RLC Bridge (2-6-2021)

Been a while since I had a go at one of these circuit problems. This was fun.

Circuit equations as per Kirchoff's laws:

$\dot{Q}_{C1} = I_{C1} \dots (1)$ $\dot{Q}_{C2} = I_{C2} \dots (2)$ $I_S = I_{L1} + I_{R1} \dots (3)$ $I_{L1} = I_{C1} + I_{C2} \dots (4)$ $I_{L2} = I_{R1} + I_{C1} \dots (5)$ $L_1 \dot{I}_{L1} + R_3 I_{C2} + \frac{ Q_{C2}}{C_2}=V_S \dots (6)$ $L_1 \dot{I}_{L1} + \frac{ Q_{C1}}{C_1} = R_1I_{R1} \dots (7)$ $R_3 I_{C2} + \frac{ Q_{C2}}{C_2} = R_2 I_{L2} + L_2 \dot{I}_{L2} + \frac{ Q_{C1}}{C_1} \dots (8)$

Now eliminating $I_{R1}$ in (7) using (3):

$L_1 \dot{I}_{L1} + \frac{ Q_{C1}}{C_1} = R_1(I_{L2} - \dot{Q}_{C1} )\dots (9)$

(2) can also be written as:

$\dot{Q}_{C1}+\dot{Q}_{C2}=I_{L1}$

Using (2), (6), (8) and (9) to construct a system of ODE's in matrix form gives:

$\left[\begin{matrix} L_1&0&0&R_3\\L_1&0&R_1&0\\0&-L_2&0&R_3\\0&0&1&1 \end{matrix} \right] \left[\begin{matrix} \dot{I}_{L1}\\\dot{I}_{L2}\\\dot{Q}_{C1}\\\dot{Q}_{C2} \end{matrix} \right] = \left[\begin{matrix} 0&0&0&-1/C_2\\0&R_1&-1/C_1&0\\0&R_2&1/C_1&1/C_2\\1&0&0&0 \end{matrix} \right] \left[\begin{matrix} I_{L1}\\I_{L2}\\Q_{C1}\\Q_{C2} \end{matrix} \right] + \left[\begin{matrix} 1\\0\\0\\0 \end{matrix} \right]V_S$ Let:

$A = \left[\begin{matrix} L_1&0&0&R_3\\L_1&0&R_1&0\\0&-L_2&0&R_3\\0&0&1&1 \end{matrix} \right]$ $B = \left[\begin{matrix} 0&0&0&-1/C_2\\0&R_1&-1/C_1&0\\0&R_2&1/C_1&1/C_2\\1&0&0&0 \end{matrix} \right]$ $C = \left[\begin{matrix} 1\\0\\0\\0 \end{matrix} \right]$ $x = \left[\begin{matrix} I_{L1}\\I_{L2}\\Q_{C1}\\Q_{C2} \end{matrix} \right]$

$\implies A \dot{x} = Bx + CV_S$ $x(0) = \left[\begin{matrix} 0\\0\\0\\0 \end{matrix} \right]$

Explicit Euler numerical integration yields:

$x_{k+1} = x_{k} + \delta t \left(A^{-1}(Bx_k + CV_S)\right)$

Iterating this in a loop using a short script of code yields the required answer. To get $I_S$ , a linear combination of appropriate states and their derivatives need to be taken keeping in mind equation (3). For a step size of $10^{-5} \ \mathrm{sec}$ , the answer I get is $\boxed{704.95}$ .