Magnetically Coupled Circuit EXAMPLE

[@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!!

[@DuneDiscord]

Thank you!!!!!

[@priyanshu2575]

Got it now for sure.Thanks you!

[@juyoungsim4557]

Thank you. It really helped me :D

[@balajisriram6363]

such a beautiful video thanks

[@rubaiyatrashid7572]

Thank you so much! ❤️

[@utkarsh_gupta_dev]

Thankyou very much helped a lot ❤️

[@BentHestad]

Very good! Thanks!

[@amanchander7837]

Please make your cursor visible and thank you so much.

[@md.shaheenalam8517]

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

[@kanthakborkar7845]

Ma'am do you still response Hike messages ??

[@amygdalabandhere]

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. - +

[@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).

[@saisudhir1765]

Mam what to do with angle??

[@bayanhassounah8258]

what's the program which you write on?

[@surendrakverma555]

Very good. Thanks 🙏🙏🙏🙏🙏

[@mjlky887]

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

[@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

[@Asifur_Rahman]

Thanks a lot Ms...

[@mahendraverma5943]

Nice explain... I am form india..

[@kalharasandaruwan9290]

Great video

[@anannabiswas9464]

Mam would you tell the books name??

[@lechbrtube]

sadiku, 5ª Ed

[@abhishekkumarsah4886]

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

[@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!!

[@kamrulhasan442]

Wonderful

[@hamadhalbahrani8781]

thank you so much

[@쿠키-j8e]

cool ❤

[@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

[@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

[@arslanafzal1265]

keep it up

[@Momen-]

Thank you very much for helping me 🌷

[@mans0ur_]

Thanks

[@liya5114]

❤️

[@yasinaltuntas763]

Mail address

[@harshvardhanpatil8057]

Solve 10.1 problem

[@miguelmendoza8895]

So dumb 🙄