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!
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.
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.
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!
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
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?
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.
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.
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?
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
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.
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
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!
Finally!! someone who did a more complex question. Thanks
Thank you for solving a difficult-ish problem, all the other videos were doing problems that were not hard enough!
You're very welcome! If you have any problems you'd like us to solve, let us know!
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.
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.
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!
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
Very helpful indeed! Thanks bro! Keep it up with these kinds of tutorials :)
Dude, you're the best, keep up the good work!
give this guy a raise
3rd one was special...superbbbbbbb!!
for anyone wanting to check:
i_1 = (-92/43) A
i_2 = (-34/43) A
i_3 = (-49/43) A
You made it crystal clear .Thanks.
This is underrated.
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?
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.
When you go from negative to postive of voltage it mean voltage rise it must be positive 20v in I1 (loop1)
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.
nice video. I have clearer insight now. Thanks
What kind of smart board is this
shouldn't the current in loop 3 go in the opposite direction? since the current is going from - to + from the 40 V source?
You prob know this already, but for other commenter, to obey sign convention, each loop must go in same direction
i found this very helpful..i understood mesh in just one vedio
Good
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?
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
ITS GOOD EFFORT..... KEEP IT UP
THANK YOU!
This is an excellent video thanks a bunch.
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.
clear and concise, thanks!
thanks a million. my lecturer sucks at explaining things
Glad you enjoyed it!
ua-cam.com/video/xDUhFxeevc4/v-deo.html
Won't 40 be negative because it shows a voltage drop?
yeah i think the third equation should be -40v too
How does he write like this? Backward!! Reverse !!
He writes normally, they flip the video during editing.
He wrote on the glass between him and camera, and they flipped the video
He writes on the glass as normal, once the video is in the editing stage, the video is then edited
@@jonathanchiii1706 thanks for pointing that out. I can now sleep peacefully 😌😌😌
I was thinking the same thing! LOL
That 1amp is the common to both of meshes thats what makes it super mesh.
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
Nice tutorial ! very helpful, thanks . :)
How do i find the next part to this video?
You're so blessed
good stuff g
Awesome vid
it was very helpful
Glad to hear that! Let us know if we can do any other problems for you ;)
At 1A why should we take I3-I1 not
I1-I3
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
جميييييل جدا👍
شكرا🌷🌷
You're the best
Nice tutoring :3
THANK YOU
Wow! INSPIRED! Super mesh! UKULELE HOLLY BLOE USA
Your country name?
I am from Bangladesh 🇧🇩
bro in 2nd equation where 6ohmscome from
Current I2 goes through the 6 ohm resistor on the top of the circuit, bro
ua-cam.com/video/xDUhFxeevc4/v-deo.html
Learn the technique and u can solve any problem
Tq , ur the best
literally taught more than my garbage online class b/c of covid
It's a bird... it's a plane... it's SUPERMESH!!!
Hahaha great super mesh!
5:12 BURP!!!!
unfortunately this is wrong. The sign notation is wayy off and you cannot get the right answer following this method.
I agree! He wrote the voltages with the wrong sign WRT the the method used for the resistor voltages.
5:13
Can you make a solution in an old fashion? Our teacher says that we need to show our solution to get a points.
Much appreciated Sir 🫡
🤭
Use matrix to solve it for 30 secs