Question

In Problems 73 and 74 use the Laplace transform to find the charge $q\left(t\right)$on the capacitor in anseries circuit subject to the given conditions. $q\left(0\right)=q_0,R=10\Omega ,C=0.1f,E\left(t\right)$given in Figure 7.3.21.

Short answer

The charge of the Laplace transform equation is

$\left(q_0-\frac{3}{2}\right)e^{-t}+\frac{3}{2}{e}^t+\frac{6e^{1.5}}{5}\left(e^{-(t-1.5)}-e^{1.5(t-1.5)}\right)U(t-1.5)$

Step-by-step solution

Definition of Laplace Transform

The integral transform of a real-valued derivative function into a complex function with variable is known as the Laplace transform.

Determine the Laplace transform equation

$E\left(t\right)$is given by

$E\left(t\right)=\left\{\begin{array}{ll}30e^t,& 0⩽t<1.5\\ 0,& t⩾1.5\end{array}\right.$

The differential equation corresponding to RC series Circuit is

$E\left(t\right)=R\frac{\text{d}q}{\text{d}t}+\frac{1}{c}q$

Given localid="1663939011275" $R=10\Omega ,C=0.1f$so the equation becomes.

$$\begin{gathered} E\left(t\right)=10\frac{\text{d}q}{\text{d}t}+\frac{1}{0.1}q \\ E\left(t\right)=10\frac{{\displaystyle \text{d}q}}{{\displaystyle \text{d}t}}+10q \end{gathered}$$

Take Laplace transform of both side of the above equation.

$$\begin{gathered} \mathcal{L}\left\{E\right(t\left)\right\}=10\mathcal{L}\left\{\frac{\text{d}q}{\text{d}t}\right\}+10\mathcal{L}\left\{q\right\} \\ =10s\mathcal{L}\left\{q\right\}-10q\left(0\right)+10\mathcal{L}\left\{q\right\}\left(\mathcal{L}\left\{\frac{\text{d}q}{\text{d}t}\right\}=s\mathcal{L}\left\{q\right\}-q\left(0\right),q\left(0\right)=q_0\right) \\ =10s\mathcal{L}\left\{q\right\}-10q_0+10\mathcal{L}\left\{q\right\}\left(1\right) \end{gathered}$$

we express E(t) in terms of unit step functions.

We know that the piecewise function

$f\left(t\right)=\left\{\begin{array}{ll}g\left(t\right),& 0⩽t<a\\ h\left(t\right)& t⩾a\end{array}\right.$

is the same as

$f\left(t\right)=g\left(t\right)-g\left(t\right)U(t-a)+h\left(t\right)U(t-a)$

Use the above formula and replace the values$a=1.5,g\left(t\right)=30e^{t}and\$h\left(t\right)=0,toobtain$

$E\left(t\right)=30e^t-30e^{t}U(t-1.5)$

Take Laplace transform of both side of the above equation.

$$\begin{gathered} \mathcal{L}\left\{E\right(t\left)\right\}=\mathcal{L}\left\{30e^t-30e^{t}U(t-1.5)\right\} \\ =\mathcal{L}\left\{30e^t\right\}-\mathcal{L}\left\{30e^{1.5}{e}^{t-1.5}U(t-1.5)\right\} \\ =30\mathcal{L}\left\{e^t\right\}-30e^{1.5}\mathcal{L}\left\{e^{t-1.5}U(t-1.5)\right\} \\ =\frac{30}{s-1}-\frac{30e^{1.5}{e}^{-.5s}}{s-1.5} \end{gathered}$$

Substitute the known values

Substitute $\mathcal{L}\left\{E\right(t\left)\right\}=\frac{30}{s-1}-\frac{30e^{1.5}{e}^{-1.5s}}{s-1.5}$in , to obtain.

