Answer to Question #344428 in Differential Equations for Zainab

Question

 An RCL circuit connected in series has R = 10 ohms, C = 10^2 farad, L = 1.5 henry, and an applied voltage E = 12 volts. Assuming no initial current and no initial charge at t = 0 when the voltage is first applied, find the subsequent current in the system.

Answer 1

$Lq''+Rq'+\dfrac{q}{C}=E$

$1.5q''+10q'+0.01q=12$

Homogeneous differential equation

$1.5q''+10q'+0.01q=0$

Corresponding (auxiliary) equation

$1.5r^2+10r+0.01=0$

$r=\dfrac{-10\pm\sqrt{(10^2)-4(1.5)(0.01)}}{2(1.5)}$

$r_1=\dfrac{-10-\sqrt{99.94}}{3}$

$r_2=\dfrac{-10+\sqrt{99.94}}{3}$

The general solution of the homogeneous differential equation is

$q=c_1 e^{-(10+\sqrt{99.94})t/3}+c_2 e^{-(10-\sqrt{99.94})t/3}$

The general solution of the non homogeneous differential equation is

$q=c_1 e^{-(10+\sqrt{99.94})t/3}+c_2 e^{-(10-\sqrt{99.94})t/3}+1200$

$q(0)=0=>c_1+c_2+1200=0$

$c_2=-1200-c_1$

$i(t)=q'=-\dfrac{10+\sqrt{99.94}}{3}(c_1 e^{-(10+\sqrt{99.94})t/3})$

$-\dfrac{10-\sqrt{99.94}}{3}(c_2 e^{-(10-\sqrt{99.94})t/3})$

$i(0)=0$

$=>-\dfrac{10+\sqrt{99.94}}{3}c_1-\dfrac{10-\sqrt{99.94}}{3}c_2=0$

$-10c_1-\sqrt{99.94}c_1+12000+10c_1$

$-1200\sqrt{99.94}-\sqrt{99.94}c_1=0$

$c_1=600(\dfrac{10-\sqrt{99.94}}{\sqrt{99.94}})$

$c_2=-600(\dfrac{10+\sqrt{99.94}}{\sqrt{99.94}})$

$i(t)=-\dfrac{12}{\sqrt{99.94}}e^{-(10+\sqrt{99.94})t/3}$

$+\dfrac{12}{\sqrt{99.94}} e^{-(10-\sqrt{99.94})t/3}$