The circuit in Figure 4 can be described by the equation [i'L v'C] = [-R2/L -1/L 1/C -1/(R1C)][iL vC] where iL is the current passing through the inductor L and vC is the voltage drop across the capacitor C. Suppose R1 is 5 ohms, R2 is .8 ohm, C is .1 farad, and L is .4 henry. Find formulas for iL

Question

The circuit in Figure 4 can be described by the equation

[latex] \left [ \begin{matrix} \acute{i}_{L} \ \acute{v}_{C} \end{matrix} ight ] = \left [ \begin{matrix} -R_{2}/L & -1/L \ 1/C & -1/(R_{1}C) \end{matrix} ight ] \left [ \begin{matrix} i_{L} \ v_{C} \end{matrix} ight ] [/latex]

where [latex] i_{L} [/latex] is the current passing through the inductor L and [latex] v_{C} [/latex] is the voltage drop across the capacitor C. Suppose [latex] R_{1} [/latex] is 5 ohms, [latex] R_{2} [/latex] is .8 ohm, C is .1 farad, and L is .4 henry. Find formulas for [latex] i_{L} [/latex] and [latex] v_{C} [/latex] , if the initial current through the inductor is 3 amperes and the initial voltage across the capacitor is 3 volts.

Accepted Answer

For the data given, [latex] A = \left [ \begin{matrix} -2 & -2.5 \ 10 & -2 \end{matrix} ight ] ext{and} x_{0} = \left [ \begin{matrix} 3 \ 3 \end{matrix} ight ] [/latex] . The method discussed in Section 5.5 produces the eigenvalue [latex] \lambda = -2 + 5i [/latex] and the corresponding eigenvector [latex] v_{1} = \left [ \begin{matrix} i \ 2 \end{matrix} ight ] [/latex] . The complex solutions of [latex] \acute{x} = Ax [/latex] are complex linear combinations of

[latex] x_{1}(t) = \left [ \begin{matrix} i \ 2 \end{matrix} ight ] e^{(-2+5i)t} ext{and} x_{2}(t) = \left [ \begin{matrix} -i \ 2 \end{matrix} ight ]e^{(-2-5i)t} [/latex]

Next, use equation (7) to write

[latex] e^{(a+bi)t} = e^{at} \cdot e^{ibt} = e^{at}(\cos bt + i \sin bt) [/latex] (7)

[latex] x_{1}(t) = \left [ \begin{matrix} i \ 2 \end{matrix} ight ] e^{-2t}(\cos 5t + i \sin 5t) [/latex]

The real and imaginary parts of [latex] x_{1} [/latex] provide real solutions:

[latex] y_{1}(t) = \left [ \begin{matrix} -\sin 5t \ 2 \cos 5t \end{matrix} ight ] e^{-2t} , y_{2}(t) = \left [ \begin{matrix} \cos 5t \ 2 \sin 5t \end{matrix} ight ] e^{-2t} [/latex]

Since [latex] y_{1} [/latex] and [latex] y_{2} [/latex] are linearly independent functions, they form a basis for the two-dimensional real vector space of solutions of [latex] \acute{x} = Ax [/latex]. Thus the general solution is

[latex] x(t) = c_{1}\left [ \begin{matrix} -\sin 5t \ 2 \cos 5t \end{matrix} ight ] e^{-2t} + c_{2}\left [ \begin{matrix} \cos 5t \ 2 \sin 5t \end{matrix} ight ] e^{-2t} [/latex]

To satisfy [latex] x(0) = \left [ \begin{matrix} 3 \ 3 \end{matrix} ight ] [/latex] , we need [latex] c_{1}\left [ \begin{matrix} 0 \ 2 \end{matrix} ight ] + c_{2}\left [ \begin{matrix} 1 \ 0 \end{matrix} ight ] = \left [ \begin{matrix} 3 \ 3 \end{matrix} ight ] [/latex] , which leads to [latex] c_{1} = 1.5 ext{and} c_{2} = 3 [/latex] . Thus

[latex] x(t) = 1.5\left [ \begin{matrix} -\sin 5t \ 2 \cos 5t \end{matrix} ight ] e^{-2t} + 3\left [ \begin{matrix} \cos 5t \ 2 \sin 5t \end{matrix} ight ] e^{-2t} [/latex]

or

[latex] \left [ \begin{matrix} i_{L}(t) \ v_{C}(t) \end{matrix} ight ] = \left [ \begin{matrix} -1.5 \sin 5t + 3 \cos 5t \ 3 \cos 5t + 6 \sin 5t \end{matrix} ight ]e^{-2t} [/latex]

See Figure 5.

Question 5.7.3: The circuit in Figure 4 can be described by the equation [i'...

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