Supermesh Explained (with examples)

[@JanFinsch]

got exam 2morw , instead of wasting time on 1hr+ lectures i stumbled upon your channel and man you are great at explaining all the complex stuff (not complex at all after watching your vids) in less than 10-15 mins! love you!!!!

[@ALLABOUTELECTRONICS]

Wish you ALL THE BEST for your Exam.

[@satyanarayansahu4957]

Sir at 3:14 in mesh 2 the value of v2 should be 3(I2-I1) BCZ we consider that current bigger which is in the mesh we are doing the operation ..

[@keyleegibbons9950]

Have not been able to understand Supermesh or Supenode from my class notes, you explained it so easy in this video and your Supernode one! Thank you!!!!!!!!!!!!!!

[@Shreyashi07]

When we are working with loop I2, why is it still I1-I2, instead of I2-I1 at the time 2:57 of the video? Any clarification is greatly appreciated

[@aadi.p4159]

Dude my teacher spent 2 hours on this garbage nd he cleared it in literally 10 mins

[@ahsansaeed4823]

he means your teacher??? hahaha or this tutor(you)

[@lombemumba4397]

THIS EXPLANATION IS SO CLEAR THANK YOU KEEP MAKING MORE VIDEOS

[@praveens4828]

simply superb.Just 10-15 mins is awesome

[@reubenwilliammpembe667]

thank you Sir you are the best!!! #RespectFromSouthAfrica

[@kalyan7589]

Will you please explain how you written KVL at Mesh 2 like -6I2+4(I3-I2) +2(I1-I2) =0 I think it was -6I2+4(I2-I3) +2(I2-I1)=0

[@ALLABOUTELECTRONICS]

In the mesh -2, the net current flowing through the 4 ohm resistor is I2 - I3. And it has been assumed that the current is flowing from right to left. So, while applying the KVL, in clock wise direction, the voltage drop acorss it would be - 4 (I2- I3). But here it has been written as 4*(I3 - I2). Same is the case for 2*(I1-I2) I hope it will clear your doubt.

[@Keshavsworld]

Please help sir You've told us in previous videos that.. when apply kvl in a mesh then the current of that should be considered as the higher current than the other mesh... But at 2:55 you write the opposite thing their..... Is it right their or am i just confused ? Please tell me sir

[@ALLABOUTELECTRONICS]

Yes, it's one of the same thing. Here, I have considered voltage V2 polarity. If you go with an earlier convention, then it would come as 3 (I2 - I1). Here it is (-3) (I1 - I2). therefore, it's the same thing. I hope it will clear your doubt.

[@Sportsman134]

for mesh 2 shouldn't it be 3*(I2-I1)

[@ashishjog]

Great video, one doubt, At 11:15, how you assumed that current flowing through 4 Ohm resistor is (I3-I2) and through 2 Ohm is (I1-I2); in this comment section actually, there are many people who asked this doubt, and you replied to them but still, I don't understand the reason behind it.

[@kedarbikkannavar5907]

You can assume anything which you want just that the signs would be reversed in that case. He is focusing on the voltage sign convention he had given to that 4ohm resistor (ie that + and -)..so whichever current enters the +ve he took it greater than the one entering the -ve.

[@dinethfonseka990]

In the last example shouldn't we consider about the 2 ohm resistor connected series to the current source

[@veeralakshmimuddana5831]

Good explanation sir,and hope u have bright future in this area

[@soul2197]

Thanks from my deepest heart possible

[@Venceroam]

Thank you Indian guy😍😍😍

[@mayurshah9131]

Superb Presentation

[@sunnyvj2636]

Sir in that super mesh technique (last sum) we have to neglect that 2ohm resistance connected to current source

[@ALLABOUTELECTRONICS]

Yes, because here we are interested in the current (I1 -I3). And current in series through the resistor will remain the same.

[@sunnyvj2636]

@@ALLABOUTELECTRONICS thanks sir

[@sachinrath219]

at 5.58 in the video instead of using v,can we use iR as v= iR,thanks.

[@soniyawawhal7234]

Does the direction of arrow change the sign of current

[@DineshGupta-fp7il]

how to determine the polarity of current in supernode or supermesh such as I2 - I1 or I1- I2. pl explain it.

[@ALLABOUTELECTRONICS]

Depending on the direction of the assumed current I1 and I2 (clockwise or the anti clock wise), it can be decided. In fact you can choose any of them. The final output will remain same. Because let's say if you choose i1 -i2, then you will get negative answer. But at the end you will get the same answer. I hope it will clear your doubt.

