4th-Order Circuit

This problem is interesting on many levels. I hope to explain it all in the following description. At a general time $t$ , say the charge on $C_1$ is $Q_1$ , that on $C_2$ is $Q_2$ . Let the current flowing through $L_1$ be $I_1$ and that through $L_2$ be $I_2$ . Since the capacitors are discharging, it can be concluded that:

$\dot{Q}_1 = -I_1$ $\dot{Q}_2 = -I_2$ Using Kirchoff's laws, the circuit equations can be written as:

$-\frac{Q_1}{C_1} + L_1 \dot{I}_1 + R(I_1 + I_2)=0$ $-\frac{Q_2}{C_1} + L_1 \dot{I}_2 + R(I_1 + I_2)=0$

Now, I often make use of linear algebra to make this system of equations more compact. The benefit of using linear algebra is that doing so enables me to code in a more compact manner. By rearranging the above equations, one can obtain the following state-space form:

$\left[\begin{matrix}\dot{Q}_1\\ \dot{Q}_2\\ \dot{I}_1\\ \dot{I}_2\end{matrix}\right] = \left[\begin{matrix}0&0&-1&0\\0&0&0&-1\\ \frac{1}{L_1C_1}&0&-R/L_1&-R/L_1\\ 0&\frac{1}{L_2C_2}&-R/L_2&-R/L_2\end{matrix}\right] \left[\begin{matrix}Q_1\\ Q_2\\ I_1\\ I_2\end{matrix}\right]$

Or:

$\dot{x} = Ax$ Where $A = \left[\begin{matrix}0&0&-1&0\\0&0&0&-1\\ \frac{1}{L_1C_1}&0&-R/L_1&-R/L_1\\ 0&\frac{1}{L_2C_2}&-R/L_2&-R/L_2\end{matrix}\right]$ $x = \left[\begin{matrix}Q_1\\ Q_2\\ I_1\\ I_2\end{matrix}\right]$

The initial conditions are stated in the problem and are written as a vector: $x(0) = \left[\begin{matrix}C_1V_{C_1}\\ C_2V_{C_2}\\ 0\\ 0\end{matrix}\right]$

Now, the total energy of the circuit is as follows:

$E = \frac{Q_1^2}{2C_1} + \frac{Q_2^2}{2C_2} + \frac{L_1I_1^2}{2} + \frac{L_2I_2^2}{2}$

This can be written in a compact form using linear algebra as follows:

$E = \frac{1}{2}x^TWx$

Where $x$ is the vector defined above and

$W = \left[\begin{matrix}\frac{1}{C_1}&0&0&0\\0&\frac{1}{C_1}&0&0\\ 0&0&L_1&0\\ 0&0&0&L_2\end{matrix}\right]$

The energy at $t=0$ is: $E_o = \frac{1}{2}x(0)^TWx(0)$

Now, I define normalized energy as follows:

$E_{R} = \frac{E}{E_o}$

The following plot shows two cases. One where $R=2$ and the other where $R=0$ . Notice how well the law of conservation of energy is demonstrated through this . It is known that a resistor is a dissipative element and without it, energy remains constant.

What also makes this problem interesting is the answer. I am lucky to have got it right cause I obtain an answer of approximately $\boxed{42.7}$ seconds and not 42.5 seconds. Subsequent tries got me to the correct ballpark. Moreover, the use of linear algebra made the numerical solution more compact in my opinion. The simulation code is below (MATLAB).

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clear all
clc

%System Parameters:
R       = 2;
L1      = 1;
L2      = 2;
C1      = 3;
C2      = 1;
VC1     = 5;
VC2     = 10;

% Definition of A and W matrices:
A       = [0 0 -1 0;0 0 0 -1;1/(L1*C1) 0 -R/L1 -R/L1;0 1/(L2*C2) -R/L2 -R/L2];
W       = diag([1/C1;1/C2;L1;L2]);

% Initial conditions and circuit energy initialisation:
xs(:,1) = [C1*VC1;C2*VC2;0;0];
E(1)    = 0.5*xs(:,1)'*W*xs(:,1);

% Time initialisation:
dt      = 1e-5;
tf      = 50;
t       = 0:dt:tf;
N       = length(t);

% Solving Linear System of Diff. eqs by Explicit Euler..
% And computing Circuit Energy:

for k = 2:N
    dxs     = A*xs(:,k-1);
    xs(:,k) = xs(:,k-1) + dt*dxs;
    E(k)    =  0.5*xs(:,k)'*W*xs(:,k);
end

% Normlaised energy:
Er      = E/E(1);

% Postprocessing:
plot(t,E/E(1),'Linewidth',1.5)
grid on
xlabel('Time')
title('Normalised Energy of Circuit vs. Time')
ylabel('E/E_o')
ylim([0 1.1])
hold on

The state variables are the capacitor voltages $(V_{C1} , V_{C2})$ and the inductor currents $(I_{L1} , I_{L2})$ . Express the rates of change of the state variables in terms of the state variables themselves.

Capacitor equations:

$-I_{L1} = C_1 \dot{V}_{C1} \\ -I_{L2} = C_2 \dot{V}_{C2}$

Inductor equations:

$V_{C1} - R(I_{L1} + I_{L2}) = L_1 \dot{I}_{L1} \\ V_{C2} - R(I_{L1} + I_{L2}) = L_2 \dot{I}_{L2}$

Isolating the derivative terms and numerically integrating yields the desired solution. Recall also the equations for the energy stored in capacitors and inductors:

$E_C = \frac{1}{2} C V_C^2 \\ E_L = \frac{1}{2} L I_L^2$

It takes about $42.5$ units of time for the total stored energy to drop to $9 \%$ of its original value. A plot of the resistor current and energy ratio is shown below: