A series circuit consists of a resistor $R = 24 \Omega$, an inductor with $L = 2 H$, a capacitor with $C = 0.005 \mathrm {~F}$, and a generator producing a voltage of $E ( t ) = 12 \cos ( 10 t )$. If the initial charge is $Q = 0.001 \mathrm { C }$ and the initial current is 0 , find the charge $Q ( t )$ at time $t$.
A) $Q ( t ) = e ^ { - 6 t } ( \cos ( 6 t ) - 0.06175 \sin ( 6 t ) )$
B)$Q ( t ) = e ^ { - 2 t } ( 0.051 \cos ( 2 t ) - 0.06175 \sin ( 2 t ) ) + \frac { 1 } { 20 } \sin ( 10 t )$
C)$Q ( t ) = e ^ { - 6 t } ( 0.001 \cos ( 8 t ) - 0.06175 \sin ( 8 t ) ) + \frac { 1 } { 20 } \sin ( 10 t )$
D) $Q ( t ) = e ^ { - 6 t } ( 0.001 \cos ( 8 t ) - 0.06175 \sin ( 8 t ) )$
E) $Q ( t ) = e ^ { - 3 t } ( \cos ( 5 t ) - 0.06175 \sin ( 5 t ) )$

Question

Quizplus · Accepted Answer

The circuit is a series circuit, so the current through all three components is the same, and we can use Kirchhoff's voltage law to write the differential equation for the charge on the capacitor:
$$
L \frac { d ^ { 2 } Q } { d t ^ { 2 } } + R \frac { d Q } { d t } + \frac { Q } { C } = E ( t )
$$
Substituting for the given values, we get:
$$
2 \frac { d ^ { 2 } Q } { d t ^ { 2 } } + 24 \frac { d Q } { d t } + 2000 Q = 12 \cos ( 10 t )
$$
This is a non-homogeneous differential equation with constant coefficients, so we can use the method of undetermined coefficients. The particular solution is
$$
Q_p ( t ) = A \cos ( 10 t ) + B \sin ( 10 t )
$$
Substituting into the differential equation yields
$$
200 A = 12 \Rightarrow A = \frac { 3 } { 50 } \quad 	ext { and } \quad B = 0
$$
The homogeneous solution is obtained by setting the right-hand side to 0:
$$
2 \frac { d ^ { 2 } Q } { d t ^ { 2 } } + 24 \frac { d Q } { d t } + 2000 Q = 0
$$
The characteristic equation is
$$
2 r ^ { 2 } + 24 r + 2000 = 0 \Rightarrow r = - 10 \pm 5 \sqrt { 6 } i
$$
So the general solution is
$$
Q ( t ) = Q_h ( t ) + Q_p ( t ) = e ^ { - 6 t } ( C _ { 1 } \cos ( 5 \sqrt { 6 } t ) + C _ { 2 } \sin ( 5 \sqrt { 6 } t ) ) + \frac { 3 } { 50 } \cos ( 10 t )
$$
We use the initial conditions to find the values of $C_1$ and $C_2$ :
\begin{align*}
Q ( 0 ) & = 0.001 \
C _ { 1 } & = 0.001 \
\frac { d Q } { d t } ( 0 ) & = 0 \
C _ { 2 } & = - 0.06175
\end{align*}
Therefore, the final answer is
$$
Q ( t ) = e ^ { - 6 t } ( 0.001 \cos ( 5 \sqrt { 6 } t ) - 0.06175 \sin ( 5 \sqrt { 6 } t ) ) + \frac { 3 } { 50 } \cos ( 10 t )
$$

A Series Circuit Consists of a Resistor R=24ΩR = 24 \OmegaR=24Ω

A Series Circuit Consists of a Resistor $R = 24 \Omega$