# understanding thevenin theorem

Discussion in 'Homework Help' started by PG1995, May 24, 2011.

1. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Hi

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 subscripts more meaningful?

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

2. ### jegues Well-Known Member

Sep 13, 2010
735
43
When you have a dependent source you need two parameters to determine the equivalent Thevenin resistance,

The open circuit voltage, $v_{oc}$ (Does that answer your second question?) and the short circuit current, $i_{sc}$.

The approach your authour takes is essentially the same thing. (This is very similiar to the problem you and I had previously worked out on the forums)

He is applying a voltage source at the a-b terminals and is going to determine Rth by seeing how much current the circuit pulls out of the source. Note that the value of the source he applies does not matter. If done correctly it should yield the same numerical answer if you were to compute $\frac{v_{oc}}{i_{sc}}$.

If he applys a 1V voltage source, and the circuit pulls a current io, then Rth will be,

$R_{th} = \frac{1V}{i_{o}}$

If I have more time I'll take a look at your other 2 questions.

Hope this helps!

Last edited: May 24, 2011
3. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Thank you, jegues. I hope you would have free time soon, or perhaps someone else would drop in to help.

Best regards
PG

4. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Hi again,

I'm trying to understand how a circuit is thevenized practically. Please have a look on the attachment. When the switch is pressed one can measure the voltage on the terminals 'A' and 'B' by connecting a voltmeter. In theory it is said to short the voltage supply and open the current source. I believe in this case it would mean that disconnect the battery's terminals from the rest of the circuit and connect the points 'a' and 'b' with a wire. Then, connect an ohmmeter to the terminals 'A' and 'B' to note the equivalent resistance. Am I thinking along the same lines? Please let me know. Thanks a lot.

Regards
PG

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5. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Would someone please comment on it? Thanks.

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Yes, you could use this procedure to obtain the Vth and Rth.
But in practice you do not have to disconnect the battery's terminals from the rest of the circuit.
You simply use voltmeter and measure the voltage between 'A' and 'B' terminals. And next you connecting the ammeter to measure the "short" current. And use Ohm's law to find Rth.
Rth = Vth/Isc

PG1995 likes this.
7. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Thank you, Jony, for the confirmation.

Let's imagine we use the procedure I described in my previous post find Thevenin equivalent. We disconnected voltage source and replaced it with a short circuit. But the question arises that why it wasn't made an open circuit once we disconnected the voltage source because we would have done that if it were current source instead of a voltage source. This post might help you to see what I think of a voltage source and current source. As I state in the linked post that a current source is in reality a voltage source with very, very high resistance in series, therefore I think that's the reason that we make it an open circuit while thevenizing when it's a current source because virtually no current can pass through that very, very high resistance. What do you think of it? Please let me know. Thanks.

Best wishes
PG

8. ### PG1995 Thread Starter Active Member

Apr 15, 2011
753
5
Would someone please comment on it? I hope it's not considered a premature bumping.

Regards
PG

9. ### crutschow Expert

Mar 14, 2008
13,472
3,359
In theory an ideal voltage source provides a voltage with no series resistance, and a current source provides a current with infinite series resistance. Thus to find the Thevenin equivalent circuit impedance you short the voltage source and open the current source.

Practical current sources are not made with a voltage source and a high series resistance. They use a voltage supply and electronic feedback to simulate the infinite output impedance (current independent of output voltage and load impedance). In practice this only works up to the voltage limit of supply.