$$\begin{gathered} \frac{30}{s-1}-\frac{30e^{1.5}{e}^{-1.5s}}{s-1.5}=10s\mathcal{L}\left\{q\right\}-10q_0+10\mathcal{L}\left\{q\right\} \\ \frac{3}{s-1}-\frac{3e^{1.5}{e}^{-1.5s}}{s-1.5}=s\mathcal{L}\left\{q\right\}-q_0+\mathcal{L}\left\{q\right\} \\ (s+1)\mathcal{L}\left\{q\right\}=\frac{3}{s-1}-\frac{3e^{1.5}{e}^{-1.5s}}{s-1.5}+q_0 \\ \mathcal{L}\left\{q\right\}=\frac{3}{(s-1)(s+1)}-\frac{3e^{1.5}{e}^{1.5s}}{(s-1.5)(s+1)}+\frac{q_0}{s+1}\left(2\right) \end{gathered}$$

Decompose the fraction$\frac{3}{(s-1)(s+1)}$using the partial fraction technique.

$$\begin{gathered} \frac{3}{(s-1)(s+1)}=\frac{A}{s-1}+\frac{B}{s+1}\left(3\right) \\ =\frac{A(s+1)+B(s-1)}{(s-1)(s+1)} \end{gathered}$$

Since the fractions in above equation have the same denominators, it takes their numerators to be equal.

role="math" $3=As+A+Bs-B$

If $s=1$then,

$$\begin{gathered} 3=A+A+\cancel{B-\cancel{bA}} \\ =\frac{3}{2} \end{gathered}$$

If S=1, then

$$\begin{gathered} 3=-A+A-B-B \\ B=-\frac{3}{2} \end{gathered}$$

$SubstitutingAandBintheequation\left(3\right),weget.$

$\frac{3}{(s-1)(s+1)}=\frac{3}{2(s-1)}-\frac{3}{2(s+1)}\left(4\right)$

Decomposition of the equation

Decompose the fraction $\frac{1}{(s-1.5)(s+1)}$using the partial fraction technique.

$$\begin{gathered} \frac{1}{(s-1.5)(s+1)}=\frac{A}{s-1.5}+\frac{B}{s+1}\left(5\right) \\ =\frac{A(s+1)+B(s-1.5)}{(s-1.5)(s+1)} \end{gathered}$$

Since the fractions in above equation have the same denominators, it takes their numerators to be equal.

$$\begin{gathered} 1=As+A+Bs-1.5B \\ Ifs=1.5,then \end{gathered}$$

$$\begin{gathered} 1=1.5A+A+1.5\cancel{B-1.5\cancel{B}} \\ A=\frac{2}{5} \end{gathered}$$

If $s=-1$, then

data-custom-editor="chemistry" $$\begin{gathered} 1=-A+A-B-1.5B \\ B=-\frac{2}{5} \end{gathered}$$

Substituting A and B in the equation (5), we get.

$\frac{1}{(s-1.5)(s+1)}=\frac{2}{5(s-1.5)}-\frac{2}{5(s+1)}$

Based on (4) and (6) the equation (2) becomes.

$\mathcal{L}\left\{q\right\}=\frac{3}{2(s-1)}-\frac{3}{2(s+1)}-3e^{1.5}{e}^{1.5s}\left[\frac{2}{5(s-1.5)}-\frac{2}{5(s+1)}\right]+\frac{q_0}{s+1}$

Take inverse Laplace transform of both side.

$$\begin{gathered} q=\frac{3}{2}{\mathcal{L}}^1\left\{\frac{1}{s-1}\right\}-\frac{3}{2}{\mathcal{L}}^{-1}\left\{\frac{1}{s-(-1)}\right\}-\frac{6e^{1.5}}{5}{\mathcal{L}}^{-1}\left\{\frac{e^{-1.5s}}{s-1.5}\right\}+\frac{6e^{1.5}}{5}{\mathcal{L}}^{-1}\left\{\frac{e^{-1.5s}}{s-(-1)}\right\}+{\mathcal{L}}^{-1}\left\{\frac{q_0}{s-(-1)}\right\} \\ =\left(q_0-\frac{3}{2}\right)e^{-t}+\frac{3}{2}{e}^t+\frac{6e^{1.5}}{5}\left(e^{-(t-1.5)}-e^{1.5(t-1.5)}\right)U(t-1.5) \end{gathered}$$

Hence, the charge of the Laplace transform equation is

$\left(q_0-\frac{3}{2}\right)e^{-t}+\frac{3}{2}{e}^t+\frac{6e^{1.5}}{5}\left(e^{-(t-1.5)}-e^{1.5(t-1.5)}\right)U(t-1.5)$