•−•  ••−•    −•−•  •−  ••−•  •
RF Cafe Morse Code >Hear It<

# Thevenin Equivalence Question - RF Cafe Forums

Because of the high maintenance needed to monitor and filter spammers from the RF Cafe Forums, I decided that it would be best to just archive the pages to make all the good information posted in the past available for review. It is unfortunate that the scumbags of the world ruin an otherwise useful venue for people wanting to exchanged useful ideas and views. It seems that the more formal social media like Facebook pretty much dominate this kind of venue anymore anyway, so if you would like to post something on RF Cafe's Facebook page, please do.

Below are all of the forum threads, including all the responses to the original posts.

themaze
Post subject: Thevenin Equivalence Question Posted: Wed Aug 24, 2005 7:50 am

Lieutenant

Joined: Wed Aug 24, 2005 7:43 am
Posts: 2
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...

Thanks

Top

john-ee
Post subject: Posted: Wed Aug 24, 2005 11:32 am

Lieutenant

Joined: Tue Aug 23, 2005 6:33 pm
Posts: 3
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.

So, Voc/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.

-John-

Top

themaze
Post subject: Thank you so much!Posted: Wed Aug 24, 2005 1:25 pm

Lieutenant

Joined: Wed Aug 24, 2005 7:43 am
Posts: 2
Great clear explanation-
Many thanks-

Neil

Posted  11/12/2012
Custom Search
More than 10,000 searchable pages indexed.

#### Your RF CafeProgenitor & Webmaster

... single-handedly redefining what an engineering website should be.

Carpe Diem!
(Seize the Day!)

5th MOB: My USAF radar shop

Airplanes and Rockets: My personal hobby website

Equine Kingdom: My daughter Sally's horse riding website