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Polar forms

  1. Nov 27, 2013 #1
    1. The problem statement, all variables and given/known data

    A=1.5495<21.0363°x(22.1009<30.3658°/69.9667<9.1884°)

    2. Relevant equations



    3. The attempt at a solution
    A=1.5495<21.0363x(22.1009/69.9667(30.3658-9.1884)=1.5495<21.0363(0.3159<21.1774)=(1.5495x0.3159)(21.0363+21.1774)=0.4895<42.2137°

    Solution above is it correct or I have to convert polar forms to complex number, and final result to convert to polar form?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Nov 27, 2013
  2. jcsd
  3. Nov 27, 2013 #2

    gneill

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    Staff: Mentor

    You can do it either way. Your solution looks fine.
     
  4. Nov 27, 2013 #3
    why result is completely different compared with MatLAB
     
  5. Nov 27, 2013 #4

    gneill

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    Staff: Mentor

    I couldn't say; I haven't seen your Matlab work.
     
  6. Nov 27, 2013 #5
    IL=IN(ZTH/ZTH+RL)
    IN=VTH/ZTH

    FOR:
    VTH=33.79310-j5.51724
    ZTH=19.06896+j11.172413
    RL=50

    Here comes the tricky part:
    I convert VTH and ZTH in polar forms I got
    VTH=34.2460<-9.32954
    ZTH=22.1009<30.3658
    and
    ZTH+RL=19.06896+j11.172413+50=69.06896+j11.172413=69.9667<9.1884

    Then

    IN=34.2460<-9.32954/22.1009<30.3658=1.1278<-39.6953

    IL=1.1278<-39.6953(22.1009<30.3658/69.9667<9.1884)=1.1278<-39.6953(0.3159<21.1774)=0.3563<-18.5169
     
  7. Nov 27, 2013 #6

    gneill

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    Staff: Mentor

    That doesn't look right. (34.2460)/(22.1009) does not yield 1.1278. Maybe you copied the wrong value down?
     
  8. Nov 27, 2013 #7
    sorry
    IN=34.2460<-9.32954/22.1009<30.3658=1.5495<-39.6953
     
  9. Nov 27, 2013 #8

    gneill

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    Staff: Mentor

    Okay, that looks better.

    I note that in your various conversions you are specifying several decimal places but the values after the first decimal place or so aren't always accurate... it looks like some truncation or rounding was done on some previous intermediate values, so the extra digits aren't helping (or useful). Make sure to keep extra digits in all intermediate values.

    For example, you converted Vth from rectangular to polar as

    33.79310-j5.51724 → 34.2460 ∠-9.32954

    If I do the same conversion, being sure to retain all digits for intermediate steps, I obtain:

    33.79310-j5.51724 → 34.24052 ∠-9.2726°

    Angles are notorious sources of rounding/truncation errors if you skimp on digits, because the functions involved are not linear; in some regions a tiny difference can blow up into a big difference.
     
  10. Nov 27, 2013 #9
    therefor

    IL=1.5495<-39.6953(22.1009<30.3658/69.9667<9.1884)=1.5495<-39.6953(0.3159<21.1774)=0.4895<-18.5179
     
  11. Nov 27, 2013 #10

    gneill

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    Staff: Mentor

    Yep. Again, rounding/truncation errors are suspect: I get 0.489 ∠-18.461° .
     
  12. Nov 27, 2013 #11
    interesting I calculate using a MatLAB, and that's what MatLAB gives me?
     
  13. Nov 27, 2013 #12
    fx>> A=33.79310-5.51724i
    fx>> magn=abs(A)
    fx>> magn=34.2405
    fx>>ANGLE=angle(A)*180/pi
    fx>>ANGLE=-9.2726
     
  14. Nov 27, 2013 #13
    Ohh, for calculation of Vth I used calculator
    sqr(33.793102+5.517242)=34.2460
    then tan-1(5.51724/33.79310)=-9.2726°

    I don't know what I have done to get -9.32954 but obviously is not right :)
     
  15. Nov 27, 2013 #14

    gneill

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    Staff: Mentor

    Right. Compare these values with the ones you provided in post #5.
     
  16. Nov 27, 2013 #15
    yes you are right angle is a bit off.And it has been corrected
     
  17. Nov 27, 2013 #16

    gneill

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    The inverse tan value looks suspect. Did you truncate a radian value before converting to degrees? My "calculator" gives 0.161837 radians for the arctan. Converting that to degrees gives 9.2726°.
     
  18. Nov 27, 2013 #17
    you are correct, I made a mistake.I copied the wrong value on the forum.I have 9.2726, and MatLAB confirmed it.Thanks
     
  19. Nov 27, 2013 #18

    gneill

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    Staff: Mentor

    Excellent :smile:
     
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