Question

L1di2dt+Ri2+Ri3=E(t)L2di3dt+Ri2+Ri3=E(t)A

(a) Show that the system of differential equations for the currentsand in the electrical network shown in Figure 7.6.7 is

$\begin{array}{l}{\mathbf{L}}_{\mathbf{1}}\frac{\mathbf{d}{\mathbf{i}}_{\mathbf{2}}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathbf{R}{\mathbf{i}}_{\mathbf{2}}\mathbf{+}\mathbf{R}{\mathbf{i}}_{\mathbf{3}}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\\ {\mathbf{L}}_{\mathbf{2}}\frac{\mathbf{d}{\mathbf{i}}_{\mathbf{3}}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathbf{R}{\mathbf{i}}_{\mathbf{2}}\mathbf{+}\mathbf{R}{\mathbf{i}}_{\mathbf{3}}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\end{array}$

(b) Solve the system in part (a) if,$\mathit{R}\mathbf{=}\mathbf{5}\mathit{\Omega }{\mathit{L}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}\mathit{h}\mathbf{,}{\mathit{L}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0125}\mathit{h}\mathbf{,}\mathit{E}\mathbf{=}\mathbf{100}\mathit{V}{\mathit{i}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}{\mathit{i}}_{\mathbf{3}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$,, .

(c) Determine the current.

Short answer

$\begin{array}{l}{L}_1\frac{d{i}_2}{dt}+R{i}_2+R{i}_3=E\left(t\right)\\ L_2\frac{d{i}_3}{dt}+R{i}_2+R{i}_3=E\left(t\right)\\ i_2\left(t\right)=\frac{100}{9}-\frac{100}{9}{e}^{-900t}\\ i_3\left(t\right)=\frac{80}{9}-\frac{80}{9}{e}^{-900t}\\ i_1\left(t\right)=20-20e^{-900t}\end{array}$

Step-by-step solution

Given Information.

$\begin{array}{l}{L}_1\frac{d{i}_2}{dt}+R{i}_2+R{i}_3=E\left(t\right)\\ L_2\frac{d{i}_3}{dt}+R{i}_2+R{i}_3=E\left(t\right)\end{array}$

Determining the differential equations for current i21 and i3t:

1. $i_1\left(t\right)=i_2\left(t\right)+i_3\left(t\right)$We can write the following using Kirchhoff's first law:

To get the result, apply Kirchhoff's second law to each loop.

$\begin{array}{l}E\left(t\right)=R{i}_1+L_1{i_2}^{\text{'}}\\ \underset{¯}{E\left(t\right)=R{i}_1+L_2{i_3}^{\text{'}}}\end{array}$

$\begin{array}{l}E\left(t\right)=L_1{i_2}^{\text{'}}+R{i}_2+R{i}_3\\ \underset{¯}{E\left(t\right)=L_1{i_3}^{\text{'}}+R{i}_2+R{i}_3}\end{array}$

As a result, the system is

$\begin{array}{l}{L}_1\frac{d{i}_2}{dt}+R{i}_2+R{i}_3=E\left(t\right)\\ L_2\frac{d{i}_3}{dt}+R{i}_2+R{i}_3=E\left(t\right)\end{array}$

Solving the system in Part(a):

$\begin{array}{l}0.01\frac{d{i}_2}{dt}+5i_2+5i_3=100\\ \underset{¯}{0.0125\frac{d{i}_3}{dt}+5i_2+5i_3=100}\end{array}$

Take the system's Laplace to transform to get

$\begin{array}{l}L\left\{{i_2}^{\text{'}}\right\}+500L\left\{i_2\right\}+500L\left\{i_3\right\}=10000L\left\{1\right\}\left(L\left\{1\right\}=\frac{1}{s}\right)\\ \underset{¯}{L\left\{{i_3}^{\text{'}}\right\}+400L\left\{i_2\right\}+400L\left\{i_3\right\}=8000L\left\{1\right\}}\left(L\left\{1\right\}=\frac{1}{s}\right)\\ L\left\{{i_2}^{\text{'}}\right\}+500L\left\{i_2\right\}+500L\left\{i_3\right\}=\frac{10000}{s}\\ \underset{¯}{L\left\{{i_3}^{\text{'}}\right\}+400L\left\{i_2\right\}+400L\left\{i_3\right\}=\frac{8000}{s}}\end{array}$

We are aware of this.

