Thank You. You can go through the following link for my Udemy course it's absolutely free and don't forget to give your feedback www.udemy.com/course/protection-used-for-power-transformer/?couponCode=5E6DA1AEACB51D48C687
Respected Sir, trasformer rating 2500 KVA 15.75 KV/575 V Dyn5 %Imp=6.23% How to perform Short circuit How much voltage inject in HV side, HV side current & LV side current Please explain with calculation. , I already applied your method for this LV current come 64 A But actual come 6.21 A(HV voltage inject 50 V, current-0. 250 A
Very good. Thanks 🙏🙏🙏🙏🙏🙏
Very well explained sir
How does u arrived to thIS VALUE?
Super brother
Thank you
Nice video
excuse sir, what handbook use it? thanks
Thank You. You can go through the following link for my Udemy course it's absolutely free and don't forget to give your feedback
www.udemy.com/course/protection-used-for-power-transformer/?couponCode=5E6DA1AEACB51D48C687
Hi Sir, i think there is wrong in 3 conditions. Idiff for condition2 should be 0.2< Idiff
Respected Sir,
trasformer rating
2500 KVA
15.75 KV/575 V
Dyn5
%Imp=6.23%
How to perform Short circuit
How much voltage inject in HV side, HV side current & LV side current
Please explain with calculation.
, I already applied your method for this
LV current come 64 A
But actual come 6.21 A(HV voltage inject 50 V, current-0. 250 A
Ok I will make an video on above topic
Sir hindi me explains kariye na sir
Ok i will make series of videos on Hindi also... please press bell button for notification