thevenin circuits

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mikezx
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thevenin circuits

Post by mikezx »

Hey guys, just have a question about thevenin equivalents. I know how to do them with dependent sources, but my question is why can't we de-activate those dependent sources like we do to independent sources when figuring out the equivalent resistance??? thanks...
russlk
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Re: thevenin circuits

Post by russlk »

I don't know what you are talking about. What dependant sources? Is this question SPICE related?
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Chris Smith
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Re: thevenin circuits

Post by Chris Smith »

Can of Worms! You Cant, Its that simple, but make your life easy, dump the "Thev" bull and take up the Metric equivalent? <p>Wish I could remember the name of that formulae?<p>After I finished two weeks of studying the Thev, the next formulae in the book [metric] was "almost" exactly alike, but none of the "Trig., Algebra, and Rocket Science, and it also works to within one percent, in five seconds or less. Dont waste your time on the Thev, unless your passing your exams.
mikezx
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Re: thevenin circuits

Post by mikezx »

Hey thanks. Can of worms...probably a good way to put it. Russ I was just considering any arbitrary circuit that has both independent and dependent sources. The Thev. procedure doesn't let you de-activate those dependent ones to figure out the Thev. equivalent circuit, but it will let you de-activate the independent sources. Instead, you can place a test voltage or current in place of the load and figure out R(th) by taking the ratio of V(test)/I(test).
russlk
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Re: thevenin circuits

Post by russlk »

Consider the case of a 1 volt fixed source having 1 ohm source resistance in series with a dependent source whose voltage in volts is equal to the output current in amps. The dependent source resistance is also 1 ohm. The open circuit voltage is one volt so the Thevenin equivalent is one volt. If we short the output terminals the current is 1 amp so the Thevenin source resistance is 1 ohm. As a test, we put a 1 ohm load on the circuit and get a current of 1/2 amp. So, it appears that it works. As a further test, reverse the polarity of the dependent generator. Now the short circuit current is .333 amp so the Thevenin source resistance must be 3 ohms. Indeed, if we load the circuit with 3 ohms, the voltage is 1/2 volt. I used SPICE to do this test.<p>[ June 07, 2002: Message edited by: Russ Kincaid ]</p>
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