# Thevenin with dependent source question

Discussion in 'Homework Help' started by andrewintejas, Oct 16, 2014.

1. ### andrewintejas Thread Starter New Member

Sep 20, 2014
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Sorry in advance for the obtuse question. This is from a textbook example. While calculating the Thevenin Resistance for this circuit, they shorted the terminals a/b as expected. They then say, "observe that when the terminals a,b are shorted together, the control voltage v is reduced to zero..."

I see that the 25 Ohm resistor branch is shunted by this short and thus no current flows across it, but I guess I don't see why the voltage on the right-hand side (when shorted like this) would be zero throughout; whereas, when it's open (as in the photo), there is a voltage across a/b. I assume this is tied to why there's no current across ix when the a/b is open.

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Surely ix is always zero!
Shorting a-b means the control parameter v is simultaneously zero. This makes the controlled source of 3*v equal to zero. This simplifies the analysis to determine the short circuit current across a-b.

Last edited: Oct 16, 2014
3. ### MrAl Well-Known Member

Jun 17, 2014
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Hi,

Yeah that's strange, why do they show ix when it is a single wire without a loop.

Also, i think the Thevenin equivalent will have to include at least one dependent source. Do you have to calculate the Thevenin equivalent?

4. ### andrewintejas Thread Starter New Member

Sep 20, 2014
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I guess what I'm trying to understand conceptually is why there would be a voltage drop on the right-side circuit (across that 25 Ohm resistor) when the a/b port is open, but none when it's shorted.

Also, the control port for the dependent sources is not shown, but I presume that there's an implicit connection between theses dependent sources and their input, i.e., there's no voltage flowing between left and right sides in that picture, yet the 3v VCVS gets voltage from the branch on the other side.

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Vab voltage is 0V because you short A with B (they are connected together). So how can you measure any voltage across A and B if no potential difference across this two points exist, because this point are join together and form one single point.

In electronic the "current" is flowing not the voltage. Voltage do not flow.
Voltage across 25 ohm resistor is a input port for VCVS. And input port don't have to be connected with the output port of a VCVS.
And no current is drawn by VCVS input ports. VCVS input behaves like an ideal voltmeter.
And we have very similarly situation for CCCS. The "input/control port" is a current which flows through 2k ohm resistor.

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6. ### andrewintejas Thread Starter New Member

Sep 20, 2014
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Thanks for the great links. Sorry, I shouldn't have written flow for the voltage! What threw me off on the shorted scenario is that fact that this is a current source (albeit dependent) and as such, we get a current flow in a loop with any voltage differences driving it. Current sources are so counter-intuitive. I don't have day-to-day examples of such things...

What remains confusing to me though is how there's a voltage drop across the 25 ohm resistor in the open a/b scenario. The VCVS takes its input from the drop, yet I don't see what potential difference exists to create it. In the absence of the VCVS, would the drop across the 25ohm be zero?

Anyway, sorry for the idiotic questions.

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
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The current source create the voltage drop across 25ohm resistor. Without VCVS the i current ( "input current" for CCCS) is equal to
5V/2Kohm = 2.5mA. So the CCCS "output current" is equal to 20*i = 50mA. And this means that Vab voltage is equal to :
Vab = -50mA * 25 ohm = - 1.25V
But if now we put VCVS into the circuit, we see that the VCVS "input voltage" will be -1.25V and the "output" 3*-1.25V = -3.75V.
But this will change i current to (5V - (-3.75V))/2k = 4.375mA and Vab = - 2.18575V but this again will give as the new i current value
i = 5.781mA
and new Vab2 = -2.890V and so on until we find the final solution (i = 10mA and Vab = -5V ).

8. ### andrewintejas Thread Starter New Member

Sep 20, 2014
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Mathematically, I found the problem quite straight-forward. KVL, NV, OL and it all works out quickly. I was just struggling with the conceptual aspect: how one can just eyeball it to reach to those conclusions without doing the math (as it was in the textbook). I guess the answer in the first case is that the two dependent sources work together to create a voltage drop. So, I will have to be satisfied with that for now.

Thanks again