AC Series Parallel ?

Discussion in 'General Electronics Chat' started by netwillie, Aug 14, 2010.

  1. netwillie

    Thread Starter New Member

    Aug 10, 2010
    11
    3
    I'm working through the Vol. II - AC
    On page 121 the answer to my calculation for Z(R//(l--C2) just do not match up to the value shown. has any one found this or am I missing something?
    Any in put will be greatly appreciated.

    Willie
     
  2. debjit625

    Well-Known Member

    Apr 17, 2010
    790
    186
    Yes that is ok, the equation is Z = C1--[R1//(L1--C2)].

    Good Luck
     
  3. netwillie

    Thread Starter New Member

    Aug 10, 2010
    11
    3
    Thank you for the reply,
    I agree with Z = C1--[R1//(L1--C2)]
    My Problem is with Z[R1//(L1--C2)] = 1/[(1/Zr) +(1/Z(L--c2))]
    My answer for Z of this component of the circuit is 246.7/_-9.32degrees or 243.44-j39.95

    Please verify
     
  4. debjit625

    Well-Known Member

    Apr 17, 2010
    790
    186
    No ,
    The values of (Using this "\" as angle symbol )
    R1 = 470 \ 0 deg
    L1 = 245.04 \ 90 deg
    C2 = 1768 \ -90 deg
    So
    1/(1/R1 + 1/(L1+C2))
    1\0 /(1\0 / 470\0) + (1\0 /(245.04 \ 90) + (1768 \ -90))
    1\0 /(0.002128\0) + (1\0 /(245.04 \ 90) + (1768 \ -90))
    1\0 /(0.002128\0) + (1\0 /(1522.96\-90))
    1\0 /(0.002128\0) + (0.0006566\90)
    1\0 /0.002227\17.14
    449.04\-17.14

    Hope this helps
     
  5. netwillie

    Thread Starter New Member

    Aug 10, 2010
    11
    3
    My Mistake - I recorded the wrong R value,
    Thank You for taking the time to look at this for me.

    Have a Great Day

    Willie
     
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