just wanted to let you know, we had the last test before the final on Thursday on thevenin, superposition and mesh. I studied your videos for about two weeks and ended up getting a perfect score and the bonus for a score of 110. these videos are the real deal!
For someone who has dealt with Kirchoff's Laws but never Thevenin's method, I have found that Jim explains and repeats till his point is driven home. Thanks for your time.
Despite having my Bachelor's in EE and now working on my Master's in EE, it took me 5 years to finally understand how Thevenin's Theorem actually works. Thank you, sir.
This is the best explanation of Thevenin's theorem I could find. I'm still confused, but you highlighted why. I got ahead of myself and didn't get a solid grasp of series/parallel resistances. Thank you for making it simple enough for me to figure out why I couldn't figure it out.
dude.. where have you been my entire semester.. my lecturer buzzed over this section in less than two lessons.. after spending an evening with this.. it's like all the wires have finally come together.. thank you..
Thank you for making this so simple and clear. This video was a welcome lifeline after grappling for two days with the same four problems. Thanks to you, I solved three of them in under 10 minutes! The third had both a voltage AND a current source, so was substantially trickier, but I still would have struggled much much more without this video.
I put your video on repeat. Watched at least 4 times, did my Thevenin Norton Quiz for 1st yr Tech/Installer, passed first time with 89%. Would have gotten 94% but the wording of 2 true false questions threw me. I hope my class mates have used the links I left for your videos. Thanks so much. Thevenin Rules!.
Dear Jim, I'd like to take the time to thank you most sincerely for the time and effort you've obviously put in to your videos. They have helped me out no end. Your style of teaching is fabulous, certainly one of the best I've come across after many many hours of searching for suitable tutorials. I am unfortunately, one of those students being forced to use nodal and mesh analysis. On top of which our professor refuses to give us quantities for voltage/current sources and resistors, which would simplify the process greatly. As I'm sure you'll understand, this leads to massive confusing equations which are so frustrating to work with. And of course at the end not having a value to cross refernce if the calculations are correct or not is so demotivating. However, with your help I have learned how to systematically work through the circuits and solve nearly any problem that comes my way. Thanks again, please keep up the excellent work. It is greatly appreciated.
sigh, I approached 5 Eth using nodal analysis and scratched my head for ever trying to figure out why I was getting 85, finally noticed battery symbol in opposite direction. nodal then resulted correct result. thanks for this video, very helpful.
This truly is the best lecture I've found on the subject. I would have liked to see your take on the importance of the ratio between output/input impedence when one circuit drives another, and how Thevenin's equivelent circuits are applied when the load drives subsequent loads. Regardless, this lecture improved my tenuous grasp on the subject immensely.
Make sure you're using the updated lecture: th-cam.com/video/Ivs5fJ4-W6E/w-d-xo.html Definitely check out the maximum power transfer theorem lecture at: th-cam.com/video/tFtFGNZTZrA/w-d-xo.html
Can someone help me out? With the solutions to the problem @ 21:18 I got the same exact answers except for the 5K ohm one, I got 28.3 V and 5mA? I'm kind of confused how he got 42.9 V as an answer when the ETH voltage is only 30 V?
This series is fantastic man, I very much appreciate the effort. It has definitely helped clarify some things in my head and I'm now far more fluent with my approach to problems. I cannot thank you enough. You know that 'acrossed' isn't a word though right Jim? :)
On ex 3: 40 ohm is in par with 50 + 75 So 40 // 125 = 30,30 ohm You still have the 30 ohm resistor from the source and the 20 ohm from the load… How did you get 55 ohms if I may ask please?
You can just do a source transformation with the current source sample. 40mA parallel with 400ohms to a 16V voltage supply series with 400ohms which is the equivalent voltage circuit and from there rth and vth can be solve easily. Since 400ohm is in series with the other 400ohms, rth will be 800 ohms while vth is already given as 16V from the source transformation
Hi Jim, any chance you can give more hints for the 5th circuit? I got RTh easily but I don't seem to get the correct Eth when applying Superposition. I tried using Norton Theorem and even Nodal analysis to no avail. I think the 500-ohm resistor in series with the 40V voltage source is throwing me off. It's actually the same issue I ran into with problem 3. Thanks for all your lectures; I am finding them very helpful in studying for my FE/EIT exam.
I know that you don't like node voltage but it makes QUICK work of problem #5. First do source transformation on the 1 amp source in parallel with the 300 ohm resister giving 300 v in series with 300 + 200 = 500 ohm. This collapses the problem to 2 loops and a single essential node at the junction of the 200 (now 500), 500 and 750 node junction. A SINGLE equation then reveals that the voltage at that node is 121.875. Voltage division reveals Vth = 121.875X(2000/3750) = 65 V.
