Instantaneous Current (Inrush) Calculation
Well, I am trying to analyze the following circuit:



simulate this circuit &ndash; Schematic created using CircuitLab
When we use and apply KCL, we can write the following set of equations:
$$\text{I}_1=\text{I}_3+\text{I}_4\tag1$$
When we use and apply Ohm's law, we can write the following set of equations:
$$
\begin{cases}
\text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\
\\
\text{I}_1=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\
\\
\text{I}_3=\frac{\text{V}_2}{\text{R}_3}\\
\\
\text{I}_4=\frac{\text{V}_2}{\text{R}_4}
\end{cases}\tag2
$$
Substitute \$(2)\$ into \$(1)\$, in order to get:
$$
\begin{cases}
\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_2}{\text{R}_3}+\frac{\text{V}_2}{\text{R}_4}\\
\\
\frac{\text{V}_1-\text{V}_2}{\text{R}_2}=\frac{\text{V}_2}{\text{R}_3}+\frac{\text{V}_2}{\text{R}_4}
\end{cases}\tag3
$$
Now, we can solve for \$\text{V}_2\$:
$$\text{V}_2=\frac{\text{R}_3\text{R}_4\text{V}_\text{i}}{\text{R}_3\left(\text{R}_1+\text{R}_2\right)+\text{R}_4\left(\text{R}_1+\text{R}_2+\text{R}_3\right)}\tag4$$
Using \$(2)\$, we can see that:
$$\text{I}_3=\frac{\text{V}_2}{\text{R}_3}=\frac{\text{R}_4\text{V}_\text{i}}{\text{R}_3\left(\text{R}_1+\text{R}_2\right)+\text{R}_4\left(\text{R}_1+\text{R}_2+\text{R}_3\right)}\tag5$$

Now, applying this to your circuit we need to use (from now on I use the lower case letters for the function in the 'complex' s-domain where I used Laplace transform):

$$\text{R}_2=\text{sL}\tag6$$
$$\text{R}_3=\frac{1}{\text{sC}}\tag7$$
The input voltage is a stable DC voltage equal to \$\hat{\text{u}}\$, so:
$$\text{v}_\text{i}\left(\text{s}\right)=\frac{\hat{\text{u}}}{\text{s}}\tag8$$

So, we get:
$$\text{i}_3\left(\text{s}\right)=\frac{\hat{\text{u}}}{\text{s}}\cdot\frac{\text{R}_4}{\frac{1}{\text{sC}}\left(\text{R}_1+\text{sL}\right)+\text{R}_4\left(\text{R}_1+\text{sL}+\frac{1}{\text{sC}}\right)}\tag9$$
Now, when we use inverse Laplace transform (using your values) we can see:
$$\text{I}_3\left(t\right)=\frac{2 e^{-\frac{5}{2} \left(\sqrt{15976001}+4001\right) t} \left(e^{5 \sqrt{15976001} t}-1\right)}{\sqrt{15976001}}\tag{10}$$
Plotting that, gives:

The maximum occurs when:
$$\hat{t}=\frac{2 \text{arccoth}\left(\frac{4001}{\sqrt{15976001}}\right)}{5 \sqrt{15976001}}\approx0.000380305\space\text{s}\tag{11}$$
And at that time \$\$ we have a current of:
$$\text{I}_3\left(\hat{t}\right)\approx0.000498226\space\text{A}\tag{12}$$

Using Mathematica I found a way of finding the time when the maximum occurs:
In[1]:=Clear[&quot;Global`*&quot;];
R2 = s*L;
R3 = 1/(s*c);
Vi = u/s;
R1 = ;
R4 = ;
u = ;
L = ;
c = ;
\[Tau] = FullSimplify[
   Solve[{D[
       FullSimplify[
        InverseLaplaceTransform[
         I3 /. FullSimplify[
             Solve[{I1 == I3 + I4, I1 == (Vi - V1)/R1, 
               I1 == (V1 - V2)/R2, I3 == V2/R3, I4 == V2/R4}, {I1, I3,
                I4, V1, V2}]][[1]][[2]], s, t]], t] == 0, t &gt; 0}, t]];
Flatten[\[Tau]] &amp;&amp; Flatten[N[\[Tau]]]