[03] Power Electronics (Mehdi Ferdowsi, Fall 2013)

[@platinumphonesandcomputers]

Thank Mr. Mehdi for these lectures am a self taught electronics technician and these lessons a very useful.

[@sdsihddsdks9760]

Perfect lecture. The voice is so clear. The explanation is amazing and creative.

[@mehdiferdowsi9085]

Shuvankar, we can only use "Iaverage" in power calculations if the system is dc. Otherwise, we have to use "Irms."

[@منوبيالفلسطي]

Dr Mehdi , I have sent you an email plz don't forget to check it

[@كيهنزرزور]

Hello, Dr. Mahdi regarding the" we can only use "Iaverage" in power calculations if the system is dc. Otherwise, we have to use "Irms." quoted, but what does the effect if you use the average current times the average voltage to calculate the absorbed power by the resistive load?

[@MrHeraclito11]

Thanks Dr. Mehdi for your lectures, I have a question, there are any method in where you use "grondwalls inequality"?, I see there are inequality differential to solve problems with dynamic load (when you use diodes) but there aren't many book about the topic

[@ghilashamaili2733]

hello at 40:00 when calculating apparent power you assumed that Vrms = Vm/sqrt(2) but when calculating Irms you assumed that Vrms = Vm/2 , is there something i missed or is it just a mistake thanks for replying

[@bbeni1988]

I have the same question here. We have calculated the Vorms at ~35:45 for a generic half bridge rectifier as Vm/2. Why to use then Vm/sqrt(2) in the numeric example for the same case?

[@pllpsy665]

It's a late response but I am going to let the answer for anyone that has the same problem in the future. Since we are looking at apparent power we look at Vsource VsourceRMS=Vm/sqrt(2) since it's a full sine signal. VsourceRMS=/=VdRMS even though IdRMS=IsRMS. It's well explained a bit later in the vid but confusing if one tries do the problem on their own before.

[@mehdiferdowsi9085]

drive.google.com/folderview?id=0BwOBDdGlM22MbGh3Xzlhdmg2RlE&usp=sharing

[@shuvankarpodder1148]

Dear Sir, I have a little question. In calculating PF you use PF=(Average Power)/( Aparent Power). In calculating "Average Power" you use "Pave=Irms^2*R". Why not use "Iaverage" instead of "Irms" in calculating "Average Power"?

[@josephattard9943]

Question: The power factor in a resistive load should be 1.. There is no lag in this case, thus, I question your calculation of the power factor.

[@mehdiferdowsi9085]

If we directly connect a resistor to an AC source, the power factor would be 1. However, in our case there is a diode involved. Even though the load is a resistor, the current that is drawn form the source is not sinusoidal. Therefore, the power factor is not 1 anymore.

[@youmah25]

what book do you use sir

[@mehdiferdowsi9085]

Power Electronics: Daniel Hart

[@youmah25]

thank you