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

bertus

Joined Apr 5, 2008
20,650
Hello,

@omar-rodriguez ,
Did you read this thread?
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Bertus
 

omar-rodriguez

Joined Jun 24, 2015
67
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"
 

omar-rodriguez

Joined Jun 24, 2015
67
thanks , but can you explain it a little bit as i dont know much about what you did after the Vab=Rth+Vth?
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)
 

MrAl

Joined Jun 17, 2014
7,752
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.
 

WBahn

Joined Mar 31, 2012
26,045
I'm still waiting for the TS to show ANY effort at all is even attempting to work their own problem even part way.
 

WBahn

Joined Mar 31, 2012
26,045
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.
 

WBahn

Joined Mar 31, 2012
26,045
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).
 

Thread Starter

safi chn

Joined Jan 30, 2016
7
Hi,

Yeah, almost like a trick question :)

Another case of the disappearing, low posts count OP ?
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
 

Thread Starter

safi chn

Joined Jan 30, 2016
7
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.
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
 
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