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Adding capacitance to an impedance

  1. Jan 20, 2009 #1
    1. The problem statement, all variables and given/known data

    If I have an impedance of: 10+j4 ohms and was told to add a capacitance of -j15 in parallel to this, would this make the impedance 10-j11 ohms??
  2. jcsd
  3. Jan 20, 2009 #2
    sorry to bump but I need to know this for my homework. I would hate to work through the whole problem incorrectly.
  4. Jan 20, 2009 #3


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    Staff: Mentor

    Expecting less than a 2 hour turnaround on a homework question is not reasonable. Do not bump like this again.

    But since you asked.... Show us your work. Show us the math where you calculated the parallel impedance based on what relevant equations....?
  5. Jan 20, 2009 #4
    It looks like you added the -j15 to the 10+j4. The problem with doing that is that 10+j4 is for a resistance of 10 in with a reactance of +j4. Your solution would be correct if you were to add the capacitor in series with the other two.

    Can you think of what you have to do now?
  6. Jan 20, 2009 #5
    I should do the combo in parallel like combining resistor in parallel but I am not sure how to do this since the capacitance does not have a magnitude. For example, would I just use
    0+j4 in calculations?
  7. Jan 20, 2009 #6
    The easy way is to find a calculator that does complex arithmetic. I use Excel but you have to add in the engineering package.

    To convert series to parallel assuming real and imaginary do the following.
    Rp = (Rs^2 + Is^2)/Rs
    Ip = (Rs^2 + Is^2)/Is

    Now add the capacitor
    Ip(total) = 1/(1/Ip + 1/Ic)

    Convert back to series
    Rs = Rp/(Rp^2 + Ip^2)
    Is = Ip(total)/((Rp^2 + Ip(total)^2)

    where Rs = series resistance and Rp = parallel resistance.

    Let me know how they tell you to work the problem. I derived these formulas myself. I don't know the official method. When I was doing bench level RF design I used to work these in my head for practice.
  8. Jan 20, 2009 #7
    The way I would do it is: ((10+j4) * (.... -j15))/((10+j4) + (...-j15)). I would change the top do polar to multiply.

    After I added the bottom, I would change it to polar and finish the division. I just do not understand what I am suppose to use for the real part of -j15. I can do the math just unsure about that.
  9. Jan 20, 2009 #8
    By the way, with a calculator that does complex arithmetic, you work it the same way you would with resistors, product over sum.
  10. Jan 20, 2009 #9
    Since the -j15 is the series value, how much is the series resistance... zero.
  11. Jan 20, 2009 #10
    Converting to polar and back again is another good way to solve it.
  12. Jan 20, 2009 #11
    Convert -j15 to polar:

    SQRT(0^2 + 15^2)
    tan-1 (15)
    = 15 angle(86.19)
    Rectangular = 15cos(86.19) = .996
    = 15sin(86.19) = j14.96
  13. Jan 20, 2009 #12
    Convert 0-j15 to polar... Mag 15, Ang -90 or I prefer 15@-90
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