Admittance

Discussion in 'General Electronics Chat' started by NewGuy01, Aug 10, 2009.

  1. NewGuy01

    Thread Starter New Member

    Aug 9, 2009
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    0
    Hello,

    I was just wondering if someone could please check my maths and show me where I went wrong...

    Im trying to find the total add admittance of this circuit:

    http://www.flickr.com/photos/41254456@N05/3807839423/



    So far I have:

    Y = (1 / 2<0°) + (1 / 2.5<-90°) + (1 / 1.41<45°)

    Multiplying across by
    (2<0°)(2.5<-90°)(1.41<45°) / (2<0°)(2.5<-90°)(1.41<45°)

    Yeilds :

    Y = (2.5<-90°)(1.41<45°) + (2<0°)(1.41<45°) + (2<0°)(2.5<-90°)

    Y = (3.53<-45°) + (2.82<45°) + (5 < -90°)

    Converting to cartesian format =

    2.5 - 2.5i + 1.99 +1.99i - 5i

    = 4.49 - 0.51i = 4.52 < -6.48° = Y.

    This is fine and grand, however if I just get the reciprocals and sum them :

    Y = (1 / 2<0°) + (1 / 2.5<-90°) + (1 / 1.41<45°)

    Y = (0.5<0°) + (0.4<90°) + (0.71<-45)
    = 0.5 + 0.4i + 0.5 -0.5i
    Y = 1 - 0.1i = 1 < 5.71°

    These two answers are completely different, shouldn't they be the same?
    Or have I gone wrong with my math somewhere?
    Ive tried this like 3 time and got the same answers each time.

    Any help is massively appreciated...
     
    Last edited: Aug 10, 2009
  2. t06afre

    AAC Fanatic!

    May 11, 2009
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  3. NewGuy01

    Thread Starter New Member

    Aug 9, 2009
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    Thanks mate but I know what admittance is...

    My problem is how I ended up with two completely different answers using two (as far as I know) valid methods...

    :confused:
     
  4. t06afre

    AAC Fanatic!

    May 11, 2009
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    The reciprocal theorem is valid also for complex admittance and impedance. So if you find it more convinient to find the complex impedance you can use the 1/x or reciprocal theorem. But as I said the law of complex math will apply...The reciprocal value of a complex number is not 1/x as it is for real numbers. Remember that ;)
     
  5. NewGuy01

    Thread Starter New Member

    Aug 9, 2009
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    0
    Im pretty sure my reciprocals are right though, isn't

    1 / (2.5 < -90°) the same as (1 < 0°) / (2.5 < -90°)?

    And phasor division is equal to (R1 / R2 < θ1 - θ2)
    So 1 / (2.5 < -90°) = (0.4 < 90°)

    I still dont see where I went wrong....
     
  6. kkazem

    Active Member

    Jul 23, 2009
    160
    26
    Hi NewGuy01,

    With that handle chosen, what are you going to do when you've been a long-time member? Change it to oldguy01?

    Anyway, enough humor. I'm a 30+ year experienced EE with my BSEE from UC, Irvine. This is a fairly simple problem, but you did make a couple of math errors that I'll point out to you.

    The first thing to do in this problem is to compute the admittance, Y = 1/Z, of the R1 + Y(C) (recall that admittances in parallel add). ALso, recall that when you invert a complex number, the polarity of the imaginary part gets inverted. Therefore, to solve this problem, first, get the admittances of R1 and C seperately, then simply add them in the complex domain as follows:
    Y(R1) = (1/2) = 0.5, Y(C) = (-1/j2.5) = +0.4j, Now add them to get:
    Y(R1 & C) = (0.5+0.4j). Note the polarity of the capacitance admittance is positive since it's the reciprocal of a negative imaginary term.

    Next, well take on R2 & L by noticing that they are in series and since series impedances add, well do that first, then invert the sum to get the admittance, then we can add to the R1&C admittance to get the final answer. Z(R2) = 1, Z(L) = +j1, we add them to get Z(R2+L) = (1+1j), and now, since Y=1/Z, we compute Y(R2&L) = 1/(1+1j) = (0.5 - 0.5j). Again, note that reversal of polarity on the imaginary term of the reciprocal.

    For the total Y, we can simply add the 2 previous admittances since they are in parallel as follows: Ytotal = Y(R1 & C) + Y(R2 & L) = (0.5+0.4j) + (0.5 - 0.5j) = (1 - 0.1j). Or in polar coordinates: Ytotal = 1.005<-0.09967) radians or in degrees: Ytotal = (1.005<-5.711) degrees.

    That's it. I hope it helps and I hope it cleared-up your math issue, which again is that when a complex number is inverted, it's imaginary part gets a sign inversion.

    Best of luck.

    Regards,
    Kamran Kazem
    kkazem
     
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