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Find the charge q(t) on the capacitor and the current i(t) in the given LRC-series circuit. L = 5 6 h, R = 10 Ω, C = 1 60 f, E(t) = 1200 V, q(0) = 0 C, i(0) = 0 A q(t) = Correct: Your answer is correct. C i(t) = Correct: Your answer is correct. A Find the maximum charge on the capacitor. (Round your answer to three decimal places.)

Respuesta :

Answer:

Explanation:

The DE corresponding to LRC series circuit is

Lq" + Rq' +1/c q = E

[tex]\frac{5}{6} q" +  10 q' + 60 q = 1200[/tex]

q" + 12q' + 72 q =  1440

considering only homegeneous part

q" + 12q'+ 72 q= 0

the auxillary equation is [tex]m^2 + 12m + 72  = 0[/tex]

[tex]m = \frac{-12 \pm \sqrt{144-288}}{2}[/tex]

[tex]m = -6 \pm 6i[/tex]

[tex]m_1 = -6 -6i[/tex]

[tex]m_2 = -6 + 6i[/tex]

the solutiion is

[tex]qh = e^{-6t} [c_1cos6t + c_2 sin 6t][/tex]

solve for particular siolution

q" + 12q'+ 72 q= 0

let qp = a be the solution

0 + ) + 72a = 1440

a = 20

general solution is

q = qh + qp

[tex]q =e^{-6t} [c_1cos6t + c_2 sin 6t] + 20[/tex][/tex]

applying initial value i(0) = q(0) = 0

also i(t) = q'(t)

[tex]i(t) = e^{-6t} [6c_2 cos6t +6 c_1 sin 6t] -  e^{-6t} [c_1 c_2 cos6t + c_2 sin 6t][/tex]

q(0) = 0

[tex]c_1 +5 = - 0[/tex]

[tex]c_1 = -5[/tex]

i(0) = 0

[tex]6c_2 - 6c_1 = 0 [/tex]

[tex]c_2 = c_1 = -5[/tex]

[tex]q(t0 = 5 - 5e^{-6t}[ coas(6t) + sin(6t)][/tex]

[tex]i(t) = e^{-6t} ( -30 cos(6t) + 30 sin(6t) - 6e^{-6t} ( -5cos(6t) - 5sin(6t))[/tex]

[tex]= e^{-6t} [60 sin (6t)][/tex]

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