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I know that the differential equation that describes this kind of circuit is $$\frac{d^2 q(t)}{dt^2} = -\omega^2 q(t) \, .$$

I was wondering how to model the case where we have dissipation of energy. I guess I should add some term of order $1$, but I don't know what to add exactly.

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    $\begingroup$ Welcome New contributor Lorenzo Benedetti! You wouldn't be asking for an RLC circuit would you? $\endgroup$ May 27, 2020 at 11:47
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    $\begingroup$ There's more than one way to add dissipation. Please include a circuit diagram. $\endgroup$
    – DanielSank
    May 27, 2020 at 14:05
  • $\begingroup$ @AlfredCentauri I thought it behaved differently from an RLC circuit $\endgroup$
    – Benny
    May 28, 2020 at 15:19

2 Answers 2

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Assuming that the loss is due to ohmic heating then a voltage term of the form $V_{\rm R} = R I = R \dfrac {dq}{dt}$ must be introduced when Kirchhoff's voltage law is being used to set up the differential equation.

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Consider a series $RLC$ circuit (one that has a resistor, an inductor and a capacitor) with a constant driving electro-motive force (emf) $E$. The current equation for the circuit is $$L\frac{di}{dt}+Ri+\frac{1}{C}\int i\;dt=E$$ $$\therefore \;\;L\frac{di}{dt}+Ri+\frac{1}{C}q=E$$

Differentiating, we have,

$$L\frac{d^2i}{dt^2}+R\frac{di}{dt}+Ci=0$$

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  • $\begingroup$ Does that work even if there is no generator and the charge is completely on the capacitor? $\endgroup$
    – Benny
    May 28, 2020 at 15:22
  • $\begingroup$ By generator, you mean battery, right? $\endgroup$
    – SarGe
    May 28, 2020 at 15:24
  • $\begingroup$ Yes I mean battery $\endgroup$
    – Benny
    May 28, 2020 at 15:26
  • $\begingroup$ The charge doesn't lie permanently on the capacitor. Instead of charge, the more precise word here is energy. The energy oscillates between inductor and capacitor and in the course it gets decayed due to the resistor. $\endgroup$
    – SarGe
    May 28, 2020 at 15:32

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