Question

(a) Use the Laplace transform to find the currentin a single-loop series circuit when $\left(0\right)=0,L=1h,R=10\Omega$and is given in Figure $7.3.22$.

(b) Use a graphing utility to graph$i\left(t\right)\text{for}0⩽t⩽6$. Use the graph to estimate $i_{max}$and $i_{min}$, the maximum and minimum values for the current.

Figure $7.3.22$in Problem 75.

Short answer

(a) The current $i\left(t\right)$in a single-loop $LR$series circuit found using Laplace transform is.

$$\begin{gathered} i\left(t\right)=\frac{1}{101}\left[10\sin t-\cos t+e^{-10t}\right] \\ +\frac{1}{101}\left[\sin \left(t-\frac{3\pi }{2}\right)+10\cos \left(t-\frac{3\pi }{2}\right)-10e^{-10(t-3\pi /2)}\right]U\left(t-\frac{3\pi }{2}\right) \end{gathered}$$

(b) The graph to estimate , the maximum and minimum values for the current is

The maximum value of current $i\left(t\right)$is $\text{0.1at}t=1.7$and the minimum value is$-0.1$at$t=4.7$.

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 transform $\mathcal{L}$of a function is 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 of equals function F of S.

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

Find the piecewise function

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

role="math" localid="1663848831743" $$\begin{gathered} \mathrm{E}\left(\mathrm{t}\right)=\left\{\mathrm{sint},0\le \mathrm{t}<\frac{3\mathrm{\pi }}{2}\right\} \\ \mathrm{E}\left(\mathrm{t}\right)=\left\{0\le \mathrm{t}{}_3<\frac{{\displaystyle 3\mathrm{\pi }}}{{\displaystyle 2}}\right\} \end{gathered}$$

We know that the piecewise function is,

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

It 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 value $a=\frac{3\pi }{2},g\left(t\right)=\sin t$and to $h\left(t\right)=0$obtain,

$E\left(t\right)=\sin t-\sin tU\left(t-\frac{3\pi }{2}\right)\left(1\right)$

The differential equation representing the circuit is

$L\frac{\text{d}i}{\text{d}t}+Ri=E\left(t\right)$

Substitute $L=1,R=10$in the above equation.

role="math" localid="1663849206226" $$\begin{gathered} \frac{\text{d}i}{\text{d}t}+10i=E\left(t\right) \\ \frac{\text{d}i}{\text{d}t}+10i\stackrel{\left(1\right)}{=}\sin t-\sin tU\left(t-\frac{3\pi }{2}\right) \end{gathered}$$

Hence, the piecewise function is$\frac{\text{d}i}{\text{d}t}+10i\stackrel{\left(1\right)}{=}\sin t-\sin tU\left(t-\frac{3\pi }{2}\right)$

Find the Laplace transform for the given equation.

Take the Laplace transform on both sides of the equation $\frac{\text{d}i}{\text{d}t}+10i\stackrel{\left(1\right)}{=}\sin t-\sin tU\left(t-\frac{3\pi }{2}\right)$

$$\begin{gathered} \mathcal{L}\left\{\frac{\text{d}i}{\text{d}t}\right\}+10\mathcal{L}\left\{i\right\}=\mathcal{L}\left\{\sin t-\sin tU\left(t-\frac{3\pi }{2}\right)\right\}\left(\mathcal{L}\left\{\frac{\text{d}i}{\text{d}t}\right\}=s\mathcal{L}\left\{i\right\}-i\left(0\right),i\left(0\right)=0\right) \\ s\mathcal{L}\left\{i\right\}+10\mathcal{L}\left\{i\right\}=\mathcal{L}\left\{\sin t\right\}-\mathcal{L}\left\{\sin tU\left(t-\frac{3\pi }{2}\right)\right\} \\ s\mathcal{L}\left\{i\right\}+10\mathcal{L}\left\{i\right\}=\mathcal{L}\left\{\sin t\right\}+\mathcal{L}\left\{\cos \left(t-\frac{3\pi }{2}\right)U\left(t-\frac{3\pi }{2}\right)\right\} \end{gathered}$$

role="math" localid="1663849566741" $$\begin{gathered} \frac{s}{s^2+k^2}=\mathcal{L}\left\{\cos kt\right\},\frac{k}{s^2+k^2}=\mathcal{L}\left\{\sin kt\right\} \\ s\mathcal{L}\left\{i\right\}+10\mathcal{L}\left\{i\right\}\stackrel{\left(**\right)}{=}\frac{1}{s^2+1}+\frac{s{e}^{-3\pi s/2}}{s^2+1} \end{gathered}$$