Can anyone help me with understanding why we leave the voltage source as it is and calculate voltage drop across those resistors in parallel with RL (I understand this point) but we measure the value of resistors close to the voltage source as a parallel connection? What is the need of measuring different values with basically two different circuits? Why the extra step of reducing the voltage source to 0?
The easiest answer is this ... that's how Thevenin's Theorem works. It's a two step method: Step 1 - remove the load resistor, determine the open circuit voltage across the terminals of interest Step 2 - remove the load resistor, determine the resistance at the terminals of interest with the source(s) removed The difference in perspective between step 1 and 2 necessitates a complete re-evalution of the circuit in question because they're not the same circuit, thus the different observed electrical properties.
"If you haven't punched your computer screen by now you are far more patient than I". Well, this made me subscribe. You are absolutely funny and you have a cool way of thinking. I appreciate your honesty too, when you tell the viewer that you can't teach us to be "flexible". I will watch the rest of the video now. Thanks for making it btw, I am struggling a lot to understand Thevenin's theorem. EDIT: Not anymore
#3 was pretty tough. I must admit that I gave up and checked the answer. Then I was focused on how to get the right answer. I tried mesh analysis but didn't succeed either. I suppose many of us got to the 40 - 30 -125 Ohm point and then got stuck. It turns out that 40 and 125 are in parallel. I could visualize it using kind of 3D imagination: assume that you look at an opened book from above. So, 40-30 is the left page, lets say, #123, 30-125 is the right page #125 and 30 ohms is the page in the middle - #124 - you are actually holding before passing. Then the result of parallel 40&125 that happens to be around 30 ohms is in series with 30 ohms placed next to the voltage source. The current of 400 mAmps then gets proportionally divided and to the 125 ohms part goes 400 mAmps * 125/165 = 96 mAmps. Then 96 mAmps * 75 ohms gives us 7.27 volts across 75 ohms resistor.
Very nice and thank you for this excellent resource. I'm not sure if it's an error but I continuously get a different result for Pl at 35:37 . I'm using I^2R to calculate power ie (11.4x10^-3)^2 x 600 = 6.84x10^-3 or 6.84mW a magnitude of 10 out? I have an exam next week stressed out hard so I could be completely ballsed up here...... =( I'm guessing I am plugging the values incorectly but when I enter them as .0114^2 x 600 I get a much closer albeit not equal 77.97x10^-3 W... Am I going insane or is this an error? kind regards from Australia, Johann
The power formula of I^2R is correct, however, the answer of 6.84mW means you're somehow not properly squaring the current. This is why I like using different variations of formulas to check the final answer.
Found you at the Columbia Gorge Community College channel! Finally i feel comfortable with Thevenins theorem! Thank you! By the way do you have any videos for Nodal Analysis ? :)
kam33p Stay in touch because "Nodal Analysis" and "Mesh Analysis" are on the production list. Spread the word to your classmates because I'm hoping to have the whole Basic Electronics 1: DC Circuit Analysis playlist done in the next couple months. I've got an interesting take on Mesh and Nodal and I'm including it in the BE1 playlist out of sheer pity for those that are enrolled in classes where an instructor forces them to use these techniques. I'm not saying they're not valid circuit analysis techniques ... but I am saying there are far more important things to learn. The Superposition Theorem and software analysis essentially render Mesh and Nodal analysis as relevant to modern society as sword fighting. It's elegant and cool ... but not something you ordinarily need to learn.
With the first 4 examples (which turned out to be the same example), was there a way to solve for the voltage and current across RL without adding the parallel resistors together to get an equivalent resistance? I feel like you were suggesting that we could, but I'm not certain
Yes, Thevenin's theorem is a more universal approach that is not dependent upon the value of the adjustable load resistor. Every time the load resistor changes one needs to perform another complete analysis of the circuit. For another explanation, I've since published an updated Thevenin's theorem lecture. Check it out at: th-cam.com/video/Ivs5fJ4-W6E/w-d-xo.html
i don't know if you reading this or not at least i gonna try to ask, at 23:30 how do you get 177.4 cause i try the parallel resistor formula which is 1/R = 1/240 + 1/680 i got 23/4080 ?
