Epic RLC circuit

Note that $V_x = V_s$ . By KCL: $2 + i_1 + i_2 - 2V_s + i_3 - i = 0$ $i_1 + i_2 + i_3 = i + 2V_s - 2$ First calculate the initial conditions. Since no independent sources are connected prior to $t = 0$ , the capacitors and inductors will have discharged. (No excitation of circuit) This means that all of the inductors have no energy ( $E = \frac{1}{2} Li^2$ ), so $i_1 = i_2 = i_3=0$ . All capacitors have discharged so all of their voltages are $0$ . Since $i_1$ , $i_2$ and $i_3$ are all in a branch containing an inductor, all of the currents must be zero immediately after the switches are closed. This is because current cannot change instantaneously across an inductor. Thus, $i_1(0^+) = i_2(0^+) = i_3(0^+) = 0$ . $V_s(0) = 0.2\sin(0) = 0$ , so the current through the dependent current source is $2V_s = 0$ . Therefore, the middle three branches all have no current after the switches are closed, so $i(0) = i(0^+) = 2\rm A$ .

Now calculate $i'(0)$ . We know that $i_1(0^+) = i_2(0^+) = i_3(0^+) = 0$ ; this forces the voltages across all of the resistors to be zero. In addition, the capacitors have zero voltage across them immediately after the switches are closed since a capacitor's voltage cannot change instantaneously. Thus, the voltage across the inductors, which can change instantaneously, must be equal to $V_s$ . However, we know $V_s(0) = 0$ so $\frac{di_1}{dt} = \frac{V_s(0)}{L} = 0$ . Same goes for $i_2$ and $i_3$ .

