A series circuit has a capacitor of .25*10^-6 F, a resistor of 5*10^3 ohms , and

Question

A series circuit has a capacitor of .25*10^-6 F, a resistor of 5*10^3 ohms , and an inductor of 1 H. The initial charge on the capacitor is zero. If a 15-volt battery is connected to the circuit and the circuit is closed at t=0 , determine the charge on the capacitor at t=0.001 seconds, at t=0.01 seconds, and at any time . Also determine the limiting charge as t approaches infinity.

So I got a= 3.75x10^-6 but yet wiley wont give me partial credit (i submitted with that value to see if it was right even turned into .00000375 and nothing). Also this is a differential equations course.

Explanation / Answer

Solution:

Voltage through the resistor Vr = R i
Voltage through the inductor Vi = L di/dt
Voltage through the capacitor Vc = q/C

15 = 1 d2q/dt2 + 5 *103 dq/dt + q/0.25 * 106

Characteristic polynomial of the homogeneous equation:

=> 2 + 5 *103 + 4 * 106 = 0
=> = {-5*103 ± (25*106 - 4 *4*106)} / {2*1} = -2.5*103 ± (6.25*106 - 4*106) = -2.5*103 ± (2.25*106)
        = (-2.5 ± 1.5) * 103
=> = -4 * 103 and -1 * 103

So a = -4 * 103 and b = -1 * 103

Particular solution: a
=> 15 = 4 * 106 a
=> a = 3.75 * 10-6

q(t) = 3.75*10-6 + Ae(-4*10^3)t + Be(-1*10^3)t
q(0) = 0 = 3.75*10-6 + A + B
i(0) = -4*103 A - 1*103 B = 0 => B = -4 A
=> 0 = 3.75*10-6 + A + (-4A) => 3.75*10-6 - 3A = 0
=> A = 1.25*10-6
=> B = -4A = -4 (1.25*10-6) = -5 * 10-6

q(t) = 3.75*10-6 + 1.25*10-6 e(-4*10^3)t - 5*10-6 e(-1*10^3)t

Vc(t) = q(t)/C = {3.75*10-6 + 1.25*10-6 e(-4*10^3)t - 5*10-6 e(-1*10^3)t} / {0.25 * 10-6)

                       = 15 + 5e(-4*10^3)t - 20e(-1*10^3)t

Vc(t = 0.001) = 15 + 5e-4 - 20e-1 = 7.734V
Vc(t = 0.01) = 15 + 5e-40 - 20e-10 = 14.999V
Vc(t = ) = 15

q(t = 0.001) = 3.75*10-6 + 1.25*10-6 e-4 - 5*10-6 e-1 = 1.9335 *10-6 C
q(t = 0.001) = 3.75*10-6 + 1.25*10-6 e-40 - 5*10-6 e-10 = 3.7498 *10-6 C
q(t = ) = 3.75*10-6 C

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