Check out the parallel resistors lecture at: th-cam.com/video/gUwoCJJgxt8/w-d-xo.html Additionally check out the DC Math lecture at th-cam.com/video/ItdTvzv32BQ/w-d-xo.html specifically the section about the scientific calculator. Sounds like you still have your calculator in EXACT mode. It needs to be in APPROXIMATE. Additionally, you've solved for 1/Rp not Rp. To solve for Rp you need to do a final inversion.
In #4 it is not necessary to do a delta-y conversion. The 180 is in parallel with the 600 giving 138.46. The 138.46 is in series with the 1000 giving 1138.46. The 1138.46 is in parallel with the 400 giving 296.
Great instruction but at 21 min. Your answer by the 5k ohm resistor is wrong. It should be 5.66 mA and 28.3V Your instruction is very good, my teacher is very difficult to understand.
Daan550 Very nice! I annotated it in the video and gave you 10k bonus points (100k if you were crazy enough to solve for it using the original series-parallel circuit). Normally I can track down my errors but I have no idea what I was smoking when I wrote down those values. How can you get a 42.9V drop from a 30V source? Thanks for letting me know. Let your classmates know this resource exists as a supporting resource for those struggling with certain subjects.
thanks a lot for the videos and great explanation i keep hearing you say that one has to view parallel circuits at a nodal level and from perspective of flowing current , can you please elaborate more on this or point me to the resources that would help me get a grasp on this ? thank you
hey i was just going through your practice questions and i am stuck finding the Vth of #3 should vth be equal to the voltage drop across the the 75 ohms resistance which i have found to be 11.61v am i missing something here ?
i still didn't get in haha but after taking a quick 10 minutes break i figured that i could use mesh analysis to find the current through 75 ohms after re-positioning the 40 ohms of course and to my surprise i found the current to be 0.096A AND times 75 ohms i got 7.2V ( i did it!!!!!) i am still having a little bit of trouble finding the Rth on the first try though in #3,4 any extra tips? i have watched all the video related to it though i cant thank you enough for the videos and your time : God bless
I think that I figured out why #3 is Vth is 7.2 v( yes I am still trying to figure it out lol ) the 30 ohms resistance has a voltage drop, therefore, we can not use the full 24 voltage to find the voltage drop across 75 , we have to find the voltage drop across the 30ohms subtract it from the total voltage and use the rest to find the voltage across the 75 ohms
Hate to bother you sir; however, I got stuck on the third exercise trying to solve for Eth. Any help would be greatly appreciated. BTW, thank you for all the lessons. They have been invaluable in developing my understanding of DC theory. Thumbs up sir!
+Rj DeSoto Thanks much, super appreciate you're making use of this material. Not sure if you're talking about the third illustrated example problem or the third challenge problem at the end. Either way, Eth is the "open circuit voltage" ... no current can flow through an open circuit ... this may fundamentally change the nature of your circuit. Try to redraw the circuit and see if a reconfiguration of elements clarifies the two point voltage measurement. I'm a visual thinker and when presented with these types of problems I keep picturing different configurations until I get it. Keep up the valiant effort!
+Jim Pytel Thank you for the reply. It was the third exercise at the end of the lesson (the challenge exercises). I know that current will not flow through the 20 ohm resistor after removing the variable load. For the life of me, I just could not figure out how the 7.2 v drop occured on the 75 ohm resistor. The 40 ohm resistor has acctually been throwing me off a bit. Thanks for the help.
+Rj DeSoto Oh yeah that's a tricky one. As you figured out the 40 ohm resistor can be repositioned such that it is in parallel with the 50 and 75 in series. Any more hints and I might give away the secret to everyone else!
I got discouraged when I couldn't calculate VL or IL correctly for ANY given value, let alone try to simplify the process. I wish this video walked through the calculations more than saying "calculate! Good luck"
There is a veritable mountain of material that precedes this lecture. Check out the "DC Circuit Analysis" playlist at: th-cam.com/play/PLdnqjKaksr8qQ9w3XY5zFXQ2H-zXQFMlI.html
At my university for whatever reason they taught us Theverin theorem and Norton theorem before teaching us nodal analysis. Honestly it made understanding them quite confusing, I felt lost the entire lecture which is why I'm here.
