Working out parallel complex impedances

Discussion in 'Homework Help' started by vane, Mar 20, 2010.

  1. vane

    Thread Starter Active Member

    Feb 28, 2007
    181
    0
    I have a question on my maths assignment at college which involves working out he total of two parallel complex impedances.

    I am given:

    Z1 = 6 + 5j &
    Z2 = -2 - 4j

    I have written notes about multiplying, dividing, adding and taking away these complex numbers i am just a bit puzzled as to which one I use to work out the combined impedance.

    Any help would be greatly appreciated as soon as possible so that I can get the assignment done and move onto the next fun installment!!!...... :D
     
  2. jlcstrat

    Active Member

    Jun 19, 2009
    58
    3
    It can be done the same way as simple resistances...product over sum, etc.
     
  3. vane

    Thread Starter Active Member

    Feb 28, 2007
    181
    0
    I think it is a specific way I am supposed to do it

    Could it be:
    (Z1 x Z2)/(Z1+Z2)?

    This is the reciprocal method is it not? I think I may have figured it out myself :)
     
  4. jlcstrat

    Active Member

    Jun 19, 2009
    58
    3
    Yeah, that's product over sum. The reciprocal would be:
    Inverse(1/z1 +1/z2)
     
  5. vane

    Thread Starter Active Member

    Feb 28, 2007
    181
    0
    thank you very much for your ultra fast help!
     
  6. jlcstrat

    Active Member

    Jun 19, 2009
    58
    3
    No problem.
     
  7. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    You are aware that you must flip back and forth between polar and rectangular coordinates depending on the operation you are performing, right?

    Multiplication and division are done in polar coordinates and addition and substraction are performed in rectangular coordinates.

    hgmjr
     
  8. Papabravo

    Expert

    Feb 24, 2006
    10,137
    1,786
    Are you sure that is necessary. I've never had a problem with multiplying complex impedances and removing the j from the denominator by multiplying top and bottom by the complex conjugate of the denominator.

    I know that it can be convenient to use polar notation, but it is not absolutely necessary. Is it?
     
  9. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    Polar and rectangular notations are entirely equivalent, it's just easier to work with one over the other sometimes.
    You can stick with one if you really want to.

    Though, adding in polar is the most counter productive thing ever...
     
    Last edited: Mar 20, 2010
  10. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    If you're going to be doing this sort of thing much, such as you will if you're studying Electrical Engineering, it would be worthwhile to get a calculator that can do complex arithmetic.

    As an alternative, I think you can find applets on the web for doing it.
     
  11. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    I am certain that your technique is a viable one. It is just that I have always tackled these problems by converting to the coordinate system suited to the math operation that I needed to perform.

    hgmjr
     
    Last edited: Mar 20, 2010
  12. Papabravo

    Expert

    Feb 24, 2006
    10,137
    1,786
    Whew -- that's a relief. I was ready to believe you had discovered some new mathematics!
     
Loading...