HELP ME PLEASE!!! 3-phase (confusion)

Discussion in 'General Electronics Chat' started by Avionicstech, May 5, 2010.

  1. Avionicstech

    Thread Starter New Member

    May 5, 2010
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    0
    Hi,
    So I'm studying for an upcoming exam and i am stuck.
    I have to calculate the line currents in different 3 phase systems. So what i will do is i will just post the example that i am having trouble with and if anyone could help me by explaining what is happening mainly in finding the neutral current that would be hugely appreciated! I just dont understand where *=(3.464 - j2) + (-3.464 - j2) + (0 + j4)* is coming from what the hell is j2, j4 etc. Oh and just so you know I am not able to find it anywhere in my text!

    thanks soo much
    Avtech
     
  2. kompteck

    New Member

    May 4, 2010
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  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    It would be of some concern that you appear to be unfamiliar with the rectangular form of a complex number. If you are about to do an exam, my expectation would be that you could routinely convert from polar to rectangular form - and vice versa. Most hand held scientific calculators provide this as a standard inbuilt function. Perhaps your text only uses the polar form of a complex number - which would be somewhat surprising.
     
  4. Avionicstech

    Thread Starter New Member

    May 5, 2010
    3
    0
    Thanks for the help guys! not sure how im going to go. i think what you may be showing me is a little in depth to what i need to know as i am just an aircraft maintenance engineer.
    If it helps in my syllabus for my exam it states for my topic on 3 phase systems:

    • In relation to star and delta connected 3 phase systems, describe, and given sufficient data, calculate line and phase voltages and currents.

    • Calculate power in a 3-phase system.

    • Describe the advantages of star and delta 3-phase systems.
    Thanks again for anymore assistance!
     
  5. kompteck

    New Member

    May 4, 2010
    8
    2
    Ok well for part A they just found the current in each where V=I*Z

    So, I = V/Z they divided the Voltage by the Impedance. They did it the shorthand way. I always convert to rectangular to do these types of problems.

    j = i, j is used by engineers i is used my mathematicians cuz i is current to us.

    What u need to know:
    convert degrees to radian: theta=deg*pi/180
    e^(j theta) = cos(theta)+j*sin(theta)

    Ex.
    V = 120<120deg Volts
    Z = 30<30deg Ohms

    V = 120<120deg = 120*e(j*120pi/180) = 120(cos(2pi/3)+j*sin(2pi/3))
    = 120(-1/2+j*sqrt(3)/2) = -60+j*60*sqrt(3) V

    Z = 30<30deg = 30*e^(j*30pi/180) = 30(cos(pi/6) + j*sin(pi/6))
    =15*sqrt(3) + j*15 Ohms

    I = V/Z = (-60+j*60*sqrt(3)) / (15*sqrt(3) + j*15) = (0+j*4) A

    you have to multiply the top and bottom buy the complex conjugate of the denominator and you get that final answer. You put the zero out from just as a place holder. There isnt a real part for this answer just an imaginary part.

    Do this for all three angles and get all your currents then add them up to get the total current which is that is done in part B...

    If you need to find power P=Vrms*Irms*cos(theta), Vrms = V/sqrt(2), Irms=I/sqrt(2)

    If you have a TI-83 or higher you can just put the calculator in rectangular mode and just plug in the e^(j*theta) part and it will just spit out the answer.

    Good luck on your test.
     
    Last edited: May 6, 2010
    Avionicstech likes this.
  6. Avionicstech

    Thread Starter New Member

    May 5, 2010
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    0
    Ok,
    Well that explanation is soo much easier too understand than my text! Thankyou.

    Just want to clarify on the quoted section. What 'e' is? (sorry pls bear with me) and is there ever an actual value for j that i would insert into the equation?

    and

    I do have a TI-83 plus but i am not fantastic at using it :S so could you give me an example of exactly how i would enter e^(j*theta) into it. Oh and also how i put it into rectangular mode? (is rectangular mode if I select 'a+bi' in the mode screen from the options of 'Real, a+bi, re^ iθ')

    Thanks so much for the help
    Sorry for my incompetence I am getting there even if it may be slowly!
     
    Last edited: May 6, 2010
  7. kompteck

    New Member

    May 4, 2010
    8
    2
    e is "e" on the calculator above LN on a TI-83.
    j = i and " i " above the period on a TI-83.

    a+bi is rectangular mode

    re^(θi) is polar mode

    this will make the calculator give you the answer in either of those forms.
     
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