Replace $\mathcal{L}\left\{i\right\}=I\left(s\right)$in the above equation

$$\begin{gathered} sI\left(s\right)+10I\left(s\right)=\frac{1}{s^2+1}+\frac{s{e}^{-3\pi s/2}}{s^2+1} \\ (s+10)I\left(s\right)=\frac{1}{s^2+1}+\frac{s{e}^{-3\pi s/2}}{s^2+1} \\ \mathrm{I}\left(s\right)=\frac{1}{\left(s^2+1\right)(s+10)}+\frac{s{e}^{-3\pi s/2}}{\left(s^2+1\right)(s+10)}\left(2\right) \end{gathered}$$

Hence, the Laplace transform is$I\left(s\right)=\frac{1}{\left(s^2+1\right)(s+10)}+\frac{s{e}^{-3\pi s/2}}{\left(s^2+1\right)(s+10)}$

Find the decomposed value of the given expression.

Decompose $\frac{1}{\left(s^2+1\right)(s+10)}$using partial fraction technique.

$$\begin{gathered} \frac{1}{\left(s^2+1\right)(s+10)}=\frac{As+B}{s^2+1}+\frac{C}{s+10}\text{(3)} \\ =\frac{(As+B)(s+10)+C\left(s^2+1\right)}{\left(s^2+1\right)(s+10)} \end{gathered}$$

Since the denominators are equal in the above equation, it takes their numerators to be equal.

$$\begin{gathered} 1=A{s}^2+10As+Bs+10B+C{s}^2+C \\ 1=(A+C)s^2+(10A+B)s+10B+C \end{gathered}$$

Comparing the coefficients we get,

$$\begin{gathered} \text{A + C = 0} \\ \text{10 A + B = 0} \\ \text{A = - C} \\ \text{B = - 10 A} \\ \text{100 C + C = 1} \\ \text{A = - (1)/(101)} \\ \text{B = (10)/(101)} \\ \text{C = (1)/(101)} \end{gathered}$$

Substituting in equation

$\frac{1}{\left(s^2+1\right)(s+10)}=\frac{10-s}{101\left(s^2+1\right)}+\frac{1}{101(s+10)}\text{(4)}$

Hence the decomposed value of is $\frac{1}{\left(s^2+1\right)(s+10)}$is.

$\frac{1}{\left(s^2+1\right)(s+10)}=\frac{10-s}{101\left(s^2+1\right)}+\frac{1}{101(s+10)}$

Find the decomposed value of the given expression.

Decompose $\frac{s{e}^{-3\pi s/2}}{\left(s^2+1\right)(s+10)}$by partial fraction technique.

$$\begin{gathered} \frac{s}{\left(s^2+1\right)(s+10)}=\frac{As+B}{s^2+1}+\frac{C}{s+10} \\ =\frac{(As+B)(s+10)+C\left(s^2+1\right)}{\left(s^2+1\right)(s+10)}\text{(5)} \end{gathered}$$

Since the denominators are equal in the above equation, it takes their numerators to be equal.

$$\begin{gathered} s=A{s}^2+10As+Bs+10B+C{s}^2+C \\ s=(A+C)s^2+(10A+B)s+10B+C \end{gathered}$$

Comparing the coefficients we get,

$$\begin{gathered} \backslash \mathrm{begingathered}\text{A + C = 0} \\ \text{10 A + B = 1} \\ \text{(10 B + C)/() = 0} \\ \text{A = - C} \\ \text{10 A + B = 1} \\ \text{B = - (C)/( 10)} \\ \text{A = - C} \\ \text{- 10 C - (C)/( 10) = 1} \\ \text{B = - (C)/( 10)} \\ \text{A = (10)/(101)} \\ \text{C = - (10)/(101)} \\ \text{B = (1)/(101)} \end{gathered}$$

Substituting in A,B,C equation

$\frac{1}{\left(s^2+1\right)(s+10)}=\frac{1+10s}{101\left(s^2+1\right)}-\frac{10}{101(s+10)}\text{(6)}$

Hence the decomposed value of $\frac{s{e}^{-3\pi s/2}}{\left(s^2+1\right)(s+10)}$is

$\frac{1}{\left(s^2+1\right)(s+10)}=\frac{1+10s}{101\left(s^2+1\right)}-\frac{10}{101(s+10)}$

Find the inverse Laplace of the given expression.

