Question

Recall that the differential equation for the instantaneous charge q(t) on the capacitor in an LRC-series circuit is given by$\mathit{L}\frac{{\mathbf{d}}^{\mathbf{2}}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\mathit{R}\frac{\mathbf{d}\mathbf{q}}{\mathbf{d}\mathbf{t}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{C}}\mathit{q}\mathbf{=}\mathit{E}\mathbf{\left(}\mathit{t}\mathbf{\right)}$,

See section 5.1. Use the Laplace transform to find q(t) when$\mathit{L}\mathbf{=}\mathbf{1}\mathit{h}\mathbf{,}\mathit{R}\mathbf{=}\mathbf{20}\mathit{\Omega }\mathbf{,}\mathit{C}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{005}\mathit{f}\mathbf{,}\mathit{E}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{150}\mathit{V}\mathbf{,}\mathit{t}\mathbf{>}\mathbf{0}\mathbf{,}\mathit{q}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{a}\mathit{n}\mathit{d}\mathit{i}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$. What is the current i(t)?

Short answer

The value of is obtained a

$Q\left(t\right)=\frac{3}{4}-\frac{3}{4}\left[e^{-10t}\cos 10t-e^{-10t}\sin 10t\right]-\frac{15}{10}{e}^{-10t}\sin 10t$

The current is obtained as $I\left(t\right)=15e^{-10t}\sin 10t$.

Step-by-step solution

Define Laplace Transform

Let f be a function defined for $t\ge 0$. Then the integral $L\left\{f\right(t\left)\right\}={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt$is said to be the Laplace transform of f, provided that the integral converges

Substitute the values

We have been given the values

$L=1h,R=20\Omega ,C=0.005f,E\left(t\right)=150V,t>0,q\left(0\right)=0,andi\left(0\right)=0$

The equation on replacing the given values will be

$\begin{array}{l}\frac{d^{2}q}{d{t}^2}+20\frac{dq}{dt}+200q=150\\ s^{2}Q\left(s\right)-sQ\left(0\right)-I\left(0\right)+20\left(sQ\right(s)-Q(0\left)\right)+200Q\left(s\right)=\frac{150}{s}\end{array}$

Evaluate and find Q(s)

We evaluate the initial values of the equation and further factorize Q(s) using partial fractions

$\begin{array}{c}Q\left(s\right)\left(s^2+20s+200\right)=\frac{150}{s}\\ Q\left(s\right)=\frac{150}{s\left(s^2+20s+200\right)}\\ \frac{A}{s}+\frac{Bs+C}{s^2+20s+200}=150\end{array}$

$\begin{array}{c}A{s}^2+20As+200A+B{s}^2+Cs=150\\ A+B=0\\ 20A+C=0\\ 200A=150\end{array}$

On replacing the values of A, B and C we get

$Q\left(S\right)=\frac{3}{4s}+\frac{-\frac{3}{4}s-15}{s^2+20s+200}$

Use the Laplace Transform

$\begin{array}{c}Q\left(s\right)=\frac{3}{4}{L}^{-1}\left\{\frac{1}{s}\right\}-\frac{3}{4}{L}^{-1}\left\{\frac{s}{s^2+20s+200}\right\}n-15L^{-1}\left\{\frac{1}{s^2+20s+200}\right\}\\ L^{-1}\left\{\frac{s}{s^2+20s+200}\right\}=\frac{s}{\left(s^2+20+100\right)+200-100}\\ \frac{(s+10)-10}{{(s+10)}^2+100}=L^{-1}\left\{-\frac{3}{4}\left[\frac{s+10}{{(s+10)}^2+100}-\frac{10}{{(s+10)}^2+100}\right]\right\}\end{array}$

The solution for the second term is obtained as

data-custom-editor="chemistry" $-\frac{3}{4}\left[e^{-10t}\cos 10t-e^{-10t}\sin 10t\right]$

On completing the square

$-15L^{-1}\left\{\frac{1}{s^2+20s+200}\right\}=-15L^{-1}\left\{\frac{1}{{(s+10)}^2+100}\right\}$

Further on multiplying and dividing by 10, we also get the solution for third term as

$\begin{array}{l}-\frac{15}{10}{L}^{-1}\left\{\frac{10}{{(s+10)}^2+100}\right\}\\ -\frac{15}{10}{e}^{-10t}\sin 10t\end{array}$

Obtain the function of time

$\begin{array}{rcl}Q\left(s\right)& =& \frac{3}{4}{L}^{-1}\left\{\frac{1}{s}\right\}-\frac{3}{4}{L}^{-1}\left\{\left[\frac{s+10}{{(s+10)}^2+100}-\frac{10}{{(s+10)}^2+100}\right]\right\}-\frac{15}{10}{L}^{-1}\left\{\frac{10}{{(s+10)}^2+100}\right\}\\ Q\left(t\right)& =& \frac{3}{4}-\frac{3}{4}\left[e^{-10t}\cos 10t-e^{-10t}\sin 10t\right]-\frac{15}{10}{e}^{-10t}\sin 10t\end{array}$

Obtain the current I(t)

When we differentiate the charge which is a function of t, we will get the current I as follow:

$\begin{array}{l}I\left(t\right)=-\frac{3}{4}\left[-10e^{-10t}\cos 10t-10e^{-10t}\sin 10t+10e^{-10t}\sin 10t\right.\left.-10e^{-10t}\cos 10t\right]\\ -\frac{15}{10}\left[-10e^{-10t}+10e^{-10t}\cos 10t\right]\end{array}$