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RCL Circuit - Problem due at 11 PM EST tonight, have been working on it for days!

  1. Mar 3, 2009 #1
    1. The problem statement, all variables and given/known data
    A series RCL circuit contains only a capacitor (C = 8.46 μF), an inductor (L = 6.01 mH), and a generator (peak voltage = 77.4 V, frequency = 2.45 x 103 Hz). When t = 0 s, the instantaneous value of the voltage is zero, and it rises to a maximum one-quarter of a period later. (a) Find the instantaneous value of the voltage across the capacitor/inductor combination when t = 4.32 x 10^-4 s. (b) What is the instantaneous value of the current when t = 4.32 x 10^-4 s?


    2. Relevant equations
    v(t) = V*sin(2pi*ft)
    z= square root(R^2 + (Xl-Xc)^2)
    I=V/Z



    3. The attempt at a solution
    the instantaneous voltage in part a = 27.77 V. (this is correct)

    my answer to part b is incorrect but I don't know why:

    X of L=2pi(2450)(.00601)=92.5 Ohms
    X of X=1/(2pi(2450)c) = 7.678 Ohms
    Z=square root(0^2 - (92.5 - 7.678)^2) = 84.822
    I0=V0/Z = 77.4/84.822 = 0.91249 A
    I=I0sin(2pi*ft + (pi/2))
    I=(0.91249)sin(2pi(2450)(4.34x10^-4) + (pi/2)) = 0.841 A
     
  2. jcsd
  3. Mar 3, 2009 #2
    Going through your answer very quickly, all I can spot is your phase angle. Since it's almost purely inductive you should have a minus where the plus now is. Hopefully I'm not too late.
     
  4. Mar 3, 2009 #3
    Oh wow... I can't believe that was it. It was too late for full credit, but now I know that for my test tomorrow. Thanks! :D
     
  5. Mar 4, 2009 #4
    To further help you determine the sign of the phase angle. If XL is greater than XC then you need to subtract the angle, if XL is smaller than XC you need to add the angle.
     
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