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L_C series circuit Problem

  1. Dec 3, 2012 #1
    1. The problem statement, all variables and given/known data
    You have a 0.400H inductor and a 6 micro-Farad capacitor. Suppose you take the inductor and capacitor and make a series circuit with a voltage source that has a voltage amplitude of 30.0V and an angular frequency of 250 rad/s. What is the phase angle (phi) of the source voltage with respect to current?


    2. Relevant equations

    tan(phi) = (V_L - V_C) / V_R = (I(X_L - X_C)) / (IR) = (X_L - X_C) / R


    3. The attempt at a solution

    First, with no resistor, all the above equations diverge. Also, I figured that since in an inductor, v leads i by pi/2, and in a capacitor, v lags i by pi/2, the answer may be zero, but Mastering Physics deemed it wrong.

    Any help greatly appreciated!
     
  2. jcsd
  3. Dec 3, 2012 #2

    gneill

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    Staff: Mentor

    The only impedances involved are reactive in nature (XL, XC). So the question becomes, which dominates the overall reactance of the circuit? What values have you calculated for XL and XC? What's their signed sum? Is the result "inductive" or "capacitive" nature?
     
  4. Dec 3, 2012 #3
    I calculated X_L = 100 and X_C = 667 and I guess the signed sum is >0. And since the capacitive reactance is larger, I assume that capacitance dominates the circuit. How does this lead to obtaining a value for R in order to get the phase angle?
     
  5. Dec 3, 2012 #4

    gneill

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    Staff: Mentor

    When series reactances are "summed", you will see the expression "XL - XC". This is a consequence of the general concept of impedance which you will no doubt cover in your courses at some point. Impedance is like resistance but uses complex numbers. The reactive "resistance" ends up in the imaginary component of the impedance, where inductive values have positive imaginary terms and capacitive values have negative imaginary terms. In this case, the individual impedances for the inductor and capacitor would look like ZL = jXL, and ZC = -jXC.

    Anyway, to make a long story short, your overall circuit will "look" like a capacitor with reactance 566.6 Ω (or impedance Z = -j566.6 Ω). With no resistance (R) in the circuit it just looks like a capacitor. How does the voltage vs current angle look for a capacitor?
     
  6. Dec 3, 2012 #5
    Thanks, but I still don't understand. There is no mention of imaginary numbers in my textbook, and no equation that involves angles other than the ones I listed initially.

    Should I simply consider reactance = resistance = R = 566.6 ohms and use it in my 2rd equation above?
     
  7. Dec 3, 2012 #6

    gneill

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    Staff: Mentor

    You'll get to complex impedances eventually, if not in this particular course then in some other follow-on. Never mind it for now.
    Nope. Reactance isn't really the same thing as resistance, despite having the same units. What you've got, without resistance in the circuit, is something that's going to "look like" either a capacitor or an inductor. Which one it "looks like" depends upon which of the reactances dominates. Don't try to use a formula that requires a resistance when none is present. Instead, consider the relationship of current and voltage in pure inductances and pure capacitances. Hint: There's only two choices of angle!
     
  8. Dec 3, 2012 #7
    OK I think I'm beginning to get it... since capacitance dominates, it won't cancel the effect of the inductance, but the circuit will "look" like a capacitor, so voltage will lag current by pi/2...?
     
  9. Dec 3, 2012 #8

    gneill

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    Staff: Mentor

    Yup. Actually, the effects of the reactances are to mutually cancel, with the "winner" being the one that's larger. Here the inductor manages to "cancel" only a portion of the capacitor's reactance.
     
  10. Dec 3, 2012 #9
    Thanks for your help, much appreciated.

    As a side note, for those of you attempting this problem, Mastering Physics does not recognize the answer if you enter 270 degrees. Use NEGATIVE 90 degrees.
     
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