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Equivalent impedance

  1. Apr 14, 2006 #1
    see diagram http://img.photobucket.com/albums/v11/biggm/z.jpg

    The question is to find the equivalent impedance for the circuit.

    I am not sure how to solve this, it is simple in the case of all three elements in series but with the capacitor in parallel i am not sure how to solve it.
    thanks for the help
     
  2. jcsd
  3. Apr 14, 2006 #2

    Andrew Mason

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    Find the impedance of the inductor and resistor in series. Then find the impedance (pure reactance) of the capacitor in parallel. Then add the impedances using:

    [tex]\frac{1}{Z_{total}} = \frac{1}{Z_1} + \frac{1}{Z_2}[/tex]

    It gets a little difficult because the impedances have phase differences.

    [tex]\frac{1}{Z_{total}} = \frac{1}{\sqrt{R^2 + \omega^2 L^2}} + \omega C[/tex]

    AM
     
    Last edited: Apr 14, 2006
  4. Apr 14, 2006 #3
    An equivalent alternative is to say,

    [tex] Z_r = R [/tex]

    [tex] Z_c = - \frac{j}{ \omega C} = \frac{1}{j \omega C } = \frac {1}{\omega C } \angle {-90^o} [/tex]

    [tex] Z_L = j \omega L = \omega L \angle {90^o} [/tex]

    Then treat the R-L in series, and call it [itex] Z_{RL} [/tex]

    and make that in parallel with [tex] Z_c[/tex],

    which gives you: [tex] Z_{eq} = (Z_c \backslash \backslash (Z_r + Z_L) ) [/tex]
     
    Last edited: Apr 14, 2006
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