I'm having trouble understanding WHY primary current flow through a transformer with an open circuit on the secondary side is negligible... Since the induced magnetic field is constantly changing, doesn't that mean there is some flow of current through the coil, a back and forth flow? I get the notion that the field gives back what is put into it as the voltage phasor rotates, and I think I understand how the inductor causes the timing of the current phasor to be shifted back in time by pi/2. So yeah, I get that the product of two vectors, phase shifted by pi/2, should be negligible. But I'm having trouble understanding that in the real world. How do you distinguish current flow through transmission lines that is producing work from the imaginary component on the phasor diagram?
No... how high current will flow... as secondary is open I2 is zero... now low voltage winding current I.e. primary current is Io + Im + I2' But I2' is zero... so only small amount of current will flow in primary side... remember in OC test high impedance is offered to the current .. So current is very low...
Vedio was very helpful sir thank you sir explained in very simple language very much convinent in understanding
Thanks a lot Sakshi..
Simply best explanation.
Raghunath chakraborty thanks... if u want any topic to be discussed then let me know..
Digital Blackboard ,sure sir
Sir make a video on single phase lnduction motor why it is not self start
Nice lecture sir .great explanation
Soo helpfull thnkq sir👌👌👌👌
Very good explanation
Thank you Ahmed ...
Good explanation sir
Thanks...
but sir explanation is really good
Thank you... Your suggestions are always welcome...
very good..keep it up
sir in oc test ckt dia supply is given to low vtg side then look at the no of turns...for 115 should be less no of turns than 230 .Am l right..
Pranali Kanase yes you r right... less no. of turns should b there for low voltage side.. I'll improve it..
I'm having trouble understanding WHY primary current flow through a transformer with an open circuit on the secondary side is negligible... Since the induced magnetic field is constantly changing, doesn't that mean there is some flow of current through the coil, a back and forth flow? I get the notion that the field gives back what is put into it as the voltage phasor rotates, and I think I understand how the inductor causes the timing of the current phasor to be shifted back in time by pi/2. So yeah, I get that the product of two vectors, phase shifted by pi/2, should be negligible. But I'm having trouble understanding that in the real world. How do you distinguish current flow through transmission lines that is producing work from the imaginary component on the phasor diagram?
Thanks for the question... I'll get back to you..
Make a video of brake test on three phase induction motor
Sir my question is ,if u connect ammeter in OC test towards LV side then there will be high current na ??how u justify this
No... how high current will flow... as secondary is open I2 is zero... now low voltage winding current I.e. primary current is Io + Im + I2'
But I2' is zero... so only small amount of current will flow in primary side... remember in OC test high impedance is offered to the current ..
So current is very low...
Sir please upload sumpners test
Thank u sir
Explanation is in clarity
thankyou sir,,,,, do some problem from jb gupta integrated course.. specially sc and oc problems
Thank you... definitely I'll upload problems on o.c & sc test from jb Gupta...
Fast explaination
nothing but