Rearranging the KCL equation and differentiating: $i = i_1+i_2+i_3 -2V_s + 2$ $\frac{di}{dt} = \frac{di_1}{dt} + \frac{di_2}{dt} + \frac{di_3}{dt} - 2\frac{dV_s}{dt}$ $i'(0) = 0 + 0 + 0 - 2\cos(5(0)) = -2$ Using node voltage $V_s$ : $V_s = i_1R + L\frac{di_1}{dt} + \frac{1}{C}\int i_1(t) dt$ $V_s = i_2R + L\frac{di_2}{dt} + \frac{1}{C}\int i_2(t) dt$ $V_s = i_3R + L\frac{di_3}{dt} + \frac{1}{C}\int i_3(t) dt$ $3V_s = R(i_1 + i_2 + i_3) + L\left(\frac{di_1}{dt} + \frac{di_2}{dt} + \frac{di_3}{dt}\right) + \frac{1}{C}\int i_1(t) + i_2(t) + i_3(t) dt$ $3V_s = R(i+2V_s - 2) + L\frac{d}{dt}(i + 2V_s - 2) + \frac{1}{C}\int (i + 2V_s - 2) dt$ Differentiating: $3\frac{dV_s}{dt} = R\frac{di}{dt} + 2R\frac{dV_s}{dt} + L\frac{d^2i}{dt^2} + 2L\frac{d^2V_s}{dt^2} + \frac{1}{C}(i+2V_s -2)$ $L\frac{d^2i}{dt^2} + R\frac{di}{dt} + \frac{1}{C}i(t) = 3\frac{dV_s}{dt} - 2R\frac{dV_s}{dt} - 2L\frac{d^2V_s}{dt^2} - \frac{2V_s}{C} + \frac{2}{C}$ Use $L = 1\rm H$ , $R = 7\rm \Omega$ , $C = 100\rm mF$ , $V_s = 0.2\sin(5t)$ . $\frac{dV_s}{dt} = 1\cos(5t) \quad \frac{d^2V_s}{dt^2} = -5\sin(5t)$ Simplify RHS: $(3-2(7))\cos(5t) - 2(-5\sin(5t)) - \frac{2}{0.1} (0.2\sin(5t)) + \frac{2}{0.1} = -11\cos(5t) + 10\sin(5t) - 4\sin(5t) + 20$ $\frac{d^2i}{dt^2} + 7\frac{di}{dt} + 10i(t) = 6\sin(5t) -11\cos(5t) + 20$ Use trial solution method to find solution to homogeneous version. $\lambda^2 + 7\lambda + 10 = (\lambda+5)(\lambda+2) =0$ $\lambda_1 = -5 \quad \lambda_2 = -2$ $i_{c}(t) = Ae^{-5t} + Be^{-2t}$ Guess for particular solution is $i_{p}(t) = a\sin(5t) + b\cos(5t) + c$ . $i_{p}'(t) = 5a\cos(5t) -5b\sin(5t)$ $i_{p}''(t) = -25a\sin(5t) -25b\cos(5t)$ Substitute into ODE: $(-25a\sin(5t) -25b\cos(5t)) + 7(5a\cos(5t) -5b\sin(5t)) + 10(a\sin(5t) + b\cos(5t) + c ) = 6\sin(5t) -11\cos(5t) + 20$ $(-15a -35b)\sin(5t) + (35a-15b)\cos(5t) + 10c = 6\sin(5t) -11\cos(5t) + 20$ Equating coefficients: $-15a - 35b = 6$ $35a - 15b = -11$ $10c = 20\longrightarrow c = 2$ $\begin{pmatrix} -15&-35 \\ 35 & -15\end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}6\\-11 \end{pmatrix}$ $5\begin{pmatrix} -3&-7 \\ 7 & -3\end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}6\\-11 \end{pmatrix}$ $\begin{pmatrix}a\\b\end{pmatrix} = \frac{1}{5(3^2 + 7^2)}\begin{pmatrix} -3&7 \\ -7 & -3\end{pmatrix}\begin{pmatrix}6\\-11 \end{pmatrix} = \frac{1}{5(3^2 + 7^2)}\begin{pmatrix} -18 -77\\ -42 + 33 \end{pmatrix} = \frac{1}{5(3^2 + 7^2)}\begin{pmatrix} -95\\-9\end{pmatrix}$ $a = \frac{-19}{3^2+7^2} \quad b = \frac{-9}{5(3^2+7^2)}$ $i(t) = Ae^{-5t} + Be^{-2t} - \frac{19}{3^2+7^2}\sin(5t)- \frac{9}{5(3^2+7^2)}\cos(5t) + 2$ $i'(t) = -5Ae^{-5t} -2Be^{-2t} - \frac{95}{ 3^2+7^2}\cos(5t) + \frac{9}{3^2+7^2}\sin(5t)$ Use initial conditions $i(0) = 2$ and $i'(0) = -2$ . $2 = A + B - \frac{9}{5(3^2+7^2)} + 2 \longrightarrow A + B = \frac{9}{290}$ $-2 = -5A - 2B - \frac{95}{3^2+7^2} \longrightarrow 5A + 2B = 2 - \frac{95}{58} = \frac{21}{58}$ $\begin{pmatrix} 1 & 1\\5 & 2\end{pmatrix}\begin{pmatrix}A\\B\end{pmatrix} = \begin{pmatrix} \frac{9}{290} \\ \frac{21}{58} \end{pmatrix}$ $\begin{pmatrix}A\\B\end{pmatrix} = \frac{1}{-3}\begin{pmatrix} 2 & -1\\-5 & 1\end{pmatrix} \begin{pmatrix} \frac{9}{290} \\ \frac{21}{58}\end{pmatrix} = \frac{-1}{3}\begin{pmatrix} \frac{18}{290} - \frac{105}{290} \\ \frac{-45}{290} + \frac{105}{290}\end{pmatrix} = \frac{-1}{3}\begin{pmatrix} \frac{-87}{290}\\ \frac{60}{290} \end{pmatrix} = \begin{pmatrix} \frac{29}{290} \\ \frac{-20}{290}\end{pmatrix}$ $A = \frac{1}{10}\quad B = \frac{-2}{29}$ $i(t) = \frac{e^{-5t}}{10} - \frac{2e^{-2t}}{29} - \frac{19}{58}\sin(5t)- \frac{9}{290}\cos(5t) + 2$ Net amount of charge is the integral over the time interval. $Q = \displaystyle \int_{0}^{2\pi}\frac{e^{-5t}}{10} - \frac{2e^{-2t}}{29} - \frac{19}{58}\sin(5t)- \frac{9}{290}\cos(5t) + 2 \ dt$ Integrating the sine and cosine terms on $[0,2\pi]$ result in zero, Thus, we just have $Q = \displaystyle \int_{0}^{2\pi}\frac{e^{-5t}}{10} - \frac{2e^{-2t}}{29} + 2 \ dt$ $= \frac{-e^{-5t}}{50}+ \frac{e^{-2t}}{29} + 2t\Big|_{0}^{2\pi}$ $= \frac{-e^{-10\pi}}{50} + \frac{e^{-4\pi}}{29} + 4\pi + \frac{1}{50} - \frac{1}{29}$ $= \boxed{\frac{e^{-4\pi}-1}{29} + \frac{1 - e^{-10\pi}}{50} + 4\pi \approx 12.55}$

