# how to solve for voltage in this circuit using thevinin's theorem... (url)

Discussion in 'Homework Help' started by safi chn, Feb 13, 2016.

Jan 30, 2016
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2. ### omar-rodriguez Member

Jun 24, 2015
51
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That's the thevenin equivalent, then just make a voltage divisor to know the voltaje over the resistor

Last edited: Feb 13, 2016
safi chn likes this.

Apr 5, 2008
15,652
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Hello,

@omar-rodriguez ,

There is also a part for the ones that give replies:

1. Treat all members with respect.

Remember the members here have different levels of knowledge and understanding; please take this into consideration when assessing the validity of their questions.

2. Authors vary in their clarity.

If you are unsure of the original question, ask for clarification. When answering a question reply in a manner that you would like someone to respond to your query. Where appropriate, cite sources where the questioner can read more about the topic.

Please do NOT give complete solutions, but hints how to get to the solution.

Bertus

4. ### safi chn Thread Starter New Member

Jan 30, 2016
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thanks , but can you explain it a little bit as i dont know much about what you did after the Vab=Rth+Vth?

5. ### omar-rodriguez Member

Jun 24, 2015
51
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I'm sorry Bertus but as you can see, I didn't give to him the complete solution at the final part of my answer, I wrote "then just make a voltage divisor"

6. ### omar-rodriguez Member

Jun 24, 2015
51
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Well, Vab=Iab*Rth+Vth is the result of the thevenin theorem, the next thing I did was KCL on the node A, so entering currents are equal to the currents leaving, the most important thing is to let Vab in terms of Iab, the coefficient of Iab is Rth (thevenin resistance) and the other term is Vth (thevenin voltage)

7. ### MrAl Well-Known Member

Jun 17, 2014
2,443
492
Hi,

Another way to do this which seems simpler to me is to use Thevenin and Norton theorms together one after the other as needed, and using Ohm's Law.

The procedure would go like this...

First convert the 1v source and it's connected series 1 ohm resistor into a current source and parallel resistance.
Then combine the parallel resistance with the 4 ohm resistor.
Then convert that parallel resistance and the current source back into a voltage source with series resistance with a value equal to the parallel resistance.
Knowing the series resistance is now the only resistance in series with the 16 amp current source, you know the voltage drop using Ohm's Law.
Knowing the transformed voltage source voltage and the voltage drop, you can add the two.
The result of this addition is the voltage U in the schematic.
Test the result by computing the sum of currents in both original resistors and the total current of 16 amps.

If you know how to parallel unlike resistances like 1 ohm and 4 ohms in your head, you can do this whole problem in your head. If you dont know how to parallel resistances like this in your head then you can use a calculator to do that part and the rest in your head. That's how easy the Thevenin and Norton theorems make these problems.

8. ### WBahn Moderator

Mar 31, 2012
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I'm still waiting for the TS to show ANY effort at all is even attempting to work their own problem even part way.

9. ### WBahn Moderator

Mar 31, 2012
17,788
4,808
Oh, and if the TS would look at the problem carefully, they'll see that there are only two sources that matter at all. Everything else is just a distraction. One node equation yields the answer and is simple enough that it can be done, with some thought, by inspection. Hint: -15 V < v < -10 V.

10. ### MrAl Well-Known Member

Jun 17, 2014
2,443
492
Hi,

Yeah, almost like a trick question

Another case of the disappearing, low posts count OP ?

11. ### WBahn Moderator

Mar 31, 2012
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4,808
I suspect is WAS a trick question intended to see if the students are looking at the problem and applying fundamental concepts instead of just throwing equations at it (or trying to get others to work the problem for them on the internet).

12. ### safi chn Thread Starter New Member

Jan 30, 2016
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i disappeared because the solutions given here are too abstract for me , no one solved it step by step in order for me to understand it ....btw electronics is not my major , i am just new to it

13. ### safi chn Thread Starter New Member

Jan 30, 2016
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well , according to what thevinin's theorem says on this site http://www.allaboutcircuits.com/textbook/direct-current/chpt-10/thevenins-theorem/ , its almost impossible for me to apply that same rules to this question

14. ### omar-rodriguez Member

Jun 24, 2015
51
2
There's nothing abstract at the Thevenin theorem that I showed to you

I gave you almost the 90% of the exercise, you only have to do this at the end

safi chn likes this.
15. ### safi chn Thread Starter New Member

Jan 30, 2016
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okay much better now