Adjustable capacitor in steady state

Discussion in 'Homework Help' started by peter_morley, Apr 10, 2011.

  1. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    Frustration is increasing exponentially...erggg well I have a copy of how I am approaching the problem as an attachment here. I get to a point at which I try to eradicate the real part but dont know how. I clearly wrote all the steps i took so I'd a appreciate if you could look at my work and steer me to a better course. thanks
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    782
    You wrote ...

    Z=j\omega L +\frac{\frac{R}{j\omega C}}{\frac{1}{(j\omega C}+R)}\frac{(R-\frac{1}{j\omega C})}{(R-\frac{1}{j\omega C})}

    which is OK

    then you wrote

    Z=\frac{\frac{R^2}{j\omega C}+\frac{1}{\omega^2C^2}}{R^2+\frac{1}{\omega^2C^2}}+j\omega L

    which is incorrect

    It should be

    Z=\frac{\frac{R^2}{j\omega C}+\frac{R}{\omega^2C^2}}{R^2+\frac{1}{\omega^2C^2}}+j\omega L

    You then equated this all to zero which is also incorrect.

    Only the imaginary part of Z should equate to zero. The impedance doesn't reduce to zero altogether. The total Z simply forces zero phase shift between voltage and current at certain values of C. It effectively becomes a purely resistive impedance for those values.

    So your next steps would involve splitting the complex rectangular form of Z into its real and imaginary parts. Then equate the imaginary part to zero to set up the second order equation with C as the unknown.
     
  3. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    Does the real part have to be just an integer by itself or can it be something like 15C. And at this point do I just cross out that real part and just solve for the variables that have a j coefficient equaling to zero?
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The only requirement is that the imaginary part is equated to zero. Doesn't matter what the real term is. Can you post what you finally come up with as the imaginary part, which should be a quadratic expression in C?
     
  5. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    I'm having trouble reducing it down from (wcR^2 +jR)/(R^2 + 1/(w^2*c^2)) - w^3*c^2*L...Ive done so much algebra today i may puke but atleast im getting better at it, any suggestions on where to go from here?
     
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    782
    This is how I would do it ...

    Step 1

    Multiply the left hand fractional term top & bottom by

    \omega^2C^2

    This gives

    Z=\frac{\frac{R^2\omega C}{j}+R}{(R^2\omega^2C^2+1)}+j\omega L

    or

    Z=\frac{-jR^2\omega C+R}{(R^2\omega^2C^2+1)}+j\omega L

    Step 2

    Separate real & imaginary terms

    Z=\frac{R}{(R^2\omega^2C^2+1)}+\frac{-jR^2\omega C}{(R^2\omega^2C^2+1)}+j\omega L

    and

    Z=\frac{R}{(R^2\omega^2C^2+1)}+j\omega L-j\frac{R^2\omega C}{(R^2\omega^2C^2+1)}

    or

    Z=\frac{R}{(R^2\omega^2C^2+1)}+j(\omega L-\frac{R^2\omega C}{(R^2\omega^2C^2+1)})

    Step 3

    Then equate the imaginary part to zero

    (\omega L-\frac{R^2\omega C}{(R^2\omega^2C^2+1)})=0

    Step 4

    Multiply everything in the equation by

    (R^2\omega^2C^2+1)

    To give

    \omega L(R^2\omega^2C^2+1)-R^2\omega C=0

    & since ω is common

    L(R^2\omega^2C^2+1)-R^2 C=0

    You should be able to simplify & solve for the two unknown C values from there ....
     
    Last edited: Apr 10, 2011
  7. peter_morley

    Thread Starter Member

    Mar 12, 2011
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    thanks for the detailed help. My problem is understanding when the equation is considered separated in real and imaginary terms. Just gotta get better at algebra ultimately
     
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