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+{{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) $
+  * $R_i = 200~\rm m\Omega$
+  * $L = 20~\rm mH$
+  * $C = 30~\rm \mu F$
+{{drawio>ee2:nYnieWAMxfShPuwt_question1.svg}}
+a) What is the resonance frequency $f_0$?
+#@HiddenBegin_HTML~nYnieWAMxfShPuwt_11,Path~@#
+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*}
+#@HiddenEnd_HTML~nYnieWAMxfShPuwt_11,Path~@#
+#@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)
+#@HiddenBegin_HTML~nYnieWAMxfShPuwt_21,Path~@#
+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*}
+#@HiddenEnd_HTML~nYnieWAMxfShPuwt_21,Path~@#
+#@HiddenBegin_HTML~nYnieWAMxfShPuwt_22,Result~@#
+$R = 6.255~\Omega$
+#@HiddenEnd_HTML~nYnieWAMxfShPuwt_22,Result~@#
+#@TaskEnd_HTML@#

resonance resonant circuit rms exam ee2 ss2021