Find V th

WBahn

Joined Mar 31, 2012
25,223
One of the beautiful things about most engineering disciplines is that, most of the time, you can determine the correctness of the answer from the answer itself.

You have come up with what you believe Vth to be.

What should the current in the 4 Ω resistor be if you connect a source equal to Vth across the terminals?

So, connect such a source across the terminals and analyze the resulting circuit and see if you get the current you expect. If you do, your answer is correct. If not, it is wrong.

Get in the habit of doing this. You will see your grades shoot through the roof while you are in school and you will be much more valuable to your eventual employers once you leave.
 

Thread Starter

Ahmed Y Osman

Joined Jan 15, 2020
3
Well the current should be zero , but I got a small current , if why say that it's negligible , Vth might be right ,,
I saw the prof's answer it is definitely not like mine lol , he just added the source connected to a terminal with the source in the first branch lol ,,, when I first saw it , I was like WOT IS THAT !!
Anyways thanks for your tip :)
Wish me luck , I have an exam tomorrow
 

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WBahn

Joined Mar 31, 2012
25,223
Well the current should be zero , but I got a small current , if why say that it's negligible , Vth might be right ,,
There's almost always going to be some round off error involved, so if the current is negligible, then that is okay. The thing that you have to be a bit careful of is what counts as "negligible". Just because it's small doesn't mean it's negligible. For instance, consider a circuit that has a Thevenin equivalent resistance of 10 MΩ and a Thevenin voltage of 10 V. If I screw up and come up with a Thevenin voltage of 110 V and plug it into my original circuit I would get a current flow in it of 10 uA, which I might be tempted to call negligible, but it certainly isn't. On the other hand, if I had a circuit that had a Thevenin equivalent resistance of 10 mΩ and the same 10 V Thevenin voltage and my answer turned out to be 10.1 V (i.e., within 1% of the correct value), then the current in the source would turn out to be 10 A and I might be tempted to declare my answer complete wrong. So "negligible" needs to be put into a meaningful context; the most useful one for a problem like this would probably be as a fraction of the short-circuit current.

I saw the prof's answer it is definitely not like mine lol , he just added the source connected to a terminal with the source in the first branch lol ,,, when I first saw it , I was like WOT IS THAT !!
You can always brute for the analysis and be guaranteed to get the correct answer (barring mistakes), but with a bit of consideration you can often come up with significantly cleaner and quicker ways to get the correct answer for the specific problem at hand. In this case you know that the current from the left hand source must all flow through the left hand branch and the voltage that is produced is then added to the voltage of the right hand source because you know that there is no current flowing in that 4 Ω resistor when the terminals are open.

Wish me luck , I have an exam tomorrow
Good luck.
 
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