I am extremely sad that I've jsut found your videos now that im studdying for my final in Circuits. You are very clear, have a fun accent, and clearly have a deep understanding of what you are talking about. Thank you very much for this sereis of videos, its been very helpful, and I hope I have trouble with another topic on which you have a series on so I can watch more.
Thank you, both Mr. Carrico and Mr. Hashioka, for your comments and, very important to me, for including the country from where you are posting. Both Brazil and Portugal are among my favourite destinations when I'm abroad.
You are really enthusiastic. I am sure there are. You just have to run the odds of finding one. The connection teacher-student is similar to the one between patient and doctor. It depends on so many apparently insignificant parameters. The Internet just makes it more likely to find the one that suits you. I am sure that my style may not suit everybody, but it does suit you. Many thanks for your enthusiasm, praise, and support.
+Vinicius Munhoz Martello Thank you kindly, Mr. Martello. I have had some Brazilian exchange students in my classroom over the years and I have learned from them to say: Obrigado!
Thank you, Mr. Caglayan, for taking the time to submit your feetback. And also, for letting me know from which country. I mention to my colleagues and students a post like this of yours: "... today, a student from Turkey, Mr.Caglayan, posted a kind note on one of the RU-vid videos ...". Thank you.
First of all thank you Sir for sharing your knowledges. Second, someone solved the first problem? If yes here are my answers: S(Source) = 0.704 - j3.874 kVA S(Factory )= 1.346 - j3.7 kVA I considered 125V and 63A as peak values.
Thank you for your kind words. AC voltmeters and AC ammeters read the RMS value of the corresponding parameters. The standard in the electric energy industry is RMS for any voltage or current (unless explicitly specified otherwise). That means that 125 V and 63 V are both RMS values. Please, try again. Observe that the active power delivered by the source cannot be less than the active power absorbed by the load.
You are welcome. Oh, Zambia, I was in Lusaka last year, in June. I was there for two weeks. I loved the place,, the atmosphere, the people and the food.
rolinychupetin oh that's great to hear, for sure we are great people here in southern Africa, I am working on a project of generating free electrical energy by converting the energy possessed by a flywheel in to electrical energy, if you anything in that line of thought that I need ,I would appreciate, thanks
Ah, I understand now ! At last ! THANK YOU ! I don't understand why books can't be written the way you explain this in your videos. If there are such books kindly tell me where to find them because the ones I read obviously lack the pedagogical approach you present here...
S, the apparent power is the product of two real numbers, the rms value V and the rms value I, so S = VI is all reals, no phasors. Now, if what we are computing is the complex power $$\overhang{S}=\overhang{V}\overhang{I}^*$$. the are all phasors (RMS phasors, a.k.a. IEEE phasors). I hope this helps. I used the LaTeX notation to represent phasors with "hats" etc.
Even if the "cable" or transmission link joining the source to the load is always an RL link, sometimes in the power industry, for very long links, there may be some series capacitive compensation and you end with the situation that you describe, which makes yours a very good question. In that case, the little triangle (RI, XI) keeps the RI side as before but its XI side points in the opposite direction. I lave it to you all developing the rest of the process of this interesting scenario.
