Circuits 1 - Mesh Analysis and Super Mesh - Example

[@edmundohernandez8464]

duuuuude this video is great!! you explained everything, even the simple stuff that someone who didn't pay attention in class at all will need. THANK YOU!

[@oluwatobiakinola25]

Finally!! someone who did a more complex question. Thanks

[@NineInchFailz]

I have been seriously struggling for my entire class understanding why these kinds of circuits behaved the way they do it and you just cleared it up so well. I was specifically curious about how a supermesh could possibly be accurate given you're essentially ignoring the middle elements, but the way you explained it made so much sense.

[@pabombs1244]

Thank you for solving a difficult-ish problem, all the other videos were doing problems that were not hard enough!

[@UConnHKN]

You're very welcome! If you have any problems you'd like us to solve, let us know!

[@scottgreen3807]

Nice, finally got to the level. I learned all this stuff at Albuquerque Technical Vocational Institute in 1981. Used it for decades, well stuff like this. These are the building blocks of the world. Where’s your op amp lectures? Great stuff.

[@Dalie3011]

give this guy a raise

[@davedarius7346]

Informative class ✌🏻 but just one thing I noticed is that, we here learned to take Mesh Analysis by the others sign i.e. 20-3(i1-i3)-5i1=0(First mesh) and in that manner! Basically the same but we flip the signs, I guess It's different conventions in different regions!

[@adhamelkasaby9423]

I think ur right, because current direction is from the +ve terminal to the -ve terminal (which is the opposite to electron direction), But in the end, both are the same because we impose all this

[@MarcoLopez-wy6zx]

Very helpful indeed! Thanks bro! Keep it up with these kinds of tutorials :)

[@xyzpg]

You made it crystal clear .Thanks.

[@awildmoosey]

for anyone wanting to check: i_1 = (-92/43) A i_2 = (-34/43) A i_3 = (-49/43) A

[@estebamp3720]

Dude, you're the best, keep up the good work!

[@fubrian2945]

This is underrated.

[@mdsiam9109]

3rd one was special...superbbbbbbb!!

[@Bass000]

When you go from negative to postive of voltage it mean voltage rise it must be positive 20v in I1 (loop1)

[@altuber99_athlete]

Ibnu Bastistah No. There are two ways to apply KVL (in resistive linear circuits). One is how you say, i.e. the voltage sign is determined by which terminal the current loop is *exiting* (and thus the voltage of the battery in that loop is positive and the voltage of the resistors is ALWAYS negative). The other way, which is the one he applis and the one I think Sadiku's book applies, is that the voltage sign is determined by which terminal the current loop is *entering* (and thus the voltage ofthe battery in that loop is negative andthe voltage of the resistors is ALWAYS positive). Now, is the equation obtained by the first methos the same or a equivalent of the one obtained by the secons method? Yes! Just multiply an equation by -1 and you'll get the other.

[@kevinmain3399]

solving for i_1 I got ... i_1=i_3-1 .... substitute i_3 - 1 for i_1 to get these equations (eq1) -3i_3 + 3 + 11i_2 - 2i_3 = 0 (eq2) 8i_3 - 8 - 5i_2 + 6i_3 = -20... simplify to get 2 unknowns... 11i_2 - 5i_3 = 0 (eq1) and -5i_2 + 14i_3 = 12 (eq2) using inverse matrix to solve... I got i2 = 0.47 A, i3 = 1.02 A, and i1 = 0.02 A.

[@HQR000]

thanks a million. my lecturer sucks at explaining things

[@UConnHKN]

Glad you enjoyed it!

[@zahidnabi21]

kzbin.info/www/bejne/rnW4mXmumsqpmZY

[@jimcoelectronicsystemsengr325]

nice video. I have clearer insight now. Thanks What kind of smart board is this

[@istainblack]

So I had a question. When you solved the super mesh circuit, you drew a new composite box to sum the voltage drops around. Was there anything special about the new box you chose? Could you have not used one of the previous box's you'd drawn to sum the voltage drops around in order to solve for your currents?

[@istainblack]

Ok so I can see now that calculating a voltage drop across smaller boxes you had previously drawn would not work as you would have to calculate a voltage drop over the ideal current provider, which wouldn't be so cool.

[@narutouchiha1946]

This is an excellent video thanks a bunch.

