Understanding thevenin theorem

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Main Question or Discussion Point

Hi, :smile:

I don't think this fits into homework section because in the first place it's not homework related problem, and secondly I'm only concerned about understanding the theorem. If you think it doesn't belong here, then please forgive me, and I request you to move it to other section. Thanks.

I'm trying to understand Thevenin's thoerem through an example problem. Given below are two scans of the example problem.

1: The author says: Because of the presence of the dependent source, however, we excite the network with a voltage source Vo...

Okay, what would the author have done if there were no dependent source?

2: Why does the author choose subscripts "o" and "oc" in Vo and Voc when the author is concerned about the terminals "a" and "b" of the circuit? Why didn't the author choose more meaningful subscripts?

3: I1 = 1/6 A, I2 = -1/18 A, I3 = -1/6 A

The above currents, I1, I2, and I3, are for three loops. The circuit consists of three loops, right? The current supplied by the 1V source is assumed to be Io. This is the current being fed to the entire circuit. The authors then says Io = -I3. Why? The I3 is the current only in ONE loop which is only one part of the circuit, then how can we equate something such as Io which is being fed into the entire circuit to something which is only being fed in a part of that entire circuit?

4: The author says: 4(I1 - I2) = Vx.
Is this correct? Previously, the author has said: -4I2 = Vx = I1 - I2.

Please help me with the above queries. It would be really kind and nice of you. Thank you

Scans
1: part one of the example:
http://img197.imageshack.us/img197/5708/theveninpart1.jpg
2: part two of the example
http://img709.imageshack.us/img709/3244/theveninpart2.jpg

Cheers
 

Answers and Replies

  • #2
Zryn
Gold Member
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1: The author says: Because of the presence of the dependent source, however, we excite the network with a voltage source Vo...

Okay, what would the author have done if there were no dependent source?
He would have skipped adding a source to excite the dependent one, Open Circuit (OC) all the current sources and Short Circuit (SC) all the voltage sources, and then calculated the resistance seen by the terminals of the network.

2: Why does the author choose subscripts "o" and "oc" in Vo and Voc when the author is concerned about the terminals "a" and "b" of the circuit? Why didn't the author choose more meaningful subscripts?
The terminals of the network are where it would traditionally connect to a load of some sort, and are generally labeled as Vout or Vo for short. Any voltage could be Vab but Vo is generally a unique term for the output voltage.

Voc is the Open Circuit Voltage seen at the output terminals while calculating the Rth

3: I1 = 1/6 A, I2 = -1/18 A, I3 = -1/6 A

The above currents, I1, I2, and I3, are for three loops. The circuit consists of three loops, right? The current supplied by the 1V source is assumed to be Io. This is the current being fed to the entire circuit. The authors then says Io = -I3. Why? The I3 is the current only in ONE loop which is only one part of the circuit, then how can we equate something such as Io which is being fed into the entire circuit to something which is only being fed in a part of that entire circuit?
Io is the only current in the I3 mesh and is going in the opposite direction to I3, hence the minus sign. If the voltage source was a current source then the mesh's current value would be that sources current value, would it not? Yes, I3 is being fed into the entire circuit, otherwise how would the I1 and I2 mesh currents have values?

4: The author says: 4(I1 - I2) = Vx. Is this correct? Previously, the author has said: -4I2 = Vx = I1 - I2.
Be careful which circuit and which set of calculations you are looking at. The values you find with the added voltage source are not compatible with the values you find in the original circuit (excluding the Vx value that gives the dependent source a static value)!
 
  • #3
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Many, many thanks, Zryn. I'm much grateful for all the help.

Cheers
 

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