Electrical Engineering: Ch 13: 3 Phase Circuit (39 of 53) How to find Source and Load Power?

[@MeridianPodcast]

I watched one of your older statistics help vids and it was a real help!! Just came to this video since it is recent and you are more likely to see it! Again, thank you so much! Keep up the great work, it is helping people! God bless.

[@altuber99_athlete]

Hehe, it was a good idea to fast-forward scenes were you only write numbers. Also, I like the fact that after solving a problem, you analyze/interpret the results, which is actually useful.

[@Zuwwar]

Hi Michel. Two remarks. As others have mentioned in the comments section below; VL needs to be Vp that is the phase voltage, and the units need to be VA instead of Watts. Please confirm or correct me if I am wrong.

[@sazib8333]

Why there is -Ve in front of the formula while calculating power supplied by source?

[@JM-ql4ok]

Supplied power is usually called negative (passive sign convention) so that the total power of the supplied and absorbed = 0. The total ABSORBED power of Zgen, ZL, and ZY is 1054.2W so the SUPPLIED power will be called -1054.2W (also in this video he is actually talking about complex power S but he is calling it P)

[@english_ielts_teacher1720]

that was amazing, thanks alot

[@MichelvanBiezen]

Thank you. Glad you liked it!

[@nyanffie9905]

isn't the given 120 voltage is still in phase voltage? and we have to convert it yet to VL before actually getting the Line Current? or do we have to assume that its already in VL.

[@halfengineer5]

That power must be VA not watt.since it is complex power.

[@romeissa5232]

Sir ,which V(L) you mean is it V(ab) or V( bc) or V (ca) ????

[@MichelvanBiezen]

V(L) = the voltage across one of the loads (from A to Neutral, or from B to Neutral, or from C to Neutral). Note that the line impedance is also included.

[@romeissa5232]

Ok sir I get it thank you

[@sangitpradhan6818]

@@MichelvanBiezen but dont we call it the phase voltage?

[@sangitpradhan6818]

all in VA

[@Jnglfvr]

Why are you converting the power to polar format only to convert it back to rectangular format? It's not like you then need to multiply or divide 2 phasor quantities. Instead just multiply the scaler quantity 3*(3.75)^2 by the impedance and just it leave it all in rectangular format since that is what you are using as the basis of comparison. I would recommend an inexpensive calculator (I use the Casio 115 ES Plus) in which you can mix polar and rectangular quantities and express the result in either format with no drama. It would dramatically reduce the time and labor of these polar/rectangular conversions by allowing everything to be done in a single step. E.g., -3*(120

[@mahdi.hasnat]

is not P = 3 * VP * IP * cos (theta) = sqrt(3) * VL* IL * cos ( theta) ?

[@nishajawahar]

In Y-Y circuits, the line current is equal to the phase current. So, P = -3 VP IP* is the same as P = -3 VP IL*.

[@mahdi.hasnat]

but why did you put negative in power equation?

[@nishajawahar]

@@mahdi.hasnat The equation I mentioned was for the power supplied. Power supplied should be taken as a negative quantity and power absorbed should be taken as a positive quantity. This comes down from the law of conservation of energy which implies that the total power of all elements in a circuit is zero.

[@LakBhatnagar-dn8mj]

​@@nishajawaharhey if you still remember these concepts can you tell me why did we take only complex power to calculate all powers? Why not total power formula?