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I've got the following problem:

We've got a RLC circuit in serie and I would like to find the function:$$\text{V}_{C}(t)$$ the voltage across the capacitor.

Given are the following things:

  • $$\text{V}_{\text{in}}(t)=20\space\text{V}$$
  • $$\text{C}=4\space\text{F}$$
  • $$\text{L}=1\space\text{H}$$
  • $$\text{R}=5\space\Omega$$
  • $$\text{I}_{C}(0)=-2\space\text{A}$$
  • $$\text{V}_{C}(0)=10\space\text{V}$$
  • $$\text{I}_{T}(t)=\text{I}_{R}(t)\space\text{is the total current}$$


My work:

$$\text{V}_{\text{in}}(t)=\text{V}_{R}(t)+\text{V}_{C}(t)+\text{V}_{L}(t)\Longleftrightarrow$$


Knowing that:

  • $$\text{V}_{L}(t)=\text{LI}'_{L}(t)=\text{LI}'_{T}(t)$$
  • $$\text{I}_{C}(t)=\text{I}_{T}(t)=\text{CV}'_{C}(t)\to\text{I}'_{C}(t)=\text{I}'_{T}(t)=\text{CV}''_{C}(t)$$

$$\text{V}_{\text{in}}(t)=\text{RI}_{T}(t)+\text{V}_{C}(t)+\text{LI}'_{T}(t)\Longleftrightarrow$$ $$\text{V}_{\text{in}}(t)=\text{R}\cdot\text{CV}'_{C}(t)+\text{V}_{C}(t)+\text{L}\cdot\text{CV}''_{C}(t)\Longleftrightarrow$$ $$\text{V}_{\text{in}}(t)=\text{CRV}'_{C}(t)+\text{V}_{C}(t)+\text{CLV}''_{C}(t)$$

So the problem becomes:

$$ \begin{cases} 20\text{V}'_{C}(t)+\text{V}_{C}(t)+4\text{V}''_{C}(t)=20\\ \text{V}_{C}(0)=10\\ 4\text{V}'_{C}(0)=-2 \end{cases} $$

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  • \$\begingroup\$ Please add a schematic for easier reference. Also, is the current \$I_T\$ a typo, maybe it should be \$I_R\$? \$\endgroup\$
    – Timo
    Jan 15 '16 at 12:42
  • \$\begingroup\$ According to given information, the ODE should be fine. An schematic would greatly improve question readability and answer quality. \$\endgroup\$
    – cjferes
    Jan 15 '16 at 14:03
  • \$\begingroup\$ oh dear, why all this \text? \$\endgroup\$ Jan 15 '16 at 20:27
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That means you are looking for state space model of R-L-C network. It would be better if you use Laplace transform.

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  • \$\begingroup\$ I know how to use Laplace Transform, but I don't know how to use it in this case! \$\endgroup\$ Jan 15 '16 at 15:25
  • \$\begingroup\$ \$V_C(t)\rightarrow V_C(s)\$; \$V'_C(t)\rightarrow sV_C(s)\$;\$V''_C(t)\rightarrow s^2V_C(s)\$; \$\endgroup\$
    – Chu
    Jan 15 '16 at 15:42
  • \$\begingroup\$ @Chu Notice that we've got other conditions! So, $$\text{V}_{C}(0)\ne 0$$ \$\endgroup\$ Jan 15 '16 at 15:43
  • \$\begingroup\$ \$V'_C(t)\rightarrow sV_C(s)-V_C(0)\$, and similar for the 2nd derivative should do it. \$\endgroup\$
    – Chu
    Jan 15 '16 at 15:49
  • \$\begingroup\$ I know how to use Laplace, but not in this case! \$\endgroup\$ Jan 15 '16 at 15:50

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