Answer to Question #140134 in Differential Equations for Longinus Lukas

Question

A series circuit contains a resistor with R = 40Ω, an inductor with L = 2 H, a capacitor with
0.0025 F, and a 12 − V battery. The initial charge is Q = 0.001 C and the initial current is
0. Using the method of undetermined coefficients, find the charge at time t A series circuit contains a resistor with R = 40Ω, an inductor with L = 2 H, a capacitor with
0.0025 F, and a 12 − V battery. The initial charge is Q = 0.001 C and the initial current is
0. Using the method of undetermined coefficients, find the charge at time t

Answer 1

Given,

L=2H, R=40$\Omega$ ,C=0.0025F

V=12 volt, initial charge $q_o=$ 0.001Columb

The classical differential equation of RLC series circuit is given by

$L\dfrac{d^2Q}{dt^2}+R\dfrac{dq}{dt}+\dfrac{Q}{C}=V$

$\to$ $2\dfrac{d^2Q}{dt^2}+40\dfrac{dq}{dt}+\dfrac{Q}{0.0025}=12$

$2\dfrac{d^2Q}{dt^2}+40\dfrac{dq}{dt}+400Q=12$

or

$\to \dfrac{d^2Q}{dt^2}+20\dfrac{dq}{dt}+200Q=6~~~~~-(1)$

Calculation of complementary part

Auxiliary equation is given by,

$m^2+20m+200=0$

$m=\dfrac{-20\pm \sqrt{400-800}}{2}$

=$\dfrac{-20\pm20i}{2}$

$=-10\pm10i$

Complementary function is given by,

C.f.=$e^{-10t}(c_1cos10t+c_2sin10t)~~~~~~~~-(1)$

Calculation of particular integral

Particular integral=A

$Q=A , Q'=0,Q''=0$

Put these value in (1)

(1)$\to 0+0+200A=6$

$\therefore A=\dfrac{6}{200}$

=$0.03$

Complete solution

C.S.=C.F.+P.I.

Q =$e^{-10t}(c_1cos10t+c_2sin10t)+0.03$

At t=0 ,q=0.001

$0.001=1(c_1cos0+c_2sin0)+0.03$

$c_1=0.001-0.03$ =$0.029$

At t=0, current I=0

So $\dfrac{dq}{dt}=0$

$\to e^{-10t}(10c_2cost-10c_1sint) -10e^{-10t}(c_1cos10t+c_2sin10t)$ =0

$\to10c_2-10c_1=0$

$\to c_2=c_1=0.029$

Therefore charge Q is

$Q=-0.029e^{-10t}(cos10t+sin10t)+0.03$

Current

I=$\dfrac{dq}{dt}$

=$-0.029(e^{-10t}(10cos10t-10sin10t)-10e^{-10t}(sin10t+cos10t))$

$=-0.029(e^{-10t}(10cos10t-10sin10t-10sin10t-10cos10t)$

=$-0.029(e^{-10t}(-20sin10t))$

$=0.58e^{-10t}sin10t$

Hence current I=$0.58e^{-10t}sin10t$