# Thevenin prob

Discussion in 'Homework Help' started by ihaveaquestion, May 5, 2009.

1. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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http://img412.imageshack.us/img412/6826/83067646.jpg
I did the problem already there... two problems actually... first one I'm pretty sure is right...

I used nodal analysis for the second one... for such a simple looking circuit it looks like I might have done more work than I needed to.. in other words maybe there is an easier way... keep in mind I know only know the basic methods... KVL and KCL, superposition, stuff like that...

Sorry for the messy handwriting on this one... For Vthev I wrote two node equations one on the left one on the right and solve for e2, which should be Vthev, correct?

2. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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Here's another one that's done already...

http://img154.imageshack.us/img154/5412/91651177.jpg

that's using superposition and I'm asked to find vout..

The only confusion I'm having is after I found the current at the bottom to be 10 A... the current is heading into v minus of where I'm finding v.. so I can think of it as a negative current (-10A) and so V = iR = -12 so by superposition 12 + (-12) so Vtotal = 0?

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790

This has an obvious shortcut to an answer without all the laborious calculation.

4. ### hgmjr Retired Moderator

Jan 28, 2005
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I believe you will find that you have arrived at an incorrect value for Vth in the first problem.

If you review Norton's Theorem you will see where you have made your error.

hgmjr

5. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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Hmm.. I did it again with node method and came up with the top node voltage to be:

e = I*(R1R2)/(R1+R2)

i(norton) = V/R = e/R1 = I*(R2/(R1+R2))

Thanks for spotting that

Last edited: May 5, 2009
6. ### hgmjr Retired Moderator

Jan 28, 2005
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Norton's Theorem allows you to combine the current source in the circuit and the resistor in parallel with that current source into an equivalent voltage source in series with the resistor. It is this voltage source that is Vth in the first problem in your initial post in this thread.

hgmjr

7. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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Oh ya I'm not sure how to convert current sources into voltage sources etc.. I forgot to mention that sorry... I'll learn it later I'm sure

Is what I did/the answer I got in my last post correct?

Last edited: May 5, 2009
8. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,711
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You got Vth for the first one wrong. It should be Vth = Is * R2.

Vth is the open circuit voltage. You can see that R1 has nothing to do with the open circuit voltage, because if the output is open circuited, no current flows in R1 and therefore there is no voltage drop in R1.

It looks like you got the second part right.

9. ### hgmjr Retired Moderator

Jan 28, 2005
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The Electrician,

Shouldn't you be able to apply Norton's to the current source in parallel with R1 and get Vth = Is * R1. (THIS IS MY
ERRONEOUS COMMENT)

Ooops! I see where I screwed up. I confused R1 and R2 in the circuit. Of course you are correct,

My apologize to ihaveaquestion as well since I may have misled him in an earlier posting to this thread.

hgmjr

Last edited: May 5, 2009
10. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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Ahh I see it now Electrician, thanks.

No problem hgmjr! Thanks for your input as well