Scroll up and hit that 👍 button if you found this video helpful! Full Circuit Analysis playlist is here: ua-cam.com/play/PLOAuB8dR35ocf9Typ1iX9NRmX0V04UYfQ.html
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
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
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.
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.
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
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??
Why does my professor only use the circle voltage sources? The four-line ones look so much better when drawing. The art classes are in another building, man.
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.
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!
Scroll up and hit that 👍 button if you found this video helpful!
Full Circuit Analysis playlist is here: ua-cam.com/play/PLOAuB8dR35ocf9Typ1iX9NRmX0V04UYfQ.html
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
100x better than my university teacher.
😁😁
😂😂😂😂😂
True brother❤
exactly
Yea 🫡
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
This one and the explanation made Nodal analysis so much more understandable
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 😬
5:05 Its supposed to be -18+6va, because negative times negative becomes a positive.
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
This guy made my day, and now i can teach my girlfriend!
You get a subscription.
Totally amazing. I really could understand thanks to you.
thankyou you are better than teach in my prof.
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
Your voice is soothing
-6 *-Va is +6va
Life made simple thank you very much.. very simplified
Much informative vid keep it up👍
Thanks for this video , my guy saving my assignment rn
TQ for this sir..I am able to understand ur calsses easily
*“We learn more by looking for the answer to a question and not finding it than we do from learning the answer itself.”*
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
you made this easy for me thank you
Thank u so soo much . this helped alot❤❤
In NIGERIA, lecturers don't teach. They just give course outlines then ask you to go study yourself with brief explanations.
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.
Thanks understood clearly ❤
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.
I like how you used the fraction multiplied by 7 for less rounding issues
May God bless you 🙏❤ bro
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
But how do you get KCL A -6 to -18 after simplify? may i know?
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??
Anyone clarify me if we consider I under 2 in opposite direction wether the equation Kcl A is messed or not
How to know the currents are out or in in direction please reply me soon
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?
what if the voltage source is flipped and negative sign is above?
at kcl b the final equation is wrong
Wow, Thank you so much
Why does my professor only use the circle voltage sources? The four-line ones look so much better when drawing. The art classes are in another building, man.
Wow 🎉 thanks so much
KCL A is wrong, it isnt 13Va but 3Va. -6Va + 4Va + 3Va = -2Va + 3Va = Va
4:11 how to you know what to multiple with like when u did with 12
Through LCM
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.
Hi how to solve if the 2 ohm's is unknown?
Hope you notice
Thanks for this video
Nice video
thank you so much !!
Thank you sir
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!
Erm how is -6x-3va still a -18va at kcl A
I love u so much dude
THANKS
-6(3-Va)=-18+6Va😢😢 4:43
Here no 13 Va but only Va 5:08
love you man
What if we have only one voltage source
why is it not -va + 7vb = 8
The results are wrong i'm afraid. You made sign destribution mistake in the sign distribution in KCL A and B.
What happened to 7Vb?
It never existed
sorry the -6Va i mean is +6Va
i think that -6Vb is +6Vb
you should use mesh hhere
minute 7:30 big mistake does not correspond to previous level throw you out of loop!!!!!!!!!!! review review
Which one of you guys student at university of lahor
Our sir is wallhi azab😶
VB should be 1.32v
Is it just me or he change the KCL at I⁵
to come up 13Va
huge
ماتقدرش تهدر عربي
ابحث عن عربي دي انجليزي.
All I’ve seen on this guys videos is that he makes mistakes in almost every single video. Can’t trust his teaching
Don't put this type of videos
Speak arabe
HOW DO YOU GET 7VB WHEN 1VB+2VB = 3VB (+)-4 YOU WOULD THINK THAT WOULD BE -1 NOT 7