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RLC Circuits - Resonant Frequency, Power

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data
    An AC circuit consists of an alternative emf of 1 V connected to a resistor of 500
    Ohms, an inductance of 0.4 mH, and two capacitors connected in parallel of 50 pF
    each, We want to find the resonance frequency of this circuit, the maximum power
    dissipated by the resistance, and at what frequencies ω will it be half as large.
    a. Draw the circuit and includes all the relevant quantities for this problem. What
    quantities do you need to find ω0, Pmax, and ω’s? (10 pts)
    b. What concepts and equations will you use to solve this problem? (5 pts)
    c. Solve for ω0, Pmax, and ω’s in term of symbols. (15 pts)
    d. Solve for ω0, Pmax, and ω’s in term of numbers. (5 pts)
    e. Verify the units, and verify if your values are plausible. (5 pts).


    2. Relevant equations
    1. omega_0 = 1/((LC)^.5) *omega_0 is resonant frequency*
    2. Power Maximum = [root mean sqaure of (V^2)]/R
    3. Power = (Power Maximum)(x^2)/(x^2 + Q^2(x^2-1)^2)
    4. x = omega / omega_0
    5. Z=(R^2 + (X_L - X_C)^2)^.5


    3. The attempt at a solution

    a. I just have a normal AC circuit with a 500 ohm resistor and .4mH inductor in series, followed by two 50pF capacitors in paralell.
    b. This is pretty much covered in my relevant equations section above.

    c. omega_0 = 1/((L)(C_1+C_2))
    I put this down because I'm given L and the total capacitance of two capacitors in series is just their sum.

    In an RLC circuit, maximum power occurs at the resonant frequency. Thats because the current is maximized when impedance is at a minimum. And the impedance is at a minimum at the resonant frequency, (X_L and X_C are equal and cancel out, leaving the impedance equal to R. So:

    Power_Maximum = (I^2)*R = Power Maximum = [root mean sqaure of (V^2)]/R

    Then to find omega, (the frequency when power is at half its maximum) I said:

    (1/2)Power_Maximum = (Power Maximum)(x^2)/(x^2 + Q^2(x^2-1)^2)
    What I want to do is solve for x in terms of everything else, then substitute x=omega/omega_0 and Q=(omega_0)(L)/R

    The problem with that is i dont know the algebra to do that, and even if i did im not sure that is correct. Part d and e follow if i can get part c. Can anyone help?
     
  2. jcsd
  3. Nov 13, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    Hi Fogel. I'm no expert on this, but am interested and may be able to help with the algebra before your post falls off the first page. It looks fairly easy to solve your equation
    (1/2)Power_Maximum = (Power Maximum)(x^2)/(x^2 + Q^2(x^2-1)^2)
    for x. If you cancel out the Power-Maximum's and multiply both sides by the denominator you soon end up with
    Q²(x²)² - (2Q²+1)x² + Q² = 0
    If you replace x² with y you just have a quadratic equation in y that can be solved with the quadratic formula. Only a positive solution is useful and only positive values of x itself are possible since x = ω/ωo.
     
  4. Nov 14, 2009 #3
    Thanks for your reply, i'd like some verification by someone on my physics if anyone can help.
     
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