Thevenin's equivalent with dependent and independent sources Problem 5.53
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- เผยแพร่เมื่อ 16 ต.ค. 2012
- example on how to obtain the Thevenin or Norton equivalent of a circuit with dependent and independent sources
For an example of circuit containing only dependent sources, check out my other video
• Thevenin Equivalent fo...
I love how you did just the right amount of math in your head. It's so refreshing to see people who can do that, as it really makes everything easier to follow. Thank you. I just failed a quiz, but now I feel confident that I'll pass the final!
I made these when I was a student. Hope they help; making them really helped me! Good luck!
I've been struggling so much in this class & finding your video gives me hope thank you!!!
you don't know how you saved my grades I mean my life right now ! THANK YOU
God bless you and your beautiful ECE knowledge! Life saver for ECE 201 exam 2!!!
Exam 2 for 209 tommorow. Hope you had more fun in ECE than I am lol
Thank you for your videos! They are very helpful.
Darren,
When looking for Voc, keep in mind the terminals of this voltage. The negative terminal is connected to the bottom of the circuit. The key to solving this lies in using KVL or KCL. There a several different approaches. You could also do KVL on the right loop. That would be Voc= -2Va + V(R2k) This solution give Voc = -16+4=-12 which is the same answer I get in the video using KVL around the outside. Hope that helps
Is the 4K ohm next to Vo the load? I did straight mesh analysis on the circuit and got half of the values for your answers for Isc and Voc.
Tobias,
The way I learned it, KVL, KCL and Ohm's law are the tripod foundation of linear circuit analysis. Any of them will always work. Developing intuition is an important part of circ analysis, especially once you know how to simplify using thevenin, norton, superposition and source transformation. To decide between KVL and KCL, I usually count the essential nodes and the mesh loops (accounting for "supermeshes"). I will often use whichever will give me the least number of equations
great advice for newcomers people watch video after video on thevenin but they need to undestand no matter how many videos youwatch if mesh and nodal are not clear you can't do anything
Thank you, now I fully understand it.
Does the method provided in this Video only work for any circuit where you are tasked to find thvenins?
Can you do thevenins if there is only dependent sources?
Thanks Nathalia. I use SmoothDraw3 which is a free download (you could use MS Paint.) I also use a Wacom Intuos pen tablet for writing. Camtasia is what I use for the screencapture. My videos are licensed under the creative commons licensed. You'll have this option from TH-cam. It means these videos can be used by anyone for educational purposes. No permission is required to use them. Best of Luck!
Hi ! tanks for this very good course
i have a question wich method can i use to find thevenin when Vos=0 and Isc= 0 ?
Hi, don't you need to have a supermesh to use KVL around the big loops?
Great video, I just have one question. Can we use this method to solve a circuit with independent sources only?
Hi there, thanks for the great videos. I teach circuits at Mason (GMU) and would love to post some examples too. Which software do you use? (if it is okay with you, I'll post links to your videos on my blackboard site - so my students look at them).
I appreciate these videos- Nathalia
please what is the load is a dependent generator do we also take it out as in the case of a resistor?
How would you find Voc if there was an independent current source in place of the independent voltage source?
@Learn First Thanks
Learn First XD yeah only 5 years late
Learn First yeahhh Xd
Thank you so much for this video
+ Axceed1 If there is not two sources then you do resistance equivalency till you have the thevinin model.
Do you have any good tips on how to know if you should use KCL or KVL? Everything seems completely random ro me.
Awesome video! Thank you!
can we just assume that voltage in parralel are the same? 12 = 12 ? for Vo
bro...you're the man!
Hey any other way to find Rth with both dependent and independent souce...more like a short cut..?
Why is it not in a loop? Just because their isn't a wire connected doesn't mean their cant be a loop, it can still be a circuit. So what is the reason?
Do you have any good sources contain dependent source that can help me out?
we can apply KVL to oly closed loops rt?? u applied fr open circuit tats fr voc!! how come?
nicely done! Thanks a lot (y)
How to find the current across the dependent source using Norton's theorem ?
Thanks man much appreciated
nice explanation
how could you employ KVL on the open circuit to calculate Voc in the first part ? isn't that wrong ?
There is still a voltage drop across an open circuit. just no current.
watch at 1.25 speed
Thank you!
perfect
watch at 2 speed
watch this video on thevenin with dependent sources
th-cam.com/video/D1QxFA9_U5w/w-d-xo.html
Thanks! I appreciate the answers- will definitely let my students know.
why is 2va not 4? are you taking into account the 2 in the va? did I miss something?
How did you get Vth to be 12? Can you just ignore the VDVS?
For V open circuit why could he do (12)(2/2+4) voltage division to find v at ab???
He didn't do that, I'm not sure what you are asking. But I see that your talking about voltage division, just not sure what you're asking
because the resistor is the load?
Excellent, thanks.
Thank you, dependent sources are no longer a problem to me. >:3
What if one of the dependencies is on the load?
Hey Matthew for V_oc isn it = to the voltage drop in 2K resistor
Ash W not if there is also a dependent source, that contributes to total resistance seen by the load
making things complex?
my teacher said, and I've tried and worked for me, if you want to find something in a resistance parallel to a current source you use kcl, else you use kvl, because most of the time you lose less time.