Substitute 4 and 6 in equation 2.

$$\begin{gathered} I\left(s\right)=\frac{10-s}{101\left(s^2+1\right)}+\frac{1}{101(s+10)}+e^{-3\pi s/2}\left[\frac{1+10s}{101\left(s^2+1\right)}-\frac{10}{101(s+10)}\right] \\ I\left(s\right)=\frac{1}{101}\left[10\frac{1}{s^2+1}-\frac{s}{s^2+1}+\frac{1}{s+10}\right]+\frac{1}{101}{e}^{-3\pi s/2}\left[\frac{1}{s^2+1}+10\frac{s}{s^2+1}-10\frac{1}{s+10}\right] \end{gathered}$$

Take inverse Laplace on the above equation.

$$\begin{gathered} {\mathcal{L}}^{-1}\left\{I\right(s\left)\right\}=\frac{1}{101}\left[10{\mathcal{L}}^{-1}\left\{\frac{1}{s^2+1}\right\}-{\mathcal{L}}^{-1}\left\{\frac{s}{s^2+1}\right\}+{\mathcal{L}}^{-1}\left\{\frac{1}{s+10}\right\}\right] \\ +\frac{1}{101}\left[{\mathcal{L}}^{-1}\left\{\frac{e^{-3\pi s/2}}{s^2+1}\right\}+10{\mathcal{L}}^{-1}\left\{\frac{e^{-3\pi s/2}s}{s^2+1}\right\}-10{\mathcal{L}}^{-1}\left\{\frac{e^{-3\pi s/2}}{s+10}\right\}\right] \end{gathered}$$

$$\begin{gathered} e^{at}={\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\},{\mathcal{L}}^{-1}\left\{\frac{k}{s^2+k^2}\right\}=\sin kt \\ {\mathcal{L}}^{-1}\left\{\frac{s}{s^2+k^2}\right\}=\cos kt \end{gathered}$$

$$\begin{gathered} =\frac{1}{101}\left[10\sin t-\cos t+e^{-10t}\right] \\ +\frac{1}{101}\left[\sin \left(t-\frac{3\pi }{2}\right)+10\cos \left(t-\frac{3\pi }{2}\right)-10e^{-10(t-3\pi /2)}\right]U\left(t-\frac{3\pi }{2}\right) \end{gathered}$$

Find the value of the current.

$$\begin{gathered} i\left(t\right)={\mathcal{L}}^{-1}\left\{I\right(s\left)\right\} \\ =\frac{1}{101}\left[10\sin t-\cos t+e^{-10t}\right] \\ +\frac{1}{101}\left[\sin \left(t-\frac{3\pi }{2}\right)+10\cos \left(t-\frac{3\pi }{2}\right)-10e^{-10(t-3\pi /2)}\right]U\left(t-\frac{3\pi }{2}\right) \end{gathered}$$

The value of the current is

$$\begin{gathered} \frac{1}{101}\left[10\sin t-\cos t+e^{-10t}\right] \\ +\frac{1}{101}\left[\sin \left(t-\frac{3\pi }{2}\right)+10\cos \left(t-\frac{3\pi }{2}\right)-10e^{-10(t-3\pi /2)}\right]U\left(t-\frac{3\pi }{2}\right) \end{gathered}$$

Graphing the minimum and maximum value of the current.

Using graphing calculator graph the function

$$\begin{gathered} \frac{1}{101}\left[10\sin t-\cos t+e^{-10t}\right] \\ +\frac{1}{101}\left[\sin \left(t-\frac{3\pi }{2}\right)+10\cos \left(t-\frac{3\pi }{2}\right)-10e^{-10(t-3\pi /2)}\right]U\left(t-\frac{3\pi }{2}\right) \end{gathered}$$

From the graph we can see the maximum value of current is at $\text{0.1at = 1.7}$and the minimum value is$\text{-0.1at =4.7}$