Here's a solution using Laplace-Transforms. We use the same currents and voltages Charley Shi defined in the other solution and assume $V_s(t)$ is defined downward. Before we begin, we normalize all currents, voltages, time and elements by the values given below: $\begin{aligned} \text{voltages:}&&&\SI{1}{V}, &&&\text{currents:}&&&\SI{1}{A},&&&\text{time:}&&&\SI{1}{s}&&& \Rightarrow &&&&R:&&&\SI{1}{\ohm}, &&& C:&&&\SI{1}{F}, &&&L:&&&\SI{1}{H},&&&Q:&&&\SI{1}{C} \end{aligned}$ The network equations remain the same in the process, but now all currents, voltages, time and elements represent their normalized dimensionless counterparts.

Before switching ( $t<0$ )

Assume the network is asymptotically stable for all $t<0$ (discussion in the appendix)

Both input sources can be omitted, as they are only connected on one side. Assuming the remaining network is asymptotically stable, it will be in steady state for all $t<0$ , because it exists for all time and has no input sources. In this steady state, all currents and voltages are zero.

After switching ( $t\geq 0$ )

All initial conditions vanish at $t=0^-$ because of the steady state for $t<0$ . We notice $V_x(t)=V_s(t)$ , so we may regard the controlled source as another independent input source. We also notice by symmetry of the elements and initial conditions, we must have $i_1(t)=i_2(t)=i_3(t)$ .

Now we compute $i(t)$ . As the initial conditions vanish, it only depends on three input sources. We use superposition and consider each input source independently. For both current sources, all impedances are shorted and can be omitted, so we directly get $i(t)$ . For the remaining input $V_s(t)$ we need a transfer function $H(s)$ :: $H(s)=\frac{I(s)}{V_s(s)}=\frac{1}{ \frac{1}{3}\left(sL+R+\frac{1}{sC}\right) } = \frac{s 3C}{s^2CL + sRC+1}=\frac{3}{L}\cdot\frac{s}{s^2+s\frac{R}{L} + \frac{1}{CL}}=\frac{3s}{s^2+ 7s + 10}= \frac{3s}{(s+2)(s+5)}$ For the input $V_s(t)$ with Laplace-Transform $V_s(s)$ , we use PFD by Covering-Method to split the single real poles: $I(s)=H(s) V_s(s)=\frac{3s}{(s+2)(s+5)}\cdot\frac{5\cdot 0.2}{s^2+25}\underset{\text{PFD}}{=}3\left[ \frac{-2}{3\cdot 29}\cdot\frac{1}{s+2} + \frac{5}{3\cdot 50}\cdot\frac{1}{s+5} + \frac{Xs+Y}{s^2+25} \right],\qquad X,\:Y\in\mathbb{R}$ We don't need to calculate the coefficients $X,\:Y$ , as that part will cancel later anyway. Via superposition of all three sources and inverse Laplace-Transform, we get $i(t)=2 - 2V_s(t) + \cancel{\frac{3}{3}}\left[ -\frac{2}{29}e^{-2t} + \frac{5}{50}e^{-5t} + K\cos(5 t+\varphi) \right],\qquad K,\:\varphi\in\mathbb{R}$ Again, we don't need to calculate the actual values of amplitude $K$ and phase $\varphi$ , as that part will cancel out anyway, just like $V_s(t)$ of the controlled source. The reason is both functions are sines/cosines with a period of $\frac{2\pi}{5}$ , and we integrate over five whole periods: $Q(2\pi)=\int_0^{2\pi}i(t)\:dt = 4\pi - 0 + \left[ \frac{1}{29}e^{-2 t} - \frac{1}{50}e^{-5 t} \right]_0^{2\pi} + 0 = 4\pi - \frac{1}{29}(1-e^{-4\pi}) + \frac{1}{50}(1 - e^{-10\pi})\approx\boxed{12.552}$

