# Thevenin's Theorem with Dependent Source

Discussion in 'Homework Help' started by francisg3, Mar 8, 2009.

1. ### francisg3 Thread Starter Member

Mar 8, 2009
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0
Hello,
I was recently assigned this problem for homework. I needed to find the Thevenin equivalent circuit, more specifically, the voltage across terminals A and B. I know with a dependent source we must excite the circuit to obtain a value for the Thevenin resistance 'Rt'. I was told to use nodal analysis along with the 'voltage open circuit and current short circuit method'. I am familiar with both mesh and nodal analysis however this problem has been giving me a hard time and any advice would be greatly appreciated. The answer I obtained for voltage at terminals A,B was 6v. Thank you.

*Dependent source is 0.5Ix where Ix is the current along the 10 ohm resistor

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Hi francisg3,

I'm not familiar with the circuit notation for the controlled voltage-to-current (?) source with respect the controlling polarity. The input arrow direction may indicate that the (+ve) polarity is sensed from the negative (right hand) end of R1 - I'm not sure. The reason I ask this is that the outcome would be different if the polarity were the opposite.

Suppose the notation implies that the current source direction is "up" when the current Ix flows from left to right through R1. In this case the current Ix in R1 would be added to the 0.5 Ix from the current source at the intersecting node. If the sensing is the opposite, then the 0.5 Ix from the current source would be subtracted from the current flowing out of R1 towards node A and into the 20Ω.

Suppose it's the first case - the currents add at their intersecting node.

The current flowing into the 20Ω would be 1.5 Ix. The voltage Vab would be 20 x 1.5 x Ix or 30 Ix.

Vab = 30 Ix .....(1)

You will find Ix from the relationship

Ix = (Vs - Vab)/R1 ......(2)

Solving equations (1) & (2) to give Vab will hopefully give

Vab = 4.5 V (your Thevenin voltage, Vt)

Use the relationship:

Rt =Vt / Isc where Isc is the short circuit current from A to B

Perhaps I could leave it to you to find Isc & thence Rt and also consider the other case of the current subtracting (if that's really what the circuit implies)

I believe Rt would be 5Ω for the first case.

3. ### The Electrician AAC Fanatic!

Oct 9, 2007
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The OP's first post would seem to indicate that it's a current controlled current source:

"Dependent source is 0.5Ix where Ix is the current along the 10 ohm resistor"

but the symbol looks like a voltage controlled current source.

Can you clarify this point for us, francisg3?

4. ### francisg3 Thread Starter Member

Mar 8, 2009
10
0
Thanks for the replies. The voltage Ix is experienced from left to right through the 10 ohm resistor. The dependent source (a current controlled current source) is 0.5Ix moving upwards. As I said, I simulated the circuit and I obtained a value of 6V for Vab, I may be wrong. This is the following nodal equations obtained:

Ix=(6-Vab)/10 (i)

-Ix-0.5Ix+(Vab/20)=0 (ii) *using a current entering as a negative sign convention*

...solving gives me an answer of 9v, which I believe to be wrong.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Hello francisg3,

I was wondering if the circuit you supply is an original question posed or your own simulation diagram.

I would expect if the controlled source is a current controlled device the connection you have shown for the drive side current path is bypassing the 10Ω resistor. The source input side should probably be in series (not parallel) with the 10Ω. Under these circumstances I believe you will have Vt = 4.5V etc.....

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Hi again,

As in the attachment this is how my simulation cct would look .... Notice how the controlled source current input loop is in series with the 10Ω.

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7. ### hgmjr Retired Moderator

Jan 28, 2005
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t.n.k.

Your diagram appears to differ from the OP's diagram in that the assumed direction of current for Ix in the OP's diagram is from the node of interest to the 6V source.

I agree that the actual current flow is opposite that assumed by the OP's diagram but that appears to be the way the OP's problem has been posed.

I do realize that the assumed direction has no bearing on the final result but I thought the difference worth mention.

hgmjr

8. ### francisg3 Thread Starter Member

Mar 8, 2009
10
0
t_n_k, you were right. a careless mistake on my part making the Ix parallel rather than in series. Problem solved. Thanks for all your help!

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Thanks for the feedback hgmjr,

It is interesting to look at the other option for the controlled source's drive side current path as attached - in this alternate case I think there is a different solution [Vt=3V, Rt=10Ω ??) - suggested in an earlier post. Hopefully the OP is comfortable with the result - I'm not sure they were given a "correct" answer for the original question.

best regards,

t_n_k

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10. ### francisg3 Thread Starter Member

Mar 8, 2009
10
0
correct, i was not given a 'correct' answer. so as far as i can see (the question is from memory) either the 3v or 4.5v are correct. at least i am familiar with the method of solving it. thanks again.

11. ### stupid Active Member

Oct 18, 2009
81
0
hi tnk,
i stumble upon this website while looking for thevenin's theroem info
i m a part time student of a long distance education.
thevenin's theroem is one of my dreaded type

may i know how u have obtained Rth = 5 ohms?