91影视

Using Kirchhoff's voltage law for loop \({\bf{1}}\)we will get \(\frac{{\bf{1}}}{{\bf{C}}}{{\bf{q}}_{\bf{1}}}{\bf{ - }}{{\bf{R}}_{\bf{2}}}{{\bf{I}}_{\bf{2}}}{\bf{ = 0}}\) where q₁ (t)=I₁ (t), and Kirchhoff's voltage law for loop \({\bf{2}}\) gives us that \({{\bf{I}}_{\bf{2}}}{{\bf{R}}_{\bf{2}}}{\bf{ - }}{{\bf{R}}_{\bf{1}}}{{\bf{I}}_{\bf{3}}}{\bf{ - L}}\frac{{{\bf{d}}{{\bf{I}}_{\bf{3}}}}}{{{\bf{dt}}}}{\bf{ = 0}}\)

At least, Kirchhoff's current law for the top juncture gives us that\({\bf{ - }}{{\bf{I}}_{\bf{1}}}{\bf{ - }}{{\bf{I}}_{\bf{2}}}{\bf{ - }}{{\bf{I}}_{\bf{3}}}{\bf{ = 0}}\), so one has to solve the following system:

\(\begin{array}{c}\frac{{\bf{1}}}{{\bf{C}}}{{\bf{q}}_{\bf{1}}}{\bf{ = }}{{\bf{R}}_{\bf{2}}}{{\bf{I}}_{\bf{2}}}\\{{\bf{I}}_{\bf{2}}}{{\bf{R}}_{\bf{2}}}{\bf{ = }}{{\bf{R}}_{\bf{1}}}{{\bf{I}}_{\bf{3}}}{\bf{ + L}}\frac{{{\bf{d}}{{\bf{I}}_{\bf{3}}}}}{{{\bf{dt}}}}\\{{\bf{I}}_{\bf{1}}}{\bf{ = - }}{{\bf{I}}_{\bf{2}}}{\bf{ - }}{{\bf{I}}_{\bf{3}}}\end{array}\)

Substituting the given values \({{\bf{R}}_{\bf{1}}}{\bf{ = }}{{\bf{R}}_{\bf{2}}}{\bf{ = 1\Omega ,C = 1\;F,L = 1H}}\), one will get

\(\begin{array}{c}{{\bf{q}}_{\bf{1}}}{\bf{ = }}{{\bf{I}}_{\bf{2}}}\\{{\bf{I}}_{\bf{2}}}{\bf{ = }}{{\bf{I}}_{\bf{3}}}{\bf{ + }}\frac{{{\bf{d}}{{\bf{I}}_{\bf{3}}}}}{{{\bf{dt}}}}{\bf{n}}\\{{\bf{I}}_{\bf{1}}}{\bf{ = - }}{{\bf{I}}_{\bf{2}}}{\bf{ - }}{{\bf{I}}_{\bf{3}}}\end{array}\)

Differentiating the third equation we will get that \({{\bf{q}}_{\bf{1}}}{\bf{ = - }}{{\bf{q}}_{\bf{2}}}{\bf{ - }}{{\bf{q}}_{\bf{3}}}\).

One will express the currents \({{\bf{I}}_{\bf{i}}}{\bf{,i = }}\overline {{\bf{1,3}}} \) as \(\frac{{{\bf{d}}{{\bf{q}}_{\bf{i}}}}}{{{\bf{dt}}}}{\bf{,i = }}\overline {{\bf{1,3}}} \) and substitute \({{\bf{q}}_{\bf{1}}}{\bf{ = - }}{{\bf{q}}_{\bf{2}}}{\bf{ - }}{{\bf{q}}_{\bf{3}}}\) into the first equation which gives us the following system:

\(\begin{array}{c}{\bf{ - }}{{\bf{q}}_{\bf{2}}}{\bf{ - }}{{\bf{q}}_{\bf{3}}}{\bf{ = }}\frac{{{\bf{d}}{{\bf{q}}_{\bf{2}}}}}{{{\bf{dt}}}}\\\frac{{{\bf{d}}{{\bf{q}}_{\bf{2}}}}}{{{\bf{dt}}}}{\bf{ = }}\frac{{{\bf{d}}{{\bf{q}}_{\bf{3}}}}}{{{\bf{dt}}}}{\bf{ + }}\frac{{{{\bf{d}}^{\bf{2}}}{{\bf{q}}_{\bf{3}}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}\end{array}\)

One will solve this system using the eliminating method. First, one will rewrite this system in operator form:

\(\begin{array}{c}{\bf{(D + 1)}}\left[ {{{\bf{q}}_{\bf{2}}}} \right]{\bf{ + }}{{\bf{q}}_{\bf{3}}}{\bf{ = 0}}\\{\bf{D}}\left[ {{{\bf{q}}_{\bf{2}}}} \right]{\bf{ - }}\left( {{{\bf{D}}^{\bf{2}}}{\bf{ + D}}} \right)\left[ {{{\bf{q}}_{\bf{3}}}} \right]{\bf{ = 0}}\end{array}\)

One can eliminate \({{\bf{q}}_{\bf{3}}}\) by multiplying the first equation by \({\bf{D}}\left( {{\bf{D + 1}}} \right)\) and then add those two equations together:

