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Series/Parallel impedance problem

  1. Oct 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Problem description:
    http://imageshack.us/a/img707/2060/beskrivelse.png [Broken]

    2. Relevant equations

    http://imageshack.us/a/img543/5302/lign1.png [Broken]

    [tex]Z = Z_1 + Z_2 + ... +Z_n (Series), [/tex]

    [tex]\frac{1}{Z}= \frac{1}{Z_1} + \frac{1}{Z_2} + ... + \frac{1}{Z_n} (parallel)[/tex]

    3. The attempt at a solution

    [tex]\frac{1}{Z}= (\frac{1}{i \omega C} + R)^{-1} + (i \omega L + R)^{-1}[/tex]

    [tex]\frac{1}{Z}= (\frac{R + i \omega C R}{i \omega C R})^{-1} + (i \omega L + R)^{-1}[/tex]

    [tex]\frac{1}{Z}= \frac{i \omega C R}{R + i \omega C R} + \frac{1}{i \omega L + R}[/tex]

    [tex]\frac{1}{Z}= \frac{i \omega C R (i \omega L + R)}{(R + i \omega C R)(i \omega L + R)} + \frac{R + i \omega C R}{(R + i \omega C R)(i \omega L + R)}[/tex]

    [tex]\frac{1}{Z}= \frac{i \omega C R (i \omega L + R) + R + i \omega C R}{(R + i \omega C R)(i \omega L + R)}[/tex]

    [tex]\frac{1}{Z}= \frac{i C L \omega^2 + C R \omega + L \omega - iR}{i C L \omega^2 + C R \omega + C \omega - i}[/tex]

    [tex]\frac{1}{Z}= \frac{i C L R \omega^2 + C R^2 \omega + L \omega R - iR^2}{i C L R \omega^2 + C R^2 \omega + C \omega R - i R}[/tex]

    [tex]R^2 = \frac{L}{C}[/tex]

    [tex]\frac{1}{Z}= \frac{i C L R \omega^2 + L \omega + L \omega R - i \frac{L}{C}}{i C L R \omega^2 + L \omega + C \omega R - i R}[/tex]

    .. and here I'm stuck. I need to eliminate those frequencies. (My approach may have been wrong all along?)
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 19, 2012 #2

    gneill

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    Staff: Mentor

    That first term, the rearrangement of the capacitor's admittance, doesn't look right. Rework the algebra.

    Regarding your approach, note that if the overall impedance Z remains constant with respect to frequency, then the overall admittance Y = 1/Z is also constant.

    What can you say about the derivative of a constant?
     
  4. Oct 19, 2012 #3
    You are right I screwed up there. I worked out a new solution by the same principle:

    [tex]\frac{1}{Z}= \frac{(i \omega C)}{1+R} + \frac{1}{i \omega L+R}[/tex]

    [tex]\frac{1}{Z}= \frac{(i \omega C)(i \omega L+R) + 1 + R}{(1+R)(i \omega L+R)}[/tex]

    [tex]Z= \frac{(1+R)(i \omega L+R)}{i \omega C(i \omega L+R)+1+R}[/tex]

    Still can't see it

    I don't get what you're saying in your second sentence though. (derivative of a constant is zero) Is that part of solving the problem?
     
  5. Oct 19, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    Still a problem with the first term. Start with the series impedance:

    ##R + \frac{1}{j ω C} = \frac{j ω R C + 1}{j ω C}##

    multiplying top and bottom by j:

    ## = \frac{ω R C - j}{ω C}##

    The admittance term is the inverse of this, so: ##\frac{ω C}{ω R C - j}## and altogether:
    $$\frac{1}{Z} = Y = \frac{ωC}{ωRC - j} + \frac{1}{R + jωL} $$

    If Y is a constant w.r.t. ω then, as you say, its derivative w.r.t. ω must be zero, right? So take the derivative and find out what the value of R2 must be that makes it so.
     
    Last edited: Oct 19, 2012
  6. Oct 19, 2012 #5
    Okay so I've taken the derivative with respect to ω

    [tex]\frac{d Y}{d \omega}= - \frac{i C}{(C R \omega -i)^2} - \frac{i L}{(R+ i L \omega)^2} = 0 [/tex]

    I'm supposed to calculate the parenthesis and isolate R2 from there?
     
  7. Oct 19, 2012 #6

    gneill

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    Staff: Mentor

    The idea is to simplify the derivative (giving it a common denominator) and then isolating R2. But your derivative does not look right. There should be three terms in the initial expansion of the derivative.
     
  8. Oct 19, 2012 #7
    My math calculator program says it's correct? What's wrong with it?

    Tried finding R2 from the derivative i calculated:

    [tex]- \frac{iC(R+iLω)^2}{(CRω−i)^2 (R+iLω)^2} - \frac{i L(CRω−i)^2}{(CRω−i)^2 (R+iLω)^2} = 0[/tex]

    [tex] \frac{- iC(R+iLω)^2 - i L(CRω−i)^2}{(CRω−i)^2 (R+iLω)^2}= 0[/tex]

    [tex] \frac{- iC (R^2+2 i L R ω-L^2 ω^2) - i L(-1-2 i C R ω+C^2 R^2 ω^2)}{(-1-2 i C R ω+C^2 R^2 ω^2) (R^2+2 i L R ω-L^2 ω^2)}= 0[/tex]

    [tex] \frac{-C i R^2-2 C i^2 L R ω+C i L^2 ω^2 + i L+2 C i^2 L R ω-C^2 i L R^2 ω^2}{(-1-2 i C R ω+C^2 R^2 ω^2) (R^2+2 i L R ω-L^2 ω^2)}= 0[/tex]

    I plotted all this into a math calculator. The denominator in the last term was incredibly huge.

    Btw thanks for the help so far. I at least got the idea behind solving the problem now.
     
  9. Oct 19, 2012 #8

    gneill

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    Staff: Mentor

    My mistake; I failed to recognize that you'd already applied a simplification.
    Since the derivative is being set to zero, you can ignore the denominator. You only need to expand the numerator and collect the R2 terms.
     
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