Nodal Analysis Example Problem #1: Two Voltage Sources
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- čas přidán 23. 07. 2024
- This tutorial works through a Nodal Analysis example problem. Nodal Analysis is a method of circuit analysis where we basically just apply Kirchhoff's Current Law (KCL) at each node with an unknown voltage. You can choose to ground any of the nodes, but generally, grounding one with the most branches (especially if they are voltage sources) will make the problem easier. We will end up with one equation per unknown node voltage, and can just solve the system of linear equations to find the branch currents. This example problem features two voltage sources.
This video is part of a full free course on electric circuits. The course covers DC circuits, circuit laws, current & voltage sources, series & parallel resistors, nodal analysis, mesh analysis, and AC circuits.
Links to the course are here:
Website: www.engineer4free.com/circuits
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Thanks for watching, I hope it helps!
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Full Circuit Analysis playlist is here: czcams.com/play/PLOAuB8dR35ocf9Typ1iX9NRmX0V04UYfQ.html
Thanks, though you have made a mistake at kcL B , because if you multiply -4(2-vB)
You are supposed to get ( -8+4vB)
So that you get =2vB-VA+vB-8+4vB
If you solve further you get vA=7vB-8
Thank God i wasn't mistaken
bro no wonder I was so confused
I hate long video tutorial and yours is just so brief and simple easy to understand and fast to learn a big topic in less then 10 minutes thank you for this so helpful
This one and the explanation made Nodal analysis so much more understandable
100x better than my university teacher.
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True brother❤
exactly
Yea 🫡
Totally amazing. I really could understand thanks to you.
You do amazing job! I am going through the playlist and smashing like on every video, cuz you deserve some positive feedback for this work
Thanks for this video , my guy saving my assignment rn
7:02 For the second KCL you did an algebraic mistake. It is not +4 but -4
Yes you are correct 😂
True, although he didn't include it in his answer, as the correct amswer is Va=7Vb-8.
Oops 😬
TQ for this sir..I am able to understand ur calsses easily
Life made simple thank you very much.. very simplified
This guy made my day, and now i can teach my girlfriend!
You get a subscription.
Much informative vid keep it up👍
Thanks understood clearly ❤
Hi how to solve if the 2 ohm's is unknown?
Hope you notice
Thanks for this video
Sir thank you so much, you saved my day....I already subscribed your channel....Thank you and God Bless
You're welcome!! Thanks for watching =)
Exellent .i visit a lot of channel bit your channel is best .you solve my confusion
thank you so much !!
Wow, Thank you so much
Wow 🎉 thanks so much
Hi,I have a problem and i can't solve it.Is there a way to send the problem so that you help me out?
But how do you get KCL A -6 to -18 after simplify? may i know?
May God bless you 🙏❤ bro
Your voice is soothing
-6 *-Va is +6va
*“We learn more by looking for the answer to a question and not finding it than we do from learning the answer itself.”*
I love u so much dude
Nice video
what if the voltage source is flipped and negative sign is above?
Thank you sir
I like how you used the fraction multiplied by 7 for less rounding issues
love you man
I dont understand why i3 on one side is traveling in and is respresented as a positive ? and on the other set on node be as a negative ??? what is going on??
What if we have only one voltage source
How to know the currents are out or in in direction please reply me soon
Can you do one based on wheatstone bridge? also it seems from other comments on other videos that professors leave specific details out to "appreciate how to get the right answer" people don't have time to mess around.
i absolutely hate professors like that.
In NIGERIA, lecturers don't teach. They just give course outlines then ask you to go study yourself with brief explanations.
Hi, I like your teaching but i have one doubt in this video. I want to know why i3 goes on right side? can't it go to left too? Please reply! much appreciated
I just assumed a direction for i3, in this case it was to the right. When I drew it, it was an unknown, and I could have assumed either direction. If i3 ended up being a negative vale, then that would have told us that it was in fact pointing the other direction. However, in this example, we calculated i3 to be positive, so our assumed direction was correct.
@@Engineer4Free Thanks a lot . Subscribed your channel. I hope your channel will help others a lot too Keep it up
Hi Merry,
If you're interested in electrical/electronics engineering, we're building a channel to help students and professionals prepare for job interviews with real world questions. We'd love it if you could swing by. Also, our LinkedIn is: www.linkedin.com/company/hardware-ninja
5:05 Its supposed to be -18+6va, because negative times negative becomes a positive.
at kcl b the final equation is wrong
4:11 how to you know what to multiple with like when u did with 12
Through LCM
Erm how is -6x-3va still a -18va at kcl A
How we find common dinominator?
it's quite easy, actually! it's elementary level maths; basically you single out all the numbers on the bottoms of the fractions, line them up and start dividing with numbers in hopes of getting each number to a 1.
for example, if you have 1, 2, 4 like in the Vb portion of the video - you ignore the 1 because it's already a one, so you have 2 and 4 left. both are divisible by 2, which - when divided - leaves you with a 1 and a 2.
then you divide them again with a 2 and the end result is three ones, with two twos you used to divide them with.
then you multiply the numbers you used while dividing (aka 2 and 2 here) with one another, and the end result is your common denominator (4 in this case).
hope i was able to help!
In KCL - A you multiplied -6 by -va which you have as -6VA which should be +6VA . If it was -6VA you would not have 13VA but instead just have 1VA.. to get 13VA it has to be positive.
Thanks i thought i was tripping
distribute the negative first then it would be -3+VA
huge
Am i the only one who noticed that he made a mistake in KCL B, while subtracting I5 from I4 and I3. He changed I5 sign from minus to plus.
why is it not -va + 7vb = 8
I'm boggled a little bit on 7:51
How is 2VB + VB - 4VB = 7VB
Shouldn't it be -VB?
VB counts as 1. Same as how X counts as 1.
What happened to 7Vb?
It never existed
to come up 13Va
you should use mesh hhere
i think that -6Vb is +6Vb
sorry the -6Va i mean is +6Va
Is it just me or he change the KCL at I⁵
The results are wrong i'm afraid. You made sign destribution mistake in the sign distribution in KCL A and B.
Which one of you guys student at university of lahor
Our sir is wallhi azab😶
VB should be 1.32v
ماتقدرش تهدر عربي
ابحث عن عربي دي انجليزي.
minute 7:30 big mistake does not correspond to previous level throw you out of loop!!!!!!!!!!! review review
Don't put this type of videos
Speak arabe