Sistemas Electricos con Laplace

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\begin{gather*} L_{1} \ \frac{diz}{dt} +Ri_{2} +Ri_{3} =E( t)\\ L_{2}\frac{di_{3}}{dt} +Ri_{2} +Ri_{3} =E( t) \end{gather*}

Buscamos: i2 = ? ; i3 = ?


\begin{gather*} \frac{di_{2}}{dt} +\frac{R}{L_{1}} i_{2} +\frac{R}{L_{1}} i_{3} =\frac{E( t)}{L_{1}}\\ \frac{di_{3}}{dt} +\frac{R}{L_{2}} i_{2} +\frac{R}{L_{2}} i_{3} =\frac{E( t)}{L_{2}} \end{gather*}


\begin{gather*} \frac{di_{2}}{dt} +\frac{5}{0.0.} i_{2} +\frac{5}{0.01} i_{3} =\frac{100}{0.01}\\ \\ \frac{di_{3}}{dt} +\frac{5}{0.0125} i_{2} +\frac{5}{0.0125} i_{3} =\frac{100}{0.0125} \end{gather*}


\begin{gather*} \frac{di_{2}}{dt} +500i_{ \begin{array}{l} 2\\ \end{array}} +500i_{3} =\ 10000\\ \frac{di3}{dt} +400i_{ \begin{array}{l} 2\\ \end{array}} +400i_{3} =\ 8000 \end{gather*}


\begin{gather*} \mathscr{L} \ \{i_{2}\} =sI_{2}( s) -i_{2}( 0)\\ =\ sI_{2}( s) -0\\ =sI_{2}( s)\\ \\ \mathscr{L} \ \{500i_{2}\} \ =\ 500\mathscr{L} \ \{i_{2}\}\\ =500\ I_{2}( s)\\ \\ \mathscr{L} \ \{400i_{3}\} =500\mathscr{L} \ \{i_{3}\}\\ =500I_{3}( s)\\ \\ \mathscr{L} \ \{10000\} \ =\ 10000\mathscr{L}\{1\}\\ =\ \frac{10000}{S} \end{gather*}


\begin{gather*} i{_{2}}^{1} +500i_{2} +500i_{3} =\ 10000\\ \Longrightarrow \ SI_{2}( s) +500I_{2}( s) +500I_{3}( s) =\frac{10000}{s} \end{gather*}


\begin{gather*} i{_{3}}^{1} +400i_{2} +400i_{3} =8000\\ \Longrightarrow \ SI_{3}( s) -0\ +400I_{2}( s) +400I_{3}( s) =\frac{8000}{s}\\ \Longrightarrow \ SI_{3}( s) +400I_{2}( s) +400I_{3}( s) =\frac{8000}{s} \end{gather*}


\begin{gather*} SI_{2}( s) +500I_{2}( s) +500I_{3}( s) =\frac{10000}{s}\\ SI_{3}( s) +400I_{2}( s) +400I_{3}( s) =\frac{8000}{s} \end{gather*}


\begin{gather*} ( S+500) I_{2}( s) +500I_{3}( s) =\frac{10000}{s}\\ 400\ I_{2}( s) +( s+400) I_{3}( s) =\ \frac{8000}{s} \end{gather*}


\begin{gather*} ( S+500) \ I_{2}( S) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \frac{10000}{s} \ -\ 500\ I_{3}( S)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I_{2}( S) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{10000}{s( s+500)} -\ \frac{5000}{( s+500)} \ \ I_{3}( S) \ \ \ \\ \ \ \ \ \ \end{gather*}


\begin{gather*} 400\ I_{2}( S) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \frac{8000}{s} \ -\ ( S+400) \ I_{3}( S)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I_{2}( S) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{8000}{400s} -\ \frac{( s+400)}{400} \ \ I_{3}( S) \ \ \ \end{gather*}

\begin{gather*} \frac{10000}{s( s+500)} -\frac{500}{s+500} \ I_{3}( s) =\frac{8000}{400s} -\frac{( s+400)}{400} \ I_{3}( s)\\ \\ \Longrightarrow \ \frac{( s+400)}{400} \ I_{3}( s) -\frac{500}{S+500} \ I_{3}( s) =\frac{8000}{400s} -\frac{10000}{s( s+500)}\\ \\ \Longrightarrow \ \left[ \ \frac{s}{400} -\frac{500}{s+500} +1\right] I_{3}( s) =\frac{20}{s} -\frac{10000}{s( s+500)}\\ \Longrightarrow \ I_{3}( s) =\frac{\frac{20}{s} -\frac{10000}{s( s+500)}}{\frac{s}{400} -\frac{500}{s+500} +1}\\ I_{3}( s) =\frac{8000\left(\frac{1}{s} -\frac{500}{s( s+500)}\right)}{\left( s-\frac{200000}{s+500} +400\right)}\\ \\ I_{3}( s) =\frac{8000\left( 1\ -\ \frac{500}{s+500}\right)}{s^{2} -\frac{200\ 000\ s}{s+500} +400s}\\ \\ I_{3}( s) =\frac{8000( s+500\ -\ 500)}{s^{2}( s+500) -200\ 000s+400s( s+500)}\\ \\ \\ =\frac{8000s}{s[ s( s+500) -\ 200\ 000\ +\ 400( s+500)]}\\ \\ =\frac{8000}{s( s+500) -\ 200\ 000\ +\ 400( s+500)}\\ \end{gather*}


