# LC circuit with dissipation [closed]

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.

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

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.
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 $$L\frac{di}{dt}+Ri+\frac{1}{C}\int i\;dt=E$$ $$\therefore \;\;L\frac{di}{dt}+Ri+\frac{1}{C}q=E$$
$$L\frac{d^2i}{dt^2}+R\frac{di}{dt}+Ci=0$$