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Below are all of the forum threads, including all
the responses to the original posts.
Post subject: Thevenin Equivalence Question Posted: Wed
Aug 24, 2005 7:50 am
Joined: Wed Aug 24, 2005
Hello all- I am studying for my PE after 20
years away from school...
In a very simple circuit of a dependent
current source (value= .75* voltage across resistor) in parallel with
a resistor of 4 ohms, how would one compute the equivalent Thevenin
resistance of that circuit (measured across the parallel combo). Yeah,
I know it's probably very simple... The answer is -2 ohms, I'm told.
Not sure how to arrive at that answer...
Post subject: Posted: Wed Aug 24, 2005 11:32 am
Joined: Tue Aug 23, 2005 6:33 pm
You need to think out of the box on this one. Since you are dealing
with a voltage-dependent current source, you need to to know what the
voltage across the 4ohm resistor is in order to get the source's current
value. Try applying a 1V independent voltage source across the output
(i.e. parallel to the 4ohm R). Then you know that Vresistor = Voc =
1V. That means that Iresistor = 1V/4ohm = 250mA. Also, Idependentsource
will be 0.75A*1V = 750mA. The remaining 500mA will flow through your
1V source at the output and be considered your Isc.
= 1V/500mA = 2ohm. This result is dependent upon the position of the
current source. If, in normal circuit convention, the arrow points up,
then the above answer is correct. However, if the arrow points down,
the current flow is reversed, and Isc becomes negative. This may be
why you are told the answer is -2ohms.
Good luck with your PE!
Post subject: Thank you
so much!Posted: Wed Aug 24, 2005 1:25 pm
Wed Aug 24, 2005 7:43 am
Great clear explanation-