# Simple Thevenin problem. (Dependant sources)

Joined Nov 20, 2013
43  Can anyone tell me if this is correct, and if not what did I do wrong? I'll really appreciate any help I could get.

Thanks

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#### WBahn

Joined Mar 31, 2012
24,700
In your first one, do you really buy that Ix is 1A?

Look at the original circuit (with R replaced by a 1A current source), if Ix is 1A, then that means that the node on the right hand side of it has 1A flowing into it from the 1A source that is part of the circuit and another 1A flowing into in from the test source, but only 1A leaving it through the 2Ω resistor.

Does that make sense?

You need to maintain the integrity of your control signals, which means that you can many any simplifications or transformations that are going to change the control signal. In this case, you've basically replaced the 2Ω resistor in the original circuit with a different resistor that just happens to also be 2Ω in your simplified circuit and the current flowing in the new resistor is NOT Ix.

You also need to learn to check your own work. In most field of engineering, the validity of an answer can usually be determined from the answer itself. Just think of the answer as a given and pick just about any other convenient parameter in the original problem as an unknown. For instance, you are claiming that Rth=4Ω. Okay, let's accept that as a given and then find the voltage of the 2V source (i.e., treat it as an unknown). Since we have 1A flowing into 4Ω, the voltage across the test source is 4V. The two 1A currents combine and go through the 2Ω resistor, so Ix=2A and Vx=4V. This makes the voltage at the negative terminal of the dependent source equal to 0V. With 2A flowing through 2Ω, we drop another 4V getting over to the voltage source which would require that it have a voltage of -4V, not the +2V that it is supposed to have. So the answer is wrong.

• Joined Nov 20, 2013
43
Yeah it doesn't make much sense. For some reason this dependant source problems always get to me.

From what I read apparently to find the Rth you need to first cut all sources (voltage->short, current->open) leave in there the depandant sources and apply a test 1A current (or voltage).Then solve for the circuit V in the Rth and you'll get the answer.

But it's wrong, so what first steps would you say can I do to solve it?

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#### WBahn

Joined Mar 31, 2012
24,700
Actually, I owe you an apology. I thought you were doing something different than you actually were. What you did in order to find Rth is just fine (except you should learn to track your units properly). Rth=4Ω.

Having recognized that, I then looked at your second image and your work for finding Vth is close, but not quite there. Take a look at Voc'' more closely.

My comments regarding whether the current of Ix=1A apply to a 1A source applied to the terminals of the entire circuit (i.e., with both independent sources on). In that case, you have a 1A current source driving current into a 4Ω resistor that is in series with a 6V supply (i.e., the Thevenin equivalent circuit -- so there's a free hint) resulting in a terminal voltage of 6V+1A*4Ω=10V.

You then have Ix=2A which makes the voltage at negative terminal of the dependent source 10V-(2V/A)*Ix=6V. You then drop another 2A*2Ω=4V going across the 2Ω resistor, leaving you with 2V at the top of the 2V source, so everything works out.

• #### anhnha

Joined Apr 19, 2012
880
WBahn:
I am wondering how could you see these tiny images in OP?

#### WBahn

Joined Mar 31, 2012
24,700
WBahn:
I am wondering how could you see these tiny images in OP?
They weren't tiny at all -- the top one is 1275 pixels wide and 1650 pixels high. The contrast wasn't too great, but it was readable.

Joined Nov 20, 2013
43
Actually, I owe you an apology. I thought you were doing something different than you actually were. What you did in order to find Rth is just fine (except you should learn to track your units properly). Rth=4Ω.

Having recognized that, I then looked at your second image and your work for finding Vth is close, but not quite there. Take a look at Voc'' more closely.

My comments regarding whether the current of Ix=1A apply to a 1A source applied to the terminals of the entire circuit (i.e., with both independent sources on). In that case, you have a 1A current source driving current into a 4Ω resistor that is in series with a 6V supply (i.e., the Thevenin equivalent circuit -- so there's a free hint) resulting in a terminal voltage of 6V+1A*4Ω=10V.

You then have Ix=2A which makes the voltage at negative terminal of the dependent source 10V-(2V/A)*Ix=6V. You then drop another 2A*2Ω=4V going across the 2Ω resistor, leaving you with 2V at the top of the 2V source, so everything works out.
That was a dumb mistake in the Vo'' part by me. I think then it should be 2V, since there's no current for the dependant source to be on. So the overall Voc would be 6V, is that correct?

#### WBahn

Joined Mar 31, 2012
24,700
That was a dumb mistake in the Vo'' part by me. I think then it should be 2V, since there's no current for the dependant source to be on. So the overall Voc would be 6V, is that correct?
Mucho correcto (IIRC).

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