Question

Consider a battery of constant voltage ${\mathit{E}}_{\mathbf{0}}$ that charges the capacitor shown in Figure 7.3.10. Divide equation (20) by L and define $\mathbf{2}\mathit{\lambda }\mathbf{=}\mathit{R}\mathbf{/}\mathit{L}$ and ${\mathit{\omega }}^{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{/}\mathit{L}\mathit{C}$.Use the Laplace transform to show that the solution $\mathit{q}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{\lambda }\mathit{q}\mathbf{\text{'}}\mathbf{+}{\mathit{\omega }}^{\mathbf{2}}\mathit{q}\mathbf{=}{\mathit{E}}_{\mathbf{0}}\mathbf{/}\mathit{L}$q(t) of subject to$\mathit{q}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{i}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ is

$q\left(t\right)=\left\{\begin{array}{l}{E}_{0}C\left[1-e^{-\lambda t}\left(\cosh \sqrt{{\lambda }^2-{\omega }^{2}t}\right.\right.\left.\left.+\frac{\lambda }{\sqrt{{\lambda }^2-{\omega }^2}}\sinh \sqrt{{\lambda }^2-{\omega }^2}t\right)\right]\lambda >\omega \\ E_{0}C\left[1-e^{-\lambda t}\left(\cos \sqrt{{\omega }^2-{\lambda }^2}t\right.\right.\left.\left.+\frac{\lambda }{\sqrt{{\omega }^2-{\lambda }^2}}\sin \sqrt{{\omega }^2-{\lambda }^2}t\right)\right]\lambda <\omega \end{array}\right.$

Short answer

The solution q(t) of$q\text{'}\text{'}+2\lambda q\text{'}+{\omega }^{2}q=\frac{E_0}{L}$ subject to $q\left(0\right)=0,i\left(0\right)=0$ is obtained as

role="math" localid="1664313108164" $q\left(t\right)=\left\{\begin{array}{c}{E}_{0}C\left(1-e^{-\lambda t}\cosh \sqrt{{\lambda }^2-{\omega }^2}t-\frac{\lambda e^{-\lambda t}\sinh \sqrt{{\lambda }^2-{\omega }^2}t}{\sqrt{{\lambda }^2-{\omega }^2}}\right),\lambda >\omega \\ E_{0}C\left[1-e^{-\lambda t}\cos \sqrt{{\omega }^2-{\lambda }^2}t-\frac{\lambda e^{-\lambda t}}{\sqrt{{\omega }^2-{\lambda }^2}}\sin \sqrt{{\omega }^2-{\lambda }^2}t\right],\lambda <\omega \\ E_{0}C\left(1-e^{-\lambda t}-\lambda t{e}^{-\lambda t}\right),\lambda =\omega \end{array}\right.$

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

We know that

$\left\{\begin{array}{c}\mathcal{L}\left\{q^{\text{'}\text{'}}\right\}=s^2\mathcal{L}\left\{q\right\}-sq\left(0\right)-q^{\text{'}}\left(0\right)\\ \mathcal{L}\left\{q^{\text{'}}\right\}=s\mathcal{L}\left\{q\right\}-q\left(0\right)\\ q\left(0\right)=0,q^{\text{'}}\left(0\right)=0\end{array}\right.$

On taking Laplace Transform, we get

$$\begin{gathered} \frac{E_0}{L}\mathcal{L}\left\{1\right\}=\mathcal{L}\left\{q^{\text{'}\text{'}}\right\}+2\lambda \mathcal{L}\left\{q^{\text{'}}\right\}+{\omega }^2\mathcal{L}\left\{q\right\} \\ \frac{E_0}{L}\frac{1^{\left(*\right)}}{s}=s^2\mathcal{L}\left\{q\right\}+2s\lambda \mathcal{L}\left\{q\right\}+{\omega }^2\mathcal{L}\left\{q\right\}\left(\mathcal{L}\left\{1\right\}=\frac{1}{s}\right) \\ \begin{array}{l}\frac{E_0}{Ls}=\left(s^2+2\lambda s+{\omega }^2\right)\mathcal{L}\left\{q\right\}\\ \mathcal{L}\left\{q\right\}=\frac{E_0}{L}\frac{1}{s\left(s^2+2\lambda s+{\omega }^2\right)}n-----\left(1\right)\end{array} \end{gathered}$$

