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Archived Series RLC circuits

  1. Mar 26, 2008 #1
    1. The problem statement, all variables and given/known data

    A series RLC circuit has a 0.19 mF capacitor, a 16 mH inductor, and a 10.0 resistor, and is connected to an ac source with amplitude 9.0 V and frequency 60 Hz.
    (a) Calculate the voltage amplitudes VL, VC, VR, and the phase angle phi .
    VL = V
    VC = V
    VR = V
    angle phi= °


    VL= IrmsXL
    VC=IrmsXC
    VR=IrmsR
    angle phi =


    3. The attempt at a solution

    I calculated XL=wL=2pi60Hz*16x10^-3 = 6.03
    XC = 1/wC = 1/2pi*60Hz*0.19*10^-3 = 13.96

    angle phi =tan-1 (XL-XC/R) which is tan-1(6.03-13.96/10 ohms) = -38.41 degrees.

    I have been unable to find VL, VC and VR. I tried translating 9V amplitude into rms and got 6.36V. I tried using Vrms=IrmsR to find Irms but have been unsuccessful. Any ideas?
     
  2. jcsd
  3. Feb 6, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    A complete solution is offered.

    Let:

    E = 9.0 V {We assume given magnitudes are RMS unless otherwise indicated}
    R = 6.0 Ω
    L = 16 mH
    C = 0.19 mF

    Calculate the reactances:

    ##XL = 2\pi f L = 6.032~Ω##
    ##XC = 1/(2 \pi f C) = 13.96~Ω##

    Calculate the current in the series circuit:

    ##I = \frac{E~∠~0°}{\sqrt{R^2 + (X_L - X_C)^2}~∠~atan \left(\frac{X_L - X_C}{R} \right) }##

    ##~~= \frac{9~V~∠ 0°}{12.76~Ω~∠~-38.41°} = 0.7052~A ~∠~38.41°##

    The phase angle ##\phi## is the angle of the current above: ##\phi = 38.41°##

    To find the individual voltages across the components we multiply the current by the reactances. Both numbers have phase angles associated with them, so:

    ##V_L = I~X_L = (0.7052~A~ ∠~ 38.41°)(6.032~Ω~∠~90°) = 4.25~V~∠~128.4°##

    ##V_C = I~X_C = (0.7052~A~∠~ 38.41°)(13.961~Ω~∠~-90°) = 9.84~V~∠~-51.59°##

    ##V_R = I~R = (0.7052~A~∠~ 38.41°)(10.0~Ω~∠~0°) = 7.05~V~∠~38.41°##
     
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