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Admittance Magnitude in a Parallel RL Circuit

Discussion in 'Electronics Homework Help' started by Cine, Sep 28, 2010.

  1. Cine

    Cine

    1
    0
    Sep 28, 2010
    Hi,

    I some have problems regarding the parallel RL circuit and it is the admittance magnitude. I have tried to apply the solution that was given to me in the text book. Here is my attempt on the question:

    "R=2 Ohms, L=19 H & e is a 300 V, 18Hz power supply. Calculate the magnitude of the admittance of the circuit?"

    The solution:

    6.28x19 H x 18Hz=214.88

    1/2=0.5

    1/214.88=0.046

    0.5^2+0.046^2 square root=502


    I'm really lost. Thanks.
     

    Attached Files:

  2. Militoy

    Militoy

    180
    0
    Aug 24, 2010
    First step - check the decimal placement on your calculation of inductive reactance.

    BTW - the methodology used in the rest looks OK to me. If your numbers above are accurate - you'll note that the inductor makes very little difference to the operation of the circuit - especially after the "first step" above.
     
    Last edited: Sep 28, 2010
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