$(*)\left\{\begin{array}{l}L\left\{{i_2}^{\text{'}}\right\}=sL\left\{{i_2}^{\text{'}}\right\}-i_2\left(0\right)\\ L\left\{{i_3}^{\text{'}}\right\}=sL\left\{{i_3}^{\text{'}}\right\}-i_3\left(0\right)\end{array}\right.$

To get the result, replace (*) in the above system.

$\begin{array}{l}sL\left\{i_2\right\}-i_2\left(0\right)+500L\left\{i_2\right\}+500L\left\{i_3\right\}=\frac{10000}{s}\left(i_2\left(0\right)=0\right)\\ s\underset{¯}{L\left\{i_3\right\}-i_3\left(0\right)+400L\left\{i_2\right\}+400L\left\{i_3\right\}=\frac{8000}{s}}\left(i_3\left(0\right)=0\right)\\ sL\left\{i_2\right\}+500L\left\{i_2\right\}+500L\left\{i_3\right\}=\frac{10000}{s}\\ s\underset{¯}{L\left\{i_3\right\}+400L\left\{i_2\right\}+400L\left\{i_3\right\}=\frac{8000}{s}}\end{array}$

In the above system, write $L\left\{i_2\right\}=I_2\left(s\right)andL\left\{i_3\right\}=I_3\left(s\right)$

$\begin{array}{l}s{I}_2\left(s\right)+500I_2\left(s\right)+500I_3\left(s\right)=\frac{10000}{s}\\ s\underset{¯}{I_3\left(s\right)+400I_2\left(s\right)+400I_3\left(s\right)=\frac{8000}{s}}\end{array}$

The system is reduced to its simplest form.

$\begin{array}{l}\left(s+500\right)I_2\left(s\right)+500I_3\left(s\right)=\frac{10000}{s}----\left(2\right)\\ 400I_2\left(s\right)+\left(s+400\right)I_3\left(s\right)=\frac{8000}{s}-----\left(3\right)\end{array}$

Equation (2) is multiplied by (s+400) while equation (3) is multiplied by -500.

$\begin{array}{l}\left(s+500\right)\left(s+400\right)I_2\left(s\right)+500\left(s+400\right)I_3\left(s\right)=\frac{10000\left(s+400\right)}{s}\\ -200000I_2\left(s\right)-500\left(s+400\right)I_3\left(s\right)=-\frac{4000000}{s}\end{array}$

To reach the answer, multiply (4)and (5) equations together

$\left(s+500\right)\left(s+400\right)I_2\left(s\right)-200000I_2\left(s\right)=\frac{10000\left(s+400\right)}{s}-\frac{4000000}{s}$

$$\begin{gathered} \left(s^2+900s+20000-20000\right)I_2\left(s\right)=\frac{10000s+400000-400000}{s} \\ \begin{array}{l}\left(s^2+900s\right)I_2\left(s\right)=10000\\ I_2\left(s\right)=\frac{10000}{s\left(s+900\right)}\end{array} \end{gathered}$$

Using the partial fraction approach, decompose the fraction $\frac{10000}{s\left(s+900\right)}$

$\begin{array}{l}*\frac{10000}{s\left(s+900\right)}=\frac{A}{s}+\frac{B}{s+900}\\ \left(s^2+900s\right)I_2\left(s\right)=10000\\ I_2\left(s\right)=\frac{10000}{s\left(s+900\right)}(*)\end{array}$

$\begin{array}{l}*\frac{10000}{s\left(s+900\right)}=\frac{A}{s}+\frac{B}{s+900}\\ =\frac{A(s+900)+Bs}{s\left(s+900\right)}\end{array}$

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

$$\begin{gathered} \begin{array}{l}10000=As+900A+Bs\\ Ifsiszero,then\\ 10000=900A\end{array} \\ A=\frac{100}{9} \\ \begin{array}{l}Ifs=-900,then\\ 10000=-900A+900A-900B\end{array} \\ B=\frac{-100}{9} \\ WegetbysubstitutingAandBinequation\left(6\right). \\ \frac{10000}{s\left(s+900\right)}=\frac{100}{9s}-\frac{100}{9\left(s+900\right)} \\ Theequation(*)istransformedinto \\ {I}_2\left(s\right)=\frac{100}{9s}-\frac{100}{9\left(s+900\right)} \end{gathered}$$

Both sides should be transformed using the inverse Laplace transform.

