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+{{tag>self:resonance impedance resonant_circuit exam_ee2_SS2024}}{{include_n>1090}}
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Magnetic Circuit \\
+<fs medium>(written test, approx. 10 % of a 120-minute written test, SS2024)</fs> #@TaskText_HTML@#
+A real capacitor behaves like an $RLC$ resonant circuit, with an equivalent series resistance $R$ and an equivalent series inductance $L$.
+{{drawio>ee2:yh4srwxu1bo1rdy4_question1.svg}}
+A given capacitor shall have the following values:
+  * $C=10 ~\rm nF$
+  * $R=20 ~\rm m\Omega$
+  * $L=1.6 ~\rm nH$
+. What is the impedance $Z_{RLC}$ of this real capacitor for $f_0=44 ~\rm MHz$? (Phase and magnitude)
+#@HiddenBegin_HTML~yh4srwxu1bo1rdy4_11,Path~@#
+The impedance is based on the resistance $R$ and the reactance $X_{LC}= {\rm j}\cdot (X_L - X_C)$:
+\begin{align*}
+\underline{Z}_{RLC} &= R + {\rm j}\cdot (X_L - X_C) \\
+                    &= R + {\rm j}\cdot (\omega L - {{1}\over{\omega C}}) \\
+                    &= R + {\rm j}\cdot (2\pi f \cdot L - {{1}\over{2\pi f \cdot C}}) \\
+\end{align*}
+The reactive part is
+\begin{align*}
+X_{LC} &= 2\pi f \cdot L - {{1}\over{2\pi f \cdot C}} \\
+       &= 2\pi 44 \cdot 10^{6} {~\rm MHz} \cdot  1.6 \cdot 10^{-9} {~\rm H} - {{1}\over{2\pi \cdot 10^{6} {~\rm MHz} \cdot 10 \cdot 10^{-9} {~\rm F}}} \\
+       &= +0.08062... ~\Omega \\
+\end{align*}
+To get the magnitude of the impedance $|\underline{Z}_{RLC}|$ one can use the Pythagorean Theorem:
+\begin{align*}
+|\underline{Z}_{RLC}| &= \sqrt{R^2 + X_{LC}^2} \\
+                      &= \sqrt{(0.020~\Omega)^2 + ( 0.08062... ~\Omega )^2} \\
+                      &= 0.0830 ... ~\Omega \\
+\end{align*}
+For the phase $\varphi$ the $\arctan$ can be applied:
+\begin{align*}
+\varphi &= \arctan \left( {{X_{LC}}\over{R}} \right) \\
+        &= \arctan \left( {{0.08062... ~\Omega}\over{0.020 ~\Omega}} \right) \\
+        &= 1.3276 ... \hat{=} +76° \\
+\end{align*}
+#@HiddenEnd_HTML~yh4srwxu1bo1rdy4_11,Path~@#
+#@HiddenBegin_HTML~yh4srwxu1bo1rdy4_12,Result~@#
+  * $|\underline{Z}_{RLC}| = 83.0  ~\rm m \Omega$
+  * $\varphi = +76°$
+#@HiddenEnd_HTML~yh4srwxu1bo1rdy4_12,Result~@#
+. What is the resonance frequency $f_r$ for the given capacitor? What is the impedance in this case?
+#@HiddenBegin_HTML~yh4srwxu1bo1rdy4_21,Path~@#
+The formula for the resonance frequency $f_r$ is:
+\begin{align*}
+f_r &= {{1}\over{2\pi \sqrt{LC}}} \\
+    &= {{1}\over{2\pi \sqrt{1.6 \cdot 10^{-9} {~\rm H} \cdot 10 \cdot 10^{-9} {~\rm F}}}} \\
+    &= 39.788... ~\rm MHz \\
+\end{align*}
+The impedance at resonance is purely the resistance.
+#@HiddenEnd_HTML~yh4srwxu1bo1rdy4_21,Path~@#
+#@HiddenBegin_HTML~yh4srwxu1bo1rdy4_22,Result~@#
+  * $f_r = 39.79 ~\rm MHz$
+  * $|\underline{Z}_{RLC}(f_r)| = 20.0  ~\rm m \Omega$
+#@HiddenEnd_HTML~yh4srwxu1bo1rdy4_22,Result~@#
+. For an application, the component shall be used in resonance on a supply of $5 ~\rm V$.
+What is the voltage on the ideal capacity $C$ in the shown circuit?
+#@HiddenBegin_HTML~yh4srwxu1bo1rdy4_31,Path~@#
+The voltage on the ideal capacity is the input voltage by the $Q$-factor increased:
+\begin{align*}
+U_C &= U_{\rm s} \cdot Q \\
+    &= U_{\rm s} \cdot \sqrt{ {{L}\over{C}} } \cdot {{1}\over{R}}\\
+    &= 5 {~\rm V} \cdot \sqrt{ {{ 1.6 \cdot 10^{-9} {~\rm H} }\over{ 10 \cdot 10^{-9} {~\rm F} }} } \cdot {{1}\over{0.020~\Omega}}\\
+    &= 100 ~\rm V
+\end{align*}
+#@HiddenEnd_HTML~yh4srwxu1bo1rdy4_31,Path~@#
+#@HiddenBegin_HTML~yh4srwxu1bo1rdy4_32,Result~@#
+$U_C = 100 ~\rm V$
+#@HiddenEnd_HTML~yh4srwxu1bo1rdy4_32,Result~@#
+#@TaskEnd_HTML@#

resonance impedance resonant circuit exam ee2 ss2024