Both solve any problem the thing is to find the one that does it faster so you won't lose time in the test.
Open terminals will still have a voltage drop/rise, just no current. So you can apply KVL through an open terminal but not KCL from an open terminal.
Thanks man!!
You made this shit look easy thanx bro & keep up.
at 2:47 why is it 4/6?
how do you identify the load?
please reply asap because i have test tomorrow
I think there's a mistake at 10:12. You acknowledge that i1= 1mA, but the only way to have isc = -3 mA is if i1= 1A. I've found this material a bit more difficult than previous material. Thank you for taking the time to make these videos!
I1 and Isc will not be the same value
Comments be like "Hey I realize that I don't know SHIT but you're doing it wrong and I don't care if you're doing it right."
When you solve for Voc what happens to the 2ohm resistor?
+Goce Dimitrov Old comment, but in case you're still wondering, he used KVL to solve for the voltage around that one loop. Remember that voltage across two parallel loads will always be equal, so the 2kohm resistor is not included in the equation. This is the implication of Kirchoff's voltage law.
Sorry, I don't see the mistake....2k(1m) =2V. Therefore -2k(Isc) = 6V. Then 6/-2k = -3mA. Hope that clarifies things.
why did you open the 4K resistor at the beginning? does the Voc depends on it?
yeah it is thanks for pointing that out to me :)))
love you
voltage across 2k resistor = Voc , as u remove load (open ci rcuit)
.so why are u calculating Box seperatly ?
@tahmidur rahman i think he was totally wrong , how could he remove the 4K resistor !!!
look at this prof. solving problem without removing the load :(21:45) th-cam.com/video/S-3NWNrAQRQ/w-d-xo.html
I don't understand how Va can be equal to 8V when we used the voltage division rule. And how it can change to 4V when we use KVL to find the short circuit current.
Isn't Va=Va so it should only have one value ?
the Va for the short circuit and the Va for the open circuit is not the same as they are different circuits (adding the jumper makes the difference) thus you cant use them interchangeably and same is true for the current. you have to solve for it again.
Thanks rik.yan. It all makes sense once you understand it hey....
why did the first equation for Va have the value of resistance 4/6 and not simply 4 x 2/4+2 =8/6 that gives 4/3 on simplification
I was wondering this as well
at 13:27 Va = 4 you said but when it was calculated earlier it was 8 V??
i think it's because Va is different for a different circuit assumption. First one had open circuit setting for load, 2nd one had short circuit setting for load.
This is a good video but the KVL for used to find the Isc was unnecessary. If you have the voltage for the dependent source, then you have the voltage for the resistor in parallel. Well, if not KVL, then KCL but I am under the impression that the KVL form is more involved than the KCL version.
How Voc= -12V I mean shouldnt it be 16V since Voc= 2Va and Va= 8V ?
Crystal clear. Thank you :)
Resistance CAN be negative, especially for a Thevenin/Norton circuit. In that case, it's an amplifier circuit. A resistOR cannot be negative.
Shahyad khani,
I added a link to another example with only dependent sources. It's in the video description of this video.
Good luck!
Thank You.
Not to quibble, but if you have the Norton circuit you don't have to back solve to find the voltage drop across the load. V_Load = I_N÷(G_N + G_Load). In this case, - 3mA ÷(1/4k + 1/4k) = - 6 volts.
How did we got i1=1???
we know that voltage across 4k ohm resistor is 4V. so current is V/R=4/4k.
use x2 or x1.5 speed when listening. very informative video, but very slowly presented
i can't understand why we removed the load ! , and there's a direct way to calculate Rth even with dependent source ,by applying test voltage source
@Kermit DuBois yea that's true , but my point is about removing the 4k resistor which i didn't understand , look at this prof solving a similar problem without removing anything :
th-cam.com/video/S-3NWNrAQRQ/w-d-xo.html
That's good thanks
whouldn't Voc be 2Va??
thanks....
i once got negative resistance in an exam and thought it was wrong, but it was actually right. You can have negative resistance according to my professor.
How?
Linear circuits dont have negative resistance
Great video can you help me to solve this problem
how do you identify a load??? he didn't explain that part at all
Mostly in question the value of load is provided
Load is that resistor across which thevenin or norton equivalent has to be drawn. In other words you could simplify the complex circuit to thevenin or norton equivalent ciruit across the load.
good
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You are making complecate make that was easy problem
Excuse me. In 2nd part of the question you easily found Va = 4V. But then you did mesh analysis which took very long. As Va=4, voltage of the node at top middle is 8V. Then we can easily determine Isc
Agreed. We can do the 4V drop divided by the 4kohm resistor and the 8V drop across the 2kohm resistor to get 1mA going through the first resistor and 4mA going through the second resistor. With 1mA going into the node and 4mA going down and out of the node, we know that 3mA must be what the 2Va source is supplying. This means Isc must be -3mA (because it is defined as the opposite direction as the current from the 2Va source
Can you explain why the first mesh is accepted (the outside mesh KVL)?
fdd
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Love the video, but use your units! Not 3m, 3mA!
Too slow dude. Speak up a notch.
Download the Video Speed Controller Chrome extension. "Speed up, slow down, advance and rewind any HTML5 video with quick shortcuts."
good