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Adding polar coordinates

  1. Apr 2, 2013 #1
    Hello all i have a question about adding 3 impedances given in polar form, must i convert them to x..y.. first or is there a quicker way on a calculator and if so can anyone give advice i have the equation Zo=√Z(oc)Z(sc) but finding it hard to understand many thanks.
  2. jcsd
  3. Apr 2, 2013 #2
    Are you sure you didn't mean multiplying two impedances? In that case, if Zoc and Zsc are two complex impedances in polar form, their product is a new complex number with magnitude |Zoc|× |Zsc| and phase angel φoc + φsc.

    It's easier to understand many thanks if you use punctuation.
  4. Apr 3, 2013 #3
    well its 3 impedance in series and parallel so if i use the formula z =z1z2/z1 + z2 and am i right that in polar form you multiply the x axis and add the y (i.e the value given as the angle)? that will sort the 2 impedances out in parallel then the remaining impedance will be in series with that and i can just add them, Thank you
  5. Apr 3, 2013 #4

    jim hardy

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    Congratulations on improved punctuation !

    Yes - you can add in rectangular,

    and you can multiply in polar as you described. I'd have said " Multiply the magnitudes and add the angles" , though. To say X and Y implies rectangular.

    I learned this in slide rule days so it's ingrained.
    I really suggest you work about ten examples on paper so it'll become habit for you.

    I was through college before pocket calculators appeared. When the HP35 came out ~1972 with its one button polar-rectangular conversion the EE world was flabbergasted.

    So there's probably an easy way to do it with a modern calculator. But you should become fluent in the method so you can spot errors coming from missed keystrokes.

    My two cents !

    old jim
  6. Apr 3, 2013 #5
    :) thanks english is not a strong point of mine, thank you for the help, im sure my calculater can add them but im not sure how, so as long as it can convert between polar and rectangular and then do what i need to do il be happy, thank you.
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