Ugh ... nodal and mesh are antiquated techniques that are no longer relevant. I feel your pain. Glad I can help. Make sure you're using the updated lectures in this playlist: th-cam.com/play/PLdnqjKaksr8qQ9w3XY5zFXQ2H-zXQFMlI.html
Check the lead in to the mesh analysis lecture for my disclaimer: th-cam.com/video/Mj5H1Y4pJno/w-d-xo.html I prefer using software circuit analysis or the superposition theorem: th-cam.com/video/q47g1Nz_4lM/w-d-xo.html
I'm having trouble to understand why the 1.3k and 2.2k are in parallel and also why when finding E th, the 1.8 is insignificant yet it was to find Rth and why the 1.1k is not significant when finding Rth but was to find Eth
Make sure you're using the updated lectures: th-cam.com/video/Ivs5fJ4-W6E/w-d-xo.html th-cam.com/video/UdEUPUn_onY/w-d-xo.html To answer your question solving for Eth and Rth are two totally separate steps and have NOTHING to do with each other. This being said both techniques require "modification" of the circuit. In the case of Eth we're solving for voltage across an open circuit. In the case of Rth the load and source are removed and we're solving for resistance. Depending on which property you're solving for modifications like shorts and opens can fundamentally change the nature of the circuit you're working with.
I mean COME ON. For anyone looking for a bare bones easy explanation of this... you set it up perfectly and then trashed it with the exercise around 5:30. Sometimes seeing it done start to finish is a GREAT teaching tool
By 02:50 I was "Primed" Out & totally mathematically Derailed from any Practical usefulness of Thevenin & his Twisted theorems....Never needed this concept in 50+ yrs electrical/ Electronics work employment.....Suppose it all in "Satisfying" the Almighty course professor's Power of making students eat "Shite" in exchange for the "Grade".....
You've got an interesting perspective of what instructors do for their students. Yes, theorems may or may not be used in certain career fields. As mentioned in the lecture Thevenin's Theorem specifically deals with frequently changing load conditions. If this is a scenario that presents itself its far simpler to use the Thevenin's equivalent circuit.
The problem is, you're explaining the basic laws of circuit analysis and I know all that stuff.. I want a proof for thevenin theorem.. I kept watching for 20 minutes and you didn't get started in proving the theorem yet!
Abdulrahman Mahdaly WTH I watched the whole video and you did not prove it! why did I waste my time watching this !! I want A PROOF .. a freaking proof!
Abdulrahman Mahdaly Sorry, proving TT wasn't the intended purpose of this lecture. The intended audience of this lecture series are those students enrolled in technician level programs and the stated objectives of this lecture are to learn to use TT. Long story short, technicians don't need proof ... they just need to know how to use a proven theorem.
I’m honestly really disappointed in the video. I need to see a full thevenins start to finish and your “start low, turn up the heat” thing really detracts from seeing it start to finish. Do it start to finish and then ask the viewer to calculate. I’m still struggling to see the big picture and no one just finishes the damn picture before tangenting or doing exercises
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just wanted to let you know, we had the last test before the final on Thursday on thevenin, superposition and mesh. I studied your videos for about two weeks and ended up getting a perfect score and the bonus for a score of 110. these videos are the real deal!
One does not simply teach Thevenins Theorem. He's a very flexible professor. Always ready and able to reach his students where they are.
For someone who has dealt with Kirchoff's Laws but never Thevenin's method, I have found that Jim explains and repeats till his point is driven home. Thanks for your time.
Despite having my Bachelor's in EE and now working on my Master's in EE, it took me 5 years to finally understand how Thevenin's Theorem actually works. Thank you, sir.
This is the only way any teacher should explain anything, especially in electronics.
This is the best explanation of Thevenin's theorem I could find. I'm still confused, but you highlighted why. I got ahead of myself and didn't get a solid grasp of series/parallel resistances. Thank you for making it simple enough for me to figure out why I couldn't figure it out.
dude.. where have you been my entire semester.. my lecturer buzzed over this section in less than two lessons.. after spending an evening with this.. it's like all the wires have finally come together.. thank you..
Glad you're making use of this material. Please let your friends know this resource is available.
I love your philosophical take on this. These lectures have been very helpful to me. Thank you.
Thank you for making this so simple and clear. This video was a welcome lifeline after grappling for two days with the same four problems. Thanks to you, I solved three of them in under 10 minutes! The third had both a voltage AND a current source, so was substantially trickier, but I still would have struggled much much more without this video.
This was really simple and truly the best video for Thevenin's theorem. Thanks much!
saying you are my LIFESAVER is an understatement best explanation ever
Best tutorial for thevénin ever. Thanks for your effort.
Truly the best explanation I have seen. I've watched it 3 times now to drill it in
Thank you so much for this lecture. You explained it so well and made it simple. The best teacher.
This is one of the best presentations on this subject I have seen. Thanks!
True to the saying...practice..practice..practice. love this video. I corrected my instructor by watching this video. Thanks..
