Question

(a) Use the Laplace transform to find the chargeon the capacitor in anseries circuit when$q\left(0\right)=0,R=50\Omega ,C=0.01f$and$E\left(t\right)$is given in Figure$7.3.23$.

(b) Assume that $E_0=100V$. Use a graphing utility to graphfor $0⩽t⩽6$. Use the graph to estimate $q_{max}$, the maximum value of the charge.

Figure 7.3.23in Problem 76.

Short answer

(a) The charge on the capacitor in an series circuit is

$\frac{E_0}{100}U(t-1)-\frac{E_0}{100}{e}^{-2(t-1)}U(t-1)-\frac{E_0}{100}U(t-3)+\frac{E_0}{100}{e}^{-2(t-3)}U(t-3)$

(b) The maximum value of charge $q\left(t\right)$ is approximately 1 at $t=3$. The graph for the charge$q\left(t\right)$is

Step-by-step solution

Define Laplace transform.

When specific initial conditions are supplied, especially when the initial values are zero, the Laplace transform is a handy method of solving certain types of differential equations. Laplace transformof a functionis defined as

$\mathcal{L}\left\{f\right(t\left)\right\}=\backslash {\mathrm{smallint}}_0^{\mathrm{\mathbb{N}}}{e}^{-st}f\left(t\right)dt$

In words we can describe this expression as the Laplace transform ofequals functionof.

$\mathcal{L}\left\{f\right(t\left)\right\}=F\left(s\right)$

Find the equation

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

$$\begin{gathered} E\left(t\right)=\left\{0,0⩽t<1\right\} \\ E\left(t\right)=\left\{E_0,1⩽t<3\right\} \\ E\left(t\right)=\left\{0,t⩾3\right\} \end{gathered}$$

The differential equation corresponding to series circuit is

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

Substitute $R=50,\text{C = 0.01 f}$ in the above equation.

Hence, the equation for$E\left(t\right)$is$E\left(t\right)=50\frac{\text{d}q}{\text{d}t}+100q$

Find the Laplace transform for the given equation.

Take Laplace transform on both sides of the equation $E\left(t\right)=50\frac{\text{d}q}{\text{d}t}+100q$

$$\begin{gathered} \mathcal{L}\left\{E\right(t\left)\right\}=50\mathcal{L}\left\{\frac{\text{d}q}{\text{d}t}\right\}+100\mathcal{L}\left\{q\right\} \\ =50s\mathcal{L}\left\{q\right\}-50q\left(0\right)+100\mathcal{L}\left\{q\right\} \\ \left(\mathcal{L}\left\{\frac{\text{d}q}{\text{d}t}\right\}=s\mathcal{L}\left\{q\right\}-q\left(0\right),q\left(0\right)=0\right)\left(1\right) \\ =50s\mathcal{L}\left\{q\right\}+100\mathcal{L}\left\{q\right\} \end{gathered}$$

Hence the Laplace transform of $E\left(t\right)=50\frac{\text{d}q}{\text{d}t}+100q\text{is}50s\mathcal{L}\left\{q\right\}+100\mathcal{L}\left\{q\right\}$

Find the unit step function.

$E\left(t\right)$ is expressed in the terms of unit step function.

$$\begin{gathered} f\left(t\right)=\left\{0,0⩽t<a\right\} \\ f\left(t\right)=\left\{g\left(t\right),a⩽t<b\right\} \\ f\left(t\right)=\left\{0,t⩾b\right\} \end{gathered}$$

It can also be expressed as

$f\left(t\right)=g\left(t\right)\left[U\right(t-a)-U(t-b\left)\right]$

Substituteandin the above equation.

$E\left(t\right)=E_0\left[U\right(t-1)-U(t-3\left)\right]$

Hence the unit step function of $E\left(t\right)$is $E\left(t\right)=E_0\left[U\right(t-1)-U(t-3\left)\right]$.

Find the Laplace transform for the given equation.

