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I've a series circuit of a coil and a capacitor, in between those components we've a switch that will close when \$t=0\$. We can write:

$$ \begin{cases} \text{U}_\text{C}\left(t\right)=-\text{U}_\text{L}\left(t\right)\\ \\ \text{I}_\text{C}\left(t\right)=\text{U}'_\text{C}\left(t\right)\cdot\text{C}\\ \\ \text{U}_\text{L}\left(t\right)=\text{I}'_\text{L}\left(t\right)\cdot\text{L}\\ \\ \text{I}\left(t\right)=\text{I}_\text{C}\left(t\right)=\text{I}_\text{L}\left(t\right)\\ \end{cases}\space\space\space\space\space\therefore\space\space\space\space\space\frac{1}{\text{C}}\cdot\text{I}\left(t\right)=-\text{L}\cdot\text{I}''\left(t\right)\tag1 $$

Using Laplace transform:

$$\begin{align} \text{I}\left(\text{s}\right)&=\frac{\text{s}\cdot\text{I}\left(0\right)+\text{I}'\left(0\right)}{\frac{1}{\text{C}}+\text{L}\cdot\text{s}}\tag2\\ \text{U}_\text{C}\left(\text{s}\right)&=\frac{1}{\text{C}\cdot\text{s}}\cdot\left\{\frac{\text{s}\cdot\text{I}\left(0\right)+\text{I}'\left(0\right)}{\frac{1}{\text{C}}+\text{L}\cdot\text{s}}+\text{C}\cdot\text{U}_\text{C}\left(0\right)\right\}\tag3\\ \text{U}_\text{L}\left(\text{s}\right)&=\text{s}\cdot\text{L}\cdot\frac{\text{s}\cdot\text{I}\left(0\right)+\text{I}'\left(0\right)}{\frac{1}{\text{C}}+\text{L}\cdot\text{s}}-\text{L}\cdot\text{I}\left(0\right)\tag4 \end{align}$$

Well, I know that:

  1. $$\text{U}_\text{C}\left(0\right)=200\tag5$$
  2. $$\pi\sqrt{\text{C}\cdot\text{L}}<10\cdot10^{-6}=10^{-5}\space\Longleftrightarrow\space\text{C}\cdot\text{L}<\frac{10^{-10}}{\pi^2}\tag6$$

How can I find the value of \$\text{C}\$ and \$\text{L}\$ using the things I know?

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Given the frequency and the initial condition left one parameter missing in the LC circuit.

The standard LC curcuit initial condition and oscilating frequency parameters are, here assuming the initial current is zero, and the voltage is maximum: $$ \omega_0=\sqrt{\frac{1}{LC}}\\ I(0)=-I_0 cos(\phi)=0\\ V(0)=\omega_0 L I_0 sin(\phi)=\omega_0 L I_0 $$

Evaluating: $$ \frac{\pi}{10^{-5}}<\sqrt{\frac{1}{LC}}\\ \frac{\pi}{10^{-5}}LI_0<\sqrt{\frac{1}{LC}}LI_0=200 $$

Hence, both the L and C values cannot be calculated for the equality case.

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  • \$\begingroup\$ First of all thanks for your answer. Second in the first line of your answer you say: RL circuit, but it is a LC circuit :). Third how did you get: $$\frac{\pi}{10^{-5}}LI_0<\sqrt{\frac{1}{LC}}LI_0=200$$ \$\endgroup\$ – user135663 Jan 16 '17 at 15:25
  • \$\begingroup\$ Lol. Well the \$U_C(0)=200\$ in the paragraph, assuming \$V=U\$. Ref. Wiki. \$\endgroup\$ – Brethlosze Jan 16 '17 at 20:42

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