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How to solve the differential equation for driven series RLC circuit?

  1. Apr 8, 2017 #1
    1. The problem statement, all variables and given/known data

    It is the driven series RLC circuit. It is given in the following images. zqTR1Ra.png KUpllgB.png

    It is from the section 12.3 in this note.

    2. Relevant equations

    The differential equation, as given by 12.3.3, is ##L \frac{d^2 Q}{d t^2} + R \frac{d Q}{d t} + \frac{Q}{C} = V_0 \sin{(\omega t)}##.

    It says that one possible solution is ##Q(t) = Q_0 \cos{(\omega t - \phi)}##, where ##Q_0 = \frac{V_0}{\omega \sqrt{R^2 + (\omega L - 1 / \omega C)^2}}## and ##\tan \phi = \frac{1}{R} \left( \omega L - \frac{1}{\omega C} \right)##.

    3. The attempt at a solution

    So, assuming ##Q(t) = Q_0 \cos{(\omega t - \phi)}##, I get$$\frac{d Q}{d t} = -Q_0 \omega \sin{(\omega t - \phi)}\\\frac{d^2 Q}{d t^2} = -Q_0 \omega^2 \cos{(\omega t - \phi)}$$
    Substituting,$$-L Q_0 \omega^2 \cos{(\omega t - \phi)} - R Q_0 \omega \sin{(\omega t - \phi)} + \frac{Q_0}{C} \cos{(\omega t - \phi)} = V_0 \sin{(\omega t - \phi)}$$

    I cannot figure out show to eliminate ##t## and ##\phi## and solve this for ##Q_0##.

    I try applying$$\cos{(\omega t - \phi)} = \cos{(\omega t)} \cos \phi + \sin{(\omega t)} \sin \phi\\\sin{(\omega t - \phi)} = \sin{(\omega t)} \cos \phi - \cos{(\omega t)} \sin \phi$$

    But still there's no way cancel ##t## and ##\phi## and solve for ##Q_0##, or cancel ##Q_0## and ##t## and solve for ##\phi##.
     
    Last edited: Apr 8, 2017
  2. jcsd
  3. Apr 8, 2017 #2

    gneill

    User Avatar

    Staff: Mentor

    The equations should hold for any value of t. Choose wisely :wink: :smile:
     
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