2 Weeks of lectures that im paying for < one Jim boi
One of the best teachers on youtube
I put your video on repeat. Watched at least 4 times, did my Thevenin Norton Quiz for 1st yr Tech/Installer, passed first time with 89%. Would have gotten 94% but the wording of 2 true false questions threw me. I hope my class mates have used the links I left for your videos. Thanks so much. Thevenin Rules!.
Very nice! Make sure to check out the updated lectures:
th-cam.com/video/Ivs5fJ4-W6E/w-d-xo.html
th-cam.com/video/UdEUPUn_onY/w-d-xo.html
You are an excellent teacher thanks for your hard work!
Dear Jim, I'd like to take the time to thank you most sincerely for the time and effort you've obviously put in to your videos. They have helped me out no end. Your style of teaching is fabulous, certainly one of the best I've come across after many many hours of searching for suitable tutorials. I am unfortunately, one of those students being forced to use nodal and mesh analysis. On top of which our professor refuses to give us quantities for voltage/current sources and resistors, which would simplify the process greatly. As I'm sure you'll understand, this leads to massive confusing equations which are so frustrating to work with. And of course at the end not having a value to cross refernce if the calculations are correct or not is so demotivating. However, with your help I have learned how to systematically work through the circuits and solve nearly any problem that comes my way. Thanks again, please keep up the excellent work. It is greatly appreciated.
Glad you’re making use of this material! Please let your friends know this resource is available.
i wish this was a bit more difficult, these example circuits are super simple compared to what brought me here :(
we have the same problem dude. i was like searching on the internet "Thevenin's Theorem difficult problems level 999999999"
thanks Jim, this was a brilliant video
sigh, I approached 5 Eth using nodal analysis and scratched my head for ever trying to figure out why I was getting 85, finally noticed battery symbol in opposite direction. nodal then resulted correct result. thanks for this video, very helpful.
@Brad Gardner Crap.... same thing
This truly is the best lecture I've found on the subject. I would have liked to see your take on the importance of the ratio between output/input impedence when one circuit drives another, and how Thevenin's equivelent circuits are applied when the load drives subsequent loads. Regardless, this lecture improved my tenuous grasp on the subject immensely.
Make sure you're using the updated lecture: th-cam.com/video/Ivs5fJ4-W6E/w-d-xo.html
Definitely check out the maximum power transfer theorem lecture at: th-cam.com/video/tFtFGNZTZrA/w-d-xo.html
I hear you on the source/load placement...electronics student here, background DC electrical.....thank you for your pity!
Can someone help me out? With the solutions to the problem @ 21:18 I got the same exact answers except for the 5K ohm one, I got 28.3 V and 5mA? I'm kind of confused how he got 42.9 V as an answer when the ETH voltage is only 30 V?
Yes, there should be an annotation that displays the correct values of 5.7mA and 28.3V
Well explained!!!!! 💪🏻
This series is fantastic man, I very much appreciate the effort. It has definitely helped clarify some things in my head and I'm now far more fluent with my approach to problems.
I cannot thank you enough.
You know that 'acrossed' isn't a word though right Jim? :)
On ex 3: 40 ohm is in par with 50 + 75
So 40 // 125 = 30,30 ohm
You still have the 30 ohm resistor from the source and the 20 ohm from the load…
How did you get 55 ohms if I may ask please?
Can you have a current source and voltage source in the same circuit as opposed to your example of 2 voltages sources
You can just do a source transformation with the current source sample. 40mA parallel with 400ohms to a 16V voltage supply series with 400ohms which is the equivalent voltage circuit and from there rth and vth can be solve easily. Since 400ohm is in series with the other 400ohms, rth will be 800 ohms while vth is already given as 16V from the source transformation
Hi Jim, any chance you can give more hints for the 5th circuit? I got RTh easily but I don't seem to get the correct Eth when applying Superposition. I tried using Norton Theorem and even Nodal analysis to no avail. I think the 500-ohm resistor in series with the 40V voltage source is throwing me off. It's actually the same issue I ran into with problem 3. Thanks for all your lectures; I am finding them very helpful in studying for my FE/EIT exam.
I'm going to say it one more time. Thank you.
Best videos ever
Damn I love how interactive this is
at least you were able to get something into my dumb little brain, thanks for that
This was an extremely good video
I know that you don't like node voltage but it makes QUICK work of problem #5. First do source transformation on the 1 amp source in parallel with the 300 ohm resister giving 300 v in series with 300 + 200 = 500 ohm. This collapses the problem to 2 loops and a single essential node at the junction of the 200 (now 500), 500 and 750 node junction. A SINGLE equation then reveals that the voltage at that node is 121.875. Voltage division reveals Vth = 121.875X(2000/3750) = 65 V.
