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Q of RLC circuit, confused with provided answer

  1. Sep 4, 2016 #1
    1. The problem statement, all variables and given/known data
    The main question is to derive the Q of the series RLC circuit. This book shows the whole solution but I have a question about a part of their answer. Basically I want to know how ##\frac{1}{2}CV_{Cmax}^2 = \frac{1}{2}C(\frac{i_{max}^2}{wC}^2)##
    to me it looks like they are saying ##V=i_{max}/Z## instead of ##V=IZ##.

    Also, at the end why is the energy dissipated per cycle ##W_d=\frac{I_{max}^2R\pi}{w}##.
    I think ##I^2R\frac{2\pi}{w}## should be dissipated the energy dissipated per cylce. Why are they leaving it as ##I^2R\frac{\pi}{w}## so that the energy dissipated per cycle is divided by 2? Also then they multiply Q by ##2\pi## at the end which I also found to be strange. Can anyone please shed some light?
    2. Relevant equations


    3. The attempt at a solution
    The book gives the solution:

    In the time domain, the instantaneous stored energy in the circuit is given by : ##W_s=\frac{1}{2}Li^2 + \frac{q^2}{2C}##
    For a maximum: ##\frac{dW_s}{dt} = Li\frac{di}{dt}+\frac{q}{c}\frac{dq}{dt} = i(L\frac{di}{dt}+\frac{q}{C})=i(V_l+V_C)=0##
    Thus, the maximum stored energy is ##W_s## at i=0 or ##W_s## at ##V_L+V_C=0##, whichever is larger. Now the capacitor voltage, and therefore the charge, lags the current by ##90^\circ##; hence, i=0 implies ##q=\pm Q_{max}## and :
    ##W_{s|i=0} = \frac{Q_{max}^2}{2C} = \frac{1}{2}CV_{Cmax}^2 = \frac{1}{2}C(\frac{I_{max}}{wC})^2=\frac{I_{max}^2}{2Cw^2}##

    On the other hand, ##V_L + V_C=0## implies ##V_L=V_C=0## and ##i=\pm I_{max}## so that ##W_{s|V_L+V_C=0}=\frac{1}{2}LI_{max}^2##
    It follows that
    ##W_{s-max}=\frac{I_{max}^2}{2Cw^2}## for ##(w\leq w_0)## and
    ##W_{s-max}=\frac{LI_{max}^2}{2}## for ##(w \geq w_0)##

    The energy dissipated per cycle (in the resistor) is ##W_d=\frac{I_{max}^2R\pi}{w}##. Consequently,
    ##Q = 2\pi\frac{W_{smax}}{W_d} = \frac{1}{wCR}, (w\leq w_0)## and ##\frac{wL}{R}, (w \geq w_0)##
     
  2. jcsd
  3. Sep 5, 2016 #2

    rude man

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    Homework Helper
    Gold Member

    Z = 1/ωC for a capacitor so what they wrote (& which you copied incorrectly BTW) is correct.
    If you integrate Rimax2sin2(ωt) dt over 1 period = 2π/ω you will get what they state, which is πRimax2/ω.
     
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