Mesh current steps 1 to 3

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We solve a circuit by writing Kirchhoff's Voltage Law in terms of "mesh currents." First three steps of four.

[@lolliequeen101]

Thank you so much, I finally understand it!

[@jedelcaldovlogs857]

A very kind information that you explain. what if 3 resistor and parallel they are the same?
thank you for your information about the mesh Current.

[@ohmakademi]

very good education. thank you so much.

[@richardhall9815]

You just saved my life! Thank you for everything you do!!

[@Micheal184]

i'm replying to you after a year, do you remember watching this video or not? tell me.

[@archiedelacroix880]

@@Micheal184 i'm replying to you after 2 year, do you remember typing this comment? tell me.

[@sotiris6116]

Why, in the 3rd mesh, R4 is solely flowed through by i3 and not by (i3-i2) ??

[@mistervoldemort7540]

Very good overall explanation although I would have appreciated to have a more detailed explanation as to how to determine whether a resistor is plus, then minus or vice versa

[@mattwells19]

The way I understand it is that you can label your resistor to have +/- on either side, aka it's up to you. When you do this, just make sure you stay consistent from mesh to mesh. For example, in mesh 1 it was -R2(i1 - i2) and in mesh 2 it was +R2(i1 - i2). As long as you stay consistent for the two meshes then you will be fine.

[@toothnailgaming42]

Resistors result in a drop in voltage (calculated via Ohm's Law) so any time your current passes through a resistor the first side becomes more positively charged while the far end becomes negatively charged. whereas we are doing mesh calculations, your point of origin is the bottom left corner of the mesh rather than the power source. This means that your superimposed current now has the ability to pass through a resistor in the opposite direction of your original current, this would result in a voltage rise rather than a voltage drop which is why some resistors result in a voltage rise when using this mesh method.

[@abdullahraja8496]

@@toothnailgaming42 then it means that mesh current and physically flowing current may be different.

[@StephenMendes]

Students might also find one (or more) of my videos helpful for Mesh and Nodal Analysis ..... I have videos for every aspect of electronics ...... but in circuit analysis Playlist I have videos for all the methods ..... supernode, supermesh, superposition, Thevenin, Norton, source transformation and etc. ..... sometimes it helps to watch several videos for the same topic to get a fuller understanding

[@jvillalobos6988]

hey, doesn't the current of the consider mesh is greater than its neighboring's? thus in the mesh 2 would yield R2(i2-i1).

[@mr.unknown9237]

Thank you this is very useful

[@Micheal184]

hello, Im replying from year 2020, are yo still alive?

[@saad5419]

Thanks a lot❤️

[@djtoddles8750]

I'm a stand-up guy, so I use the mensch current method

[@sunilkumarsingh9673]

well done gru

[@murtez22]

thanks

[@redblack8766]

Why is i3 equal to negative I only? What about the current through R4?

[@BlueyMcPhluey]

Red Black I think because of the constant current source?

[@hallfighter7]

If you are analyzing the mesh in the middle, the current through R4 is i2-i3. If you are analyzing the mesh on the far right, the current through R4 is i3-i2. However, you don't need to set up a KVL equation for the mesh on the far right as you can tell by inspection of the current source that i3 = -I.
Remember, a "mesh current" is just an analysis tool that allows you to analyze a circuit in a systematic and repeatable manner. Don't confuse the mesh currents flowing clockwise through each loop with the ACTUAL currents flowing through each circuit element.

[@mennatallahmamdouh6058]

I still don't understand

[@joachimjorrydeleon3868]

you are finding the current in mesh 3 and there is already the current source given

[@craigjones6457]

I dont understand the logic behind i3 = -I either. All other meshes (i1 and i2) took into account a full mesh/loop from one node and back to that node. So if you follow that principle in the i3 mesh you would get i3 = -I - R4 (i2-i3).
If I applied the same logic for i3=-I to mesh i1, then I would assume that i1 = v/R1.
The only thing that I can tell is different is that i3 is a current source and i1 a voltage source, but why would that make any difference to the way you calculate the mesh current value?
If someone explain why that would be really helpful!
Would be great if someone could explain why I

[@vishaldhiman1625]

what happened if resistor is connected in series with I current ?

[@emirgo6825]

It would be the same.

[@olaayman1900]

thank you

[@justridemtbofficial9483]

thx

[@shahriartanvir977]

Why R2 is a voltage rise in second equation? Can someone explain?

[@hebam.3318]

in the first mesh, we moved along the circuit as the voltage goes from positive to negative from up to down, so when in the second mesh as we are going from the opposite direction(down to up) we can't change the voltage direction already assumed in mesh 1.

[@aayushthakur1079]

Is mesh rule is different from loop law or if it then which is imp for cbse 12 th board

[@ALEX-gr7dx]

Mesh current method is like an extension of KVL.

[@PunkHippie1971]

Step 2 didn't make sense to me because of the junction the current goes through.

[@aaronreybonaobra9109]

At 3:18, is the direction of element current of R2 assumed?

[@JonasVercoutere]

no, look at your voltage. The current goes from the positive side through the circuit to the negative side. So at the voltagepoint the current goes up, goes through the circuit in meets back 'under' the voltagepoint.

[@ShabDhdgdh]

what if i dont have a "easy" mesh?

[@lot7887]

Why didn't i3 mesh current include R4?

[@xyronnedavidyabut394]

π lot i think it is from kichhoffs current law which states that current in is equal to current out

[@kyleeething]

Thanks Sal; great material as usual. Keep up the good work brother

[@aselim20.]

I wrote it.

[@albertrenshaw4252]

In the second equation shouldn't it be i1+i2? You wrote i1-i2 because i2 goes in reverse, but wouldn't that mean the i2 value is negative, the equation would still be i1+i2 with i2 being a negative value? For example if I have f(x) = x+y and I know y is -3 I don't rewrite it as f(x)x=-y because y is negative ... otherwise I'd end up with x-(-3).

[@wealthy_concept1313]

was thinking same thing

[@adils3196]

hello i think at 7:38 it should be i2-i1, not i1-i2

[@vincentlouis622]

Adil Shareef you are right

[@shahwali4291]

You can write it as well both are one and the same thing!

[@emirgo6825]

@@shahwali4291 Can you please elaborate how? +R2*I1 would mean voltage increase according to the KVL.

[@shahwali4291]

@@emirgo6825 it depends upon in which direction are you going i mean conventional current direction!

[@jcamilorg7420]

Not really

[@rey_lone9655]

Finally

[@Nshiime]

you are the promised saviour

[@curtpiazza1688]

🎉

[@alhasan6751]

5:33 Why not i3=+I

[@alaa2069]

because both are in opposite direction

[@alhasan6751]

@@alaa2069 Thank you😃

[@djphylearninglake7172]

not good