A series circuit has a capacitor of {\textstyle\frac{1}{48}} \times 10^{-6} F and an inductor of 3 H. If the initial charge on the capacitor is 2 \times 10^{-2} C and there is no initial current, find the charge on the capacitor Q(t).


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)

Print this page