Answer to Question #106232 in Differential Equations for Gayatri Yadav

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

Kirchhoff's voltage law:

$u_R+u_L+u_C=0$

where $u_R,u_L,u_C$ are the voltage across $R,L,C$ 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: 

$\frac{d^2i(t)}{dt^2}+2 \alpha\frac{di(t)}{dt}+\omega^2_0i(t)=0\\ \alpha=\frac{R}{2L}, \omega_0=\frac{1}{\sqrt{LC}}$

The differential equation has the characteristic equation:

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

The roots of the equation in $s$ are: 

$s_{1,2}=-\alpha \pm\sqrt{\alpha^2-\omega_0^2}$

The general solution of the differential equation is an

exponential in either root or a linear superposition of both

$i(t)=Ae^{s_1t}+Be^{s_2t}$

The initial current is zero:

$i(0)=Ae^{0}+Be^{0}=0$

Therefore: $A=-B$  

$i(t)=A(e^{s_1t}-e^{s_2t})=\\ =Ae^{-\alpha t}(e^{\sqrt{\alpha^2-\omega_0^2}t}-e^{-\sqrt{\alpha^2-\omega_0^2}t})\\ \sqrt{\alpha^2-\omega_0^2}=\sqrt{(30)^2-10\cdot 50}=\\ =\sqrt{400}=20, \alpha=30$

therefore

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

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

$L\frac{di(t)}{dt}|_{t=0}+\frac{q_0}{C}=0\\ \frac{di(t)}{dt}|_{t=0}=-500q_0\\ \frac{di(t)}{dt}=A(-10e^{-10t}+50e^{-50t)}\\ \frac{di(t)}{dt}|_{t=0}=40A\\ 40A=-500q_0\\ A=-12.5q_0$

Therefore

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