# Finding the Thevenin voltage with a dependent source circuit

Discussion in 'Homework Help' started by Grossenhahn, Feb 16, 2015.

1. ### Grossenhahn Thread Starter New Member

Feb 16, 2015
5
0
Hello,

I am to analyse this circuit: http://i.imgur.com/49oZf5F.png

The main goal is to find the Thevenin voltage across the open port, as well as the Thevenin resistance.

I'm having a hard time with the dependent source, as I'm trying to use the node method but cannot figure out my equations.

Can anyone suggest a good place to start? Thanks!

2. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Presumably you learned how to analyze circuits with dependent sources before being told about this think called a Thevenin equivalent. If so, then just treat this like any old circuit and find the voltage at the output. Sometimes people forget what they already know because they think that a Thevenin equivalent is somehow new and different.

If that isn't the issue, then pick an analysis method and set up your equations as best you can. Show them here and we can use that as a starting point.

3. ### Grossenhahn Thread Starter New Member

Feb 16, 2015
5
0
Ok, so I want to use the node method. I'm setting up nodes e_1 between the current source and the dependent source, e_2 between the dependent source and resistor R_2, and my ground node at the bottom.

For node e_1, I know i + (current through dependent source) - I_0 = 0

For node e_2, I know e_2 / R_2 - (current though dependent source) = 0

Presumably, if I know e_2, then I know the Thevenin voltage, however I'm unsure as to what the current through my dependent source is. Is it just i?

4. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
A picture's worth a thousand words and a well-annotated picture embedded in the post is worth at least ten thousand.

I used slightly different labels -- what you called node e_1 I called Node A and what you called node e_2 I called Node B. What you called the (current through dependent source) I called current Ix.

Notice also that (current through the dependent source) says nothing about what direction that current is flowing in. Don't leave that undefined. It doesn't matter how you define it, but you MUST define it. I've defined it according to the passive sign convention, which seems to also be how you meant for it to be defined as well.

So you are doing good so far. You are summing the current out of the node, so you have

Node A: i + Ix - Io = 0
Node B: Vb/R2 - Ix = 0

So you have two equations and four unknowns, {i, Ix, Va, and Vb}

But what do you know about the current i and the voltage Va?
What do you know about the constraint placed on Va and Vb by the dependent voltage source?

5. ### Grossenhahn Thread Starter New Member

Feb 16, 2015
5
0
Thanks for that, I'm not much of a forum user!

Would I be correct in saying that Va - Vb = Zi, and Va = i R1 ?

6. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
No and yes (but possibly by coincidence).

Look at the dependent voltage source. It is saying that the voltage at the + node is equal to Z·i relative to the voltage at the - node. That means that

(V+) - (V-) = Z·i

Since V+ = Vb and V- = Va, what does that mean about your equation above?

As for Va = i·R1, that is correct, but had the current i been defined as going upward, that would have been wrong. Do you understand why?

For a resistor, the current through the resistor from node 1 to node 2 is equal to the voltage at node 1 (V1) relative to the voltage at node 2 (V2) divided by the resistance (R).

So

i = (V1-V2)/R

In this case V1 is Node A and V2 is the ground node, which is 0V, so you have

i = Va/R1

7. ### Grossenhahn Thread Starter New Member

Feb 16, 2015
5
0
Ah yes I see you mean about my first equation, I'm so used to seeing the positive terminal on the left.
So Vb - Va = Zi ?

Now, I can write my two node equations in terms of Va and Ix which gives me

Va / R1 + Ix - Io = 0, and
(Va*(1+Z/R1))/R2 - Ix = 0

Is this correct?

Edit: it seems to be correct since I got the correct answer for Vb = Vth

8. ### Grossenhahn Thread Starter New Member

Feb 16, 2015
5
0
How do I go about determining the Thevenin resistance?

I open the circuit across Io, and try to find the resistance equivalent as looking in from the port, yet this isn't the right answer. What am I doing wrong? I am assuming the value of Z is 2 ohm as this is the value given in the problem.

9. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Yes, as you have surmised, that is correct. It's not the prettiest way to do it, but it certainly works. I'll throw this out for your consideration. Starting with:

Node A: i + Ix - Io = 0
Node B: Vb/R2 - Ix = 0

The second equation let's you immediately combine these to get:

i + Vb/R2 - Io = 0

You then know that

i = Va/R1, so you have:

(Va/R1) + (Vb/R2) - Io = 0

Now you have Vb-Va = Z·i, which can be used to eliminate Va from the above equation to leave you with an equation for Vb, which is what you want.

But any way that is valid is good.

10. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
With a dependent source you can't just turn off all of the supplies and look at the resistance seen looking into the port. You have two options. The first is to take the original circuit and short the output terminals and find the short circuit current (also known as the Norton equivalent current). You can then use this and the Thevenin equivalent voltage to find the equivalent resistance. The other way is to turn off all of the independent sources and apply a text voltage to the outputs and find the text current that flows as a result and, from these two values, fine the equivalent resistance.