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RCL Circuits

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data
    The AC power supply (maxium voltage 20.0V) in RCL circuit is run at resonance for the following parameters:

    R = 20.0 Ω
    L = 88.0 mH
    C = 79.9 μF

    When an iron core is added to the inductor, the self-inductance is doubled

    A) Find the frequency of the generator
    B) After the core had been added, find the phase angle and impedance in the circuit.
    C) Draw phasor diagram at time 4.1267 ms
    D) Find the voltages across each circuit element at this time and show that they follow Kirchoff's voltage rule.


    2. Relevant equations

    σ = tan-1 ([itex]\frac{X_{L}-X_{C}}{R}[/itex])

    XL = ωL
    XC = [itex]\frac{1}{ωC}[/itex]

    @ Resonance XL = XC
    frequency (f) = [itex]\frac{ω}{2\pi}[/itex]
    impedance (z) = [itex]\sqrt{(R)^2+(X_{L}-X_{C})^2}[/itex]

    3. The attempt at a solution

    A)

    ω= [itex]\frac{1}{\sqrt{LC}}[/itex] = 377 rad/s
    frequency (f) = [itex]\frac{ω}{2\pi}[/itex] = [itex]\frac{377 rad/s}{2\pi}[/itex] = 592 Hz

    B)

    Assuming ω stays the same, L is now 2*88.0 mH = 176 mH



    XL = ωL = (377 rad/s)(176 mH) = 66.3

    XC = [itex]\frac{1}{ωC}[/itex] = [itex]\frac{1}{(377 rad/s)(79.9 μF)}[/itex] = 33.2

    σ = tan-1 ([itex]\frac{X_{L}-X_{C}}{R}[/itex]) = tan-1 ([itex]\frac{66.3-33.2}{20 Ω}[/itex]) = 58.9 degrees

    impedance (z) = [itex]\sqrt{(R)^2+(X_{L}-X_{C})^2}[/itex] = [itex]\sqrt{(20 Ω)^2+(66.3-33.2)^2}[/itex] = 38.7 Ω

    C)

    E = Emax sin (ωt) = 20.0 v sin(377 rad/s 4.167 ms) ≈ 20 v

    Thus Emax should be vertical, and the phase angle of 58.9 degrees is between Vr (also Imax) and Emax

    qRfNN.png

    D)

    This is where I am stuck...

    I know the vertical components should all equal 0.

    Emax = 20 v
    VL = XL Sin(180-59.8) = 66.3 Sin(120.2) = 57.3 v


    VC = XC Sin(0-59.8) = 33.2 Sin(-59.8) = -28.7 v


    VR = R sin(90-59.8) = 20 Ω sin(30.2) = 10.1 v

    -20 + 57.3 - 28.7 + 10.1 ≠ 0








    OK, so this isn't the right way... maybe using Imax..

    Imax = [itex]\frac{E_{max}}{z}[/itex] = [itex]\frac{20.0 v}{38.7 Ω}[/itex] = 0.517 A

    Emax = 20 v
    VL = XL Imax = 66.3 * 0.517 = 34.3 v


    VC = XC Imax = 33.2 * 0.517 = 17.2 v


    VR = R Imax = 20 Ω * 0.517 = 10.3 v

    -20 + 34.3 - 17.2 + 10.3 ≠ 0
     
    Last edited: Apr 25, 2012
  2. jcsd
  3. Apr 25, 2012 #2
    I realized my mistake. Combination of both attempts really...

    I was assuming since the graph of reactance was similar to voltages that it would make sense to use them in the calculation but now I see I should first find the max voltage using the impedance multipled by max current, and then getting max voltage use it with trig.

    For future googler's, here is how it's done.

    Imax = [itex]\frac{E_{max}}{z}[/itex] = [itex]\frac{20.0 v}{38.7 Ω}[/itex] = 0.517 A

    Emax= 20 v
    VLmax = XL Imax = 66.3 * 0.517 = 34.3 v


    VCmax = XC Imax = 33.2 * 0.517 = 17.2 v


    VRmax = R Imax = 20 Ω * 0.517 = 10.3 v


    vertical components should all equal 0.

    E = EMax Sin(0) = 20.0 Sin(0) =20.0 v
    VL = VLmax Sin(180-59.8) = 34.3 v Sin(120.2) = 29.6 v


    VC = VCmax Sin(0-59.8) = 17.2 v Sin(-59.8) = -14.9 v


    VR = VRmax sin(90-59.8) = 10.3 v sin(30.2) = 5.18 v

    -20 + 29.6 - 14.9 + 5.18 = -0.12 (Due to significant figures and rounding) ≈ 0
     
    Last edited: Apr 25, 2012
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