Question

A series RLC circuit with R = 6 ohm, C = 0.02 Farad and L = 0.1 has no applied voltage. Find the subsequent current in the circuit if the initial charge, on the capacitor is q0 and the initial current is zero.

Answer 1

A series $RLC$ circuit with  R=6 ohm, C=0.02 Farad and L=0.1 has no applied voltage. Find the subsequent current in  the circuit if the initial charge, on the capacitor is $q_0$ and the initial current is zero. 

Kirchhoff's voltage law:

$u_R+u_L+u_C=0$

where $u_R,u_L,u_C$ are the voltages across R,L and C respectively.

Substituting in the constitutive equations:

$Ri(t)+L\frac{di(t)}{dt}+\frac{1}{C}\int \limits_{-\infty}^{t}i(\tau)d\tau=0$

Differentiating and dividing by L :

$\frac{d^2i(t)}{dt^2}+\frac{R}{L}\frac{di(t)}{dt}+\frac{1}{LC}i(t)=0$

This can usefully be expressed in a more generally applicable form:

$\frac{d^2i(t)}{dt^2}+2\alpha\frac{di(t)}{dt}+\omega_0^2i(t)=0\\ \alpha=\frac{R}{2L}, \omega_0=\frac{1}{\sqrt{LC}}$

The differential equation has the characteristic equation: 

$s^2+2\alpha s+\omega_0^2=0$

The roots of the equation in s are:

$i(t)=A(e^{s_1t}-e^{s_2t})=\\ =Ae^{-\alpha t}(e^{\sqrt{\alpha^2-\omega_0^2}t}-e^{-\sqrt{\alpha^2-\omega_0^2}t})\\ \alpha=\frac{6}{0.2}=30, \omega_0=\frac{1}{\sqrt{0.002}}\\ \sqrt{\alpha^2-\omega_0^2}=\sqrt{900-\frac{1}{0.002}}=\sqrt{400}=20$

therefore

$i(t)=Ae^{-30 t}(e^{20t}-e^{-20t})=A(e^{-10t}-e^{-50t})$

The initial charge on the capacitor is $q_0$ and initial current is zero: 

$L\frac{di(t)}{dt}|_{t=0}+\frac{q_0}{C}=0\\ \frac{di(t)}{dt}|_{t=0}=A(-10e^{-10t}+50e^{-50t})\\ \frac{di(t)}{dt}|_{t=0}=40A\\ \frac{q_0}{LC}=-40A\\ A=-\frac{q_0}{40LC}=-12.5q_0$

Therefore:

$i(t)=-12.5q_0(e^{-10t}-e^{-50t})$