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Homework Help: AC Circuits: Phasors/Polar to rectangular transformation

  1. Jun 27, 2010 #1
    40<50degrees + 20<-30 degrees,
    I get how to convert to rectangular,

    I got 43.03 + j20.64, but converting it back to polar.... how exactly do you do so? The answer is 47.72<25.63degrees..... my book doesn't explain it.

    Also with
    (2+j4)(3-j5).... how exactly do you tackle this sort of problem? I tried the FOIL method and it didn't exactly work.


    Thank you!
     
  2. jcsd
  3. Jun 27, 2010 #2
    1. The problem statement, all variables and given/known data

    40<50degrees + 20<-30 degrees,
    I get how to convert to rectangular,

    I got 43.03 + j20.64, but converting it back to polar.... how exactly do you do so? The answer is 47.72<25.63degrees..... my book doesn't explain it.

    Also with
    (2+j4)(3-j5).... how exactly do you tackle this sort of problem? I tried the FOIL method and it didn't exactly work.


    Thank you!
     
  4. Jun 27, 2010 #3
    Length: |I|= sqrt(Re2+Im2)

    sqrt(432+212)=48

    Angle: < = arctan (Im/Re)

    arctan (21/43) = 26 (deg)

    Draw a phasor diagram and you see it easily. Simple geo/trig.

    Part 2:
    Either convert to polar form and multiply the length and add the angles, or just multiply out the two parenthesis inn a normal fashion. Remember j*j=-1
     
  5. Jun 27, 2010 #4

    Zryn

    User Avatar
    Gold Member

    Rectangular [tex]\rightarrow[/tex] Polar

    x + jy [tex]\rightarrow [/tex] [tex]\sqrt{x^{2} + y^{2}}[/tex] [tex]\angle[/tex] [tex]tan^{-1}(\frac{y}{x})[/tex]

    There are alot of calculators which can do this for you too!


    (2+j4)(3-j5) ... FOIL is the way, expand the brackets first

    2*3 + 2*-j5 + j4*3 + j4*-j5 ... simplify

    6 - 10j + 12j - 20j*j ... keep in mind j = [tex]\sqrt{-1}[/tex], so [tex]j^{2}[/tex] = -1

    6 +2j + 20

    26 + 2j
     
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