Thevenin Voltage & Impedance with Dependent Source: Homework & Solution"

[eehelp150]

Homework Statement

Homework Equations

The Attempt at a Solution

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

 

[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.

 

[eehelp150]

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.

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?

 

[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.

 

[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.

 

[The Electrician]

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?

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

 

[gneill]

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

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

 

[eehelp150]

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

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

 

[gneill]

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

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

 

[eehelp150]

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

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

 

[gneill]

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

And that would be a much better result!

 

[eehelp150]

And that would be a much better result!

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?

 

[gneill]

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?

Right.

 

[eehelp150]

Right.

How would I find Zth?

 

[gneill]

How would I find Zth?

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.

 

[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.

 

[eehelp150]

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.

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?

 

[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?

 

[eehelp150]

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?

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

 

[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:

 

[eehelp150]

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:

View attachment 107413

redid with wolfram and got it. Thanks!