Find a system of differential equations and initial conditions for the currents in the networks given in the schematic diagrams (Figures \({\bf{5}}.{\bf{39}} - {\bf{5}}.{\bf{42}}\) on pages\({\bf{294}} - {\bf{295}}\)). Assume that all initial currents are zero. Solve for the currents in each branch of the network.

One has that\({L_1} = 0.02H\),\({L_2} = 0.025H\), \(R = 10\Omega \)and\(E\left( t \right) = 10\;V\). Applying the

Kirchhoff's voltage law on the first loop one will get\(E\left( t \right) = {L_1}\frac{{d{I_2}}}{{dt}} + R{I_1}\)and from the second loop, one has that\(0 = - {L_1}\frac{{d{I_2}}}{{dt}} + {L_2}\frac{{d{I_3}}}{{dt}}\).

Applying Kirchhoff's current law at the point \(A\)one will get \(0{\bf{ }} = - {I_1}{\bf{ }} + {I_2}{\bf{ }} + {I_3}\)and the point \(B\)gives us that\(0 = {I_1} - {I_2} - {I_3}\), so in both cases, one has that\({I_1} = {I_2} + {I_3}\)

One has to solve the following system:

\(\begin{aligned}{c}E\left( t \right) = {L_1}\frac{{d{I_2}}}{{dt}} + R{I_1}\\0 = - {L_1}\frac{{d{I_2}}}{{dt}} + {L_2}\frac{{d{I_3}}}{{dt}}\end{aligned}\)

Expressing \({I_1}\)in terms of \({I_2}\)and \({I_3}\)in the first two equations of the system one will get;

\(\begin{aligned}{c}E\left( t \right) = {L_1}\frac{{d{I_2}}}{{dt}} + R{I_2} + R{I_3},\\0 = - {L_1}\frac{{d{I_2}}}{{dt}} + {L_2}\frac{{d{I_3}}}{{dt}}\end{aligned}\)

Substituting the values for \(E,{\bf{ }}{L_1},{\bf{ }}{L_2},{\bf{ }}{R_1}\)and \({R_2}\)our system becomes:

\(\begin{aligned}{c}10 = 0.02\frac{{d{I_2}}}{{dt}} + 10{I_2} + 10{I_3}{\bf{ }}\\0 = - 0.02\frac{{d{I_2}}}{{dt}} + 0.025\frac{{d{I_3}}}{{dt}}\end{aligned}\)

Dividing the first equation by \(10\) and multiplying the second by \(1000\)one will get:

\(\begin{aligned}{c}1 = 0.002\frac{{d{I_2}}}{{dt}} + {I_2} +

One will solve this by the elimination method. First, one needs to rewrite the system in operator form:

\(\begin{aligned}{c}0.002D\left( {{I_2}} \right) + {I_2} + {I_3} = 1{\bf{ }}\\ - 20D\left( {{I_2}} \right) + 25D\left( {{I_3}} \right) = 0\end{aligned}\)

From the first equation, one has that\({I_3} = 1 - {I_2} - 0.002D{I_2}\), so substituting this into the second equation one has,

\(\begin{aligned}{c} - 20D\left( {{I_2}} \right) + 25D\left( {1 - {I_2} - 0.002D\left( {{I_2}} \right)} \right) = 0\\ - 20D\left( {{I_2}} \right) - 25D\left( {{I_2}} \right) - 0.05{D^2}\left( {{I_2}} \right) = 0\\0.05{D^2}\left( {{I_2}} \right) + 45D\left( {{I_2}} \right) = 0\end{aligned}\)

The auxiliary equation is \(0.05{r^2} + 45r = r\left( {0.05r + 45} \right) = 0\) and its roots are:

\(\begin{aligned}{c}{r_1} = 0,{r_2} = - \frac{{45}}{{0.05}}\\{r_1} = 0,\;\;\;{r_2} = - 900\end{aligned}\)

So, the general solution for \({I_2}\)is\({I_2}\left( t \right) = {c_1} + {c_2}{e^{ - 900t}}\)

Now, one can find the current \({I_3}\)from \({I_3} = 1 - {I_2} - 0.002D\left( {{I_2}} \right){\rm{:}}\)

\(D\left( {{I_2}} \right) = - 900{c_2}{e^{ - 900t}}\)

\(\begin{aligned}{c}{I_3} = 1 - {c_1} - {c_2}{e^{ - 900t}} - 0.002 \times \left( { - 900{c_2}{e^{ - 900t}}} \right)\\ = 1 - {c_1} - {c_2}{e^{ - 900t}} + 1.8{c_2}{e^{ - 900t}}\\\;{I_3}\left( t \right) = 1 - {c_1} + 0.8{c_2}{e^{ - 900t}}\end{aligned}\)

One will find the current \({I_1}\)from the relation\({I_1} = {I_2} + {I_3}\):

\(\begin{aligned}{c}{I_1} = {c_1} + {c_2}{e^{ - 900t}} + 1 - {c_1} + 0.8{c_2}{e^{ - 900t}}{\bf{ }}\\{I_1}\left( t \right) = 1 + 1.8{c_2}{e^{ - 900t}}\end{aligned}\)

One will find the constant \({c_1},{\bf{ }}{c_2}\)from the initial conditions which are:

\({I_1}\left( 0 \right) = {I_2}\left( 0 \right) = {I_3}\left( 0 \right) = 0,\)So, one has a system

\(\begin{aligned}{c}{I_1}\left( 0 \right) = 1 + 1.8{\bf{ }}{c_2}\\ = 0\end{aligned}\)

\(\begin{aligned}{c}{I_2}\left( 0 \right) = {c_1} + {c_2}\\ = 0\end{aligned}\)

\(\begin{aligned}{c}{I_3}\left( 0 \right) = 1 - {c_1} + 0.8{\bf{ }}{c_2}\\ = 0\end{aligned}\)

From the first equation, one has that\({c_2} = \frac{{ - 1}}{{1.8}} = - \frac{{10}}{{18}}\), from the second one has that \({c_1} = - {c_2}\),and therefore\({c_1} = \frac{{10}}{{18}}\), so the currents in the given network are:

\(\begin{aligned}{c}{{\bf{I}}_{\bf{1}}}{\bf{(t) = 1 - }}{{\bf{e}}^{{\bf{ - 900t}}}}\\{{\bf{I}}_{\bf{2}}}{\bf{(t) = }}\frac{{{\bf{10}}}}{{{\bf{18}}}}{\bf{ - }}\frac{{{\bf{10}}}}{{{\bf{18}}}}{{\bf{e}}^{{\bf{ - 900t}}}}\\{{\bf{I}}_{\bf{3}}}{\bf{(t) = }}\frac{{\bf{4}}}{{\bf{9}}}{\bf{ - }}\frac{{\bf{4}}}{{\bf{9}}}{{\bf{e}}^{{\bf{ - 900t}}}}\end{aligned}\)