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Answers without the blur. Sign up and see all textbooks for free! Q 3.5-2E

Expert-verified Found in: Page 121 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # 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.

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

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

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

See the step by step solution

## 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 $\frac{dq}{dt}+\frac{q}{CR}=\frac{E}{R} ......\mathbf{\left(}\mathbf{1}\mathbf{\right)}\phantom{\rule{0ex}{0ex}}$.

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

Put these values in equation (1) then

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

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

Now the equation is

${e}^{{10}^{6}t}q=\int {e}^{{10}^{6}t}\mathrm{sin}100tdt\phantom{\rule{0ex}{0ex}}{e}^{{10}^{6}t}q=-{10}^{-10}{e}^{{10}^{6}t}\mathrm{cos}100t+{10}^{-6}{e}^{{10}^{6}t}\mathrm{sin}100t+C\phantom{\rule{0ex}{0ex}}q\left(t\right)={10}^{-10}\mathrm{cos}100t+{10}^{-6}\mathrm{sin}100t+C{e}^{-{10}^{6t}}\phantom{\rule{0ex}{0ex}}$

Now, find the value of done

$I=\int {e}^{{10}^{6}t}\mathrm{sin}100tdt\phantom{\rule{0ex}{0ex}}=-\frac{1}{100}{e}^{{10}^{6}t}\mathrm{cos}100t+\frac{{10}^{6}}{100}\int {e}^{{10}^{6}t}\mathrm{cos}100tdt\phantom{\rule{0ex}{0ex}}=-\frac{1}{100}\mathrm{cos}100t+1000\int {e}^{{10}^{6}t}\mathrm{cos}100tdt\phantom{\rule{0ex}{0ex}}=-\frac{1}{100}{e}^{{10}^{6}t}\mathrm{cos}100t+1000\left(\frac{1}{100}{e}^{{10}^{6}t}\mathrm{sin}100t-\frac{{10}^{6}}{100}\int {e}^{{10}^{6}t}\mathrm{sin}100tdt\right)\phantom{\rule{0ex}{0ex}}=-\frac{1}{100}{e}^{{10}^{6}t}\mathrm{cos}100t+100\left(\frac{1}{100}{e}^{{10}^{6}t}\mathrm{sin}100t-\frac{{10}^{6}}{100}-10000I\phantom{\rule{0ex}{0ex}}=-\frac{1}{100}{e}^{{10}^{6}t}\mathrm{cos}100t+100{e}^{{10}^{6}t}\mathrm{sin}100t-100000000I+C\phantom{\rule{0ex}{0ex}}I=-{10}^{-10}{e}^{{10}^{6}}\mathrm{cos}100t+{10}^{-6}\mathrm{sin}100t+C$

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

$I=-{10}^{-10}{e}^{{10}^{6}}\mathrm{cos}100t+{10}^{-6}\mathrm{sin}100t+-{10}^{-10}\phantom{\rule{0ex}{0ex}}q\left(t\right)={10}^{-10}\mathrm{cos}100t+{10}^{-6}\mathrm{sin}100t-{10}^{-10}{e}^{-{10}^{6}t}\phantom{\rule{0ex}{0ex}}$

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

## Step 3: evaluate capacitor voltage.

Now, find the value of capacitor voltage.

${E}_{C}=\frac{q\left(t\right)}{C}\phantom{\rule{0ex}{0ex}}{E}_{C}=\frac{{10}^{-10}\mathrm{cos}100t+{10}^{-6}\mathrm{sin}100t-{10}^{-10}{e}^{-{10}^{6}t}}{{10}^{-10}}\phantom{\rule{0ex}{0ex}}{E}_{C}=-{10}^{-4}\mathrm{cos}100t+\mathrm{sin}100t+\text{'}{10}^{-4}{e}^{-{10}^{6}t}\phantom{\rule{0ex}{0ex}}$

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

## Step 4: Determine electric resistor.

Using Kirchhoff’s voltage law to the RC circuit

${E}_{R}={E}_{C}-E\left(t\right)\phantom{\rule{0ex}{0ex}}{E}_{R}=-{10}^{-4}\mathrm{cos}100t+{10}^{-8}\mathrm{sin}100t-{10}^{-4}{e}^{-{10}^{6}t}\phantom{\rule{0ex}{0ex}}$

## Step 5: find the value of current.

Now, find the value of current.

$I=\frac{{E}_{R}}{R}\phantom{\rule{0ex}{0ex}}I=-{10}^{-4}\mathrm{cos}100t+\mathrm{sin}100t+{10}^{-4}{e}^{-{10}^{6}t}$

Therefore, the subsequent current is $I=-{10}^{-4}\mathrm{cos}100t+\mathrm{sin}100t+{10}^{-4}{e}^{-{10}^{6}t}$. ### Want to see more solutions like these? 