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Voltage across R,L,C

  1. Nov 19, 2011 #1
    I am trying to find the effective voltage across each element (R,L,C) of a series RLC circuit like the one below. I am given the following values: Ieff = 1 Amp, Veff = 1 V, i(t) lags v(t) by 45°, L = 1, f = 2/2∏ Hz, L = 1 H
    RLC_series_circuit.png

    What I have done so far:
    Veff = Vm/√2
    => Vm = √2
    Ieff = Im/√2
    => Im = √2
    XL = ωL = 2

    v(t) = √2*sin(2t)
    i(t) = √2*sin(2t - 45°)

    i know the formulas for the peak voltages for the individual elements are
    VR = Im*R
    VL = Im*XL
    VC = Im*XC

    i'm not sure if those correspond to the effective voltages or if i have to use the equation Veff = Vm/√2
    i am also having trouble finding R and XC from the information provided..
     
    Last edited: Nov 19, 2011
  2. jcsd
  3. Nov 19, 2011 #2
    I agree with your value for Xl.
    If the current lags the supply voltage then the supply voltage LEADS the voltage across R because the voltage across R and the current are in phase.
    Are you familiar with showing the voltages Vl, Vr, and Vc on a vector (phasor) diagram. Vl is drawn on the =+y axis, Vr is drawn on the +X axis (horizontal axis ) and Vc is drawn on the -y axis.
    This means, from the information in your question, that the supply voltage (1volt) will be a vector from the origin at 45 degrees to the X axis.
    Hope this is of some help
    I think that you have not been given enough information! Do you know anything about R or C?
     
  4. Nov 19, 2011 #3
    Sorry....you do have enough information,I had missed the current
     
  5. Nov 19, 2011 #4
    Imax = Vmax/Z
    => Z = 1
    (1) Z = sqrt(R2 + (XL - XC)2)
    cosθ = R/Z, θ = 45°
    1/√2 = R/1
    R = 1/√2
    From (1) I get C = 1/(4-√2)

    I am not 100% sure this is correct
     
  6. Nov 19, 2011 #5
    I got those values for R and C
     
  7. Nov 19, 2011 #6
    thanks for the help. i got
    VR = Im*R = 1
    VL = Im*XL = 2√2
    VC = Im*XC = 2√2 - 1
     
  8. Nov 19, 2011 #7
    I got
    Vr = 1 x Cos45 =0.7V
    R = 0.7/1 = 0.7Ω
    Vc = 1.3V
    Xc = 1.3 (gives C= 0.38F)

    I think the differences are due to the fact that I used the voltage and current values in the question. I assume these are rms values and I think you have converted these to peak values.

    A good result !!
    Cheers
     
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