Magnetically Coupled Circuit EXAMPLE

In this video I have solved an example about magnetically coupled circuit using dependent sources.
The book that I am referring to in this video is:
Fundamentals of Electric Circuits 5th Edition
by Charles Alexander (Author), Matthew Sadiku (Author)

[@ultralaggerREV1]

Ok, you got me lost at the I_1 going through the J2 inductor.
I_1 is ENTERING the dot, so that should make the voltage source at the J6 inductor as POSITIVE? You have negative on top and positive below of the J6 voltage source, isn’t it supposed to be reversed?

[@marcos3865]

Nice video! Alternately you could solve the circuit in less time by replaicing a star of decoupled inductors instead of the 2 coupled inductors. These are the relations that allow you to make the transformation: L1=Lm-L11, L2=Lm-L22, L3=Lm. This operation is lawful because the 2 coupled inductors ha a common terminal, therefore we do not modify the topology of the circuit. By :)

[@marcos3865]

@Bahaa Al DeenYes exactly

[@ujjawaljain4848]

Absolutely amazing!! Cleared all my doubts!!

[@priyanshu2575]

Got it now for sure.Thanks you!

[@balajisriram6363]

such a beautiful video thanks

[@utkarsh_gupta_dev]

Thankyou very much helped a lot ❤️

[@juyoungsim4557]

Thank you. It really helped me :D

[@rubaiyatrashid7572]

Thank you so much! ❤️

[@DuneDiscord]

Thank you!!!!!

[@BentHestad]

Very good! Thanks!

[@saisudhir1765]

Mam what to do with angle??

[@surendrakverma555]

Very good. Thanks 🙏🙏🙏🙏🙏

[@kanthakborkar7845]

Ma'am do you still response Hike messages ??

[@GreatnessUrge]

🤔 In the second loop equation. Why is it 3ji1 for the dependent source and not 3j(i2-i1) ?

[@a.abolhassan2003]

Cause this is the generated voltage at coil (j6) is affected by flux of coil (j2).

[@bayanhassounah8258]

what's the program which you write on?

[@md.shaheenalam8517]

please make a tutorial of exercise problem 13.15 norton equivalent for the circuit

[@mahendraverma5943]

Nice explain... I am form india..

[@Asifur_Rahman]

Thanks a lot Ms...

[@mjlky887]

ma'am could you please upload/include the matlab script here in the description box. thank you

[@egygyerekegycsontjusztlasz3040]

I still don’t understand how can i tell the polarity of the dependent source

[@diwalli]

If the current enters the dotted terminal of coil1, it will induce a voltage with positive polarity at dotted terminal of coil 2.
If currents the coil1, it will induce voltage with negative polarity at dotted terminal of coil2.
In coil 2j , the current enters, so positive polarity at dotted end and negative polarity at undotted end. So. - +

[@mostainbillahrobi1968]

thank you, so much mam, but some problem we can't see your mouse pointer

[@abhishekkumarsah4886]

It is good,it will be more helpful if you solve another problem same like this,madam

[@abhishekkumarsah4886]

It is good,it will be more helpful if you solve another problem from Alexander same like this,madam

[@amanchander7837]

Please make your cursor visible and thank you so much.

[@hamadhalbahrani8781]

thank you so much

[@kalharasandaruwan9290]

Great video

[@cristinejoyalmazan1143]

Hello po. Hoping you'll notice my reply po. I just want to ask if why didn't you transpose the 100angle of 60 po?

[@marcalbrechtmanuelparayno5859]

I think because of ohm's law, V=IR

[@kamrulhasan442]

Wonderful

[@anannabiswas9464]

Mam would you tell the books name??

[@lechbrtube]

sadiku, 5ª Ed

[@Momen-]

Thank you very much for helping me 🌷

[@mans0ur_]

Thanks

[@user-xm2og6rl5r]

cool
❤

[@craze1238]

The current going through j2 is entering the dot and because of this, you said the charge at the dot of the j6 inductor is negative. So this tells me when the current is entering the dot of one inductor it means the dot of the other inductor is negative. Later in the video, you said because the current is leaving the dot on the j6 inductor the dot on the j2 inductor must be negative. From this, I gathered that when current is leaving the dot on one inductor it means the dot on the other inductor is negative. These can't both be true so I was wondering if you could let me know what I'm misunderstanding?

[@gauravwani453]

Something's wrong

[@tiagomenezes6698]

If the current enters the dot on inductor A, then the dot on inductor B will have a + sign. If the current leaves dot A, then dot B will be - sign.
It also helps to draw the dependent source in the same direction as the inductor to see which terminal of the source the dot represents.
When current enters dot in j2, it means dot in j6, which is at the bottom terminal of the inductor, will be +, so the upper terminal of the source will be -.
When current leaves dot on j6, the dot in j2, representing the left terminal of the source, will be - and the right terminal of the source will be +.

[@ktspeedenergy]

Same problem

[@farhatullahkhan7690]

Why the sign is -3j in first loop

[@ayanchakraborty2011]

Same doubt..I think it is wrong

[@jsha6604]

Same here

[@dukedaffy5457]

The current enter through dot in one inductor and enter through non-dot in other inductor...that's why!

[@mateoslab]

@@dukedaffy5457 but then shouldn't it be negative when it leaves the dot not enters?

[@dukedaffy5457]

@@mateoslab when it enters and leaves through dot or enters and leaves through non-dot it's taken as +ve other cases -ve ..simple!!

[@arslanafzal1265]

keep it up

[@liya5114]

❤️

[@harshvardhanpatil8057]

Solve 10.1 problem

[@yasinaltuntas763]

Mail address

[@miguelmendoza8895]

So dumb 🙄