RLC with DC current

Let us call $V(t)$ the voltage supply, a step in this case. The differential equation will be:

$\large \displaystyle V(t) = Ri(t) + L \frac{di(t)}{dt} + \frac{1}{C} \int_0^t i(\tau) d \tau$

Assuming that the amplitude of the voltage step is $V_0$ and applying Laplace Transform :

$\large \displaystyle \frac{V_0}{s} = \left ( R + Ls + \frac{1}{Cs} \right ) I(s)$

$\large \displaystyle I(s) = \frac{ \frac{V_0}{L} } {s^2 + \frac{R}{L} s + \frac{1}{LC} }$

For the given values:

$\large \displaystyle I(s) = \frac{0.5V_0}{s^2 + 0.5s + 0.125}$

The roots of $s^2 + 0.5s + 0.125$ are actually complex, which leads to a sine:

$\large \displaystyle I(s) = \frac{0.5V_0}{(s + 0.25)^2 + 0.0625}$

$\large \displaystyle I(s) = 2V_0 \cdot \frac{0.25}{(s + 0.25)^2 + 0.0625}$

Since $\large \displaystyle \mathcal{L}(e^{-at} \sin(\omega t) ) = \frac{\omega}{(s+a)^2 + \omega^2}$

$\color{#20A900}\boxed{\large \displaystyle i(t) = 2V_0e^{-0.25t}\sin(0.25t) }$

Which acutally makes sense. After the unit step, the voltage is DC, and current must tend to $0$ , since capacitors in DC current are an open circuit. So, there shall be no constant term in the current value. Or, $k=0$ .

Buf, if we consider the complex values of s:

$\large \displaystyle I(s) = \frac{0.5V_0}{s^2 + 0.5s + 0.125}$

$j$ is the imaginary unit:

$\large \displaystyle I(s) = \frac{jV_0}{s + 0.25 + 0.25j} + \frac{-jV_0}{s + 0.25 - 0.25j}$

$\color{#20A900}\boxed{\large \displaystyle i(t) = jV_0 e^{-(0.25 + 0.25j)t } - jV_0 e^{-(0.25 - 0.25j)t } }$

Which is the same as the expression above, with the sine and the exponential, just written in another form. So, in this case:

$\large \displaystyle \color{#3D99F6} k = 0, \lambda_1 = -0.25 - 0.25j, \lambda_2 = -0.25 + 0.25j, \boxed{\large \displaystyle \frac{1}{2 \lambda_1 \times \lambda_2} = 4}$

We want to find both natural frequencies $\lambda_{1,2}$ of the circuit. Calculate a transfer function, e.g. per voltage divider: $H(s)=\frac{U_c(s)}{V(s)}=\frac{ \pm\frac{1}{sC} }{ sL + R + \frac{1}{sC} } = \frac{\pm 1 }{s^2 CL + s RC + 1} = \frac{\pm 1}{CL}\cdot \frac{1}{s^2 + s\frac{R}{L} + \frac{1}{CL}} \overset{!}{=}\frac{\pm 1}{CL}\cdot\frac{1}{(s-\lambda_1)(s-\lambda_2)}$ Numerator and denominator of $H(s)$ are relatively prime, so the denominator zeroes are all natural frequencies. Comparing coefficients, we get $\frac{1}{2\lambda_1\lambda_2}=\frac{CL}{2}=\boxed{\SI{4}{s^2}}$

$\text{DSolve}\left[\left\{2 i'(t)+\frac{1}{4} \int_0^t i(x) \, dx+i(t)=1,i(0)=0\right\},i,t\right] \Rightarrow i(t)=2 e^{-\frac{t}{4}} \sin \left(\frac{t}{4}\right) \Rightarrow \mathbb{i}\ e^{\left(-\frac{1}{4}-\frac{\mathbb{i}}{4}\right) t}-\mathbb{i}\ e^{\left(-\frac{1}{4}+\frac{\mathbb{i}}{4}\right) t}$

$\frac{1}{2\frac18}\Rightarrow 4$