\begin{gather*} I_{3}( s) =\frac{8000}{s( s+500) -200\ 000\ +\ 400s\ +\ 200\ 000}\\ \\ I_{3}( s) =\frac{8000}{s( s+500) +\ 400s\ }\\ I_{3}( s) =\frac{8000}{s^{2} +500s+\ 400s\ }\\ I_{3}( s) =\frac{8000}{s^{2} +900s\ }\\ \end{gather*}


\begin{gather*} \frac{8000}{s^{2} +900s} =\frac{8000}{s( s+900)} =\frac{A}{S} +\frac{B}{S+900}\\ \\ \Longrightarrow \ 8000\ =\ A( s+900) +Bs\\ \ \ \ \ \ \ \ \ \ 8000\ =\ As+900A+Bs\\ \ \ \ \ \ \ \ \ \ 8000\ =\ ( A+B) s+900A \end{gather*}


\begin{gather*} 8000=900A\ \Longrightarrow \ A=\frac{8000}{900} =\frac{80}{9}\\ 0\ =\ A+B\ \Longrightarrow \ B=\ -A\\ \Longrightarrow \ B=\ -\frac{80}{9} \end{gather*}


\begin{gather*} I_{3}( s) =\frac{8000}{s^{2} +900s} =\frac{80}{9s} +\frac{( -) 80}{9( s+900)}\\ =\frac{80}{9s} -\frac{80}{9( s+900)} \end{gather*}


\begin{gather*} I_{2}( s) =\frac{8000}{400s} -\frac{( s+400)}{400} I_{3}( s)\\ \\ \Longrightarrow \ I_{2}( s) =\frac{8000}{400s} -\frac{( s+400)}{400}\left(\frac{80}{9s} -\frac{80}{9( s+900}\right)\\ \Longrightarrow \ I_{2}( s) =\frac{8000}{400s} -\frac{80( s+400)}{( 400)( 9s)} +\frac{80( s+400)}{( 400)( 9( s+900))}\\ \Longrightarrow \ I_{2}( s) \ =\ \frac{20}{s} -\frac{2( s+400)}{90s} +\frac{2( s+400)}{90( s+900)}\\ =\frac{20}{s} -\frac{2}{90} -\frac{800}{90s} +\frac{2s}{90( s+900)} +\frac{800}{90( s+900)}\\ \\ =\left(\frac{20}{s} -\frac{80}{9s}\right) +\frac{2s}{90( s+900)} +\frac{800}{90( s+900)} -\frac{2}{90}\\ =\frac{180-80}{9s} +\frac{2s+800}{90( s+900)} -\frac{2}{90}\\ =\frac{100}{9s} +\frac{2s+800-2( s+900)}{90( s+900)}\\ =\frac{100}{9s} +\frac{2s+800-2s-1800}{90( s+900)}\\ =\frac{100}{9s} -\frac{1000}{90( s+900)}\\ I_{2}( s) =\frac{100}{9s} -\frac{100}{9( s+900)} \end{gather*}


\begin{gather*} I_{3}( s) =\frac{80}{9s} -\frac{80}{9( s+900)}\\ I_{2}( s) =\frac{100}{9s} -\frac{100}{9( s+900)}\\ \\ \\ I_{3}( s) =\frac{80}{9s} -\frac{80}{9( s+900)}\\ \\ \mathscr{L}^{-1}\left\{\frac{80}{9s}\right\} =\frac{80}{9}\mathscr{L}^{-1}\left\{\frac{1}{s}\right\}\\ =\frac{80}{9} -( 1)\\ =\frac{80}{9}\\ \\ \mathscr{L}^{-1}\left\{-\frac{80}{9( s+900)}\right\} =\frac{80}{9}\mathscr{L}^{-1}\left\{\frac{1}{s+900}\right\}\\ =-\frac{80}{9} e^{-900t}\\ \\ i_{3}( t) =\frac{80}{9} -\frac{80}{9} e^{-900t} \end{gather*}


\begin{gather*} i_{2}( t) =\frac{100}{9}\mathscr{L}^{-1}\left\{\frac{1}{s}\right\} -\frac{100}{9}\mathscr{L}^{-1}\left\{\frac{1}{s+900}\right\}\\ \\ \Longrightarrow \ i_{2}( t) =\frac{100}{9} -\frac{100}{9} e^{-900t}\\ \end{gather*}


\begin{gather*} i_{3} =\frac{80}{9} -\frac{80}{9} e^{-900t}\\ i_{2} =\frac{100}{9} -\frac{100}{9} e^{-900t} \end{gather*}