[@014arunkaruppaiah.n2]

Excellent explanation

[@yking9889]

Can we take i2-i3 current in 4ohm resistor in last question

[@ishitapinto5089]

What I do when voltage is connected in series with resistor should I do source transformation or just continue with is as it is given

[@ALLABOUTELECTRONICS]

It depends on the circuit. Sometimes it is better to use source transformation to simplify the circuit, while sometimes it is good to leave it as it is.

[@puttapallisatishyadav222]

Y it should be I3-I2..The current through 4ohm resistor? It may be I2-I3

[@harshkunvariya1714]

At (11:23) in equation you put -6i2+4(i3 - i2)*+2(i1 - i2)* is it right?

[@maniyadav3863]

Same doubt

[@sleephourofmotivatwakeupas6512]

The explanation was really awesome.. Thank you so much... But can you explain about selection of sign of the nodes please.

[@ALLABOUTELECTRONICS]

Do you mean the sign of the current in the mesh?

[@physicsdemonstrations2426]

@@ALLABOUTELECTRONICS yes

[@diannahelgeemaesampilo9984]

this is really really helpful! thanks.

[@thalapathybalaji23]

But how did u get I1=4 Ampere without simplifying?

[@SriramB-]

super explanation bro

[@nageshreddy4974]

u r great man.....ty

[@sssenterprises7497]

Nice explanation, thanku so much

[@kajalgoyal9544]

Sir super mesh ke 2nd question mei 2ohm resistor ko eqn. Mei kyu nahi liya

[@ALLABOUTELECTRONICS]

Because the current through 2 ohm resistor is 3A. Sine the current source is attached at one end, it means the current entering into 2 ohm resistor is 3A and the same current is also leaving through it. Therefore it is not considered in calculation. I hope it will clear your doubt.

[@factknowledge3297]

sir can you please do this sum for me the question is using mesh analysis find the v1 of the circuit shown in the diagram

[@parag1583]

Badiya bhai, keep it up

[@gandretiganesh412]

good explanation sir thank you.....

[@emirk8]

the last example's solution should be i1=6a i2=11/3 and i3=3a

[@sru.jaan.ah14]

Sir,when both the currents i1 and 3A are going in the same direction,why did you take it as i1-i3=3A at 10:56?

[@ALLABOUTELECTRONICS]

Actually, this equation or relation is not based on KCL. But just by inspection, you can observe that current I1 is flowing in the clockwise direction while current I3 is flowing in the counterclockwise direction. So, we can say that net current flowing through the 2-ohm resistor is I1- I3, which is flowing in the downward direction. And current through 3A source is also flowing in the downward direction. So, we can say that I1-I3 should be equal to 3A. I hope it will clear your doubt. And if you still have any confusion then do let me know here.

[@sru.jaan.ah14]

Thank you sir

[@pallab10]

great explanation.....!!

[@PRAKASH-hp6eg]

At 10:52 doesn't the 2ohm resistor make any effect?? Any one who know can please reply me🤝

[@ALLABOUTELECTRONICS]

Since it is in series with a current source, it will not have any impact on the currents.

[@souvikpramanik5183]

LOVE U SIR❤️❤️

[@koyasrinath3283]

Is 2ohm resistor series with current source can be neglected??

[@ALLABOUTELECTRONICS]

No need to neglect 2 ohm resistor which is in series with current source.

[@shayansiddiqui1931]

why u put negative sign before 2 at 9:56 ?? since the current is coming from positive terminal

[@ALLABOUTELECTRONICS]

Here, we are applying KVL. And 2(I1-I2) is the voltage across 2-ohm resistor. Now, while applying KVL we are moving from higher to the lower potential across 2-ohm resistor in the supermesh. So, the negative sign has been taken for the voltage across 2-ohm resistor. I hope it will clear your doubt. But if you still have any doubt then do let me know here.

[@kasiergwapo]

I think the equation at Mesh 2 is wrong. It is supposed to be -6I2 - 4(I3-I2) - 2(I1-I2) = 0 -6I2 - 4I3 + 4I2 - 2I1 + 2I2 = 0 -4I3-2I1 = 0 -2I3 - I1 = 0 I got my equation from the direction of the current in each resistors. Correct me, If I'm wrong thanks.

