Question

In a series LCR circuit with L = 5/3 H, R = 10 Ω and C = 1/30 F and a source of emf
E = 600 V, determine the charge on the capacitor as a function of time. It is given that
the initial charge on the capacitor is zero and the initial current in the circuit is 9A.

Answer 1

The RLC circuit is descrided by the initial value problem

$$L\frac{d^2 q}{dt^2}+R\frac{d q}{dt}+\frac{1}{C}q=E$$

$$q(0)=q_0,\quad I(0)=q'(0)=I_0$$

where $q(t)$ is a charge on the capacitor.

In our case we have

$$\frac{5}{3}\frac{d^2 q}{dt^2}+10\frac{d q}{dt}+30q=600$$

or

$$\frac{d^2 q}{dt^2}+6\frac{d q}{dt}+18q=360$$

and

$$q(0)=0,\quad I(0)=q'(0)=9$$

Solution of the IVP

$$q(t)=e^{-3t}(20e^{3t}-20\cos(3t)-17\sin(3t))$$