Solve the equation using partial fraction technique

On decomposing the fractional part, we get

$\frac{1}{s\left(s^2+2\lambda s+{\omega }^2\right)}=\frac{A}{s}+\frac{Bs+C}{s^2+2\lambda s+{\omega }^2}=\frac{A\left(s^2+2\lambda s+{\omega }^2\right)+(Bs+C)s}{s\left(s^2+2\lambda s+{\omega }^2\right)}----\left(2\right)$

We can observe that the denominators are equal, so we can equate the numerators as well.

$\begin{array}{l}1=A{s}^2+2\lambda sA+{\omega }^{2}A+B{s}^2+Cs\\ 1=(A+B)s^2+(2\lambda A+C)s+{\omega }^{2}A\end{array}$

On comparing the coefficients, we get

$\begin{array}{l}2\lambda A+C=0\\ \underset{¯}{{\omega }^{2}A=1}\\ B=-\frac{1}{{\omega }^2}11\end{array}$

$$\begin{gathered} B=-\frac{1}{{\omega }^2} \\ C=-\frac{2\lambda }{{\omega }^2} \\ A=\frac{1}{{\omega }^2} \end{gathered}$$

Thus, on substituting the values of A, B and C we get

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

Evaluate the equation

Now, the equation (1) will become

$\begin{array}{l}\mathcal{L}\left\{q\right\}=\frac{E_0}{L{\omega }^2}\left[\frac{1}{s}-\frac{s+2\lambda }{\left(s^2+2\lambda s+{\omega }^2\right)}\right]\\ Weknow,{\omega }^2=\frac{1}{LC}\iff \frac{1}{L{\omega }^2}=C\\ \mathcal{L}\left\{q\right\}=E_{0}C\left[\frac{1}{s}-\frac{s+2\lambda }{\left(s^2+2\lambda s+{\omega }^2\right)}\right]\end{array}$

$\begin{array}{l}When,\lambda >\omega \\ s^2+2\lambda s+{\omega }^2=(s+\lambda {)}^2-\left({\lambda }^2-{\omega }^2\right)\\ \mathcal{L}\left\{q\right\}=E_{0}C\left[\frac{1}{s}-\frac{s+2\lambda }{{(s+\lambda )}^2-\left({\lambda }^2-{\omega }^2\right)}\right]\end{array}$

Take inverse Laplace transform on both sides

$\begin{array}{l}q=E_{0}C{\mathcal{L}}^{-1}\left\{\frac{1}{s}-\frac{s+2\lambda }{{(s+\lambda )}^2-\left({\lambda }^2-{\omega }^2\right)}\right\}\\ q=E_{0}C{\mathcal{L}}^{-1}\left\{\frac{1}{s}-\frac{s+\lambda }{{(s+\lambda )}^2-\left({\lambda }^2-{\omega }^2\right)}-\frac{\lambda }{\sqrt{{\lambda }^2-{\omega }^2}}\frac{\sqrt{{\lambda }^2-{\omega }^2}}{{(s+\lambda )}^2-\left({\lambda }^2-{\omega }^2\right)}\right\}\\ q=E_{0}C\left(1-e^{-\lambda t}\cosh \sqrt{{\lambda }^2-{\omega }^2}t-\frac{\lambda e^{-\lambda t}\sinh \sqrt{{\lambda }^2-{\omega }^2}t}{\sqrt{{\lambda }^2-{\omega }^2}}\right)\\ \\ \left(e^{at}\sinh kt\right.\left.={\mathcal{L}}^{-1}\left\{\frac{k}{{(s-a)}^2-k^2}\right\},e^{at}\cosh kt={\mathcal{L}}^{-1}\left\{\frac{s-a}{{(s-a)}^2-k^2}\right\}\right)\end{array}$