Thank you
Can anyone help me with understanding why we leave the voltage source as it is and calculate voltage drop across those resistors in parallel with RL (I understand this point) but we measure the value of resistors close to the voltage source as a parallel connection? What is the need of measuring different values with basically two different circuits? Why the extra step of reducing the voltage source to 0?
The easiest answer is this ... that's how Thevenin's Theorem works. It's a two step method:
Step 1 - remove the load resistor, determine the open circuit voltage across the terminals of interest
Step 2 - remove the load resistor, determine the resistance at the terminals of interest with the source(s) removed
The difference in perspective between step 1 and 2 necessitates a complete re-evalution of the circuit in question because they're not the same circuit, thus the different observed electrical properties.
"If you haven't punched your computer screen by now you are far more patient than I". Well, this made me subscribe. You are absolutely funny and you have a cool way of thinking. I appreciate your honesty too, when you tell the viewer that you can't teach us to be "flexible". I will watch the rest of the video now. Thanks for making it btw, I am struggling a lot to understand Thevenin's theorem.
EDIT: Not anymore
#3 was pretty tough. I must admit that I gave up and checked the answer. Then I was focused on how to get the right answer. I tried mesh analysis but didn't succeed either. I suppose many of us got to the 40 - 30 -125 Ohm point and then got stuck.
It turns out that 40 and 125 are in parallel. I could visualize it using kind of 3D imagination: assume that you look at an opened book from above. So, 40-30 is the left page, lets say, #123, 30-125 is the right page #125 and 30 ohms is the page in the middle - #124 - you are actually holding before passing. Then the result of parallel 40&125 that happens to be around 30 ohms is in series with 30 ohms placed next to the voltage source. The current of 400 mAmps then gets proportionally divided and to the 125 ohms part goes 400 mAmps * 125/165 = 96 mAmps. Then 96 mAmps * 75 ohms gives us 7.27 volts across 75 ohms resistor.
Ok but how do you get 55ohms for rth?
Very nice and thank you for this excellent resource. I'm not sure if it's an error but I continuously get a different result for Pl at 35:37 . I'm using I^2R to calculate power ie (11.4x10^-3)^2 x 600 = 6.84x10^-3 or 6.84mW a magnitude of 10 out? I have an exam next week stressed out hard so I could be completely ballsed up here...... =( I'm guessing I am plugging the values incorectly but when I enter them as .0114^2 x 600 I get a much closer albeit not equal 77.97x10^-3 W... Am I going insane or is this an error? kind regards from Australia, Johann
The power formula of I^2R is correct, however, the answer of 6.84mW means you're somehow not properly squaring the current. This is why I like using different variations of formulas to check the final answer.
Thank you so much sir
What do you use to draw all these circuits on the PC?
Most are hand drawn but some are cut from NI MultiSim
Perfect
Found you at the Columbia Gorge Community College channel! Finally i feel comfortable with Thevenins theorem!
Thank you! By the way do you have any videos for Nodal Analysis ? :)
kam33p Stay in touch because "Nodal Analysis" and "Mesh Analysis" are on the production list. Spread the word to your classmates because I'm hoping to have the whole Basic Electronics 1: DC Circuit Analysis playlist done in the next couple months. I've got an interesting take on Mesh and Nodal and I'm including it in the BE1 playlist out of sheer pity for those that are enrolled in classes where an instructor forces them to use these techniques. I'm not saying they're not valid circuit analysis techniques ... but I am saying there are far more important things to learn. The Superposition Theorem and software analysis essentially render Mesh and Nodal analysis as relevant to modern society as sword fighting. It's elegant and cool ... but not something you ordinarily need to learn.
Jim Pytel Awesome i will! Thank you once again! :)
kam33p As you wish sir! I present the new Nodal Analysis lecture ... th-cam.com/video/QbPq-ZHyfG4/w-d-xo.html
With the first 4 examples (which turned out to be the same example), was there a way to solve for the voltage and current across RL without adding the parallel resistors together to get an equivalent resistance? I feel like you were suggesting that we could, but I'm not certain
Yes, Thevenin's theorem is a more universal approach that is not dependent upon the value of the adjustable load resistor. Every time the load resistor changes one needs to perform another complete analysis of the circuit. For another explanation, I've since published an updated Thevenin's theorem lecture. Check it out at: th-cam.com/video/Ivs5fJ4-W6E/w-d-xo.html
@@bigbadtech Thanks Jim
14:57 Is that the current division rule just modified or another formula ?
