Admittance Magnitude in a Parallel RC Circuit

Discussion in 'Homework Help' started by Ryzing, Sep 22, 2010.

  1. Ryzing

    Thread Starter New Member

    Sep 22, 2010
    4
    0
    Hi,

    I some have problems regarding the parallel RC 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=14 Ohms, C=260µF & e is a 990 V, 250Hz power supply. Calculate the magnitude of the admittance of the circuit?"

    The solution:

    1/6.28x260x 250=2449

    990/2449=40424

    990/14=70.714

    40424^2+70.714^2 square root=339


    I need someone to correct me on this. Thanks.
     
    Last edited: Sep 22, 2010
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Admittance [Y] is simply .....

    Y=1/R+jωC

    So magnitude, |Y|=sqrt((1/R)^2+(ωC)^2)
     
    Last edited: Sep 22, 2010
  3. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    Your arithmetic here is wrong. The result is not 2449, but rather 2.449E-6. And besides that you haven't done the arithmetic right, you have used a value of 260 farads for the capacitor rather than the correct value of 260 microfarads.

    You don't need to involve the value of the applied voltage, 990 volts, because the admittance has nothing to do with the applied voltage.
     
  4. Ryzing

    Thread Starter New Member

    Sep 22, 2010
    4
    0
    Thanks. I got it now.
     
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