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Impedance of circuit with capacitor

  1. Mar 17, 2004 #1
    You are given a capacitor( impedance 100 ohms), an inductor (impedance 300 ohms) and a resistor (400 ohms), all connected in series. Determine the impedance( in ohms) of this circuit.
  2. jcsd
  3. Mar 17, 2004 #2
    This sounds as if you are trying to get someone to do your homework. Can you give us a little more detail? Before we give you an answer, can you at least give us several possible outcomes of the problem?
  4. Mar 17, 2004 #3
    (Assuming your EMF is of AC:)

    [tex]{V_r}_{max} = {I_s}_{max}r[/tex]
    [tex]{V_L}_{max} = {I_s}_{max}X_L[/tex]
    [tex]{V_c}_{max} = {I_s}_{max}X_c[/tex]

    I'm not going to explain the whole theory here, but you should know that in this kind of circuit the [tex]V_c[/tex] precedes the current by 90 degrees, [tex]V_L[/tex] is lagging by 90 degrees, and [tex]V_r[/tex] corresponds to the current exactly. (I'm probably using the wrong terms and not explaining this very well ).

    Therefore you can draw a diagram of [tex]{V_c}_{max}[/tex], [tex]{V_L}_{max}[/tex], and [tex]{V_r}_{max}[/tex], whereby [tex]{V_c}_{max}[/tex] is 90 degrees ahead of [tex]{V_r}_{max}[/tex] and [tex]{V_L}_{max}[/tex] is 90 degrees behind it.

    But you should also know that all the time:

    [tex]V_r + V_L + V_c = V_s[/tex]

    So to find [tex]V_s[/tex] at any moment you need to sum up all three V's, while treating them as vectors. On the axis that connects [tex]{V_c}_{max}[/tex] and [tex]{V_L}_{max}[/tex] the sum would be [tex]{V_c}_{max} - {V_L}_{max}[/tex], and on the other axis the sum would simply be [tex]{V_r}_{max}[/tex]. Using pythagoras you can show that:

    [tex]V_s = \sqrt{({V_c}_{max} - {V_L}_{max})^2 + {V_r}_{max}^2}[/tex]

    Which becomes:

    [tex]{I_s}_{max}Z = \sqrt{({I_s}_{max}X_c - {I_s}_{max}X_L)^2 + ({I_s}_{max}r)^2}[/tex]

    And if you divide by [tex]{I_s}_{max}[/tex]:

    [tex]Z = \sqrt{(X_c - X_L)^2 + r^2}[/tex]

    So Z, or the impedance of this circuit, is [tex]\sqrt{200k\Omega}[/tex]
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