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A series circuit has a capacitor of 0.04×10−6 F and an inductor of 1 H. If the initial charge on the capacitor is 0.18×10−6C and there is no initial current, find the charge Q on the capacitor at any time t. Enter an exact answer. Do not use thousands separator in the answer field. Enclose arguments of functions in parentheses. For example, sin(2x).

Respuesta :

Answer:

[tex]Q(t) = 0.18\times 10^{-6} cos(5\times 10^3 t)[/tex]

Explanation:

We knwo that Kirchoff law

[tex]L\frac{DI(t)}{dt} + \frac{1}{C}Q = 0[/tex]

where

[tex]I(t) = \frac{dQ}{dt}[/tex]

hence

[tex]LQ" + \frac{Q}{C} = 0[/tex]

C is given as 0.04\times 10^{6} F

L= 1 H , so we have

[tex]Q" +  25\times 10^6Q = 0[/tex]

the characteristic equation of this differential equation is

[tex]r^2 + 25\times 10^6 = 0[/tex]

[tex]r = \pm 5\times 10^3 i[/tex]

Therefore differential equation is

[tex]Q(t) =  c_1 cos(5000t) + c_2sin(5000t)[/tex]

we know initial value if capacitor is given as [tex]0.18\times 10^{-6} C[/tex]

Therefore

[tex]0.18\times 10^{-6} = Q(0) =  c_1 cos(0) + c_2sin(0)[/tex]

[tex]c_1 = 0.18\times 10^{-6}[/tex]

if no inital current is present then we hvae I(0) = Q'(0) = 0

[tex]Q'(T) = -5000 C_1 sin(5000t) +  5000 c_2cos(5000t)[/tex]

[tex]0 = Q'(0) = - 5000c_1sin(0) + 5000c_2cos(0)[/tex] therefre

[tex]c_2 =0 [/tex]

hence charge is

[tex]Q(t) = 0.18\times 10^{-6} cos(5\times 10^3 t)[/tex]

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