[@240mains]

missed it ! 3 books you should read ( at least read the DC chapters of each book ) Engineering Circuit Analysis 5th edition Hayt and Kemmerly Electric Circuits ,...Nilson 2nd edition Basic Engineering Circuit Analysis Irwin , 4th edition or newer editions

[@muhammadsaqlainishtiaq4467]

That 1amp is the common to both of meshes thats what makes it super mesh.

[@adivhahofrenenetshimbupfe4124]

i found this very helpful..i understood mesh in just one vedio

[@anilkumarraju1898]

Good

[@materialsky313]

clear and concise, thanks!

[@iruettka]

shouldn't the current in loop 3 go in the opposite direction? since the current is going from - to + from the 40 V source?

[@psychoturtle509]

You prob know this already, but for other commenter, to obey sign convention, each loop must go in same direction

[@EpicNostra]

How does he write like this? Backward!! Reverse !!

[@DeanOckenden8]

He writes normally, they flip the video during editing.

[@jonathanchiii1706]

He wrote on the glass between him and camera, and they flipped the video

[@MrSmallPet]

He writes on the glass as normal, once the video is in the editing stage, the video is then edited

[@shanny3667]

@@jonathanchiii1706 thanks for pointing that out. I can now sleep peacefully 😌😌😌

[@shanny3667]

I was thinking the same thing! LOL

[@arslannazir1468]

ITS GOOD EFFORT..... KEEP IT UP

[@UConnHKN]

THANK YOU!

[@waroftheworlds2008]

4:30 so on passive elements (RLC) the passive sign convention is always positive with this method? and it's only the sources that you decide on positive or negative?

[@UConnHKN]

The sign convention is just that, a convention. The easiest thing to do is the write the positive and negative signs on the passive elements at the beginning of the problem and them as you traverse the KVL loop, use the sign for the element that you encounter first. ~Andrew F

[@uzzalh.mohammad1329]

Nice tutorial ! very helpful, thanks . :)

[@b62_shriharishatrughan79]

good stuff g

[@UnoDuosTre]

Won't 40 be negative because it shows a voltage drop?

[@fanpeter-z3c]

yeah i think the third equation should be -40v too

[@frankvalenti8350]

Awesome vid

[@safoma]

You're so blessed

[@alishajiwa884]

How do i find the next part to this video?

[@khalid98]

جميييييل جدا👍 شكرا🌷🌷

[@mdnazmulhuda2555]

it was very helpful

[@UConnHKN]

Glad to hear that! Let us know if we can do any other problems for you ;)

[@Dalie3011]

You're the best

[@richardhall9815]

It's a bird... it's a plane... it's SUPERMESH!!!

[@hollysukulelemunford2432]

Hahaha great super mesh!

[@hajer5477]

THANK YOU

[@zacharyschneider6040]

literally taught more than my garbage online class b/c of covid

[@denji3397]

Nice tutoring :3

[@hollysukulelemunford2432]

Wow! INSPIRED! Super mesh! UKULELE HOLLY BLOE USA

[@pls_letmewin9548]

Tq , ur the best

[@nithishkumar9283]

At 1A why should we take I3-I1 not I1-I3

[@carsonprokopuik5101]

KCL... sum of currents at a node is zero so if I1 is assumed as CW and I3 is also CW then I3 = I1 + 1A... which => 1A = I3-I1

[@support.2023]

Your country name? I am from Bangladesh 🇧🇩

[@rabbipaalan7586]

bro in 2nd equation where 6ohmscome from

[@UConnHKN]

Current I2 goes through the 6 ohm resistor on the top of the circuit, bro

[@zahidnabi21]

kzbin.info/www/bejne/rnW4mXmumsqpmZY Learn the technique and u can solve any problem

[@faisalaldubaikhi2357]

5:13

[@leftenantthunder]

unfortunately this is wrong. The sign notation is wayy off and you cannot get the right answer following this method.

[@dalenassar9152]

I agree! He wrote the voltages with the wrong sign WRT the the method used for the resistor voltages.

[@ralphwv]

5:12 BURP!!!!

[@bobbytaraantino]

Can you make a solution in an old fashion? Our teacher says that we need to show our solution to get a points.

[@waferlayout]

Much appreciated Sir 🫡 🤭 Use matrix to solve it for 30 secs