It's actually the DC Voltage Divider Rule. Check it out at: th-cam.com/video/-cJk1Us0Fx4/w-d-xo.html
THIS IS SOOOOO COOL
i don't know if you reading this or not at least i gonna try to ask, at 23:30 how do you get 177.4 cause i try the parallel resistor formula which is 1/R = 1/240 + 1/680 i got 23/4080 ?
Check out the parallel resistors lecture at: th-cam.com/video/gUwoCJJgxt8/w-d-xo.html Additionally check out the DC Math lecture at th-cam.com/video/ItdTvzv32BQ/w-d-xo.html specifically the section about the scientific calculator. Sounds like you still have your calculator in EXACT mode. It needs to be in APPROXIMATE. Additionally, you've solved for 1/Rp not Rp. To solve for Rp you need to do a final inversion.
@@bigbadtech Thank you
Resonance with your take on Textbook examples @ 4:25
In #4 it is not necessary to do a delta-y conversion. The 180 is in parallel with the 600 giving 138.46. The 138.46 is in series with the 1000 giving 1138.46. The 1138.46 is in parallel with the 400 giving 296.
LN2 + Jell-O + Tree = Fun!
Great instruction but at 21 min. Your answer by the 5k ohm resistor is wrong. It should be 5.66 mA and 28.3V
Your instruction is very good, my teacher is very difficult to understand.
Daan550 Very nice! I annotated it in the video and gave you 10k bonus points (100k if you were crazy enough to solve for it using the original series-parallel circuit). Normally I can track down my errors but I have no idea what I was smoking when I wrote down those values. How can you get a 42.9V drop from a 30V source? Thanks for letting me know. Let your classmates know this resource exists as a supporting resource for those struggling with certain subjects.
thanks a lot for the videos and great explanation i keep hearing you say that one has to view parallel circuits at a nodal level and from perspective of flowing current , can you please elaborate more on this or point me to the resources that would help me get a grasp on this ? thank you
Check out the "DC Kirchhoff's Current Law" lecture at:
th-cam.com/video/Wryk_WGXjRU/w-d-xo.html
hey i was just going through your practice questions and i am stuck finding the Vth of #3 should vth be equal to the voltage drop across the the 75 ohms resistance which i have found to be 11.61v am i missing something here ?
Check out the advice I give to RJ DeSoto in the comment section below.
i still didn't get in haha but after taking a quick 10 minutes break i figured that i could use mesh analysis to find the current through 75 ohms after re-positioning the 40 ohms of course and to my surprise i found the current to be 0.096A AND times 75 ohms i got 7.2V ( i did it!!!!!) i am still having a little bit of trouble finding the Rth on the first try though in #3,4 any extra tips? i have watched all the video related to it though i cant thank you enough for the videos and your time : God bless
I think that I figured out why #3 is Vth is 7.2 v( yes I am still trying to figure it out lol ) the 30 ohms resistance has a voltage drop, therefore, we can not use the full 24 voltage to find the voltage drop across 75 , we have to find the voltage drop across the 30ohms subtract it from the total voltage and use the rest to find the voltage across the 75 ohms
Hate to bother you sir; however, I got stuck on the third exercise trying to solve for Eth. Any help would be greatly appreciated. BTW, thank you for all the lessons. They have been invaluable in developing my understanding of DC theory. Thumbs up sir!
+Rj DeSoto Thanks much, super appreciate you're making use of this material. Not sure if you're talking about the third illustrated example problem or the third challenge problem at the end. Either way, Eth is the "open circuit voltage" ... no current can flow through an open circuit ... this may fundamentally change the nature of your circuit. Try to redraw the circuit and see if a reconfiguration of elements clarifies the two point voltage measurement. I'm a visual thinker and when presented with these types of problems I keep picturing different configurations until I get it. Keep up the valiant effort!
+Jim Pytel Thank you for the reply. It was the third exercise at the end of the lesson (the challenge exercises). I know that current will not flow through the 20 ohm resistor after removing the variable load. For the life of me, I just could not figure out how the 7.2 v drop occured on the 75 ohm resistor. The 40 ohm resistor has acctually been throwing me off a bit. Thanks for the help.
+Rj DeSoto Oh yeah that's a tricky one. As you figured out the 40 ohm resistor can be repositioned such that it is in parallel with the 50 and 75 in series. Any more hints and I might give away the secret to everyone else!
