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Determining Current and Voltage Drop in RLC Circuit

  1. Jul 9, 2013 #1
    1. The problem statement, all variables and given/known data
    A circuit contains a 100Ω resistor, a .0500H inductor, a .400μF capacitor, and a source of time-varying emf connected in series. The time-varying emf is 50.0 V at a frequency of 2000 Hz.

    a) Determine the current in the circuit.
    b) Determine the voltage drop across each component of the circuit.

    2. Relevant equations
    I = V / Z
    Z = sqrt (R2+ (XL - XC)2)
    XL = ωL
    XC = 1 / (ωC)
    ω = 2∏f
    Vemf = VR + VC + VL (Kirchoff's Loop Rule)
    V = IX, IR

    3. The attempt at a solution

    Using the above equations, I got XL to be 628.3 and XC to be 198. I got Z to be 441, and solved for I: I = 50/441 → .113 A to get the current. The issue is, when I get the voltage drops across the components, they don't add up to 50 V. For the inductor alone, I got V = IXL, which ended up being 71 V. I'm not sure where I'm going wrong with this problem.
     
  2. jcsd
  3. Jul 9, 2013 #2

    mfb

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    What do you get if you add the components? Did you consider the phase shift between the currents and voltages?
     
  4. Jul 9, 2013 #3
    I got around 105 V when I add the components. As for the phase shift, I guess I didn't, but I'm not entirely sure what you mean by that.
     
  5. Jul 9, 2013 #4

    mfb

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    When the capacitor has its maximal voltage, the current is zero, and the resistor has no voltage drop. When the resistor has its maximal voltage, the current is maximal, and the capacitor has no voltage drop.
    And similar for the coil.

    For every moment in time, the voltages have to add up to the source voltage. The maximal (or RMS) voltages of the individual components do not have to do that.
     
  6. Jul 9, 2013 #5
    Ohh okay I get it. They all reach maximal points at different times, so I'm calculating each component's maximal voltage drop? Still, it wouldn't make sense for the inductor to have a maximal voltage that goes over 50 Volts.
     
  7. Jul 10, 2013 #6

    mfb

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    Oh, that can make sense. You have a resonant circuit.
     
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