Question

The current i(t) in an RC-series circuit can be determined from the integral equation.

$\mathbf{R}\mathbf{i}\mathbf{+}\frac{\mathbf{1}}{\mathbf{C}}\underset{\mathbf{0}}{\overset{\mathbf{t}}{\mathbf{\int }}}\mathbf{i}\left(\tau \right)\mathbf{d}\mathbf{\tau }\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$,

where E(t) is the impressed voltage. Determine when

$\mathit{R}\mathbf{=}\mathbf{10}\mathit{\Omega }\mathbf{,}\mathit{C}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathit{f}\mathbf{,}$and$\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{(}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{t}\mathbf{)}$

Short answer

The solution is

$i\left(t\right)=-9+2t+9e^{-\frac{t}{5}}$

Step-by-step solution

given information

$Ri+\frac{1}{C}\underset{0}{\overset{t}{\int }}i\left(\tau \right)d\tau =E\left(t\right)$

$R=10\Omega ,C=0.5f,$and$E\left(t\right)=2(t^2+t)$

In the above equation, replace the supplied values

$\begin{array}{rcl}10i+\frac{1}{0.5}\underset{0}{\overset{t}{\int }}i\left(\tau \right)d\tau & =& 2(t^2+t)\\ 10i+2\underset{0}{\overset{t}{\int }}i\left(\tau \right)d\tau & =& 2t^2+2t\end{array}$

Take laplace transform of both side

$\begin{array}{rcl}10L\left\{i\right\}+2L\left\{\underset{0}{\overset{t}{\int }}i\left(\tau \right)d\tau \right\}& =& L\left\{2t^2+2t\right\}\\ 10L\left\{i\right\}+2L\left\{\underset{0}{\overset{t}{\int }}i\left(\tau \right)d\tau \right\}& =& 2L\left\{t^2\right\}+2L\left\{t\right\}=\frac{4}{s^3}+\frac{2}{s^2}...............\left(1\right)\end{array}$

Since, $\left(L\left\{t^n\right\}=\frac{n!}{s^{n+1}}\right)$

Now we calculate $L\left\{\underset{0}{\overset{t}{\int }}i\left(\tau \right)d\tau \right\}$

$L\left\{\underset{0}{\overset{t}{\int }}i\left(\tau \right)d\tau \right\}=L\left\{i\left(t\right)*1\right\}$ $\left(\underset{0}{\overset{t}{\int }}y\left(\tau \right)d\tau =y\left(t\right)*1\right)$

$=L\left\{i\left(t\right)\right\}L\left\{1\right\} \left(L\left\{f*g\right\}=L\left\{f\right\}L\left\{g\right\}\right)$

$=L\left\{i\left(t\right)\right\}.\frac{1}{s}$

 use partial fraction technique

Substitute in (1), to get

$\begin{array}{rcl}10L\left\{i\right\}+2L\left\{i\right\}\frac{1}{s}& =& \frac{4}{s^3}+\frac{2}{s^2}\\ 5L\left\{i\right\}+L\left\{i\right\}\frac{1}{s}& =& \frac{2}{s^3}+\frac{1}{s^2}\\ \left(\frac{5s+1}{s}\right)L\left\{i\right\}& =& \frac{2+s}{s^3}\end{array}$

$L\left\{i\right\}=\frac{2+s}{s^2\left(5s+1\right)}...........\left(2\right)$

Decompose the fraction$\frac{2+s}{s^2\left(5s+1\right)}$

Using the partial fraction technique

$\frac{2+s}{s^2\left(5s+1\right)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{\left(5s+1\right)}............\left(3\right)$

$=\frac{As\left(5s+1\right)+B\left(5s+1\right)+c{s}^2}{s^2\left(5s+1\right)}$

Because the denominators of the fractions in the previous equation are the same, their numerators must be the same.

$$\begin{gathered} 2+s=5A{s}^2+As+5Bs+B+c{s}^2 \\ 2+s=\left(5B+A\right)s+B+\left(5A+C\right)s^2 \end{gathered}$$

comparing coefficients

comparing coefficients on both sides, we get

$\begin{array}{rcl}5B+A& =& 1\\ 5A+C& =& 0\\ B& =& 2\end{array}$

$$\begin{gathered} 10+A=1 \\ 5A+C \\ B=2 \end{gathered}$$

$$\begin{gathered} A=-9 \\ C=45 \\ B=2 \end{gathered}$$

Substituting A and B in the equation (3), we get

$\frac{2+s}{s^2\left(5s+1\right)}=-\frac{9}{s}+\frac{2}{s^2}+\frac{45}{\left(5s+1\right)}$

The equation (2) becomes,

$L\left\{i\right\}=-\frac{9}{s}+\frac{2}{s^2}+\frac{45}{\left(5s+1\right)}$

$=-9\frac{1}{s}+2\frac{1}{s^2}+\frac{45}{5\left(s+\frac{1}{5}\right)}$

Take inverse laplace transform of both side.

$$\begin{gathered} \begin{array}{rcl}i\left(t\right)& =& -9L^{-1}\left\{\frac{1}{s}\right\}+2L^{-1}\left\{\frac{1}{s^2}\right\}+9L^{-1}\left\{\frac{1}{\left(s+\frac{1}{5}\right)}\right\}\\ & =& -9+2t+9e^{-\frac{t}{5}}\\ & & \left(**\right)\left\{e^{at}=L^{-1}\left\{\frac{1}{s-a}\right\} \\ {t}^n=L^{-1}\left\{\frac{n!}{s^{n+1}}\right\} \\ 1=L^{-1}\left\{\frac{1}{s}\right\}\right.\\ i\left(t\right)& =& -9+2t+9e^{-\frac{t}{5}}\\ & & \end{array} \end{gathered}$$

conclusion

The final solution is

$i\left(t\right)=-9+2t+9e^{-\frac{t}{5}}$