Missing units and stability

Rem.: The controlled current source and $V_s(t)$ (not yet normalized) might be missing some units: $\begin{aligned} V_s(t)&=\SI{0.2}{V}\sin(\num{5}\red{\si{s^{-1}}}t), &&& i_x(t)&=\num{2}\red{\si{\ohm^{-1}}}V_x(t) \end{aligned}$

Rem.: Let's think about asymptotic stability for $t<0$ . As the controlled current source is oriented against the voltage $V_x(t)$ , the resulting element equation is equivalent to a negative resistance $R_x=\SI{-0.5}{\ohm}$ : It seems highly unlikely that the resulting network could be asymptotically stable.

However, the state space model for $t<0$ actually yields all six poles with negative real part, so the network is asymptotically stable despite the controlled current source. I suspect the other resistances are large enough so that they can compensate the destabilizing negative resistance of the controlled current source.

Applying Kirchoff's current law:

$I = I_1 + I_2$ $I_3 = I_X + I_1$ $I_3 = I_4 + I_5$ $I_5 = 2 + I_6$

$\implies I_3 = I_4 + 2 + I_6$ $\implies I_X + I_1 = I_4 + 2 + I_6$ $\implies I_X + I - I_2 = = I_4 + 2 + I_6$ $\implies \boxed{I = I_2 + I_4 + I_6 - I_X +2}$

Now, applying Kirchoff's voltage law to various loops gives:

$-V_S + V_X=0$ $\implies V_X = V_S \implies I_X = 2V_S$ $L\dot{I}_2 + \frac{Q_2}{C} + I_2R = 0$ $L\dot{I}_4 + \frac{Q_4}{C} + I_4R = 0$ $L\dot{I}_6 + \frac{Q_6}{C} + I_6R = 0$ $\dot{Q}_2 = I_2$ $\dot{Q}_4 = I_4$ $\dot{Q}_6 = I_6$

The initial conditions can be deciphered from the fact that at when the switches are open, no current flows and that all capacitors discharge over a long period of time. Therefore:

The current through each RLC branch is governed by:

$L\dot{I}_b + \frac{Q_b}{C} + I_bR = 0$ $\dot{Q}_b = I_b$ $Q_b(0)=I_b(0) = 0$ $b = 2,4,6$

The above is solved numerically as follows:

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clear all
clc

% Parameters:
R          = 7;
C          = 100/1000;
L          = 1;

% Time step and initialisation
dt         = pi/10000;
tf         = 2*pi;
t          = 0:dt:tf;

% Initial conditions:
Qb(1)      = 0;
Ib(1)      = 0;

% Loop to simulate branch current evolution:
for k = 1:length(t)-1

    % Source voltage:
    Vs(k)       = 0.2*sin(5*t(k));

    % Governing differential equation:
    dIb         = (Vs(k) - Qb(k)/C - Ib(k)*R)/L;

    % Numerical integration: Semi implicit Euler:
    Ib(k+1)     = Ib(k) + dIb*dt;
    Qb(k+1)     = Qb(k) + Ib(k+1)*dt;

    if k == length(t)-1
        Vs(k+1) = 0.2*sin(5*t(k+1));
    end
end

% Calculating source current:
I1     = Ib;
I4     = Ib;
I6     = Ib;
Ix     = 2*Vs;
I      = 3*Ib - Ix + 2;

% Computing required integral:
Q      = 0;
for k = 1:length(I)

    Q    = Q + I(k)*dt;

end

ANSWER = Q % ~ 12.5525