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