Find the inrush current in the capacitor [closed]

How can I find the inrush current value of the capacitor in the following circuit:

simulate this circuit – Schematic created using CircuitLab

When $$\R_1=R_2=10\$$ and $$\L=10mH\$$ and $$\C=1\mu F\$$ and $$\V_1=1\$$.

• I’m voting to close this question because it's homework with no attempt to solve. Dec 22 '20 at 15:17
• I’m closing this question because it is a homework/assignment type of question with no attempt at a solution or explanation of what the OP understands. Homework type questions are permitted but need to show some effort. Dec 28 '20 at 22:45

Well, we are trying to analyze the following circuit:

simulate this circuit – 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:

$$\text{I}_3=\frac{\text{V}_2}{\text{R}_3}=\frac{\text{R}_4\text{V}_\text{i}}{\text{R}_3\left(\text{R}_1+\text{R}_2\right)+\text{R}_4\left(\text{R}_1+\text{R}_2+\text{R}_3\right)}\tag5$$

When we want to apply the derivation from above to your circuit we need to use Laplace transform (I will use lower case function names for the functions that are in the (complex) s-domain, so $$\\text{y}\left(\text{s}\right)\$$ is the Laplace transform of the function $$\\text{Y}\left(t\right)\$$):

• $$\text{R}_2=\text{sL}\tag6$$
• $$\text{R}_3=\frac{1}{\text{sC}}\tag7$$
• 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:

$$\text{i}_3\left(\text{s}\right)=\frac{\hat{\text{u}}}{\text{s}}\cdot\frac{\text{R}_4}{\frac{1}{\text{sC}}\left(\text{R}_1+\text{sL}\right)+\text{R}_4\left(\text{R}_1+\text{sL}+\frac{1}{\text{sC}}\right)}=$$ $$\frac{\hat{\text{u}}\text{R}_4\text{C}}{\text{CL}\text{R}_4\text{s}^2+\left(\text{L}+\text{C}\text{R}_1\text{R}_4\right)\text{s}+\text{R}_1+\text{R}_4}\tag9$$

It is not hard to show that when $$\x:=\left(\text{L}+\text{C}\text{R}_1\text{R}_4\right)^2-4\text{C}\text{L}\text{R}_4\left(\text{R}_1+\text{R}_4\right)>0\$$ we get the behavior that you're talking about (with just one maximum).

Now, when we use inverse Laplace transform we can see:

$$\text{I}_3\left(t\right)=\frac{2\hat{\text{u}}\text{CR}_4\exp\left(-\frac{\left(\text{L}+\text{CR}_1\text{R}_4\right)t}{2\text{CLR}_4}\right)\sinh\left(\frac{t\sqrt{x}}{2\text{CLR}_4}\right)}{\sqrt{\text{L}^2+\left(\text{CR}_1\text{R}_4\right)^2-2\text{CLR}_4\left(\text{R}_1+2\text{R}_4\right)}}\tag{10}$$

The maximum occurs when (if we obey the condition $$\0):

$$\hat{t}=\frac{\text{CLR}_4\ln\left(\frac{\text{L}+\text{CR}_1\text{R}_4+\sqrt{x}}{\text{L}+\text{CR}_1\text{R}_4-\sqrt{x}}\right)}{\sqrt{x}}\tag{11}$$

And at that time $$\\hat{t}\$$ we have a current of:

$$\text{I}_3\left(\hat{t}\right)=\frac{2\hat{\text{u}}\text{CR}_4\exp\left(-\frac{\left(\text{L}+\text{CR}_1\text{R}_4\right)\hat{t}}{2\text{CLR}_4}\right)\sinh\left(\frac{\hat{t}\sqrt{x}}{2\text{CLR}_4}\right)}{\sqrt{\text{L}^2+\left(\text{CR}_1\text{R}_4\right)^2-2\text{CLR}_4\left(\text{R}_1+2\text{R}_4\right)}}\tag{12}$$

$$\hat{t}=\frac{\text{arccoth}\left(\frac{101}{\sqrt{9401}}\right)}{500 \sqrt{9401}}\approx0.0000401357\space\text{s}\tag{13}$$

And:

$$\text{I}_3\left(\hat{t}\right)=\frac{\left(\sqrt{9401}+101\right) e^{-\left(1+\frac{101}{\sqrt{9401}}\right)\text{arccoth}\left(\frac{101}{\sqrt{9401}}\right)}}{4000}\approx0.000931611\space\text{A}\tag{14}$$

With:

$$x=\frac{9401}{100000000}\approx0.00009401>0\tag{15}$$

If you're a user of Mathematica, you can use the codes that I wrote to solve this question:

In[1]:=R2 = s*L;
R3 = 1/(s*c);
Vi = u/s;
FullSimplify[
InverseLaplaceTransform[(R4 Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4),
s, t]]

Out[1]=(2 c E^(-(((L + c R1 R4) t)/(2 c L R4))) R4 u Sinh[(
Sqrt[-4 c L R4 (R1 + R4) + (L + c R1 R4)^2] t)/(
2 c L R4)])/Sqrt[L^2 + c^2 R1^2 R4^2 - 2 c L R4 (R1 + 2 R4)]

In[2]:=FullSimplify[
Solve[{I1 == I3 + I4, I1 == (Vi - V1)/R1, I1 == (V1 - V2)/R2,
I3 == V2/R3, I4 == V2/R4}, {I1, I3, I4, V1, V2}]]

Out[2]={{I1 -> ((R3 + R4) Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4),
I3 -> (R4 Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4),
I4 -> (R3 Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4),
V1 -> ((R3 R4 +
R2 (R3 + R4)) Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4),
V2 -> (R3 R4 Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4)}}

In[3]:=R2 = s*L;
R3 = 1/(s*c);
Vi = u/s;
L = 10*10^(-3);
c = 1*10^(-6);
u = 1;
R1 = 10;
R4 = 10;
FullSimp
FullSimplify[
InverseLaplaceTransform[(R4 Vi)/((R1 + R2) R3 + (R1 + R2 + R3) R4),
s, t]]

Out[3]=(E^(-500 (101 + Sqrt[9401]) t) (-1 + E^(
1000 Sqrt[9401] t)))/(10 Sqrt[9401])

In[4]:=FullSimplify[Solve[{D[%3, t] == 0, t > 0}, t]]

Out[4]=ArcCoth[101/Sqrt[9401]]/(500 Sqrt[9401])

In[5]:=FullSimplify[Limit[%3, t ->ArcCoth[101/Sqrt[9401]]/(500 Sqrt[9401])]]

Out[5]=((101 + Sqrt[9401]) E^(-(1 + 101/Sqrt[9401]) ArcCoth[101/Sqrt[
9401]]))/4000

In[6]:=Plot[%3, {t, 0,1/1000}]

Out[6]=(*this code will plot the picture above*)

• Thank you so so much, I will study your answer in more detail. Dec 20 '20 at 14:12
• @Jan you have done this before - you should not hand homework questions on a plate to anyone - that isn't what this site is about. Dec 22 '20 at 15:18