1550078481-Ordinary_Differential_Equations__Roberts_

288 Ordinary Differential Equations

EXAMPLE 3 Finding the Equation for the Charge on a Capacitor

Find the equation for the charge, q, on the capacitor and the current flowing

in the RLC series circuit of Figure 6.3, if E = 0 V (volts), L = .04 H (Henry),

R = 20D (Ohms), C = 4x 10-^4 p (Farad), ifthe initial charge on the capacitor

is q(O) = 2 x 10-^3 F , and if the initial current flowing in the circuit is i(O) =

q'(O) = OA (amps).

SOLUTION

To find the equation for the charge on the capacitor, we must solve the fol-

lowing initial value problem which results from equations (7)- (8) by replacing

E, L , R, C, q 0 , and io by the values given in the problem.

. 04q" + 20q' + 2500q = O; q(O) = 2 x 10-^3 F , q' (0) = 0 A.

Using POLYRTS, we find the roots of the auxiliary equation .04r^2 + 20r +

2500 = 0 are r 1 = r 2 = -250. Since the roots are real and equal , this system

provides an example of critically damped motion and the equation for the

charge on the capacitor (the solution of the differential equation) is

q(t) = (A+ Bt)e-25ot

where A and B must be chosen to satisfy the given initial conditions. Differ-

entiating, we find

q' (t) = i(t) = (-250A + B - 250Bt)e-^250 t.

To satisfy the given initial condit ions, A and B must be chosen to simultane-

ously satisfy

q(O) = 2 x 10-^3 =A and q'(O) = i(O) = 0 = -250A + B.

So A = 2 x 10-^3 F, B = 250 A= .5 F /s, and the charge on the capacitor as

a function of time after the switch is closed is

q(t) = (.002 + .5t)e-^250 tcoulombs.

And the current flowing in the circuit is

i(t) = q'(t) = -125e-^250 tamps.

The negative sign in the equation for the current indicates that the current is
flowing in the direction opposite of the arrow in Figure 6.3.