Supernode Analysis: Dependent Source

[@EEAcademy2022]

Reference For this Problem: Engineering Circuit Analysis - Kemmerly & Hayt, Chapter - 4 , Exercise - 4.17 (Fig - 4.47) , 7th Edition There is an error in writing equation (iii), I skipped a minus sign.. Please use following correction - 0.5 V1 + 0.7 V2 = 8 ---- (iii)

[@johnv0625]

Hi I've been wondering in your eq. 2 the given is 5=8+(v1-v2 over 2)+(v3 over 2.5) right? then it's lcd is 10 right?then 10 divided by 2 is 5, 5 times v1-v2 is 5v1 - 5v2, then you will do the same process to the v3 over 2.5 right? then the answer is 5V1-5V2+4V3=-3 right? how come that your answer become 0.5v1 - 0.5v2 + 0.4v3 = -3?

[@tanzimislam6845]

That moaning in the beginning 0:01 when you do circuit maths 😂

[@LeBravoSpike]

thanks for the iii correction--I was following along and was wondering how I could have gotten a different answer.

[@dq.16]

we have to take Va as v2/5 ryt? because Va is flowing through 5ohm resistor ?

[@laysalhadi9186]

which software did you used to make circuit solution tutorial...? software name please...

[@josephsmth646]

How did you take 8 amps positive in kcl eqn at node 2

[@md.mahbuburrahman4379]

Vi, eta kon software use kore circuit gula draw kortechen janale khub upokrito hotam.

[@tonoyahaque]

how 8A current source is outgoing...I dont understand..

[@EEAcademy2022]

Look at the direction of current source at node 3.... You will understand

[@TheArnoldification]

Old comment, but leaving this here for reference: I believe the 8A current source is outgoing because it's leaving v3 and going into v2, and that current will split into both the bottom-middle and middle-left branches (meaning whatever current through v2 into v1 is definitely not 8 amps)

[@ambaradhikari7425]

If you are saying about the middle part so why are you not considering it in supernode???? Actually in supernode the only 2 nodes will act and for that we don't consider the middle node and so the 8A will be incoming

[@HS-fb9tw]

Is it okay that you assumed the current direction between V1 and V2 to the right at equation 2 , and then you assumed the current direction at equation 3 on the same spot to the left between V2 and V1 ?

[@EEAcademy2022]

When you apply KCL in a node, you have to consider each of them as an independent node. When you apply KCL in node 2, forget about node 1. In the nodal analysis, all unknown currents must be outgoing from the node and that node will be higher than other node voltages. So current in the second case through that path is V2-V1/R. If you are still confused, check the steps of nodal analysis. kzbin.info/www/bejne/bHWxiqOEiq9jZpo

[@md.shihabahmed7355]

I was getting a lot of benefit from your tutorial, I hope everyone will like. Thank you very much

[@md.rezuwanhassan5484]

Thank u brother for explaining the whole thing so easily! Really loved it and helped me a lot to understand the concept! But, I think u've had made a little calculation mistake while calculating the V-2 Node's incoming and outgoing current! u've had written the equation as "0.5V-1 + 0.7V-2 = 8" but, I think it will be "0.7V-2 - 0.5V-1 = 8" please, pardon me if I am wrong! U are surely senior than me and I know that u're a very brilliant student as u are studying at a very reputed public university. But, nobody is perfect... And, we humans are allowed and meant to make mistakes!

[@EEAcademy2022]

Hm... I skipped a minus sign in front of V1 in equation (iii)... Thanks for the correction

[@hemayetullah8269]

V1 = 4.28 V2= 8.36 V3= -2.40

[@ahmadsadman8023]

nice vaiya. carry on...

[@evanp.1315]

Great video when played at 0. 75 speed.

[@mahadishakkhor123]

nice one

[@ganeshpradhan9756]

It will be -0.5V1 in eq 3

[@EEAcademy2022]

I put that in the description box

[@wasiqsiddiqui3033]

where is the explanation? How did you solved the equation 1 2 and 3??

[@PriyanshuPradhan-sz7jw]

put on calculator get the value

[@antoniorobles9034]

Great explanation!!

[@ralphmorales3692]

nobody ever wants to solve the system of equations SMH

[@hasanmohammadkhan1524]

Thank you

[@md.jashimuddin7118]

Thanks....

[@dineshprasad6983]

It's "v" not "Bhi"😂😂😂..