A series circuit contains a resistor with R = 24 , an inductor with L = 2 H, a capacitor with C = 0.005 F, and

Question

A series circuit contains a resistor with R = 24 ohms, an inductor with L = 2 H, a capacitor with C = 0.005 F, and a generator producing a voltage of E(t) = 12 sin(10t). The initial charge is Q = 0.001 C and the initial current is 0.
Find the charge at time t.

Q(t)= ??

Answer 1

First thing to realize is that this circuit is in resonance. The characteristic frequency is equal to 1/sqrt(LC)
=1/sqrt(2*.005)
=10
This means the driving source won't provide net new charge to the capacitor and we can look at this in two pieces
so for the output across the capacitor we
have
v-out
=1/jωC/(R+jωL+1/(jωC))*v-in
=1/(1-ω^2LC+jωRC)

When we plug in values we get

v-out=1/(1-10^2*(2*0.005)+j10*24*.005)*v-in
v-out=1/(1-1+1.2j)*v-in
=-j/1.2*12sin(10t)
=10sin(10t-π)
so the change in Q across the capacitor from the driven source will be

Q(t)-driven
=0.005*10sin(10t-π)
=0.0510sin(10t-π) C
now we have to look at the initial conditions with a voltage of
0.2 V stored in the capacitor.

We'll short out the driven source and solve for the decay in voltage through the resistor and inductor

Because there is in essence a DC voltage stored on the capacitor and discharged through the resistor, the inductor has no role. The voltage discharge is

0.2exp(-t/RC)=0.2exp(-8.33t). The charge is just this multiplied by C.
or 0.2*0.005exp(-8.33t)=0.001exp(-8.33t)
the total will be
Q(t)=0.0510sin(10t-π)+0.001exp(-8.33t)

Answer 2

The differential equation for your circuit is:

Lq'' + Rq' + q/C = 12sin(10t).............(1)

If you assume that sin(10t) = 0 for t<0 you can find solution using one sided Laplace transform. I have assumed that sin(10t) is valid for -inf < t < inf .
The general solution of non-homogeneous differential equation (1) consists of the sum of the particular (any) solution of the equation (1) and the general solution of the homogeneous equation (2) below:

Lq'' + Rq' + q/C = 0.............(2)

Particular solution of (1) is assumed as q = Qsin(10t + fi). If you plug it in the solution is:

q = -0.05cos(10t).............(3)

The general solution of (2) is:

q = R*(e^at)*sin(bt + fi),....(4)
where:
a = -R/2L = -6 and b = sqrt(4L/C-R^2)/2L = 8.

Plugging (4) into (2) you get R = 0.051/0.8, fi = 0.923.

Finally your solution is the sum of (3) and (4):

q = - 0.05cos(10t) + (0.051/0.8)*(e^(-6t))*sin(8t + 0.923).

You can verify that qo = 0.001, and io = q'(0) = 0.

Solution (2) is so called stationary solution. In this case LC is in resonance at frequency omega = 10 and their serial impedance is 0. Therefore the stationary current is E/R = 12/24 = 0.5 A, which you get as the first derivative of q.
Solution (4) is the transient solution caused by initial charge q0.

Answer 3

The differential equation is in the type LQ'' + RQ' + (a million/C)Q = 12sin(10t) the place L = 2, R = 24, and (a million/C) = 2 hundred, so 2Q'' + 24Q' + 200Q = 12sin(10t). We first discover the respond to the homogeneous area, Q(t)[h], and then the answer for the nonhomogeneous area, Q(t)[nh], and then upload them jointly to get Q(t). The function equation is: 2r² + 24r + 2 hundred = 0 fixing this, we get r = (-6 + 8i),(-6 - 8i), and since the suggestions are complicated numbers in the type (? ± ?i), the answer to the homogeneous area of the equation is Q(t)[h] = e^?t * (c1*sin(?t) + c2*cos(?t)) = e^-6t * (c1*sin(8t) + c2*cos(8t)). Now we make a gamble for Q(t)[nh], in an identical sort because of the fact the spectacular fringe of the unique equation 12*sin(10t), so we wager Asin(10t) + Bcos(10t). Now we differentiate this two times: Q(t)[nh] = Asin(10t) + Bcos(10t) Q'(t)[nh] = 10Acos(10t) - 10sin(10t) Q''(t)[nh] = -100Asin(10t) - 100cos(10t) Then we replace that back into the unique equation: 2*(Asin(10t) + Bcos(10t)) + 24*(10Acos(10t) - 10sin(10t)) + 2 hundred*(-100Asin(10t) - 100cos(10t)) =?= 12*sin(10t) Which equals 240*A*cos(10*t)-240*B*sin(10*t) =?= 12*sin(10t) We set 240A = 0 because of the fact there are not any cosines on the spectacular part and -240B = 12 because of the fact there are 12 sines. fixing those mini equations, we get A = 0 and B = -a million/20. No all of us know the two out Q(t)[h] and Q(t)[nh]: Q(t)[h] = e^-6t *(c1*sin(8t) + c2*cos(8t)) Q(t)[nh] = 0*sin(10t) - (a million/20)*cos(10t) so Q(t) = e^-6t *(c1*sin(8t) + c2*cos(8t)) - (a million/20)*cos(10t). Now we are able to apply our preliminary values (Q(0) = 0.001 and Q'(0) = 0) to discover c1 and c2: Q(t) = e^-6t *(c1*sin(8t) + c2*cos(8t)) - (a million/20)*cos(10t). Q'(t) = -6e^(-6t) * (c1sin(8t) + c2cos(8t)) + e^(-6t) * (8c1cos(8t) - 8c2sin(8t)) + 10Acos(10t) - 10Bsin(10t) Q(0) = a million*(c1*0 + c2*a million) - (a million/20)*a million = c2 - (a million/20) = 0.001 --->> c2 = 0.051 Q'(0) = -6c2 + 8c1 = 0 --->> c1 = 0.03825 So: A = 0 B = (-a million/20) c1 = 0.03825 c2 = 0.051 and the terrific equation is: Q(t) = e^-6t * (0.03825*sin(8t) + 0.051*cos(8t)) - (a million/20)cos(10t) (that's real in accordance to Webassign.) desire that facilitates!