Question

Solveequation(10)subjectto $\mathit{i}\left(0\right)\mathbf{=}\mathbf{0}$WithR, L, CandE (t)asgiven. Useagraphingutilitytographthesolutionfor$\mathbf{0}\mathbf{⩽}\mathit{t}\mathbf{⩽}\mathbf{3}$.

$$\begin{gathered} \mathbf{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{h}\mathbf{,}\mathbf{R}\mathbf{=}\mathbf{3}\Omega ,\mathbf{C}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{f}\mathbf{,} \\ \mathbf{E}\left(\mathbf{t}\right)\mathbf{=}\mathbf{100}\left[\mathbf{u}\left(\mathbf{t}\mathbf{-}\mathbf{1}\right)\mathbf{-}\mathbf{u}\left(\mathbf{t}\mathbf{-}\mathbf{2}\right)\right] \end{gathered}$$

Short answer

The solution of equation is

$i\left(t\right)=100\left(e^{-10\left(t-1\right)}-e^{-20\left(t-1\right)}\right)u\left(t-2\right)-100\left(e^{-10\left(t-2\right)}-e^{-20\left(t-2\right)}\right)u\left(t-2\right)$

Step-by-step solution

Step 1:

Accordingtoconvolution 7.4.2theorem:

If $f\left(t\right)$and$g\left(t\right)$ are piecewise continuous on$[0,\infty )$ and of exponential order, then

$L\left\{f.g\right\}=L\left\{f\left(t\right)\right\}.L\left\{g\left(t\right)\right\}=F\left(s\right).G\left(s\right)$

Weusethefollowingformula:

$L^{-1}\left\{\frac{1}{s+a}\right\}=e^{-at}$

Step 2:

Given:

$$\begin{gathered} L=0.1h,R=3\Omega ,C=0.05f, \\ E\left(t\right)=100\left[u\left(t-1\right)-u\left(t-2\right)\right] \end{gathered}$$

From equation (10) we can write.

$L\frac{di}{dt}+Ri\left(t\right)+\frac{1}{C}\left\{{\int }_0^{t}i\left(T\right)dT\right\}=100\left[u\left(t-1\right)-u\left(t-2\right)\right]$

Substitutingthevalues, we get

$0.1\frac{di}{dt}+3i\left(t\right)+\frac{1}{0.05}\left\{{\int }_0^{t}i\left(T\right)dT\right\}=100\left[u\left(t-1\right)-u\left(t-2\right)\right]$

$\frac{di}{dt}+30i\left(t\right)+200\left\{{\int }_0^{t}i\left(T\right)dT\right\}=1000\left[u\left(t-1\right)-u\left(t-2\right)\right]$

We take Laplace transform of each term and by using theorem 7.4.2 we can write.

$$\begin{gathered} L\left\{\frac{di}{dt}\right\}+30L\left\{i\left(t\right)\right\}+200L\left\{{\int }_0^{t}i\left(T\right)dT\right\}=L\left\{1000\left[u\left(t-1\right)-u\left(t-2\right)\right]\right\} \\ sI\left(s\right)+30I\left(s\right)+\frac{200}{s}I\left(s\right)=\frac{1000}{s}\left(e^{-s}-e^{-2s}\right) \end{gathered}$$

Solving for $I\left(s\right)$:

$I\left(s\right)=\frac{1000e^{-s}-1000e^{-2s}}{s^2+30s+200}$

Now we simplify it by using partial fraction

$$\begin{gathered} I\left\{s\right\}=\frac{1}{s^2+30s+200}=\frac{A}{s+10}+\frac{B}{s+20} \\ 1=A\left(s+20\right)+B\left(s+10\right) \end{gathered}$$

We put $s=-20,ands=-10$alternatively and comparing constants on both sides, we get

$$\begin{gathered} A=\frac{1}{10},B=-\frac{1}{10} \\ I\left\{s\right\}=\frac{1}{s^2+30s+200}=\frac{1/10}{s+10}+\frac{{\displaystyle -1/10}}{{\displaystyle s+20}} \end{gathered}$$

Step 3:

Nowwetakeinverse Laplacetransformof $I\left\{s\right\}$to get the required answer:

$$\begin{gathered} I\left\{s\right\}=L^{-1}\left\{\left(\frac{1000}{s+10}-\frac{1000}{s+20}\right)\left(e^{-s}-e^{-2s}\right)\right\}=\frac{1}{10}{L}^{-1}\left\{\frac{1}{s+10}\right\}-\frac{1}{10}{L}^{-1}\left\{\frac{1}{s+20}\right\} \\ I\left\{s\right\}=\left(\frac{100}{s+10}-\frac{100}{s+20}\right)\left(e^{-s}-e^{-2s}\right) \\ i\left(t\right)=100\left(e^{-10\left(t-1\right)}-e^{-20\left(t-1\right)}\right)u\left(t-2\right)-100\left(e^{-10\left(t-2\right)}-e^{-20\left(t-2\right)}\right)u\left(t-2\right) \end{gathered}$$