Take Laplace transform on both sides of the equation$E\left(t\right)=E_0\left[U\right(t-1)-U(t-3\left)\right]$

$$\begin{gathered} \mathcal{L}\left\{E\right(t\left)\right\}=\mathcal{L}\left\{E_0\left[U\right(t-1)-U(t-3\left)\right]\right\} \\ =E_0\frac{e^{-s}}{s}-E_0\frac{e^{-3s}}{s}\left(\mathcal{L}\left\{U\right(t-a\left)\right\}=\frac{e^{-as}}{s}\right) \end{gathered}$$

Substitute$\mathcal{L}\left\{E\right(t\left)\right\}=E_0\frac{e^{-s}}{s}-E_0\frac{e^{-3s}}{s}$ in equation.

$$\begin{gathered} E_0\frac{e^{-s}}{s}-E_0\frac{e^{-3s}}{s}=50s\mathcal{L}\left\{q\right\}+100\mathcal{L}\left\{q\right\} \\ \frac{E_0{e}^{-s}-E_0{e}^{-3s}}{s}=(50s+100)\mathcal{L}\left\{q\right\} \\ \mathcal{L}\left\{q\right\}=\frac{E_0{e}^{-s}-E_0{e}^{-3s}}{50s(s+2)}\text{(2)} \end{gathered}$$

Hence the Laplace transform for$E\left(t\right)=E\left(t\right)=E_0\left[U\right(t-1)-U(t-3\left)\right]is\mathcal{L}\left\{q\right\}=\frac{E_0{e}^{-s}-E_0{e}^{-3s}}{50s(s+2)}$ is .

Find the decomposed value of the given fraction.

Decompose the fraction $\frac{1}{s(s+2)}$ using partial fraction technique.

$$\begin{gathered} \frac{1}{s(s+2)}=\frac{A}{s}+\frac{B}{s+2} \\ =\frac{A(s+2)+Bs}{s(s+2)}\text{(3)} \end{gathered}$$

Since the fractions have same denominator, it takes their numerators are equal.

$$\begin{gathered} 1=As+2A+Bs \\ 1=(A+B)s+2A \end{gathered}$$

Comparing the coefficients.

$$\begin{gathered} A+B=0 \\ \underset{¯}{2A}=1 \\ B=-A \\ A=\frac{1}{2} \\ B=-\frac{1}{2} \\ A=\frac{1}{2} \end{gathered}$$

Substitute A, B in equation 3 .

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

The decomposed value is$\frac{1}{2s}-\frac{1}{2(s+2)}$ .

Find the inverse Laplace value.

Substitute $\frac{1}{s(s+2)}=\frac{1}{2s}-\frac{1}{2(s+2)}$in equation 2 .

$\mathcal{L}\left\{q\right\}=\frac{E_0{e}^{-s}-E_0{e}^{-3s}}{50}\left[\frac{1}{2s}-\frac{1}{2(s+2)}\right]$

Take inverse Laplace on both sides.

$$\begin{gathered} U(t-a)={\mathcal{L}}^{-1}\left\{\frac{e^{-as}}{s}\right\} \\ {e}^{at}={\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\} \end{gathered}$$

$$\begin{gathered} q=\frac{E_0}{100}{\mathcal{L}}^{-1}\left\{\frac{e^{-s}}{s}\right\}-\frac{E_0}{100}{\mathcal{L}}^{-1}\left\{\frac{e^{-s}}{s+2}\right\}-\frac{E_0}{100}{\mathcal{L}}^{-1}\left\{\frac{e^{-3s}}{s}\right\}+\frac{E_0}{100}{\mathcal{L}}^{-1}\left\{\frac{e^{-3s}}{s+2}\right\} \\ =\frac{E_0}{100}U(t-1)-\frac{E_0}{100}{e}^{-2(t-1)}U(t-1)-\frac{E_0}{100}U(t-3)+\frac{E_0}{100}{e}^{-2(t-3)}U(t-3) \end{gathered}$$

So, the value of charge,

$q\left(t\right)=\frac{E_0}{100}U(t-1)-\frac{E_0}{100}{e}^{-2(t-1)}U(t-1)-\frac{E_0}{100}U(t-3)+\frac{E_0}{100}{e}^{-2(t-3)}U(t-3)$

Graphing the minimum and maximum value of the current.

Using graphing calculator graph the function.

The graph shows that the maximum charge q(t) is at$t=3$ .