{{tag>resonance resonant_circuit RMS exam_ee2_SS2021}}{{include_n>1270}}
 
#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Resonant Circuit \\
<fs medium>(written test, approx. 4 % of a 120-minute written test, SS2021)</fs> #@TaskText_HTML@#

Given a resonant circuit, that is fed by a linear voltage source (circuit diagram on the right). \\
The inductance $L$ and capacitance $C$ are fixed. The resistance $R$ can be varied. 
  * $u_{\rm s} = 12{~\rm V} \cdot \sin (2 \pi \cdot f_0 \cdot t) $
  * $R_i = 200~\rm m\Omega$
  * $L = 20~\rm mH$
  * $C = 30~\rm \mu F$

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a) What is the resonance frequency $f_0$?
  
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The resonant frequency $f_0$ is given as 
\begin{align*}
f_0 = {{1}\over{ 2\pi \sqrt{LC} }}
\end{align*}

With the values:
\begin{align*}
f_0 &= {{1}\over{ 2\pi \sqrt{20 \cdot 10^{-3} ~\rm H \cdot 30 \cdot 10^{-6} ~\rm F} }} \\
    &= 205.4681... \rm Hz
\end{align*}

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#@HiddenBegin_HTML~nYnieWAMxfShPuwt_12,Result~@#
$f_0 = 205.5 \rm Hz$
#@HiddenEnd_HTML~nYnieWAMxfShPuwt_12,Result~@#

b) How large must $R$ be, so that at resonance the voltage across the capacitor is $U_C = 4 \cdot U_{\rm s}$?  (independent)

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For the following calculation, the internal resistance $R_i$ and the resistance $R$ have to be combined:
\begin{align*}
R_\Sigma = R_i + R \\
\end{align*}

Here, either one knows that the gain factor $Q$ stands for $Q={{U_C}\over{U_{\rm s}}}$ and therefore can directly use the following formula:
\begin{align*}
Q = {{U_C}\over{U_{\rm s}}} &= {{1}\over{R_\Sigma}}    \sqrt{ {{L}\over{C}} } \\
    R_\Sigma                &= {{U_{\rm s}}\over{U_C}} \sqrt{ {{L}\over{C}} } \\
\end{align*}

<callout>
When the gain factor is not known, one has to derive it: \\
The voltage $I$ at resonance is only given by the total ohmic resistance $R_\Sigma$ and the source voltage $U_{\rm s}$:
\begin{align*}
I = {{U_{\rm s}}\over{R_\Sigma}}
\end{align*}

This current flow also through the impedance of the capacitor
\begin{align*}
U_C &= Z_C \cdot I \\
    &= {{1}\over{\omega C}} \cdot I \\
    &= {{U_{\rm s}}\over{\omega C R_\Sigma }}  \\
\end{align*}

At resonance, the angular frequency $\omega$ is given by $\omega= {{1}\over{\sqrt{LC}}}$. \\
\begin{align*}
U_C &= {{U_{\rm s}}\over{{{1}\over{\sqrt{LC}}} C R_\Sigma }}  \\
    &= {{U_{\rm s}}\over{\sqrt{{{C}\over{L}}} R_\Sigma }}  \\
    &= {{U_{\rm s}}\over{R_\Sigma }} \sqrt{{{L}\over{C}}}  \\

\end{align*}
</callout>

In both cases, we end up with the same formula, where we have to insert the physical values: 
\begin{align*}
R_\Sigma  &= {{U_{\rm s}}\over{U_C}} \sqrt{ {{L}\over{C}} } \\
          &= {{1}\over{4}} \sqrt{ {{20\cdot 10^{-3} ~\rm H}\over{30\cdot 10^{-6} ~\rm C}} } \\
          &= 6.4549...~\Omega \\
\end{align*}

And so, the resistance $R$ is:
\begin{align*}
R  &= R_\Sigma - R_i \\
   &= 6.4549...~\Omega - 0.2~\Omega \\
   &= 6.2549...~\Omega
\end{align*}

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#@HiddenBegin_HTML~nYnieWAMxfShPuwt_22,Result~@#
$R = 6.255~\Omega$
#@HiddenEnd_HTML~nYnieWAMxfShPuwt_22,Result~@#

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