Question

Use the Laplace Transform to find the charge q(t) in an RC series circuit when$\mathit{q}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ and$\mathit{E}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{E}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{-}\mathbf{kt}}\mathbf{,}\mathit{k}\mathbf{>}\mathbf{0}$

Consider two cases:$\mathit{k}\mathbf{\ne }\mathbf{1}\mathbf{/}\mathit{R}\mathit{C}$ and$\mathit{k}\mathbf{\ne }\mathbf{1}\mathbf{/}\mathit{R}\mathit{C}$.

Short answer

The charge q(t) in an RC series circuit is

$q\left(t\right)=\left\{\begin{array}{l}\frac{E_{0}C}{1-kRC}{e}^{-kt}+\frac{E_{0}C}{kRC-1}{e}^{-t/RC},\frac{1}{RC}\ne k\\ \frac{E_0}{R}t{e}^{-t/RC},\frac{1}{RC}=k\end{array}\right.$

Step-by-step solution

Define Laplace Transform

Let f be a function defined for$t\ge 0$. Then the integral

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt$

is said to be the Laplace transform of f, provided that the integral converges.

Apply Laplace Transform

The RC circuit of which the corresponding the differential equation is

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

We are given . $E\left(t\right)=E_0{e}^{-kt}$.,so the equation becomes

$E_0{e}^{-kt}=R\frac{\mathrm{d}q}{\mathrm{d}t}+\frac{1}{C}q$

Taking the Laplace transform,

$\begin{array}{l}\mathcal{L}\left\{E_0{e}^{-kt}\right\}=R\mathcal{L}\left\{\frac{\mathrm{d}q}{\mathrm{d}t}\right\}+\frac{1}{C}\mathcal{L}\left\{q\right\}\\ \mathcal{L}\left\{E_0{e}^{-kt}\right\}=Rs\mathcal{L}\left\{q\right\}-Rq\left(0\right)+\frac{1}{C}\mathcal{L}\left\{q\right\}\left(\mathcal{L}\left\{\frac{\mathrm{d}q}{\mathrm{d}t}\right\}=s\mathcal{L}\left\{q\right\}-q\left(0\right),q\left(0\right)=0\right)\\ \mathcal{L}\left\{E_0{e}^{-kt}\right\}=Rs\mathcal{L}\left\{q\right\}+\frac{1}{C}\mathcal{L}\left\{q\right\}\end{array}$

$\begin{array}{l}{E}_0\mathcal{L}\left\{e^{-kt}\right\}=Rs\mathcal{L}\left\{q\right\}+\frac{1}{C}\mathcal{L}\left\{q\right\}\left(\mathcal{L}\left\{e^{at}\right\}=\frac{1}{s-a}\right)\\ E_0\frac{1}{s+k}=\left(Rs+\frac{1}{C}\right)\mathcal{L}\left\{q\right\}\\ \mathcal{L}\left\{q\right\}=\frac{E_0}{(s+k)\left(Rs+\frac{1}{C}\right)}\\ \mathcal{L}\left\{q\right\}=\frac{E_0}{R(s+k)\left(s+\frac{1}{RC}\right)}----\left(1\right)\end{array}$

Consider the given conditions

$\frac{1}{RC}=k$

The fraction of the equation (1) is evaluated as

$\frac{1}{(s+k)\left(s+\frac{1}{RC}\right)}=\frac{A}{s+k}+\frac{B}{s+\frac{1}{RC}}=\frac{A\left(s+\frac{1}{RC}\right)+B(s+k)}{(s+k)\left(s+\frac{1}{RC}\right)}$

As the denominators are same, we equate the numerators,

$\begin{array}{l}1=A\left(s+\frac{1}{RC}\right)+B(s+k)\\ Ifs=-\frac{1}{RC},then\\ 1=B\left(\frac{kRC-1}{RC}\right)B=\frac{RC}{kRC-1}\\ Ifs=-k,then\\ 1=A\left(\frac{1-kRC}{RC}\right)\\ A=\frac{RC}{1-kRC}\end{array}$

Substituting the values of A, B and C we get

$\begin{array}{l}\frac{1}{(s+k)\left(s+\frac{1}{RC}\right)}=\frac{RC}{1-kRC}\frac{1}{s+k}+\frac{RC}{kRC-1}\frac{1}{s+\frac{1}{RC}}\\ \mathcal{L}\left\{q\right\}=\frac{E_0}{R}\left[\frac{RC}{1-kRC}\frac{1}{s+k}+\frac{RC}{kRC-1}\frac{1}{s+\frac{1}{RC}}\right]\\ \mathcal{L}\left\{q\right\}=\frac{E_{0}C}{1-kRC}\left[\frac{1}{s-(-k)}\right]+\frac{E_{0}C}{kRC-1}\left[\frac{1}{s-\left(-\frac{1}{RC}\right)}\right]\end{array}$

Taking the inverse Laplace transform,

$\begin{array}{l}q=\frac{E_{0}C}{1-kRC}{\mathcal{L}}^{-1}\left\{\frac{1}{s-(-k)}\right\}+\frac{E_{0}C}{kRC-1}{\mathcal{L}}^{-1}\left\{\frac{1}{s-\left(-\frac{1}{RC}\right)}\right\}\\ q=\frac{E_{0}C}{1-kRC}{e}^{-kt}+\frac{E_{0}C}{kRC-1}{e}^{-t/RC}\\ \\ \left(e^{at}={\mathcal{L}}^{-1}\left\{\frac{1}{s-a}\right\}\right)\end{array}$

Consider the second condition

$\frac{1}{RC}=k$

We have

$\begin{array}{l}\mathcal{L}\left\{q\right\}=\frac{E_0}{R{(s+k)}^2}\\ \mathcal{L}\left\{q\right\}=\frac{E_0}{R}\frac{1}{{(s+k)}^2}\end{array}$

Thus, taking inverse Laplace transform we get

$\begin{array}{l}q=\frac{E_0}{R}{\mathcal{L}}^{-1}\left\{{\left.\frac{1}{s^2}\right|}_{s\to s-(-k)}\right\}\\ q=\frac{E_0}{R}t{e}^{-kt}\\ q=\frac{E_0}{R}t{e}^{-t/RC}\end{array}$

Therefore, the q(t) is obtained as

$q\left(t\right)=\left\{\begin{array}{l}\frac{E_{0}C}{1-kRC}{e}^{-kt}+\frac{E_{0}C}{kRC-1}{e}^{-t/RC},\frac{1}{RC}\ne k\\ \frac{E_0}{R}t{e}^{-t/RC},\frac{1}{RC}=k\end{array}\right.$