\(\begin{array}{c}{\bf{D(D + 1)(D + 1)}}\left[ {{{\bf{q}}_{\bf{2}}}} \right]{\bf{ + D(D + 1)}}\left[ {{{\bf{q}}_{\bf{3}}}} \right]{\bf{ = 0}}\\{\bf{D}}\left[ {{{\bf{q}}_{\bf{2}}}} \right]{\bf{ - D(D + 1)}}\left[ {{{\bf{q}}_{\bf{3}}}} \right]{\bf{ = 0}}\\{\bf{(D(D + 1)(D + 1) + D)}}\left[ {{{\bf{q}}_{\bf{2}}}} \right]{\bf{ = 0}}\\{\bf{D}}\left( {{{\bf{D}}^{\bf{2}}}{\bf{ + 2D + 2}}} \right)\left[ {{{\bf{q}}_{\bf{2}}}} \right]{\bf{ = 0}}\end{array}\)

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

\(\begin{array}{c}{{\bf{r}}_{{\bf{1,2}}}}{\bf{ = }}\frac{{{\bf{ - 2 \pm }}\sqrt {{\bf{4 - 8}}} }}{{\bf{2}}}{\bf{,}}\;\;\;{{\bf{r}}_{\bf{3}}}{\bf{ = 0}}\\{{\bf{r}}_{{\bf{1,2}}}}{\bf{ = - 1 \pm i,}}\;\;\;{{\bf{r}}_{\bf{3}}}{\bf{ = 0}}\end{array}\)

So, the general solution for \({{\bf{q}}_{\bf{2}}}\) is \({{\bf{q}}_{\bf{2}}}{\bf{(t) = }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost + }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint + }}{{\bf{c}}_{\bf{3}}}\)

Now we can find the current \({{\bf{I}}_{\bf{2}}}\) :

\(\begin{array}{c}{{\bf{I}}_{\bf{2}}}{\bf{(t) = }}\frac{{{\bf{d}}{{\bf{q}}_{\bf{2}}}}}{{{\bf{dt}}}}{\bf{ = }}\frac{{\bf{d}}}{{{\bf{dt}}}}\left( {{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost + }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint + }}{{\bf{c}}_{\bf{3}}}} \right)\\{\bf{ = - }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost - }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint - }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint + }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost}}\\{\bf{ = - }}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ - }}{{\bf{c}}_{\bf{2}}}} \right){{\bf{e}}^{{\bf{ - t}}}}{\bf{cost - }}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right){{\bf{e}}^{{\bf{ - t}}}}{\bf{sint}}\end{array}\)

One can now find the charge \({{\bf{q}}_{\bf{3}}}\) from \({{\bf{q}}_{\bf{3}}}{\bf{ = - }}{{\bf{q}}_{\bf{2}}}{\bf{ - }}\frac{{{\bf{d}}{{\bf{q}}_{\bf{2}}}}}{{{\bf{dt}}}}{\bf{ = - }}{{\bf{q}}_{\bf{2}}}{\bf{ - }}{{\bf{I}}_{\bf{2}}}\)

\(\begin{array}{c}{{\bf{q}}_{\bf{3}}}{\bf{(t) = - }}{{\bf{q}}_{\bf{2}}}{\bf{(t) - }}{{\bf{I}}_{\bf{2}}}{\bf{(t)}}\\{\bf{ = - }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost - }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint - }}{{\bf{c}}_{\bf{3}}}{\bf{ + }}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ - }}{{\bf{c}}_{\bf{2}}}} \right){{\bf{e}}^{{\bf{ - t}}}}{\bf{cost + }}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right){{\bf{e}}^{{\bf{ - t}}}}{\bf{sint}}\\{\bf{ = - }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost + }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint - }}{{\bf{c}}_{\bf{3}}}\end{array}\)

So, the current \({{\bf{I}}_{\bf{3}}}\) is

\(\begin{array}{c}{{\bf{I}}_{\bf{3}}}{\bf{(t) = }}\frac{{{\bf{d}}{{\bf{q}}_{\bf{3}}}}}{{{\bf{dt}}}}{\bf{ = }}\frac{{\bf{d}}}{{{\bf{dt}}}}\left( {{\bf{ - }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost + }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint - }}{{\bf{c}}_{\bf{3}}}} \right)\\{\bf{ = }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost + }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint - }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint + }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost}}\\{\bf{ = }}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right){{\bf{e}}^{{\bf{ - t}}}}{\bf{cost - }}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ - }}{{\bf{c}}_{\bf{2}}}} \right){{\bf{e}}^{{\bf{ - t}}}}{\bf{sint}}\end{array}\)

One can introduce new constants \({\bf{A = }}{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}\)and\({\bf{B = - }}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ - }}{{\bf{c}}_{\bf{2}}}} \right)\), so now one has the solutions for the currents \({{\bf{I}}_{\bf{2}}}\) , and \({{\bf{I}}_{\bf{3}}}\) are:

\(\begin{array}{c}{{\bf{I}}_{\bf{2}}}{\bf{(t) = B}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost - A}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint}}\\{{\bf{I}}_{\bf{3}}}{\bf{(t) = A}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost + B}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint}}\end{array}\)

Now, one can find the current\({{\bf{I}}_{\bf{1}}}\):

\(\begin{array}{c}{{\bf{I}}_{\bf{1}}}{\bf{(t) = - }}{{\bf{I}}_{\bf{2}}}{\bf{(t) - }}{{\bf{I}}_{\bf{3}}}{\bf{(t)}}\\{\bf{ = - B}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost + A}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint - A}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost - B}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint}}\\{\bf{ = - (A + B)}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost + (A - B)}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint}}\end{array}\)

The general solution for the given network is:

\(\begin{array}{c}{{\bf{I}}_{\bf{1}}}{\bf{(t) = - (A + B)}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost + (A - B)}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint}}\\{{\bf{I}}_{\bf{2}}}{\bf{(t) = B}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost - A}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint}}\\{{\bf{I}}_{\bf{3}}}{\bf{(t) = A}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cost + B}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sint}}\end{array}\)