 13, 2016

### Staff: Mentor

And that would be a much better result!

12. Oct 13, 2016

### eehelp150

So now I have the voltage across the capacitor and the value of the dependent source (which is simply 2*Vcapacitor). The voltage across the 300ohm resistor is Voc which would be: (Vcapacitor + 2Vcapacitor) for a total of 3 * Vcapacitor right?

13. Oct 13, 2016

### Staff: Mentor

Right.

14. Oct 13, 2016

### eehelp150

How would I find Zth?

15. Oct 13, 2016

### Staff: Mentor

One method is to find the short circuit current across the output (i.e. the Norton equivalent current). The impedance is the ratio of the Thevenin voltage to the Norton current.

16. Oct 13, 2016

### The Electrician

Replace the 9 volt with a short and add a 1 amp current source at the output. Solve the network for V again. The voltage at the A-B terminals will be 3*V. The value of that voltage will be equal to the resistance at the A-B terminals which will be the desired Rth.

Since you already have an equation for V due to the 9 volt source, it will be a small change to add the 1 amp current source.

17. Oct 13, 2016

### eehelp150

I ended up getting I1 = -0.10975609756098 +0.01219512195122i
and I2 = -0.13414634146341 -0.20731707317073i

Did I do it right?
3*V = 197.56 -21.95i

V300ohm = (I2+1)*300 = 259.756-62.19i

Which value is right?

18. Oct 13, 2016

### The Electrician

Neither is right. Let's see the equation you used.

Your original equations was (V-9)/600 + V/(-j300) + (V+2V)/300 = 0

What would it be if you short the 9 volt source and apply 1 amp to the A-B terminals?

19. Oct 13, 2016

### eehelp150

V/600 + V/(-j300) + 3V/300 - 1 =0

20. Oct 13, 2016

### The Electrician

So is this the equation that gave you the values in post #17? If so, something's wrong with the solution because that's the right equation. Did you use MyAlgebra to solve it?

Here's what I get: