Expert: Paul Klarreich - 3/29/2008

Question
hye.. i really hope that you can answer this question.i'm studying ode now for my degree in engineering. this is the first order of differential equation, but it involve the application in electric circuit. this make me confius on how to solve it. the question is:
given equation governing the charge q(t) on the capacitor or a circuit is: L d^2/dt^2 + R dq/dt + 1/C q = E (t)
let L=2 henrys, R=4 ohms, C = 0.05 farads and E=100volts.solve for q(t) subject to the initial condiion q(0)=i(0) = 0
identify the steady-state solution.give a plot of the solution q(t) over a sufficiently long time period to clearly show the approach of q to its steady-state.

Answer
Questioner:   fiza
Category:  Calculus
Private:  No
 
Subject:  ode
Question:  hye.. i really hope that you can answer this question.i'm studying ode now for my degree in engineering. this is the first order of differential equation, but it involve the application in electric circuit. this make me confius on how to solve it. the question is:

given equation governing the charge q(t) on the capacitor or a circuit is: L d^2/dt^2 + R dq/dt + 1/C q = E (t)
let L =  2 henrys, R = 4 ohms, C = 0.05 farads and E=100volts.solve for q(t) subject to the initial condiion q(0)=i(0) = 0
identify the steady-state solution.give a plot of the solution q(t) over a sufficiently long time period to clearly show the approach of q to its steady-state.

It has been a LONG TIME since I did any of this, so you are free to treat my answer with the disrespect it deserves.

L d^2q/dt^2 + R dq/dt + 1/C q = E (t)

As I recall, you start with the 'homogeneous equation':

L d^2q/dt^2 + R dq/dt + 1/C q = 0

Assuming q = A e^kt

L k^2 + R k + 1/C k = 0

   -R +- sqrt(R^2 - 4L/C)
k = -------------------------
            2L

   -4 +- sqrt(16 - 8/0.05)
k = -------------------------
            4

   -4 +- sqrt(16 - 160)
k = -------------------------
            4

   -4 +- sqrt(- 144)
k = ------------------
            4

k = -1 +- 12i

And the solution is:

q = A e^(-t) sin (12t + phi)  Plus a constant, of course.

Since this makes the left side zero, we need + 100 as the constant, so

q = A e^(-t) sin (12t + phi) + 100

and all that is left is to find A and phi.

Now q(0) = 0 means

A sin (phi) + 100 = 0, so

A sin (phi) = -100

And i(t), as I recall, is  q'(t)

q'(t) = 12 A e^(-t) cos (12t + phi) - A e^(-t) sin (12t + phi)

if q'(0) = 0,

12 A cos (phi) - A sin (phi) = 0

12 cos (phi) - sin (phi) = 0

--> sin(phi) = 12 cos(phi)

tan(phi) = 12

So phi = arctan(12) and

sin phi = 12/sqrt(145) << from right triangles.

Going back:

A sin (phi) = -100

A (12/sqrt(145))= -100

A = -100(12/sqrt(145))

So your graph is:

q = -100(12/sqrt(145) e^(-t) sin(12t + arctan(12)) + 100
q = -99.65 e^(-t) sin(12t + arctan(12)) + 100

and I think that should about do it.


[See attached image.]