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Complex-Phasor (Moved from physics)

  1. Jul 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Two alternating currents are applied out of phase with each other. One is 4 volts and the other is 3 volts 27° out of phase.

    These can be written as (4, 0°) and (3, 27°)


    2. Relevant equations

    a)draw both phasors
    b)draw the combined phasor between (0 degree and 360 degree)
    c)label peak voltage and state its value



    3. The attempt at a solution

    a)A rough drawing on paint http://imageshack.us/photo/my-images/3/plotb.jpg/

    b) I know A= 6.8106 volts by using this formula

    r=√a2+b2

    r=3cos27 + 4i + i(3sin 27)

    r=√(3cos27 + 9)2 + (3sin27)2

    =6.81v

    angle=0.2014


    Image 2 attached.


    Please look at both images, which are correct
     

    Attached Files:

  2. jcsd
  3. Jul 16, 2012 #2

    LCKurtz

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    Is there a question here?
     
  4. Jul 16, 2012 #3
    Yeh

    (4,0°) and (3,27°)

    a)draw both phasors
    b)draw the combined phasor between (0 degree and 360 degree)
    c)label peak voltage and state its value
     
  5. Jul 16, 2012 #4

    LCKurtz

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    Well, yes, those are the questions you are working on. I knew that. And I see you have worked on them. What is your question that you are seeking help on here?
     
  6. Jul 17, 2012 #5
    I have drawn two images. Also done a calculation.

    I was wondering if the images and calculation are correct?
     
  7. Jul 17, 2012 #6

    LCKurtz

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    Your work looks OK, but I think your answer to (b) should be a picture of a phasor (which I don't see anywhere) and your answer to (c) should just be a number.
     
  8. Jul 18, 2012 #7
    b)How should I draw a phasor?

    c) Again any help on what number ?
     
  9. Jul 18, 2012 #8
    try sketching both graphs on the same axis... then imagine what the sum might look like.
     
  10. Jul 18, 2012 #9

    LCKurtz

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    The same way you drew phasors for part (a).
    You already calculated it, didn't you?
     
  11. Jul 18, 2012 #10
    LCKurtz You have confused me.

    Let's rewind back- I have drawn two images

    i) (http://imageshack.us/photo/my-images/3/plotb.jpg

    ii) The attached image



    Then you said

    "
    Your work looks OK, but I think your answer to (b) should be a picture of a phasor (which I don't see anywhere) and your answer to (c) should just be a number."



    My reply to you question:

    For the phasor image would that not be image (i) and for c the number (=6.81v ).

    And last but not least should I label (6.81v) on graph B.


    Thanks again
     
  12. Jul 18, 2012 #11

    LCKurtz

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    Part (a) Said draw both phasors. That's apparently what you did in image 1, although I would put a little arrowhead on the tips of both.

    Part (b) asks for the "combined phasor". That is another phasor like in part (a). It is just the result of adding the two given phasors and would be easy to include in your image 1. But when it asks for the combined phasor "between 0 and 360" maybe it means the corresponding sine wave like in your other image. Either way the 6.81 is the amplitude of the sine wave you have drawn and is also the length of the combined phasor, which you have not drawn yet.
     
  13. Jul 18, 2012 #12
    Thank you!

    One last question for the combined phasor. How should I be able to draw that?

    Would you please be able to attach simple exam on how to draw one? As I do not where to start.
     
  14. Jul 18, 2012 #13

    LCKurtz

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    You calculated its length and angle (in radians) in your first post. Just draw it like you did the others. Or you could geometrically use the parallelogram rule for adding the two vectors (phasors) in your first picture. Or if it's the sine wave they want, you have already drawn it.
     
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