4th Order RLC (3-20-2020)

Let's use the impedances in the Laplace form. Since the ratio voltage-current in an inductor with indutance $L$ is: $V(t) = L\frac{di(t)}{dt}$ , in the laplace domain the ratio $\frac{V(s)}{I(s)}$ is $L s$ . Likewise this ratio for a capacitor with capacitance $C$ is $\frac{1}{Cs}$ and for a resistor with resistance $R$ is just $R$ . So, the total impedance of this circuit is ( $//$ means impedances in parallel):

$\large \displaystyle Z(s) = \left ( \frac{1}{s} // 1 \right ) + s + \frac{1}{s} + ( s//1 )$

$\large \displaystyle Z(s) = \frac{\frac{1}{s} \cdot 1}{\frac{1}{s} + 1} + s + \frac{1}{s} + \frac{s \cdot 1}{s + 1}$

$\large \displaystyle Z(s) = \frac{1}{s+ 1} + s + \frac{1}{s} + \frac{s}{s + 1}$

$\large \displaystyle Z(s) = \frac{s^2 + s + 1}{s}$

The current in the laplace domain will be:

$\large \displaystyle I_S(s) = \frac{V_S(s)}{Z(s)}$

Since the input voltage is a step of amplitude $10$ , $V_S(s) = \frac{10}{s}$ . So:

$\large \displaystyle I_S(s) = \frac{10}{s^2+s+1}$

$\large \displaystyle I_S(s) = \frac{20}{\sqrt{3}} \cdot \frac{\sqrt{\frac34}}{\left (s+\frac12 \right )^2 + \frac34}$

In the time domain:

$\large \displaystyle I_S(t) = \frac{20}{\sqrt{3}} e^{- \frac{t}{2} } \sin \left ( \frac{t \sqrt{3}}{2} \right )$

Differentiating and making in equal to 0, one gets that the solutions are in the form:

$\large \displaystyle \tan \left ( \frac{t \sqrt{3}}{2} \right ) = \sqrt{3}$

Or:

$\large \displaystyle \frac{t \sqrt{3}}{2} = \frac{\pi}{3} + k \pi$

$\large \displaystyle t = \frac{2 \pi}{3 \sqrt{3}} (1 + 3k)$

Where k is an integer. Our maximum $I_{Smax}$ occurs when $k = 0$ and our minimum $I_{Smin}$ occurs when $k = 1$ . So:

$\color{#20A900} \boxed{ \large \displaystyle I_{Smax} = 10 e^{- \frac{\pi}{3 \sqrt{3}} } \approx 5.46293 }$

$\color{#20A900} \boxed{ \large \displaystyle I_{Smin} = - 10 e^{- \frac{4 \pi}{3 \sqrt{3}} } \approx -0.89064}$

Our quantity $E_s(t)$ will be:

$\large \displaystyle E_s(t) = \frac{200}{\sqrt{3}} \int_0^t e^{- \frac{t}{2} } \sin \left ( \frac{t \sqrt{3}}{2} \right ) dt$

After integration by parts twice:

$\large \displaystyle E_s(t) = 100 - \frac{200}{\sqrt{3}} e^{- \frac{t}{2} } \sin \left ( \frac{t \sqrt{3}}{2} + \frac{\pi}{3} \right )$

Since the exponential will go to $0$ as $t$ goes to $\infty$ , it's easy to see that:

$\color{#20A900} \boxed{ \large E_{S \infty} = 100}$

For $E_{Smax}$ , differentiating and making it equal to $0$ , one gets that the solutions are in the form:

$\large \displaystyle \tan \left ( \frac{t \sqrt{3}}{2} + \frac{\pi}{3} \right ) = \sqrt{3}$

Or:

$\large \displaystyle \frac{t \sqrt{3}}{2} + \frac{\pi}{3} = \frac{\pi}{3} + k \pi$

$\large \displaystyle t = \frac{2k \pi}{\sqrt{3}}$

At $k=0$ we will actually have $E_{Smin} = 0$ , and $E_{Smax}$ will be attained at $k=1$ :

$\color{#20A900} \boxed{ \large \displaystyle E_{Smax} = 100 ( 1 + e^{- \frac{\pi}{\sqrt{3}} } ) \approx 116.30335}$

Calculating $Q$ :

$\color{#3D99F6} \boxed { \large \displaystyle Q = -100 \left ( 1 + e^{- \frac{\pi}{\sqrt{3}}} \right ) e^{- \frac{5\pi}{\sqrt{3}}} \approx - 5.65875 }$

@Guilherme Niedu has given a very nice Laplace solution. I will show the state-space formulation that can be used in conjunction with numerical integration. Express the time derivatives of state variables $(V_{C1}, V_{C2}, I_{L1}, I_{L2})$ in terms of the state variables and the forcing function (the source). Once the solution routine is written, it is trivial to find energies, maximum quantities, etc.

For capacitor 1:

$I_{R1} = \frac{V_{C1}}{R_1} \\ I_{C1} = I_{L1} - I_{R1} \\ \dot{V}_{C1} = \frac{I_{C1}}{C_1}$

For capacitor 2:

$I_{C2} = I_{L1} \\ \dot{V}_{C2} = \frac{I_{C2}}{C_2}$

For inductor 1:

$I_{R2} = I_{L1} - I_{L2} \\ V_{R2} = R_2 I_{R2} \\ V_{L1} = (V_S - V_{C1}) - (V_{R2} + V_{C2}) \\ \dot{I}_{L1} = \frac{V_{L1}}{L_1}$

For inductor 2:

$V_{L2} = V_{R2} \\ \dot{I}_{L2} = \frac{V_{L2}}{L_2}$

Source Current:
$I_S = I_{L1}$

Numerical integration (Euler, for example):

$V_{C1 \, (k)} = V_{C1 \, (k-1)} + \dot{V}_{C1 \, (k-1)} \, \Delta t \\ V_{C2 \, (k)} = V_{C2 \, (k-1)} + \dot{V}_{C2 \, (k-1)} \, \Delta t \\ I_{L1 \, (k)} = I_{L1 \, (k-1)} + \dot{I}_{L1 \, (k-1)} \, \Delta t \\ I_{L2 \, (k)} = I_{L2 \, (k-1)} + \dot{I}_{L2 \, (k-1)} \, \Delta t$