Balanced Wye-Wye Connection || Y-Y Connection || Example 12.2 || Practice Problem 12.2 || ENA 12.3

[@naumankhalid2513]

outstading sir, the way u teach u hav made this tough subject as a piece of cake MASHA ALLAH JAZAK ALLAH may ALLAH bless u with His countless blessings ameen

[@ElectricalEngineeringAcademy]

Thanks Nauman for your kind appreciation. Best wishes

[@KunalKumar-qt3nt]

THANK YOU SIR. BHOT ACHCHA SAMJHAYA APNE. ALL DOUBTS CLEARED

[@vianeypadilla2499]

thank you so much!!!!! I WAS SO LOST IN POWER ENGINEERING AND COMING ACROSS THIS VIDEO HELPED ME A LOT

[@ElectricalEngineeringAcademy]

U R Welcome my dear Vianey. Best Wishes.

[@drdudleydragon9671]

Wow great explanation and layout. Loved it. Helped me moving ahead in this topic.

[@ElectricalEngineeringAcademy]

Thanks Danny. Good Luck.

[@Zeeshan12368]

Sir you're my hero!

[@ziaullahzia6896]

Great sir on 'very easy method you solve the question. Thanks sir I gain something more from you...

[@ElectricalEngineeringAcademy]

Thanks dear. Best wishes.

[@tahsinbhuiya2458]

Thank you sir.May Allah bless you.

[@ElectricalEngineeringAcademy]

Dhannobad. Jazak Allah Khairan

[@rahafal-zahrani5939]

Great Explanation. Thank you

[@ElectricalEngineeringAcademy]

Welcome dear. Best wishes.

[@HARVINDERSINGH-lx9vy]

Awesome lecture sir

[@ElectricalEngineeringAcademy]

Thanks Harvinder Jee. Best wishes

[@HARVINDERSINGH-lx9vy]

Sir aap bahut acha samjaate ho plz aur videos upload kro

[@ahmednazir3645]

Great explanation sir👍

[@ElectricalEngineeringAcademy]

Thanks dear. Best wishes.

[@nomihaider96]

thank u sir great video lecture

[@ElectricalEngineeringAcademy]

Most welcome dear.

[@Zubairkhan-rb1fx]

nice work....sir keep making videos...😘

[@ElectricalEngineeringAcademy]

Thanks Dear. Best Wishes.

[@cuisinepersan3450]

mashallah

[@namoonay6942]

alot of love sir.........

[@ElectricalEngineeringAcademy]

Thanks Nouman. Best of Luck.

[@aditit306]

Thank you sir 🙏

[@sohailjanjua123]

Salam I like your video

[@ElectricalEngineeringAcademy]

Walaikum Assalam. Thanks.

[@azlanikram9393]

thank you sir

[@ElectricalEngineeringAcademy]

U R Welcome. Best wishes.

[@arrabisjamesb.8656]

how did you get Vca = 207.8 /_-180?

[@ElectricalEngineeringAcademy]

Please listen carefully from video time 16:15. Once we know angle of Vab, then Vbc angle is(Vab angle -120) and Vca angle is (Vab angle -240) or (Vab angle +120)

[@arrabisjamesb.8656]

​@@ElectricalEngineeringAcademy Thanks man, i was confused about the why in my sci. cal it says Vca = 207.8/_ 180. 180 and -180 is the same pala. They have the same magnitude.

[@arrabisjamesb.8656]

@@ElectricalEngineeringAcademy Sorry for because i did not comment my real problem ;0