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Impedance as a function of angular frequency

  1. Apr 24, 2014 #1
    1. The problem statement, all variables and given/known data
    r0YSYfV.png


    2. Relevant equations
    Z_L = jωL
    Z_C = 1/(jωC) = -j/(ωC)

    3. The attempt at a solution

    This is my attempt for the series combination:

    Z = jωL + 1/(jωC)
    Z = j0.02ω - j20000/ω

    Is there a way to simplify this further? What would a graph look like, if the function has imaginary parts?

    And also, to find the frequency for an equivalent open circuit, I would have to set the impedance to zero right? What would it be for a short circuit?
    ^EDIT: Actually I just realized that I would set Z equal to zero for a short circuit, not an open circuit.
    For an open circuit, the impedance should be infinite, but how would I find the angular frequency?
    ∞ = j0.02ω - j20000/ω does not seem like a solvable equation.
     
    Last edited: Apr 24, 2014
  2. jcsd
  3. Apr 24, 2014 #2

    ehild

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    What happens if ω=0 (direct current)?


    ehild
     
  4. Apr 25, 2014 #3
    1. The problem statement, all variables and given/known data

    r0YSYfV.png

    2. Relevant equations

    Z_L = jωL
    Z_C = 1/(jωC) = -j/(ωC)


    3. The attempt at a solution

    This is my attempt for the series combination:

    Z = jωL + 1/(jωC)
    Z = j0.02ω - j20000/ω = j(0.02ω - 20000/ω)

    Is there a way to simplify this further? What would a graph look like, if the function has imaginary parts?
    This is what I got for the graph of Z = (0.02ω - 20000/ω) (I am not sure how to use imaginary numbers in winplot):

    FnUwcWk.png

    Does this seem correct?

    Also, to find the ω for a short circuit, I have to set Z equal to zero right?
    For an open circuit, Z=∞, so how would I solve for ω in that case?
     
  5. Apr 25, 2014 #4
    I am not sure how I would solve for that.
    Z = j0.02(0) - j20000/0

    I have a zero in the denominator.

    I know from calculus that the limit of 1/∞ = 0, so would that then mean that 1/0 = ∞?
    This wouldn't apply to real numbers, but maybe it works here because it is imaginary?

    If that is the case, then that would mean that the ω for an open circuit is equal to 0.
     
  6. Apr 25, 2014 #5
    Your are asked to plot the impedance magnitude--that is the square root of the sum of the squares of the real and imaginary parts. You have no real part, so you just plot the absolute value of the impedance.

    Also, no mention is made of the type of plot, but it's traditional to plot impedances on logarithmic vertical and horizontal scales. If you do this, your plot will look more like what you'll find in a textbook.
     
  7. Apr 25, 2014 #6

    ehild

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    ω=0 means that the voltage/current does not change with time. It is direct current/voltage. Is it any current if you connect a simple DC voltage source ( a battery) across the circuit?

    You can consider also the limit ω tends to infinity. What happens to the impedance then?

    ehild
     
  8. Apr 25, 2014 #7
    The limit as ω approaches infinity is infinity.
    So does that mean that for an open circuit (infinite impedance), ω=∞?
    That doesn't seem right, I may be misunderstanding.
     
  9. Apr 25, 2014 #8
    Okay I just realized I did my limit calculations wrong.

    Setting Z to 0 for a short circuit, and taking the limit of ω as Z approaches ∞ for an open circuit, I got these values:

    Series Circuit:
    ω = 1000 rad/s for a short circuit
    ω = 0 for an open circuit

    Parallel Circuit:
    ω = 0 for a short circuit
    ω = 1000 rad/s for an open circuit

    Do these values make sense?
     
  10. Apr 25, 2014 #9

    ehild

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    Gold Member

    Yes. You can add what happens in both cases at the limit when ω tends to infinity.

    ehild
     
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