Electrical Engineering: Ch 13: 3 Phase Circuit (34 of 53) Average Power per Phase?
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- Опубліковано 6 вер 2024
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In this video I will further explain the power in the 3-phase circuit in relation to what is the average power per phase. In the previous example we found the equation of the power as a function of phase voltage, phase current, and the cosine of the phase angle. Since phase current, phase current, and phase angle re constants then power=constant. Therefore for a single phase then Pp=(1/3)P.
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Your formula in 3:13 should have the complex conjugate of the impedance in the denominator. S = V^2/Z*.
in the rule S=Z times the magnitude of current ^2 you didnt mention of its the line current or the phase current which one is it?
Thank you so much sir. One doubt sir, (line voltage is =square root
of 3 × Vp ×angle 30°) the angle 30° is missing sir....
Professor, in 3:10 , it should be Z* than Z.
I think you forgot to say about the differences in phase voltage and line voltage angle. phase voltage will lag line voltage by 30 degrees (or pi/6 radian).
Thanks Prof.
You are welcome.
why u say that n a delta connected load vl = vp we learned that in a y-delta connection is vl = sqrt3vp with phasor 30 and il = ipsqrt3phasor -30
Nice explanation, Thank you very much
You are welcome
Sir but in Y delta 3phase circuit which u said its the common, both the linecurrent=√3Iphase and linevoltage=√3phase voltage then would the total powe of Y-Delta circuit be P=VLILcosΦ ?
The exact reason why i came back to this video. by all means please correct me if i'm wrong, but since we are only looking at the load in this stage. if you would draw a Y or delta load with the outgoing wires, not connected to any source. And deem the voltage between te lines the linevoltage, you will be able to see that in a delta circuit, the line voltage equals the phase voltaqge ( the voltage over the impedance), and the the linecurrent will be a facter sqrt(3) more than the current through the phase ( through the impedance). You will find the reverse being true for a y connected load. For me the confusing thing was looking at the possible source connections, for definitions of phase and line voltage. But once you look at the load it becomes alot clearer, hope this helps
Hi. Shouldn't you have a Z* there?
Sir, I do recommend the proper usage of the terms complex power and apparent power.
Feel free to add any comment.
yeah you are right
@@MichelvanBiezen in the rule S=Z times the magnitude of current ^2 you didnt mention of its the line current or the phase current which one is it?
Why dont you multiple the line voltage with 3 for total power?
Just like with any RCL circuit, the power generated on each element (resitance, inductanve, capacitance) it will be out of phase. When you add the voltage drops across each element, it will always be higher than the total input voltage, but when you account for the phase difference they will add up correctly. Here with a 3-phase circuit you cannot add them up algebraically because of the phase difference.
@@MichelvanBiezen thank you for your explanation, sir!
Very good lecture Sir. Thanks 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
Thank you. Glad you found our videos.
thank u sir for great explanation.❣❣❣❣
You are welcome. Glad you found our videos. 🙂
You are genius
Thank you very much
You are welcome
Many thanks sir
Hello sir. Thank you very much for the explanation. I'm currently confusing wether we need different formula for delta or star connection. Regards
The general equation for the average power per phase is as shown in the video. The impendance of the circuit needs to be calculated as shown in the previous videos depending on the type of circuit you are dealing with.
Thankអរគុណ👏
You are welcome. 🙂
I'm not going to school anymore.