Category: Test #3 Review (Page 4 of 4)

Capacitance, C = ${\textstyle\frac{1}{48}} \times 10^{-6}$

Inductance, L = $3$ H

Initial Charge, $Q(0) = Q_0$ = $2 \times 10^{-2}$ C

Initial Current, $Q'(0) = Q’_0$ or $I(0)$ = $I_0$ = 0 A

Since a resistor was not mentioned to be a part of the circuit, we can assume that $R = 0 \Omega$.

Also, there’s no impressed voltage mentioned, therefore, we can assume that $E(t) = 0$.

Using the form

$LQ”+RQ’+\frac{1}{C}Q = E(t)$

The differential equation for this problem will be

$3Q”+(\frac{1}{({\textstyle\frac{1}{48}} \times 10^{-6})})Q = 0$

The characteristic equation will then be

$3r^2 + (\frac{1}{({\textstyle\frac{1}{48}} \times 10^{-6})}) = 0$

Using the quadratic formula, the roots of the characteristic equation is

$r = 0 \pm 4000i$

Since the roots are complex imaginary numbers, the general equation for the amount of charge in the circuit will then be

$Q(t) = e^{0t}(c_{1}sin(4000t)+c_{2}cos(4000t))$

$= c_{1}sin(4000t)+c_{2}cos(4000t)$

Also note that, since the roots are imaginary, the oscillation of the circuit is underdamped.

To find the function for current, differentiate the previous equation, yielding

$I(t) = Q'(t) = 4000(c_{1}cos(4000t)-c_{2}sin(4000t))$

Next, substitute the initial conditions to solve for the undetermined coefficients

$ c_{1}sin(4000(0))+c_{2}cos(4000(0)) = 2 \times 10^{-2}$

$c_{2} = 2 \times 10^{-2}$

$4000(c_{1}cos(4000(0))-c_{2}sin(4000(0))) = 0$

$c_{1} = 0$

Finally, the equation for the amount of charge in the given circuit is

$Q(t) = 2 \times 10^{-2}cos(4000t)$