Question

Use the Laplace transform to find the charge $q\left(t\right)$on the capacitor in an RC series circuit subject to the given conditions.

$q\left(0\right)=0,R=2.5\Omega ,C=0.08f,E\left(t\right)$given in figure.

Short answer

The charge of the Laplace transform equation is$q=\frac{2}{5}U(t-3)-\frac{2}{5}{e}^{(-5(}t-3\left)\right)U(t-3)$

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

$E\left(t\right)=\left\{\begin{array}{ll}0,& 0⩽t<3\\ 5,& t⩾3\end{array}\right.$

The differential equation corresponding to series Circuit is

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

Given $R=2.5\Omega ,C=0.08f$ so the equation becomes.

$E\left(t\right)=2.5\frac{\text{d}q}{\text{d}t}+\frac{1}{0.08}q$

Take Laplace transform o both side of the equation:

$$\begin{gathered} \mathcal{L}\left\{E\right(t\left)\right\}=2.5\mathcal{L}\left\{\frac{\text{d}q}{\text{d}t}\right\}+\frac{1}{0.08}\mathcal{L}\left\{q\right\} \\ =2.5s\mathcal{L}\left\{q\right\}-2.5q\left(0\right)+\frac{1}{0.08}\mathcal{L}\left\{q\right\}\left(\mathcal{L}\left\{\frac{\text{d}q}{\text{d}t}\right\}-s\mathcal{L}\left\{q\right\}-q\left(0\right),q\left(0\right)-0\right) \\ =2.5s\mathcal{L}\left\{q\right\}+\frac{1}{0.08}\mathcal{L}\left\{q\right\}\left(1\right) \end{gathered}$$

To find $\mathcal{L}\left\{f\right(t\left)\right\}$we express role="math" localid="1663945361375" $E\left(t\right)$ in terms of unit step functions.

A piecewise function defined as:

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

Can be expressed 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 $a-3,g\left(t\right)-0andh\left(t\right)-5$to obtain

$E\left(t\right)=5U(t-3)$

Take Laplace transform of both side of the above equation

$$\begin{gathered} \mathcal{L}\left\{E\right(t\left)\right\}=\mathcal{L}\left\{5U\right(t-3\left)\right\} \\ =\frac{5e^{-ss}}{s}\left(\mathcal{L}\left\{U\right(t-a\left)\right\}-\frac{e^{-as}}{s}\right) \end{gathered}$$

Substitute the known values.

Substitute $\mathcal{L}\left\{E\right(t\left)\right\}-\frac{5e^{-3s}}{s}$in (1), to obtain.

$$\begin{gathered} \frac{5e^{-is}}{s}=2.5s\mathcal{L}\left\{q\right\}+\frac{1}{0.08}\mathcal{L}\left\{q\right\} \\ \frac{2e^{-is}}{s}=(s+5)\mathcal{L}\left\{q\right\} \\ \mathcal{L}\left\{q\right\}=\frac{2e^{-is}}{s(s+5)}\left(2\right) \end{gathered}$$

Decompose the fraction $\frac{2}{s(s+5)}$using the partial fraction technique.

$$\begin{gathered} \frac{2}{s(s+5)}=\frac{A}{s}+\frac{B}{s+5}\left(3\right) \\ =\frac{A(s+5)+Bs}{s(s+5)} \end{gathered}$$

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

$$\begin{gathered} 2=As+5A+Bs \\ 2=(A+B)s+5A \end{gathered}$$

Compared to the coefficient of both sides, we get

$$\begin{gathered} A+B=0 \\ 5A=2 \\ B=-A \\ A=\frac{2}{5} \\ B=-\frac{2}{5} \\ A=\frac{2}{5} \end{gathered}$$

Find the charge equation for the Laplace transform.

Substituting A and B in the equation (2), we get.

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

The equation (3) becomes

$$\begin{gathered} q=\frac{2}{5}{\mathcal{L}}^{-1}\left\{\frac{e^{-8s}}{s}\right\}-\frac{2}{5}{\mathcal{L}}^{-1}\left\{\frac{e^{-8s}}{s+5}\right\} \\ =\frac{2}{5}U(t-3)-\frac{2}{5}{e}^{-5(t-3)}U(t-3) \end{gathered}$$

Hence, the charge of the Laplace transform equation is$\frac{2}{5}U(t-3)-\frac{2}{5}{e}^{-5(t-3)}U(t-3)$ .