Chelapati book

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frecoder78

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I'm trying to figure out why the author uses S conjugate on pg 2-55 instead of just S. Please help! Thanks!

 
I'm trying to figure out why the author uses S conjugate on pg 2-55 instead of just S. Please help! Thanks!
possibly dumb question here, what is this chelapati book you're referring to? I don't think I've heard of them before.

 
I'm referring to the Electrical Book
I did a bunch of digging, and it seems like people have said in the past this was a great book for the PE electrical power exam review. Do you think that even with the new format change of the test this is still a valuable study tool? I mean, obviously you must otherwise you wouldn't be using it. Do you have any thoughts on this you can share?

 
I'm referring to the Electrical Book
I did a bunch of digging, and it seems like people have said in the past this was a great book for the PE electrical power exam review. Do you think that even with the new format change of the test this is still a valuable study tool? I mean, obviously you must otherwise you wouldn't be using it. Do you have any thoughts on this you can share?
I really do think this book is a very useful tool. Some of the parts I find pretty difficult to follow since the author skips a few steps. This is my second time taking the exam and I didn't use this book as much as I should have. The great thing about this book is the industrial type of applications mixed in with power engineering concepts. I would highly recommend it to anyone (well let's see if it helps me pass this time)

 
If you're interested in any of Dr. Chelapati's books, you can go to www.irvine-institute.org. There are tables of content up for each book so you can see what's inside. If you have any questions about any of the material, you can always email either myself or Dr. Chelapati directly at [email protected].

 
I'm trying to figure out why the author uses S conjugate on pg 2-55 instead of just S. Please help! Thanks!
S= VI*

V=IZ

S=|I|^2 (absolute I squared)

This is what our instructor wrote -- CVC

Hi,

I'm not sure I understand:

S=|I|^2 (absolute I squared)

wouldn't there still be a Z in the product with the magnitude? like below:

S=|I|^2 * Z

If this is too obvious to point out, sorry, otherwise I need to figure out some things that I thought I understood.

 
I'm trying to figure out why the author uses S conjugate on pg 2-55 instead of just S. Please help! Thanks!
S= VI*

V=IZ

S=|I|^2 (absolute I squared)

This is what our instructor wrote -- CVC
I think the question was why S conjugate was used rather than reciting what the author wrote down.

I'm not familiar with S*, but I've seen different texts use S=VI or S=VI*. We would use the I* since we are dealing with complex power. I* would indicate we would have both real and imaginary components which, if put in polar form, would mean we would have magnitude and direction. In this case, the apparent power should lie in either the 1st or 4th quadrant. I don't have my copy of the book with me, but depending on if it were either leading or lagging, would tell us which direction S would be going. I think the problem also provided a power factor parameter so that would help you out knowing if the system was leading or lagging.

I think the Chelapati book is a decent reference but I have noticed that there were several steps that were skipped in a majority of the problems. I wish more textbooks would show each step rather than consolidating a lot of the work and just arriving at the answer.

 
Where can I get a copy of this book? I can't search it anywhere, not even the online pdf version.

 
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