$\begin{array}{l}When,\lambda <\omega \\ s^2+2\lambda s+{\omega }^2=(s+\lambda {)}^2-\left({\omega }^2-{\lambda }^2\right)\\ \mathcal{L}\left\{q\right\}=E_{0}C\left[\frac{1}{s}-\frac{s+2\lambda }{{(s+\lambda )}^2-\left({\omega }^2-{\lambda }^2\right)}\right]\end{array}$

Take inverse Laplace transform on both sides

$\begin{array}{l}q=E_{0}C{\mathcal{L}}^{-1}\left\{\frac{1}{s}-\frac{s+2\lambda }{{(s+\lambda )}^2-\left({\lambda }^2-{\omega }^2\right)}\right\}\\ q=E_{0}C{\mathcal{L}}^{-1}\left\{\frac{1}{s}-\frac{s+\lambda }{{(s+\lambda )}^2-\left({\lambda }^2-{\omega }^2\right)}-\frac{\lambda }{\sqrt{{\lambda }^2-{\omega }^2}}\frac{\sqrt{{\lambda }^2-{\omega }^2}}{{(s+\lambda )}^2-\left({\lambda }^2-{\omega }^2\right)}\right\}\\ q=E_{0}C\left(1-e^{-\lambda t}\cosh \sqrt{{\lambda }^2-{\omega }^2}t-\frac{\lambda e^{-\lambda t}\sinh \sqrt{{\lambda }^2-{\omega }^2}t}{\sqrt{{\lambda }^2-{\omega }^2}}\right)\\ \\ \left(e^{at}\sinh kt\right.\left.={\mathcal{L}}^{-1}\left\{\frac{k}{{(s-a)}^2-k^2}\right\},e^{at}\cosh kt={\mathcal{L}}^{-1}\left\{\frac{s-a}{{(s-a)}^2-k^2}\right\}\right)\end{array}$

$\begin{array}{l}When,\lambda =\omega \\ s^2+2\lambda s+{\omega }^2=(s+\lambda {)}^2\\ \mathcal{L}\left\{q\right\}=E_{0}C\left[\frac{1}{s}-\frac{s+2\lambda }{{(s+\lambda )}^2}\right]\end{array}$

Take inverse Laplace transform on both sides

$\begin{array}{l}q=E_{0}C{\mathcal{L}}^{-1}\left\{\frac{1}{s}-\frac{s+2\lambda }{{(s+\lambda )}^2}\right\}\\ q=E_{0}C{\mathcal{L}}^{-1}\left\{\frac{1}{s}-\frac{1}{s+\lambda }-\frac{\lambda }{{(s+\lambda )}^2}\right\}\\ q=E_{0}C\left(1-e^{-\lambda t}-\lambda t{e}^{-\lambda t}\right)\end{array}$

Step 5:

Therefore, q(t) is obtained as

$q\left(t\right)=\left\{\begin{array}{c}{E}_{0}C\left(1-e^{-\lambda t}\cosh \sqrt{{\lambda }^2-{\omega }^2}t-\frac{\lambda e^{-\lambda t}\sinh \sqrt{{\lambda }^2-{\omega }^2}t}{\sqrt{{\lambda }^2-{\omega }^2}}\right),\lambda >\omega \\ E_{0}C\left[1-e^{-\lambda t}\cos \sqrt{{\omega }^2-{\lambda }^2}t-\frac{\lambda e^{-\lambda t}}{\sqrt{{\omega }^2-{\lambda }^2}}\sin \sqrt{{\omega }^2-{\lambda }^2}t\right],\lambda <\omega \\ E_{0}C\left(1-e^{-\lambda t}-\lambda t{e}^{-\lambda t}\right),\lambda =\omega \end{array}\right.$