DC Current Splitting with Transient

You can use the final limit theorem to avoid inverse Laplace-Transforms and integration at the same time! 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.

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After switching ( $t\geq 0$ )

The network was de-energized at $t=0^-$ , so all initial conditions at $t=0^-$ vanish and $i_{R_{1,2}}(t)$ only depend on the input sources $v_s(t),\:i_s(t)$ . To calculate the two currents efficiently with Laplace-Transforms, we define the voltage $u_C(t)$ over $C$ from top to bottom. With KVL, we get $\begin{aligned} V_s(s)&=\frac{10}{s},& I_s(s)&=\frac{5}{s}&&&\Rightarrow &&&& I_{R_1}(s)&=\frac{U_C(s)-V_s(s)}{R_1} = \frac{U_C(s)}{2} - \frac{5}{s},&&& I_{R_2}(s)&=\frac{U_C(s)}{sL + R_2}=\frac{U_C(s)}{50s+3} \end{aligned}$ To calculate $U_C(s)$ , we combine $R_1,\:V_s(s)$ into an equivalent current source and use superposition afterwards: $\begin{aligned} U_C(s) &= \left((sL+R_2)\|R_1\|\frac{1}{sC}\right)\cdot \left( I_s(s) + \frac{V_s(s)}{R_1} \right)=\frac{1}{ s + \frac{1}{2}+\frac{1}{50s+3} } \cdot \left(\frac{5}{s}+\frac{5}{s}\right)=\frac{s + 0.06}{s^2+0.56s+0.05}\cdot\frac{10}{s} \end{aligned}$ We calculate the Laplace-Transforms of both integrands. The poles at $s=0$ cancel out, i.e. $I_0,\:4I_0$ are the steady-state currents $i_{R_{1,2}}(t)$ will converge to: $\left.\begin{aligned} I_{R_1}(s)-\frac{I_0}{s}&=\frac{U_C(s)}{2} - \frac{6}{s} =\frac{-6s^2 + 1.64s}{(s^2+0.56s+0.05)s} = \frac{-6s+1.64}{s^2+0.56s+0.05} \\[.5em] \frac{4I_0}{s} - I_{R_2}(s) &=\frac{4}{s} - \frac{U_C(s)}{50s+3} = \frac{4s^2+2.24s}{(s^2+0.56s+0.05)s} = \frac{4s+2.24}{s^2+0.56s+0.05} \end{aligned}\qquad\right|s_{1,2}=-\frac{7}{25}\pm\sqrt{\frac{49}{625}-\frac{1}{20}}=-\frac{7}{25}\pm\frac{\sqrt{71}}{50}<0$ Both remaining poles $s_{1,2}$ lie in the open left plane, so we may use the Final Value Theorem to calculate both parts of $Q$ together: $Q = \lim_{s\rightarrow 0} s\cdot\frac{1}{s}\left( I_{R_1}(s) - \frac{I_0}{s} + \frac{4I_0}{s} -I_{R_2}(s) \right)=\lim_{s\rightarrow 0}\frac{-2s + 3.88}{s^2+0.56s+0.05}=\frac{3.88}{0.05}=\boxed{77.6}$

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Rem.: This way is efficient if you're only interested in the value of $Q$ , but not the transients $i_{R_{1,2}}(t)$ For the transients, there is no way around partial fraction decomposition (PFD) and inverse Lapace-Transform :(

Let $I_1= I_{R1}$ and $I_2 = I_{R2}$ . Let the charge on the capacitor be $Q_c$ and the current through it be $I_c$ at a general time $t$ . Applying Kirchoff's laws gives the circuit equations:

$-10 -2I_1 + 3I_2 + 50\dot{I}_2 = 0$ $50\dot{I}_2 +3I_2 = Q_c$ $I_S = 5 = I_1 + I_2 + \dot{Q}_c$

Simplifying the equations a bit more leads to the system of ODEs:

$2\dot{Q}_c + 50\dot{I}_2 = -5I_2 + 20$ $50\dot{I}_2=-3I_2 + Q_c$

$I_2(0) = Q_c(0) = 0$

Taking Laplace transform on both sides and solving for $I_2(s)$ and $I_c(s)$ knowing that $I_c(s) = sQ_c(s)$ :

$I_2(s) = \dfrac{20}{s\left(100s^2+56s+5\right)}$ $I_c(s) = \dfrac{20\left(50s+3\right)}{\left(100s^2+56s+5\right)}$ $\because 5 = I_1 + I_2 + \dot{Q}_c$

Now computing the inverse Laplace transform using Wolfram leads to the following. Let:

$a = -\frac{7}{25} - \frac{\sqrt{71}}{50}$ $b = -\frac{7}{25} + \frac{\sqrt{71}}{50}$

$I_2(t) = \frac{2}{71} \left((-71+14\sqrt{71}) \mathrm{e}^{at} -(71 + 14\sqrt{71})\mathrm{e}^{bt} +142\right)$ $I_c(t) = \frac{5}{\sqrt{71}} \left((11 + \sqrt{71}) \mathrm{e}^{at} + (\sqrt{71}-11)\mathrm{e}^{bt}\right)$ $\because 5 = I_1 + I_2 + \dot{Q}_c$ $\therefore I_1(t) = 5 - I_2(t) - I_c(t)$

After having obtained these closed form expressions, the next step is to evaluate the required integrals which also have a closed form:

$\int_{0}^{\infty} \left( 4-I_2(t)\right) \ dt = \int_{0}^{\infty} \left( 4 - \frac{2}{71} \left((-71+14\sqrt{71}) \mathrm{e}^{at} -(71 + 14\sqrt{71})\mathrm{e}^{bt} +142\right) \right) \ dt$ $\int_{0}^{\infty} \left( I_1(t)-1\right) \ dt = \int_{0}^{\infty} \left(5 - \frac{2}{71} \left((-71+14\sqrt{71}) \mathrm{e}^{at} -(71 + 14\sqrt{71})\mathrm{e}^{bt} +142\right) - \frac{5}{\sqrt{71}} \left((11 + \sqrt{71}) \mathrm{e}^{at} + (\sqrt{71}-11)\mathrm{e}^{bt}\right) - 1 \right) \ dt$

I did not even consider touching these manageable but very tedious monsters. Numerically evaluating leads to:

$Q = \int_{0}^{\infty} \left( I_1(t)-1\right) \ dt + \int_{0}^{\infty} \left( 4-I_2(t)\right) \ dt \approx 77.6$