Answer to Question #140134 in Differential Equations for Longinus Lukas

Question #140134
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
1
Expert's answer
2020-10-26T19:19:06-0400

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)\n-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"








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