Excellent presentation. I have a variable autotransformer, cheap Chinese "Variac" but Variac is a brand. Thinking about the common leg currents is important not to exceed the current rating of the windings.

Helped me a lot, the only thing I don't understand here is conversion of a mutual transformer to an auto transformer. Do you mean an 1:1 isolation transformer? Since using any standard step down transformer may mean that the primary and secondary may have different current ratings, not only that, the reduced secondary voltage common on most transformers may only allow for a small change in the auto transformers voltage. did you mean a 1:1 transformer. I'm new to all this, could you please explain. Also, how could I tap of midway on the secondary side?

Hi Zac, i need help how did you get the Vtotal? please i tried doing pythagoras like you said but i didnt get the value

This is a very good video. explained very well. I have the same machine , I finished fixing. I had to buy a expensive multimeter because all the multimeters I have couldn't measure capacitors.. But I learned with this video witch multimeter I had to buy... You saved me hundreds of dollars..

V leads by 30 deg..........A lags by 30 deg. are they the same ?

Please help me understand better. with bldc motor I read it on internet Wye 1.7X less KV as Delta. Delta 1.7X higher top-speed than wye, more current than WYE at all RPMs. I assum delta config draws 3A, and Wye also draws 3A, then will Wye give higher Rpm than Delta? Or both configurations have the same rpm, which one will draw more amps?

Thank yu so much ..l am watching from Zimbabwe

What if the Main inputs power supply or Main is 600 Volts Tap accidentally to 480 Volts Transformer Primary windings. What will be the Voltage and Current of a Particular 230/240Volts rated Secondary transformer. What woul be the results.

Nice work. Straight to the point, this is the exact same as the NEC except we switched to 125% of continuous loads vs the .8

It would be good to mention how the magnetic fields for the shaded and non-shaded pole portions do not occur at the same physical location and the armature is attracted to the core irrespective of the direction of the magnetic flux. It is the attractive forces on the armature that add. Explaining WHY the magnetic flux at the shaded pole lags the flux at the non-shaded pole by 90° (induced current in the shorting ring producing its own magnetic flux) would also be helpful.

Cool dea!; Thanks 👍 for that explanation

I love it. Thanks for making my life easier to calculate

Zack, I'm a Chartered Electrical Engineer and Technical Director for my company providing HV designs up to 132kV in the UK. I also provide training. Can I just say that your video is excellent. I was going to produce my own you tube presentation on this for internal training but I think I will just refer people to yours. No wasted words, no BS, no patronising explanations. 10/10. well done. I'll look at your other videos.

VERY INSIGHTFUL..... Asante Sana (Thanks very much)

Omg thank you so much I was trying to learn this from a textbook my brain couldn’t hack it

Bruv are you writing backwards ?

Thank you for the video 🙂

What if my circuit is being supplied by a transformer? Do my calculations change due to the inductance of the transformer?

Thank you Pablo. You are a man of the people

Great video sir ! Thank you

Great video

Thanks sir

Excellent. thanks

Could you explain where you got 20 volts. Seems like you just picked a random voltage? Did I miss something?

I am very grateful to you for this explanation...I am from Iraq

Thank you!

Thank you very much !

Hands down the very best illustration of how to do this process. Thank you so much!

i finally got the polarity down, thank you

Thank you!

oh sir you are little bit crazy.

you're a life saver

Great stuff! Thank you

I like the part where the numbers come out of nowhere. How in the world did you get 5.45? or 2.056??

In 3rd year currently and am re doing the class due to a failed re write on the gov exam. Wish I found this channel the first go around😂 explained very well and simple

How was the max 70 amp OCPD determined for the first tap example?

230 v primary howmany turn required And 60 v secondary which point turn should i tap

Such a satisfying and helpful video, thank you so much!

Thank you sir

Great

Your explanation is incomplete. First, the practical phase shift is more like 45°, not 90° (if you got anything near 90° by having the resistance of the shading coil very low, the amplitude would be hopelessly weak). Second, no matter what the phase shift is, adding 2 sinewaves will ALWAYS have 2 zero crossings per cycle. There are 2 things you miss. First, the 2 sinewaves each have their own SEPARATE magnetic paths in the iron, one passing through the shading coil, the other not. THEY ARE NOT ADDED TOGETHER. Secondly, the magnetic force is proportional to the SQUARE of the wave amplitude. So EACH wave is SQUARED, SEPARATELY, then the forces are added together mechanically. The sin² forces are always positive, each becoming zero twice per cycle. But the zero crossings of the 2 phase-shifted waves are misaligned in time, causing there to be a positive pulling force that NEVER becomes zero.

Thank you very much for the great explanation.

🎯 ty just what I needed

🎯ty

🎯 ty

please do more code calculations this might have saved my apprenticeship

I hope this channel grows. You saved my ass.

Thank you kind Sir

thanks a million

Fuck the Career center and those idiots