Electrical Engineering: Ch 12 AC Power (46 of 58) Total Complex, Real, & Reactive Power Ex***

[@zuhairalsaffar7001]

I like your way of teaching you are an excellent teacher thanks

[@affection231]

thank you sir, the S of load will have angle -38.7 instead of -33.7

[@MichelvanBiezen]

The one angle is "a poorly written" -22.8 degrees, see the lines above it. -22.8 + (-10.9) = - 33.7

[@danielcornejo1171]

I have watched your previous videos. Why do we take the conjugate of Irms to find complex power in this question?

[@MichelvanBiezen]

Power is P = I^2 R In the case of phasors, we need to square the current, but in order to eliminate an imaginary component, we multiply the current by its conjugate.

[@antoniosalcuni6943]

Always useful!! Many thanks

[@anasasim3856]

That was a very good video professor. But I want to know, why is it that you used the Irms(I=Irms

[@MichelvanBiezen]

It depends on how the information is given in the problem. Sometimes the information is given in terms of the maximum voltage and current (the usual way), but sometimes a problem will be stated with the RMS voltage and current given. The problem is solved in the same way.

[@ohgoditsjames94]

@@MichelvanBiezen RMS values make the most sense because in the world of transmission and distribution, we almost always use RMS values.

[@tashinversionkgf5675]

irms conjugate angle will be positive 22.8 degree

[@MichelvanBiezen]

Since I has a positive angle, I conjugate will have a negative angle.

[@tashinversionkgf5675]

@@MichelvanBiezen sorry sir, i had a mistake,when i note down,really you are a altruist.may i know are you a professor or electtrical engineer

[@tashinversionkgf5675]

@@MichelvanBiezen sir i would like to be your Facebook friend

[@toufikabdoallah7961]

Thank you very much

[@tragicvision775]

Danke