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I've a series R and C circuit, with a DC voltage source. I want to find the charging function of the capacitor. My question is: is the method I use right?


My work (using Laplace transform):

$$ \begin{cases} \text{U}_{\text{in}}(t)=\text{U}_{\text{C}}(t)+\text{U}_{\text{R}}(t)\\ \\ \text{U}_{\text{R}}(t)=\text{I}_{\text{R}}(t)\cdot\text{R}\\ \\ \text{I}_{\text{C}}(t)=\text{U}_{\text{C}}'(t)\cdot\text{C}\\ \\ \text{I}_{\text{in}}(t)=\text{I}_{\text{C}}(t)=\text{I}_{\text{R}}(t) \end{cases}\space\space\space\space\space\space\Longrightarrow^{\mathcal{L}}\space\space\space\space\space\space \begin{cases} \text{U}_{\text{in}}(\text{s})=\text{U}_{\text{C}}(\text{s})+\text{U}_{\text{R}}(\text{s})\\ \\ \text{U}_{\text{R}}(\text{s})=\text{I}_{\text{R}}(\text{s})\cdot\text{R}\\ \\ \text{I}_{\text{C}}(\text{s})=\text{C}\cdot\text{s}\cdot\text{U}_{\text{C}}(\text{s})-\text{C}\cdot\text{U}_{\text{C}}(0)\\ \\ \text{I}_{\text{in}}(\text{s})=\text{I}_{\text{C}}(\text{s})=\text{I}_{\text{R}}(\text{s}) \end{cases} $$

So, we get:

$$\text{U}_{\text{in}}(\text{s})=\frac{\text{I}_{\text{in}}(\text{s})+\text{C}\cdot\text{U}_{\text{C}}(0)}{\text{C}\cdot\text{s}}+\text{I}_{\text{in}}(\text{s})\cdot\text{R}\Longleftrightarrow\text{I}_{\text{in}}(\text{s})=\frac{\text{U}_{\text{in}}(\text{s})-\frac{\text{U}_{\text{C}}(0)}{\text{s}}}{\text{R}+\frac{1}{\text{C}\cdot\text{s}}}$$

So, when I want to find U_c(s):

$$\text{I}_{\text{in}}(t)=\text{U}_{\text{C}}'(t)\cdot\text{C}\space\space\space\space\space\space\Longrightarrow^{\mathcal{L}}\space\space\space\space\space\space\frac{\text{U}_{\text{in}}(\text{s})-\frac{\text{U}_{\text{C}}(0)}{\text{s}}}{\text{R}+\frac{1}{\text{C}\cdot\text{s}}}=\text{C}\cdot\text{s}\cdot\text{U}_{\text{C}}(\text{s})-\text{C}\cdot\text{U}_{\text{C}}(0)$$

Solving U_c(s), gives me:

$$\text{U}_{\text{C}}(\text{s})=\frac{\frac{\text{U}_{\text{in}}(\text{s})-\frac{\text{U}_{\text{C}}(0)}{\text{s}}}{\text{R}+\frac{1}{\text{C}\cdot\text{s}}}+\text{C}\cdot\text{U}_{\text{C}}(0)}{\text{C}\cdot\text{s}}$$

Knowing that the voltage source is DC:

$$\text{U}_{\text{in}}(\text{s})=\frac{\text{U}_{\text{in}}}{\text{s}}$$

So:

$$\color{red}{\text{U}_{\text{C}}(\text{s})=\frac{\frac{\frac{\text{U}_{\text{in}}}{\text{s}}-\frac{\text{U}_{\text{C}}(0)}{\text{s}}}{\text{R}+\frac{1}{\text{C}\cdot\text{s}}}+\text{C}\cdot\text{U}_{\text{C}}(0)}{\text{C}\cdot\text{s}}=\frac{\frac{\text{U}_{\text{in}}-\text{U}_{\text{C}}(0)}{\text{R}\cdot\text{s}+\frac{1}{\text{C}}}+\text{C}\cdot\text{U}_{\text{C}}(0)}{\text{C}\cdot\text{s}}}$$

Using inverse Laplace transform, I found:

$$\text{U}_{\text{C}}(t)=\text{U}_{\text{in}}+e^{-\frac{t}{\text{CR}}}\left(\text{U}_{\text{C}}(0)-\text{U}_{\text{in}}\right)$$

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  • \$\begingroup\$ That is one way to do it. I find just making a first order constant coefficient differential equation just as easy for a simple RC circuit using KVL. An inverse Laplace transform is not a trivial thing thats why we use tables or computers to do them for us. Whereas a first order diff eq is actually straight forward. But if your task was to use Laplace then good job. \$\endgroup\$
    – crowie
    Commented Oct 28, 2016 at 12:19
  • \$\begingroup\$ As a student, this all made sense to me... \$\endgroup\$
    – Rev
    Commented Oct 28, 2016 at 12:27
  • \$\begingroup\$ it's correct, but clearer if you express it as Uin[1-exp(-t/RC) + IC's, so the rising exponential nature is absolutely apparent. \$\endgroup\$ Commented Oct 28, 2016 at 12:43

1 Answer 1

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Yes, your method and result are correct. A quick way to check is to see if your result gives the correct result at t=0 and infinity while assuming 0 initial condition. At t=0 the the exponential term is 1 and you have Uc=Vin-Vin=0, good. At infinity, the exponential term is 0 and you have Uc=Vin. These two values are connected by an exponential function which we know is correct for a first order RC.

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