Q 3.5-2E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

An RC circuit with a $\mathbf{1}\mathbf{\Omega }$ resistor and a $\mathbf{0}\mathbf{.}\mathbf{000001}\mathbf{-}\mathbf{F}$ capacitor is driven by a voltage $\mathbf{E}\left(t\right)\mathbf{=}\mathbf{sin}\mathbf{100}\mathbf{tV}$. If the initial capacitor voltage is zero, determine the subsequent resistor and capacitor voltages and the current.

Step-by-Step Solution

Answer

The subsequent resistor is $E_R=-{10}^{-4}\cos 100t+{10}^{-8}\sin 100t-{10}^{-4}{e}^{-{10}^{6}t}$.

The subsequent capacitor voltage is $E_C=-{10}^{-4}\cos 100t+\sin 100t+{10}^{-4}{e}^{-{10}^{6}t}$.

The subsequent current is $I=-{10}^{-4}\cos 100t+\sin 100t+{10}^{-4}{e}^{-{10}^{6}t}$.

Step 1: Important formula.

The governing differential equation for RC circuit is $\frac{dq}{dt}+\frac{q}{CR}=\frac{E}{R}$.

Step 2: Determine the subsequent resistor.

The governing differential equation for RC circuit is $\frac{dq}{dt}+\frac{q}{CR}=\frac{E}{R} ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}$.

Here $q\left(0\right)=0,C={10}^{-6},E\left(t\right)=\sin 100 t$.

Put these values in equation (1) then 

$\frac{dq}{dt}+{10}^{-6}q=\sin 100 t$

The integrating factor is $e^{{10}^{6}t}$.

Now the equation is 

$$\begin{gathered} e^{{10}^{6}t}q=\int e^{{10}^{6}t}\sin 100tdt \\ {e}^{{10}^{6}t}q=-{10}^{-10}{e}^{{10}^{6}t}\cos 100t+{10}^{-6}{e}^{{10}^{6}t}\sin 100t+C \\ q\left(t\right)={10}^{-10}\cos 100t+{10}^{-6}\sin 100t+C{e}^{-{10}^{6t}} \end{gathered}$$

Now, find the value of done

$$\begin{gathered} I=\int e^{{10}^{6}t}\sin 100tdt \\ =-\frac{1}{100}{e}^{{10}^{6}t}\cos 100t+\frac{{10}^6}{100}\int e^{{10}^{6}t}\cos 100tdt \\ =-\frac{1}{100}\cos 100t+1000\int e^{{10}^{6}t}\cos 100tdt \\ =-\frac{1}{100}{e}^{{10}^{6}t}\cos 100t+1000(\frac{1}{100}{e}^{{10}^{6}t}\sin 100t-\frac{{10}^6}{100}\int e^{{10}^{6}t}\sin 100tdt) \\ =-\frac{1}{100}{e}^{{10}^{6}t}\cos 100t+100(\frac{1}{100}{e}^{{10}^{6}t}\sin 100t-\frac{{10}^6}{100}-10000I \\ =-\frac{1}{100}{e}^{{10}^{6}t}\cos 100t+100e^{{10}^{6}t}\sin 100t-100000000I+C \\ I=-{10}^{-10}{e}^{{10}^6}\cos 100t+{10}^{-6}\sin 100t+C \end{gathered}$$

Apply the initial conditions then $C=-{10}^{-10}$.

$$\begin{gathered} I=-{10}^{-10}{e}^{{10}^6}\cos 100t+{10}^{-6}\sin 100t+-{10}^{-10} \\ q\left(t\right)={10}^{-10}\cos 100t+{10}^{-6}\sin 100t-{10}^{-10}{e}^{-{10}^{6}t} \end{gathered}$$

 The subsequent resistor is $E_C=-{10}^{-4}\cos 100t+\sin 100t+\text{'}{10}^{-4}{e}^{-{10}^{6}t}$.

Step 3: evaluate capacitor voltage.

Now, find the value of capacitor voltage.

$$\begin{gathered} E_C=\frac{q\left(t\right)}{C} \\ {E}_C=\frac{{10}^{-10}\cos 100t+{10}^{-6}\sin 100t-{10}^{-10}{e}^{-{10}^{6}t}}{{10}^{-10}} \\ {E}_C=-{10}^{-4}\cos 100t+\sin 100t+\text{'}{10}^{-4}{e}^{-{10}^{6}t} \end{gathered}$$

The subsequent capacitor voltage is $E_C=-{10}^{-4}\cos 100t+\sin 100t+\text{'}{10}^{-4}{e}^{-{10}^{6}t}$

Step 4: Determine electric resistor .

Using Kirchhoff’s voltage law to the RC circuit 

$$\begin{gathered} E_R=E_C-E\left(t\right) \\ {E}_R=-{10}^{-4}\cos 100t+{10}^{-8}\sin 100t-{10}^{-4}{e}^{-{10}^{6}t} \end{gathered}$$

Step 5: find the value of current.

Now, find the value of current.

$$\begin{gathered} I=\frac{E_R}{R} \\ I=-{10}^{-4}\cos 100t+\sin 100t+{10}^{-4}{e}^{-{10}^{6}t} \end{gathered}$$

Therefore, the subsequent current is $I=-{10}^{-4}\cos 100t+\sin 100t+{10}^{-4}{e}^{-{10}^{6}t}$.