I got discouraged when I couldn't calculate VL or IL correctly for ANY given value, let alone try to simplify the process. I wish this video walked through the calculations more than saying "calculate! Good luck"
There is a veritable mountain of material that precedes this lecture. Check out the "DC Circuit Analysis" playlist at: th-cam.com/play/PLdnqjKaksr8qQ9w3XY5zFXQ2H-zXQFMlI.html
At my university for whatever reason they taught us Theverin theorem and Norton theorem before teaching us nodal analysis. Honestly it made understanding them quite confusing, I felt lost the entire lecture which is why I'm here.
Ugh ... nodal and mesh are antiquated techniques that are no longer relevant. I feel your pain. Glad I can help.
Make sure you're using the updated lectures in this playlist: th-cam.com/play/PLdnqjKaksr8qQ9w3XY5zFXQ2H-zXQFMlI.html
@@bigbadtech Do you know what the name of the modern technique is? I'm really struggling to grasp nodal and mesh.
Check the lead in to the mesh analysis lecture for my disclaimer: th-cam.com/video/Mj5H1Y4pJno/w-d-xo.html
I prefer using software circuit analysis or the superposition theorem: th-cam.com/video/q47g1Nz_4lM/w-d-xo.html
Acrosst?
how did he get 860ohms for RL?
this is what I want to know too
I'm having trouble to understand why the 1.3k and 2.2k are in parallel and also why when finding E th, the 1.8 is insignificant yet it was to find Rth and why the 1.1k is not significant when finding Rth but was to find Eth
for the 3rd problem i believe
Make sure you're using the updated lectures:
th-cam.com/video/Ivs5fJ4-W6E/w-d-xo.html
th-cam.com/video/UdEUPUn_onY/w-d-xo.html
To answer your question solving for Eth and Rth are two totally separate steps and have NOTHING to do with each other. This being said both techniques require "modification" of the circuit. In the case of Eth we're solving for voltage across an open circuit. In the case of Rth the load and source are removed and we're solving for resistance. Depending on which property you're solving for modifications like shorts and opens can fundamentally change the nature of the circuit you're working with.
Who else loves the intro song? :)
I mean COME ON. For anyone looking for a bare bones easy explanation of this... you set it up perfectly and then trashed it with the exercise around 5:30.
Sometimes seeing it done start to finish is a GREAT teaching tool
By 02:50 I was "Primed" Out & totally mathematically Derailed from any Practical usefulness of Thevenin & his Twisted theorems....Never needed this concept in 50+ yrs electrical/ Electronics work employment.....Suppose it all in "Satisfying" the Almighty course professor's Power of making students eat "Shite" in exchange for the "Grade".....
You've got an interesting perspective of what instructors do for their students. Yes, theorems may or may not be used in certain career fields. As mentioned in the lecture Thevenin's Theorem specifically deals with frequently changing load conditions. If this is a scenario that presents itself its far simpler to use the Thevenin's equivalent circuit.
This is good for the method
But thats still not what i wanted
I wanted intuition
Why does this work? Kinda like a proof
This guy takes his circuits very seriously..
Very.
@@bigbadtech lol thanks for the reply and this wonderful elaborate tutorial.
Got pretty dramatic for the first 10 mins, "circuit analysis is an art form"
Great video! Can you stop writing your 2s like Ls please, confusing!
The problem is, you're explaining the basic laws of circuit analysis and I know all that stuff.. I want a proof for thevenin theorem.. I kept watching for 20 minutes and you didn't get started in proving the theorem yet!
Abdulrahman Mahdaly WTH I watched the whole video and you did not prove it! why did I waste my time watching this !! I want A PROOF .. a freaking proof!
Abdulrahman Mahdaly Sorry, proving TT wasn't the intended purpose of this lecture. The intended audience of this lecture series are those students enrolled in technician level programs and the stated objectives of this lecture are to learn to use TT. Long story short, technicians don't need proof ... they just need to know how to use a proven theorem.
My problem is with f***ing dependent sources
I feel your pain. I need to publish a lecture on dependent sources for all those feeling the same way.
14 hours remaining for my exam, but do it I’m sure that it will be so beneficial for some one.. Good luck doing it my legend 👍🏻
Redo this video. And make your 2 look like a 2 not an L.
For R2 looks like RL
th-cam.com/video/e0KxcU_-ggY/w-d-xo.html
I’m honestly really disappointed in the video.
I need to see a full thevenins start to finish and your “start low, turn up the heat” thing really detracts from seeing it start to finish.
Do it start to finish and then ask the viewer to calculate.
I’m still struggling to see the big picture and no one just finishes the damn picture before tangenting or doing exercises