# Circuit Analysis finding Voc for circuit

Discussion in 'Homework Help' started by damm1371, Feb 23, 2012.

1. ### damm1371 Thread Starter New Member

Feb 22, 2012
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The object is to find Voc. pictured is the circuit in question. This is for a basic circuit analysis so the techniques available include, ohms law, KVL,KCL, lakeville, source transformation (Norton/Thevenin), mesh analysis, gaussian bubble rule(current into bubble is current out) and superposition. So IT almost seems possible by doing KVL and doing a determinant, however I keep running into problems. Next I have tried super position which seems to just make everything more complicated. I have a feeling that it could be a combo between lakeville and KVL and maybe a gaussian bubble around the upper right portion of the circuit. but I have yet to find something that works. I am interested in the process to get Voc. However if plugging in figures would help you explain the process to me so be it. I would like it to be step by step, just giving me the answer doesn't help learn at all. Thank You to everyone who takes the time to help.

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I've attached an approach using superposition. I've omitted some parts of the solution so that you can fill in the gaps yourself if you favour this approach over other alternatives.

• ###### Superposition Method.pdf
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3. ### damm1371 Thread Starter New Member

Feb 22, 2012
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Awesome Thank you. I had actually just come across solving it this method as well. You have been a big help thanks.

4. ### damm1371 Thread Starter New Member

Feb 22, 2012
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Okay so after looking more carefully I have some more questions. first off I am not familiar with the term "galvanic connection." mabey there is a way you can dumb it down for me.

I guess I will go down the list with my questions.

In source I1
why is I1 negative?
SO R2 is shorted via V1 to ground which is why it doesn't come into play (typo maybe?) I think you have have typed R3 while meaning R2. But then
Why is there no current in R4?

I would have thought Voc[I1] = I1 * (R3+R4)

conclusions
Voc[I2] = I2 * R4
Voc[I3] = I3 * R3 (or R3+R4) *as per previous question i didn't understand why there was no current in R4 before*

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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A galvanic connection simply means there is a means for current conduction. One may therefore measure an electric potential through a galvanic connection whether current is flowing or not. People are sometimes confused that one could measure a voltage at the unterminated end of a resistor when the other end is connected to a measurable potential.

Why is I1 negative?

It's not - the voltage it produces at Voc is negative by virtue of the direction in which it flows in R3.

SO R2 is shorted via V1 to ground which is why it doesn't come into play (typo maybe?) I think you have have typed R3 while meaning R2.

For I1 in isolation current indeed flows through R1, R2 & R3. The total value I1 passes through R3, whereas it would divide between R1 & R2. Anyway it suffices to know that all of I1 flows in R3. With one end of R3 grounded this means only the voltage across R3 contributes to the value of Voc.

Why is there no current in R4?

With I1 being considered in isolation (for superposition method) other current sources including I2 are viewed as an being open circuit. With I2 being viewed as open circuit, R4 has no means of returning current to any other part of the circuit. It simply terminates at the Voc node.

I would have thought Voc[I1] = I1 * (R3+R4)

That's incorrect for the preceding reason.

Voc[I2] = I2 * R4 [Correct]
Voc[I3]=-I3*R3

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