Transient vs. Steady State (4-16-2021)

Wow, what a great question! :D I'm learning transient analysis currently in my electrical engineering course. This was really fun to solve. Here, I labelled the currents as follows.

By KCL, $i_1 + i_2 + i_3 = 0$ . We can also formulate equations based on the voltages in the circuit. Firstly, $V_{R1} = i_1R_1 \quad V_{R2} = i_2R_2 \quad V_{L1} = L_1\frac{di_1}{dt} \quad V_{L2} = L_2\frac{di_2}{dt} \quad V_{L3} = L_3\frac{di_3}{dt}$ Using the node voltage $V_x$ : $V_x = V_1 + i_1R_1 + L_1\frac{di_1}{dt} \ \ (1)$ $V_x = V_2 + i_2R_2 + L_2\frac{di_2}{dt} \ \ (2)$ $V_x = L_3\frac{di_3}{dt} \ \ (3)$ Adding these three equations while substituting in the resistances and inductances gives: $3V_x = V_1 + V_2 + 0.25\underbrace{(i_1+i_2)}_{-i_3} + \frac{di_1}{dt} + \frac{di_2}{dt} + \frac{di_3}{dt}$ Due to the linearity of the derivative: $\frac{di_1}{dt} + \frac{di_2}{dt} + \frac{di_3}{dt} = \frac{d}{dt}\underbrace{(i_1+i_2+i_3)}_0 = 0$ Furthermore, the $V_x$ in our equation is simply $\frac{di_3}{dt}$ via equation $(3)$ . Thus, $3\frac{di_3}{dt} = V_1 + V_2 - 0.25i_3$ $3\frac{di_3}{dt} + 0.25i_3 = V_1+V_2$ This is a first order, linear, non-homogeneous ODE. Let's first find the homogeneous part of the solution. Using the trial solution method: $3\lambda + 0.25 = 0 \longrightarrow \lambda = \frac{-0.25}{3} = \frac{-1}{12}$ So the complementary solution is $i_{3c}(t) = Ae^{-\frac{t}{12}}$ .

Now we need to find the particular solution, $i_{3p}(t)$ . To do this, first simplify the RHS. $\sin\left(t+\frac{\pi}{4}\right) = \sin(t)\cos\left(\frac{\pi}{4}\right) + \cos(t)\sin\left(\frac{\pi}{4}\right) = \frac{\sin(t)+\cos(t)}{\sqrt{2}}$ $V_1(t) + V_2(t) = 10(\sin(t)+\cos(t)) + 7\sqrt{2}\sin(t) = (10+7\sqrt{2})\sin(t) + 10\cos(t)$ This is in the form $a\sin(t) + b\cos(t)$ , so this will be our 'guess' for $i_{3p}(t)$ . Then $i_{3p}'(t) = a\cos(t) - b\sin(t)$ . Substituting this into our ODE: $3(a\cos(t) - b\sin(t)) + \frac{a\sin(t) + b\cos(t)}{4} = (10+7\sqrt{2})\sin(t) + 10\cos(t)$ Equating coefficients: $\frac{a}{4} - 3b = 10+7\sqrt{2}$ $3a + \frac{b}{4} = 10$ $\begin{pmatrix} \frac{1}{4} & -3 \\ 3 & \frac{1}{4} \end{pmatrix} \begin{pmatrix} a\\b \end{pmatrix} = \begin{pmatrix} 10+7\sqrt{2} \\ 10 \end{pmatrix}$ $\begin{pmatrix} a\\b \end{pmatrix} = \begin{pmatrix} \frac{1}{4} & -3 \\ 3 & \frac{1}{4} \end{pmatrix}^{-1}\begin{pmatrix} 10+7\sqrt{2} \\ 10 \end{pmatrix} = \frac{1}{9+\frac{1}{16}} \begin{pmatrix} \frac{1}{4} & 3 \\ -3 & \frac{1}{4} \end{pmatrix}\begin{pmatrix} 10+7\sqrt{2} \\ 10 \end{pmatrix} = \begin{pmatrix} 3.859 \\ -6.312\end{pmatrix}$ Therefore, $i_3(t) = Ae^{-\frac{t}{12}} + 3.859\sin(t) -6.312\cos(t)$ The energy stored in the inductors is modelled by $E = \frac{1}{2}Li^2$ where $i$ is the current through it. This means that when the inductors are de-energized at $t = 0$ , $i_1 = i_2 = i_3 = 0$ since $L \neq 0$ . Consequently, $i_3(0) = 0$ . $0 = Ae^0 + 3.859\sin(0) - 6.312\cos(0) \longrightarrow A = 6.312$ $i_3(t) = \underbrace{6.312e^{-\frac{t}{12}}}_{\textrm{transient}} + \underbrace{3.859\sin(t) -6.312\cos(t)}_{\textrm{'steady state'}}$ The transient part is as such because as $t\rightarrow \infty$ , it goes to zero, while the steady state part doesn't. This makes sense because the homogeneous solution is always the transient while the steady state part the particular solution. The steady state part is precisely $I_{3SS}(t)$ .

Now, we are finally ready to solve the integral. The integrand is simply the complementary solution $i_{3c}(t) = 6.312e^{-\frac{t}{12}}$ . $Q = \int_{0}^{\infty} 6.312e^{-\frac{t}{12}}dt$ $= 6.312\times-12\times e^{-\frac{t}{12}}\Big|_{0}^{\infty}$ $= 12\times6.312 = \boxed{75.74}$