Second-Order RL Circuit

Let the current flowing through $R_1$ be $I$ and that flowing through $L_2$ be $I_1$ . Based on this, the equations for the circuit are derived using Kirchoff's laws and rearranged into a state-space form as shown below:

$\left[\begin{matrix}\dot{I}\\\dot{I}_1\end{matrix}\right] = \left[\begin{matrix}-2&1\\1&-1\end{matrix}\right]\left[\begin{matrix}I\\I_1\end{matrix}\right] + \left[\begin{matrix}1\\0\end{matrix}\right]V_s$

Current through the resistors $R_1$ and $R_2$ are respectively:

$I_{R1} =\left[\begin{matrix}1&0\end{matrix}\right]\left[\begin{matrix}I\\I_1\end{matrix}\right]$

$I_{R2} =\left[\begin{matrix}1&-1\end{matrix}\right]\left[\begin{matrix}I\\I_1\end{matrix}\right]$

Introducing shorthand notation. Let:

$x = \left[\begin{matrix}I\\I_1\end{matrix}\right]$ $A = \left[\begin{matrix}-2&1\\1&-1\end{matrix}\right]$ $B = \left[\begin{matrix}1\\0\end{matrix}\right]$ $u = V_s$ $C_1 = \left[\begin{matrix}1&0\end{matrix}\right]$ $C_2 = \left[\begin{matrix}1&-1\end{matrix}\right]$

So the system of governing equations becomes:

$\dot{x} = Ax + Bu$ $I_{R1} = C_1 x$ $I_{R2} = C_2 x$

Now, The first equation is solved as such:

$\dot{x} = Ax + Bu \implies e^{-At}\dot{x} - e^{-At}Ax = e^{-At}Bu$ $\implies \frac{d}{dt}\left(e^{-At}x\right) = e^{-At}Bu$

Separating variables and integrating and applying the initial condition that at $t=0$ , both $I$ and $I_1$ are equal to zero:

$x = e^{At}\left(\int_{0}^{t}e^{-As}ds\right)Bu\implies x = \left(I_{2} - e^{-At}\right)A^{-1}Bu$

Note that here $I_2$ is the identity matrix of two rows and columns. Not to be confused with the quantity of current. Finally, we get:

$I_{R1} = C_1\left(I_{2} - e^{-At}\right)A^{-1}Bu$ $I_{R2} = C_2\left(I_{2} - e^{-At}\right)A^{-1}Bu$

Now the matrix exponentials can be computed by using the eigenvalues and eigen vectors of the matrix $A$ . By substituting the matrix exponentials and other quantities and simplifying the expressions for currents gives two functions of time describing the currents $I$ and $I_1$ . A good way to check whether the answer is correct is by taking the limiting value of currents as time tends to infinity. The limiting value of $I$ is 10 amps while that for $I_1$ is zero. From here the following integrals can be computed easily but the evaluation requires some patience. I have left out most of the number-crunching and have focussed on the concepts.

$E_{R1} = \int_{0}^{10}I_{R1}^2 R_1 dt$ $E_{R2} = \int_{0}^{10}I_{R2}^2 R_2 dt$

The answer is $\boxed{\frac{E_{R1}}{E_{R2}} \approx 41.528}$

The recommended approach to this problem is numerical but I have presented an analytical route.