{{tag>resonant_circuit exam_ee2_SS2022}}{{include_n>1130}}
 
#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Series Resonant Circuit \\
<fs medium>(written test, approx. 10 % of a 120-minute written test, SS2022)</fs> #@TaskText_HTML@#

A real capacitor behaves like an RLC resonant circuit, with an equivalent series resistance $R$ and an equivalent series inductance $L$. \\ 
A capacitor shall be given with the following values: 
  * $C=10 ~\rm nF$
  * $R=88 ~\rm m\Omega$
  * $L=60 ~\rm pH$

{{drawio>ee2:7El8zLZJglAazXTw_question1.svg}}

1. What is the impedance $\underline{Z}_{RLC}$ of this real capacitor for $f_0=100 ~\rm MHz$? (Phase and magnitude)

#@HiddenBegin_HTML~7El8zLZJglAazXTw_11,Path~@#
The impedance $\underline{Z}_{RLC}$ is given by:
\begin{align*}
\underline{Z}_{RLC} &= R + \underline{X}_{L} + \underline{X}_{C} \\
                    &= R + {\rm j}\omega L - {{\rm j}\over{\omega C}} \\
                    &= R + {\rm j}\cdot \left(\omega L - {{1}\over{\omega C}} \right)\\
                    &= R + {\rm j}\cdot {X}_{LC} \\
\end{align*}

Putting in the numbers, only for the reactive part ${X}_{LC}$:
\begin{align*}
{X}_{LC} &= 2\pi\cdot    \quad \quad f_0  \quad  \quad \cdot  \quad L \quad \quad \quad \; - {{1}\over{2\pi  \cdot  \quad  \quad f_0  \quad \quad \cdot  \quad \quad C \quad\quad}}  \\
         &= 2\pi \cdot 100 \cdot 10^{6}{~\rm Hz} \cdot 60 \cdot 10^{-12}{~\rm H} - {{1}\over{2\pi \cdot 100 \cdot 10^{6}{~\rm Hz} \cdot 10 \cdot 10^{-9}{~\rm F}}}  \\
         &= -121.45... ~\rm m\Omega \\
\end{align*}

With the real and imaginary parts, we can derive the magnitude and phase:
\begin{align*}
Z_{RLC}  &=\sqrt{R^2 + {X}_{LC}^2}  \\
         &=\sqrt{(88 ~\rm m\Omega)^2 + (-121.45 ~\rm m\Omega)^2}  \\
         &= 150.0... ~\rm m\Omega \\
\end{align*}

\begin{align*}
\varphi  &=\arctan \left( { {{{X}_{LC}}\over{R}}} \right)\\
         &=\arctan \left( { {{-121.45 ~\rm m\Omega}\over{88 ~\rm m\Omega}}} \right)\\
         &= -0.9437... = -54.07...° \\
\end{align*}
#@HiddenEnd_HTML~7El8zLZJglAazXTw_11,Path~@#

#@HiddenBegin_HTML~7El8zLZJglAazXTw_12,Result~@#
  * $Z_{RLC} = 150.0~\rm m\Omega $
  * $\varphi = -54.07° $
#@HiddenEnd_HTML~7El8zLZJglAazXTw_12,Result~@#

2. The impedance magnitude of task 1. can be interpreted as the impedance magnitude of an effective ideal capacity $C_0$. \\ 
In this case, the magnitude of the impedance of $C_0$ would be $X_{C0}=Z_{RLC}$. \\
Which value would $C_0$ have for the given $f_0$?

#@HiddenBegin_HTML~7El8zLZJglAazXTw_21,Path~@#
The calculated impedance of $Z_{RLC}$ has to be set equal to $X_{C0}$
\begin{align*}
Z_{RLC} &= X_{C0}  \\
        &= {{1}\over{2\pi f \cdot C}} \\
\rightarrow C &= {{1}\over{2\pi f \cdot Z_{RLC}}}
\end{align*}

With values: 
\begin{align*}
 C &= {{1}\over{2\pi 100 \cdot 10^{6} {~\rm MHz}\cdot 0.1500... {~\rm \Omega} }} \\
   &= 10.6... ~\rm nF
\end{align*}

#@HiddenEnd_HTML~7El8zLZJglAazXTw_21,Path~@#

#@HiddenBegin_HTML~7El8zLZJglAazXTw_22,Result~@#
$ C = 10.6 ~\rm nF $
#@HiddenEnd_HTML~7El8zLZJglAazXTw_22,Result~@#

3. What is the resonance frequency $f_{\rm r}$ for the given capacitor? 
What is the impedance in this case?

#@HiddenBegin_HTML~7El8zLZJglAazXTw_31,Path~@#
The resonance frequency is given as
\begin{align*}
f_{\rm r} &= {{1}\over{2\pi\sqrt{LC}}} \\
  &= {{1}\over{2\pi\sqrt{60 \cdot 10^{-12}{~\rm H} \cdot 10 \cdot 10^{-9}{~\rm F}}}} \\ 
  &= 205.5... ~\rm MHz
\end{align*}

At resonance, the impedance is given purely by the resistor.
#@HiddenEnd_HTML~7El8zLZJglAazXTw_31,Path~@#

#@HiddenBegin_HTML~7El8zLZJglAazXTw_32,Result~@#
  * $f_{\rm r} = 205.5 ~\rm MHz$
  * $88~\rm m\Omega$
#@HiddenEnd_HTML~7El8zLZJglAazXTw_32,Result~@#

#@TaskEnd_HTML@#