[@kasiergwapo]

11:35

[@ALLABOUTELECTRONICS]

Actually, the equation depends on how you take the sign convention. Here it has been assumed that in mesh 2, the net current is equal to I3-I2. Which is flowing from right to left. And accordingly, the voltage drop is shown across the 4-ohm resistor. Now, if you apply the KVL in the mesh and move around the mesh in a clockwise direction then the voltage across the 4-ohm resistor is increasing as we are moving from lower to the higher potential. So, the positive sign is taken for the voltage drop across the 4-ohm resistor. The same is true for 2-ohm resistor. I hope it will clear your doubt. And stil lif you have any doubt then do let me know here.

[@kasiergwapo]

ALL ABOUT ELECTRONICS Thank you :D

[@reetikjoshi1087]

Does 2 ohm resistance have no voltage drop near 3A current source un last eg.?

[@ALLABOUTELECTRONICS]

Of course, there is a voltage drop across the 2-ohm resistor.

[@MrOptimusPrime.]

@@ALLABOUTELECTRONICS you missed to add it to equation 2, i mean 2 ohm resistance in series with the current source.

[@madhanbk2002]

In 2nd eg mesh 2 resistance always voltage drop not voltage gain ypu did wrong sir

[@kieroribello4684]

Sir how did you come up with -6(I2)+4(I3-I2)+2(I1-I2)=0 at 11:22? I'm a little bit confuse with the signs.

[@ALLABOUTELECTRONICS]

Better you assume the voltage across each element first. Let's assume it as V1, V2, V3. Where V1, V2, and V3 are the voltage drop across 6 ohm, 4 ohm and 2 ohm resistors respectively. And if you apply the KVL with the assumed signs, then it can be written as, -V1 + V2 + V3 = 0 Now, V1 = 6*I2, V2 = 4 (I3- I2) and V3 = 2 (I1 - I2) (Here, it has been assumed that the net current I3- I2 and I1 -I2 is larger. ) Now, if you put the value of V1, V2 and V3, then you will get that expression.

[@kushalgarg7367]

@@ALLABOUTELECTRONICS NO SIR V2 = 4 (I2 -I3) and V3 = 2(I2-I1) THIS IS CORRECT AND IT DOESNOT DEPEND ON WHICH IS BIGGER , DIRECTION MATTERS

[@saeedabdullah1734]

THE 3rd EQUITION in a SUPERMESH is wrong YOU MUST START WIT 6(I2)+4(I3-I2)+2(I1-I2)=0 WHY You started with -6 answer please

[@TheBigWins]

exactly

[@aaronjoshuasarmiento6539]

The sign is negative if it is a voltage drop and it's positive if it is a voltage gain. I don't think it's wrong. It's just a different approach from yours where you consider the polarity of the terminal encountered first as you travel around the loop. Whatever approach you use in deciding which sign to put, you will arrive with the same answers anyway.

[@hemanthkumar5259]

What about the 2ohms resistor placed below the 3A

[@ALLABOUTELECTRONICS]

As, it is in series with 3A source, the same current will also flow through 2 ohm resistor.

[@apoorva158]

Nice sir

[@jayeshsonawane7397]

In supermesh section there is a resistance of 2ohm below the current source why we had not calculated it??

[@ALLABOUTELECTRONICS]

You mean to say, while writing I1-I3 = 3A right? Well, actually it is not required. Because current through that 2-ohm resistor will be 3A only. And we are only looking at the current. If we are calculating the voltage then we need to consider the voltage drop across that 2-ohm resistor. I hope it will clear your doubt. If you still have any doubt then do let me know here.

[@jayeshsonawane7397]

thanx

[@arshadahmed8605]

ALL ABOUT ELECTRONICS sir if we are calculating voltage what ill be the equation with resistor consideration..

[@ALLABOUTELECTRONICS]

well, you can find the voltages too across each element, once you know all the currents in the circuit. Using KVL, it's easy to find the voltages across each element. I hope you understood.

[@SURAJKUMAR-fp6zb]

I got a question where there was a current source in place of the 8 v voltage source .how to solve it??

[@ALLABOUTELECTRONICS]

Do you mean to say, additional current source along with 3A current source? Or without 3A current source.

[@SURAJKUMAR-fp6zb]

Additional source along with 3 A current source

[@KaveriJayan]

U r taking opposite sign convention for currents in equation..

