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

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

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

$$c_2=-1200-c_1$$

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

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

$$i(0)=0$$

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

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

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

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

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

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

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