Analysis of a Voltage Resistor Inductor Capacitor Circuit

Consider the circuit.

resistor capacitor inductor voltage circuit

The voltage drop across a resistor is  
The voltage drop across a capacitor is  
\[V_C=\frac{Q}{C}= \frac{1}{C} \int I dt\]
The voltage drop across an inductor is  
\[V_L=L \frac{dI}{dt}\]
Using Kirchoff's Law for the left hand loop gives  
. (1)
The voltages across the resistor and inductor respectively are  
\[V_{R_1}=R_1(I_1-I_2), , \: V_L=L \frac{dI_1}{dt}\]
. Substituting these into (1) gives  
\[E- L \frac{dI_1}{dt}-R(I_1-I_2)=0\]
\[50- 0.5 \frac{dI_1}{dt}-200(I_1-I_2)=0\]
. (2)
Applying Kirchoff's Voltage Law to the right hand loop  
\[\frac{1}{C} \int I_2 dt+I_2R_2+R_1(I_1-I_2)=0 \rightarrow \frac{1}{C} \int I_2 dt-R_1I_1+I_2(R_1+R_2)=0\]
\[ \frac{Q}{C}-R_1I_1+\frac{dQ}{dt}(R_1+R_2)=0\]
Using the values from the diagram gives
\[ \frac{1}{50 \times 10^{-6}} \int I_2 dt-200I_1+500 \frac{dQ}{dt}=0 \rightarrow \frac{dQ}{dt}-0.4I_1+40Q=0 \rightarrow \frac{dQ}{dt}=-40Q+0.4I_1\]
. (3)
  in (2) to give  
\[50- 0.5 \frac{dI_1}{dt}-200(I_1-\frac{dQ}{dt})=0 \rightarrow 50-0.5 \frac{dI_1}{dt}-200I_1+200 \frac{dQ}{dt}=0\]
Substitution of (3) into the last equation results in the equation  
\[50-0.5 \frac{dI_1}{dt}-120I_1+80Q=0 \rightarrow \frac{dI_1}{dt}=-240I_1+1600Q+100\]
We now have the equations


We can write this in matrix form as  
\[\begin{pmatrix} \frac{dI_1}{dt}\\ \frac{dQ}{dt}\end{pmatrix} = \left( \begin{array}{ccc} -240 & -1600 \\ 0.4 & -40 \end{array} \right) \begin{pmatrix} I_1\\ Q \end{pmatrix} + \begin{pmatrix} 100 \\ 0\end{pmatrix}\]

The eigenvalues and eigenvectors are  
\[\lambda_1=-80, \: \lambda_2=-200\]
\[\mathbf{v}_1 = \begin{pmatrix}-100\\1\end{pmatrix}, \: \mathbf{v}_2 = \begin{pmatrix}-400\\1\end{pmatrix}\]
The solution to the homogeneous system is then  
\[\begin{pmatrix} I_1\\ Q \end{pmatrix}=Ae^{-80t}\begin{pmatrix}-100\\1\end{pmatrix}+Be^{200t}\begin{pmatrix}-400\\1\end{pmatrix} \]
To find a particular solution, assume  
. Then

We get
\[I_1=1/4, \: Q=1/400\]

The general solution is then  
\[\begin{pmatrix} I_1\\ Q \end{pmatrix}=Ae^{-80t}\begin{pmatrix}-100\\1\end{pmatrix}+Be^{200t}\begin{pmatrix}-400\\1\end{pmatrix}+\begin{pmatrix}1/4\\1/400\end{pmatrix} \]
This can be fitted to any initial conditions.

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