[@overlordbeast1792]

Thank You Sir👍🙏

[@aravindreddy1792]

super explanation

[@kushalgarg7367]

Thanks but i think last question 3rd equation is wrong please verify

[@olaayman1900]

thank you

[@priyankak6748]

Thanks alot👏👏👏👏💪

[@SanjitKumar-kh1hj]

Really helped me. ;)

[@sudeepsaha452]

Nice thanks.

[@milliepeg01]

Thank you!!!

[@pamulaparthivishal2995]

Bro this is Vishal I have a suggestion can you plzzz Remove the sub titles so,that the diagram is visible

[@ALLABOUTELECTRONICS]

If you are watching on PC then you can simply drag the substitle anywhere on the screen. On smartphone, for some period you can turn of the subtitles in the video setting. I hope it will solve your problem.

[@parmod7506]

2ohm resistor ko kvl of supermesh mai use nhi kiya Why

[@ALLABOUTELECTRONICS]

Because, the current through 2 ohm resistor will be 3 A only. In short, anything in series with current source, won't affect the supermesh equation.

[@prajwalpatil792]

please explain me HOW 3( I1 - I2 ) COMES......?

[@ALLABOUTELECTRONICS]

Would you please exactly tell me the timing in the video where you are having doubt?

[@prajwalpatil792]

ALL ABOUT ELECTRONICS 2:03

[@ALLABOUTELECTRONICS]

In that example, the direction of mesh currents I1 and I2 were assumed in the clockwise direction. So, the effective current through 3-ohm resistor will be I1 - I2. And voltage across it will be 3(I1-I2). I hope, it will clear your doubt.

[@prajwalpatil792]

thnk i got it

[@naderaljorf2835]

what is the name of intro music ??

[@ALLABOUTELECTRONICS]

I have given a link in the description. The exact name of the song is buddy I think.

[@Sk-ew7cn]

नमस्कार @10.12 how can we say current flowing through 4 ohm resistor is I3 - I2. Plz. Tell me

[@ALLABOUTELECTRONICS]

We have assumed that the I3 is greater than I2. So, net current flowing through will be equal to I3-I2. And hence drop across 4-ohm resistor is from left to right. You can also assume that I2 is greater than I3. (i.e drop across 4-ohm resistor is from right to left, and can get the same result.) It's just sign-convention that we have assumed for the calculation.

[@Sk-ew7cn]

ALL ABOUT ELECTRONICS Thanks for your response. @10.05 we take I1-I2 for resistor 2 ohm resistor. Hence same current will flow through that branch till it diverts to 4 ohm resistor. So current in 4 ohm should be less than current in 2ohm resistor. Therefore (I1-I2) - I3 is to be consider. If I3 is flowing through current source and 2 ohm resistor. I'M NEW IN THIS. PLZ CLEAR MY DOUBT शुक्रिया 🙇‍♂️

[@ALLABOUTELECTRONICS]

Here, current I1, I2, and I3 are the mesh currents. And here we are assuming that they are flowing in particular mesh only. So, if you consider the 2-ohm resistor at the top, it will interact with mesh current I1 and I2. While the 2-ohm resistor at the bottom will interact with mesh current I1 and I3 only. I hope it will clear your doubt.

[@Sk-ew7cn]

ALL ABOUT ELECTRONICS I understood. बहुत बहुत धन्यवाद

[@trinadharaoakula2448]

Super mess some more examples kavali

[@alyattalla8520]

Theres a mistake in the second problem

[@mraju2357]

Decrease ur speed sir

[@muhammadijlal3216]

I would not recommend his videos. They might get you the answers but will surely destroy your concepts. He is making the equations all in the opposite direction of the currents he is taking.

[@aaronjoshuasarmiento6539]

As I've replied to the other comment, "The sign is negative if it is a voltage drop and it's positive if it is a voltage gain. I don't think it's wrong. It's just a different approach from yours where you consider the polarity of the terminal encountered first as you travel around the loop. Whatever approach you use in deciding which sign to put, you will arrive with the same answers anyway." The approach in this video will not destroy your concepts. In my opinion, his approach illustrates the concept of KVL better than the approach you are used to.

[@factknowledge3297]

sir can you send me your facebook page link

[@ALLABOUTELECTRONICS]

facebook.com/ALLABOUTELECRONICS/

[@chiduralasunandha3353]

I had a doute

[@ALLABOUTELECTRONICS]

Yes, please let me know here.

[@AmitKumar-cx3ps]

Who asked

[@krishnateja972]

its a bad presentation i didnt understand