Thevenin Equivalent Example Problem
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- Опубліковано 22 сер 2024
- This example problem uses Thevenin's theorem to determine a Thevenin equivalent circuit for a DC electrical circuit with two voltage sources and a few resistors. Note that this video solves the example problem, but does not give a detailed explanation for Thevenin's theorem.
This problem comes from question 81, section 6.9 of James Fiore's online DC Electrical Circuit Analysis text ((tinyurl.com/dc.... )
00:13 - Problem definition
00:19 - Step 1: Identify part of circuit to Thevenize
00:35 - Step 2: Label port
00:53 - Step 3a: Set voltage sources to 0V
01:53 - Step 3b: Calculate RTh (Thevenin resistance)
02:42 - Step 4: Determine VTh (Thevenin Voltage)
03:03 - : Use superposition
03:24 - : 10V source VTh calculation
04:40 - : 24V source VTh calculation
05:40 - : Put 10V and 24V source calculations together
06:25 - Thevenin equivalent circuit
07:01 - Calculate V(load) and I(load)
08:33 - Power of Thevenin
This video (and the online textbook) have the creative commons license: CC-BY-NC-SA (tinyurl.com/cc...)
All over youtube, this is the only channel whose explanation is beginner friendly.
I remember when I was in college (about 35 years ago) attending the Electricity 101 class, all seemed so cryptic and confusing but you explain just in ten minutes what took me days to understand, and everything is so clear that I wish I had a teacher like you. Awesome. Thank you.
I'm glad to have helped, sometimes just seeing things the second (or third) time around makes things make sense.
Wait for the RTH step how is 10k in parallel with 2k?
Because there is node there.
I love how you explain your answer for this problem. I hope you can create more example problems concerning different theorems (Branch current, maxwell, kcl/kvl, nodal analysis, superposition, thevenin and norton's).
thanks alot this video is so much easier to understand than most
Glad it helped!
no one gonna talk about how he's writing backwards?
I wish I had the skill to write backwards like that. I'm just using software to mirror the video
@@ElectronXLab That's genius!!
I was watching it thinking, "wow, pretty darn good, writing backwards behind plexiglass," and then, "ooh, he's lefty too. That's rare!" and then, "oh. Flipped the video."
Regardless, nice explanation of the concepts and showing how to "see" into the circuit from the Thevenin check point perspective.
@@ElectronXLab 👍👍👍
I'm lost and I've been an engineer 40 years... never needed this cause I never could believe it
Thank you so much I understood perfectly
Thank you for this precious video Now I understand how to slove complex circuit thevnin theorem
Excellent explanations. Knowing that there are any number of ways to solve the ckt, the simpler (thus faster) method instead of Superposition would have been to use “Special Thevenin” without the 4K and then to stick it back in to come up with a final Vth answer. He could even have left the 4K in and used Millman for a quicker solution.
Great tip!
How is the 2k in parallel? It contadicts other videos. If its a speci direction due to removing Vs, thsy should be noted
when you short the voltage sources, the 2k is in parallel with the 10k and the 4k resistors
How????????????????????????@@ElectronXLab
@@ElectronXLabsuppose the 2k was shifted so that it was lying up and to the right of the 4k. Would it now be in series to the other 2 resistors?
Great video as always! Thank you
Thanks again! I'm really glad my videos are continuing to be helpful
Wow... I wish UA-cam (and ElectronX Lab) would have been around when I was in EE (40 years back).
Thank you sir for your help!!!
This was life saving, Thanks!
Thankyou, that was brilliantly explained!
Great video!! Helped so much.
New subscriber unlocked
So i am confused about how the thevenin voltage is between terminals A and B, but when you re-insert the load resistor in between these two terminals, the thevenin voltage does not drop across the load resistor. May you explain this to me? thanks
Great Teacher
Thank you Sir
I see why 4k||10k but wouldn't the result be in series with the 2k resistor? ((4||10)+2)
this was helpful 🤜🤛
Great vid legend
Well explained
What will be the Vth if it was between 10k ohm ?
Thank you
You're welcome
Sir, what happened 1.33 Kilo ohm?
YOU ARE GOATED
4:05 How did you get 2.857K as the combonation of the 10K and 4K?
(10000^-1 + 4000^-1)^-1 = 2857
@@ElectronXLabI got another question do you have a video that is solely for thevenin?
@@foryourculture_ this one maybe: ua-cam.com/video/hfyfThbrqos/v-deo.htmlsi=fVvYrxVFgQKT_5sh
Is the VAb solved by using superposition theorem?
Yes!But it is just only use simple node voltage way to find out Eab !
hi professor, now I am a student. In an exam, I solved a problem like you in the video, but my professor think that I am wrong. I am not sure if I am wrong or my prefessor is wrong. May I send you my solution and you check it if my solution is ok or not? thanks
I'm sorry, I won't be able to help you with your assignment. Hopefully you can find a classmate to work with.
@@ElectronXLab it's not my assignment, just a question in an exam, my professor's answer is different from mine. So I wanted you to see my solution and find my errors. ok forget it. Thanks anyway.
Thank sir but u have to be slow and clear when lecturing please
How did you get the 2.85?
It's the parallel combination of the 10k and 4k resistors
Why is it -10 and not +10 at 4:22 but it is +24 at 5.23 when the current is going the same way on both?
For the initial circuit, I put the 10V source "upside down" - positive at the bottom and negative at the top.
Thanks!
You bet! Thank You!
Why did you do 2k plus 4k instead of 10k+4k?
Because the 10k is in series with the voltage source and since we want to make the resistors in series to the voltage source, we can just pick up the 2k and 4k that are in parallel, sum them up (i.e 2k//4k), the result will be in series with both the 10k and the voltage source (which is what we intended) so by finding the voltage drop across the 1.333 (i.e 2k//4k) is the same as Vab