Electrical Engineering: Ch 12 AC Power (50 of 58) Power Factor Correction: Example Part 1

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In this video I will find the power-factor correction by finding the correct capacitor placed in parallel to the original load to maximize the impedance forcing a decrease in the current provided to the entire circuit. (Part 1)
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Опубликовано:

29 авг 2024

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Комментарии : 11

@lrobertiii 5 лет назад

I have wanted to ask; what are your particular credentials? Are you an engineer? A physicist? A mathematician? All of them? How did you manage to come about learning all of these subjects in the sciences and maths?

@MichelvanBiezen 5 лет назад

I am a physicist, worked as an engineer, taught classes in radar, math, physics, chemistry, astronomy, and physical science at work and multiple colleges and a university, and kept studying and learning on my own. I believe the key is to keep learning through life and not stop once you leave the university.

@RaviarMhamad Месяц назад

What if the load was capacitive reactance? so the correction will be inductive reactance?

@davidtrudeau-D.T. Год назад

There is a mistake at the unit of mesure for the angular velocity, should be; radian per second "R^-1". Not Hertz "Hz".

@MichelvanBiezen Год назад

You are correct. Even though both rad/sec and Hz have the same units, (1/sec), they have different meanings. Hz = cycles/sec

@davidtrudeau-D.T. Год назад

@@MichelvanBiezen 2πf is not equal to something in Hz. f, frequency (cycles/sec) is mesure in Hz).

@MichelvanBiezen Год назад

Yes, I wrote you were correct. The units of radians per second = 1/sec and the units of Hz = 1/sec, so the units are the same, but the meaning is different. One is 2 pi times the other as you indicated.