# Help Applying method to find Thevenin and Norton Equivalents

Discussion in 'Homework Help' started by IronBrain, Mar 3, 2010.

1. ### IronBrain Thread Starter New Member

Mar 3, 2010
6
0
I am to find the Thevin and Norton equivalents of the following circuits, my problem being I am new to these method and any help is appreciated also is there any way to simplify this circuit properly?

2. ### LoganFife New Member

Feb 7, 2010
13
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what have you tried? do you know the basic steps you should be taking?

... is that thing in the middle a current source?

3. ### IronBrain Thread Starter New Member

Mar 3, 2010
6
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Yes, it is a current source.
I haven't a good idea of how to approach this problem.
Is there anything I can simplify in the circuit?
If, so how would I procede on accurately?

4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
The symbol as drawn is somewhat ambiguous. The + and - labels makes it appear to be a current-dependent-voltage-source. The arrow inside the diamond makes it appear to be a current-controlled-current-source.

It makes a significant difference which one it happens to be. Can you clarify the symbol for us?

hgmjr

5. ### IronBrain Thread Starter New Member

Mar 3, 2010
6
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....I stated the clarifaction above.
It is a dependent current controlled current source, the indication of the polarties indicate the direction of the current source, and it is dependent on the current Ix I indicated above.

6. ### hgmjr Moderator

Jan 28, 2005
9,030
214
The direction of the arrow (inside the symbol) is all that is needed to know what direction the current is flowing in. The + and - sign are redundant.

hgmjr

7. ### Heavydoody Active Member

Jul 31, 2009
140
11
I assume you are looking in from the right side. Find the open circuit voltage between those two ends. Next, short out those two ends and determine your short circuit current (the current through your imaginary short). The open circuit voltage is the Thevenin voltage. That voltage divided by the short circuit current is the Thevenin resistance. Together, they make up the Thevenin equivalent (the resistance in series with the voltage source). Perform source transformation on that and you wind up with a Norton equivalent (the current source in parallel with the resistance).

8. ### IronBrain Thread Starter New Member

Mar 3, 2010
6
0
Thanks for the feedback.
Ill try that out and report back if any problems, so I assume there is nothing I can simplify in the circuit right now?

Does the dependent source conflict with me applying these methods?

9. ### Heavydoody Active Member

Jul 31, 2009
140
11
Sorry I disappeared last night. It was a long day. So, I am back from school and I solved this problem.

First, to answer your last question, no further simplification is necessary. The dependent source only adds a third equation to your two nodal equations (three unknowns, three equations).

The numbers here seemed a bit awkward, so if anyone else can confirm this, it would be cool. I came up with -5 volts and 11/2 ohms for the Thevenin and -10/11 amps with the 11/2 ohms for the Norton.

10. ### hgmjr Moderator

Jan 28, 2005
9,030
214
I calculated Vth = -5 Volts and Rth = 11/2 Ohms.

hgmjr

11. ### Heavydoody Active Member

Jul 31, 2009
140
11
Thanks for the confirmation. The numbers had me wondering.