Great Explanation, Better than my professor for sure !!! Amazing , thanks !
@isjoker2211 жыл бұрын
You should work with Khan Academy to have your videos added to an electrical engineer playlist on his site. Very clear, very easy to follow, you made the concepts of power factor less complex, no pun intended.
@KillaKam92211 жыл бұрын
Best video Ive seen on youtube .. Good lookin out man
@ChristianKrause8911 жыл бұрын
Awesome video! Very useful
@ayiti1010 жыл бұрын
would you append to have a video talking about the power factor correction since a useful power factor is around 0.9 or 0.95 ?
@rastarecords310 жыл бұрын
Great video thanks
@Sam-dc9bg8 жыл бұрын
Thanks for the lesson.
@crashonthehumble10 жыл бұрын
Thank you Professor
@supra332028 жыл бұрын
Could someone help me out? @5:50 - How do we draw the graph of voltage and current? How do you know the theta is negative? Can I draw the current vector on the x-axis and the voltage vector at an angle with respect with the current vector? I guess another way to ask the question is - how come vector current and voltage is in graph quadrant IV not in quadrant I?
@RianWardana8 жыл бұрын
+Supra T The current vector is in quadrant IV because the load power factor is lagging. The angle is arc cos of power factor, which is 0.83
@supra332028 жыл бұрын
+Rian Wardana Thanks! I was told that since the problem doesn't specified the voltage angle, we can assume the voltage angle is at 0 degree. And lagging PF = current lags voltage, so current vector was drawn in quadrant IV (CCW rotation). And the angle theta between voltage and current vectors is -33.9 degree.
@crashonthehumble10 жыл бұрын
Sir I have a few questions and I have a midterm in a few days are you open for questions? Have a great day.
@sharonrosedemeterio29808 жыл бұрын
uhm, i just want to ask . why did you used .83 pf in solving complex power ? why not use the 33.9 ?
@worldnetinternetcafe77698 жыл бұрын
He was using Cos of angle which also happens to be the power factor. Cos of angle is adjcnt over hyptns
@TechTins_Projects5 жыл бұрын
Don't you only need to solve to 2 or 3 significant figures given by the initial values?
@TechTins_Projects5 жыл бұрын
eg 439
@ayiti1010 жыл бұрын
Thanks man !
@d3rddrj10 жыл бұрын
How do we know that the current we found at 12:21 is RMS? Thanks
@AraujoMatt10 жыл бұрын
It's a convention that, unless specified otherwise, all values are RMS. Perhaps I should have mentioned that at the beginning. RMS is really the only thing that makes sense with problems involving power because the average current is 0 and the amplitude of the current would be misleading because it doesn't relate to power
@d3rddrj10 жыл бұрын
Matthew Araujo Thanks for the reply, and thanks for Great Video!!! I told my friends that I should have attended "KZbin University" and saved a few million dollars. LOL
@MrMineHeads.5 жыл бұрын
@@AraujoMatt Does that mean that the final voltage will also be in RMS because current was taken to be at RMS?
@TechTins_Projects5 жыл бұрын
Yes the answer is in RMS. RMS should always be used in AC circuits as the average power is the only one that makes sense. Using an instantaneous peak power does not show the average heating effect that current will have on the circuit's components. Whereas RMS does and is used for that reason.
@shivpalpatine8 жыл бұрын
In equation S= VI the V and the I are NOT in RMS right? But you used it with RMS values?
@vandaliztik92667 жыл бұрын
THAT V I SHOULD BE RMS VALUE
@johanariff14606 жыл бұрын
If you want to use peak values for V and I then the formula to use is S = VI*/2.
@Andrew-sz1iq10 жыл бұрын
How exactly do you know that the reactive power will be positive if the current if lagging? Could anyone elaborate on this?
@johanariff14606 жыл бұрын
Just calculate the complex power absorbed by the load (S = VI*) using the passive sign convention. You'll find that the reactive power will be positive for a lagging pf (inductive load) and negative for a leading pf (capacitive load). Hope that helps.
@IM-br1eb9 жыл бұрын
When you solved the equation for I , you didn't transform the resistance from Ohms to Kilo Ohms .... you left the power values in KW , but resistance in Ohms instead of Kilo-Ohms