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!
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
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
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.
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 UA-cam. It means these videos can be used by anyone for educational purposes. No permission is required to use them. Best of Luck!
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.
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!
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.
+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.
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.
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.
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?
@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) ua-cam.com/video/S-3NWNrAQRQ/v-deo.html
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.
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?
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
@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 : ua-cam.com/video/S-3NWNrAQRQ/v-deo.html
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.
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
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!!!
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
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
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
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
you don't know how you saved my grades I mean my life right now ! THANK YOU
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.
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 UA-cam. It means these videos can be used by anyone for educational purposes. No permission is required to use them. Best of Luck!
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
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.
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.
+ Axceed1 If there is not two sources then you do resistance equivalency till you have the thevinin model.
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
at 2:47 why is it 4/6?
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.
Hey any other way to find Rth with both dependent and independent souce...more like a short cut..?
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."
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.
Thank you for your videos! They are very helpful.
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.
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
Shahyad khani,
I added a link to another example with only dependent sources. It's in the video description of this video.
Good luck!
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.
we can apply KVL to oly closed loops rt?? u applied fr open circuit tats fr voc!! how come?
please what is the load is a dependent generator do we also take it out as in the case of a resistor?
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....
Hi, don't you need to have a supermesh to use KVL around the big loops?
Do you have any good sources contain dependent source that can help me out?
Sorry, I don't see the mistake....2k(1m) =2V. Therefore -2k(Isc) = 6V. Then 6/-2k = -3mA. Hope that clarifies things.
Resistance CAN be negative, especially for a Thevenin/Norton circuit. In that case, it's an amplifier circuit. A resistOR cannot be negative.
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
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
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?
How to find the current across the dependent source using Norton's theorem ?
Do you have any good tips on how to know if you should use KCL or KVL? Everything seems completely random ro me.
Thank you, dependent sources are no longer a problem to me. >:3
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) ua-cam.com/video/S-3NWNrAQRQ/v-deo.html
Thank you, now I fully understand it.
How did we got i1=1???
we know that voltage across 4k ohm resistor is 4V. so current is V/R=4/4k.
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.
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?
making things complex?
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 ?
because the resistor is the load?
can we just assume that voltage in parralel are the same? 12 = 12 ? for Vo
use x2 or x1.5 speed when listening. very informative video, but very slowly presented
how do you identify the load?
please reply asap because i have test tomorrow
Thanks! I appreciate the answers- will definitely let my students know.
Great video, I just have one question. Can we use this method to solve a circuit with independent sources only?
Thank you so much for this video
How Voc= -12V I mean shouldnt it be 16V since Voc= 2Va and Va= 8V ?
How did you get Vth to be 12? Can you just ignore the VDVS?
why is 2va not 4? are you taking into account the 2 in the va? did I miss something?
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
What if one of the dependencies is on the load?
nice explanation
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 :
ua-cam.com/video/S-3NWNrAQRQ/v-deo.html
Awesome video! Thank you!
You made this shit look easy thanx bro & keep up.
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.
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
Great video can you help me to solve this problem
nicely done! Thanks a lot (y)
whouldn't Voc be 2Va??
Thanks man much appreciated
You are making complecate make that was easy problem
bro...you're the man!
Excellent, thanks.
Crystal clear. Thank you :)
Thanks man!!
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)?
Thank You.
That's good thanks
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thanks....
fdd
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