$\begin{array}{l}{L}^{-1}\left\{I_2\left(s\right)\right\}=L^{-1}\left\{\frac{100}{9s}-\frac{100}{9\left(s+900\right)}\right\}\\ =\frac{100}{9}{L}^{-1}\left\{\frac{1}{s}\right\}-\frac{100}{9}{L}^{-1}\left\{\frac{1}{s-\left(-900\right)}\right\}\\ (**)=\frac{100}{9}-\frac{100}{9}{e}^{-900t}\\ (**)\left\{\begin{array}{l}{e}^{at}=L^{-1}\left\{\frac{1}{s-a}\right\}\\ 1=L^{-1}\left\{\frac{1}{s}\right\}\end{array}\right.\end{array}$

$\begin{array}{l}{i}_2\left(t\right)=L^{-1}\left\{I_2\left(s\right)\right\}=\frac{100}{9}-\frac{100}{9}{e}^{-900t}\\ Equation\left(2\right)ismultipliedby-400,whileequation\left(3\right)ismultipliedby\left(s+500\right).\\ -400\left(s+500\right)I_2\left(s\right)-200000I_3\left(s\right)=-\frac{4000000}{s}\\ 400\left(s+500\right)I_2\left(s\right)+\left(s+500\right)\left(s+400\right)I_3\left(s\right)=\frac{8000\left(s+500\right)}{s}\\ Addthe\left(7\right)and\left(8\right)equationstogethertoget\\ \left(s+500\right)\left(s+400\right)I_3\left(s\right)-200000I_3\left(s\right)=\frac{8000\left(s+500\right)}{s}-\frac{4000000}{s}\\ \left(s+500\right)\left(s+400\right)I_3\left(s\right)-200000I_3\left(s\right)=\frac{8000s+4000000}{s}-\frac{4000000}{s}\end{array}$

$\begin{array}{l}\left(s^2+900s\right)I_3\left(s\right)=8000\\ I_3\left(s\right)=\frac{8000}{s\left(s+900\right)}\\ Usingthepartialfractionapproach,decomposethefraction.\\ \frac{8000}{s\left(s+900\right)}=\frac{A}{s}+\frac{B}{s+900}\\ =\frac{A\left(s+900\right)+Bs}{s\left(s+900\right)}\\ Becausethedenominatorsofthefractionsinthepreviousequationarethesame,theirnumeratorsmustbeequal\\ 8000=As+900A+Bs\\ IfS=0,then\\ 8000=900A\\ A=\frac{80}{9}\\ IfS=-900,then\\ 8000=-900A+900A-900B\\ B=\frac{-80}{9}\\ WegetbysubstitutingAandBinequation.\\ \frac{8000}{s\left(s+900\right)}=\frac{80}{9s}-\frac{80}{9\left(s+900\right)}\end{array}$

$\begin{array}{l}(**)becomestheequation\\ I_3\left(s\right)=\frac{80}{9s}-\frac{80}{9\left(s+900\right)}\\ BothsidesshouldbetransformedusingtheinverseLaplacetransform.\\ L^{-1}\left\{I_3\left(s\right)\right\}=L^{-1}\left\{\frac{80}{9s}-\frac{80}{9\left(s+900\right)}\right\}\\ =\frac{80}{9}{L}^{-1}\left\{\frac{1}{s}\right\}-\frac{80}{9}{L}^{-1}\left\{\frac{1}{s-\left(-900\right)}\right\}\\ (**)\frac{80}{9}-\frac{80}{9}{e}^{-900t}\\ \left(**\right)\left\{\begin{array}{l}{e}^{at}=L^{-1}\left\{\frac{1}{s-a}\right\}\\ 1=L^{-1}\left\{\frac{1}{s}\right\}\end{array}\right.\\ i_3\left(t\right)=L^{-1}\left\{I_3\left(s\right)\right\}=\frac{80}{9}-\frac{80}{9}{e}^{-900t}\end{array}$

Determining the current i1t

$\begin{array}{l}{i}_1\left(t\right)=i_2\left(t\right)+i_3\left(t\right)\\ =\frac{100}{9}-\frac{100}{9}{e}^{-900t}+\frac{80}{9}-\frac{80}{9}{e}^{-900t}\\ =20-20e^{-900t}\end{array}$

Determining the Result

$\begin{array}{l}{L}_1\frac{d{i}_2}{dt}+R{i}_2+R{i}_3=E\left(t\right)\\ L_2\frac{d{i}_3}{dt}+R{i}_2+R{i}_3=E\left(t\right)\\ i_2\left(t\right)=\frac{100}{9}-\frac{100}{9}{e}^{-900t}\\ i_3\left(t\right)=\frac{80}{9}-\frac{80}{9}{e}^{-900t}\\ i_1\left(t\right)=20-20e^{-900t}\end{array}$