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+{{tag>impedance inductor exam_ee2_SS2021}}{{include_n>1230}}
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Impedance Characteristics \\
+<fs medium>(written test, approx. 6 % of a 120-minute written test, SS2021)</fs> #@TaskText_HTML@#
+A coil has an inductive reactance of $X_0 = X(f_0) = 80~\rm \Omega$ at a frequency $f_0 = 60 ~\rm kHz$. \\
+Calculate the frequencies $f_1$, $f_2$, $f_3$ at which the following reactances are measured: \\
+  * $X_1 = 50 ~\rm \Omega$
+  * $X_2 = 121 ~\rm \Omega$
+  * $X_3 = 147 ~\rm \Omega$
+#@HiddenBegin_HTML~OkzNHLJJycUQKBSh_11,Path~@#
+There are multiple ways to solve this question. \\ One way would be, to calculate the inductance $L$ first by rearranging $X(f) = 2\pi \cdot f \cdot L$. \\ \\
+Another way uses ratios (or "rule of three"), since $X(f) = f \cdot k$ with a constant $k$. \\
+Therefore one can set up two formulas $X_n = f_n \cdot k$, $X_0 = f_0 \cdot k$, and divide the formulae by each other. \\
+This leads to:
+\begin{align*}
+{{X_n}\over{X_0}} &= {{f_n}\over{f_0}} \\
+    f_n           &= {{X_n}\over{X_0}}\cdot f_0
+                   = {{f_0}\over{X_0}}\cdot X_n \\
+\end{align*}
+Putting in the numbers:
+\begin{align*}
+    f_n              &= {{60 ~\rm kHz}\over{80~\rm \Omega}}\cdot X_n \\
+                     &= 0.75 {{\rm \Omega}\over{\rm kHz}}\cdot X_n \\
+\end{align*}
+#@HiddenEnd_HTML~OkzNHLJJycUQKBSh_11,Path~@#
+#@HiddenBegin_HTML~OkzNHLJJycUQKBSh_12,Result~@#
+  * $f_1 = 37.5~\rm kHz$
+  * $f_2 = 90.75~\rm kHz$
+  * $f_3 = 110.25~\rm kHz$
+#@HiddenEnd_HTML~OkzNHLJJycUQKBSh_12,Result~@#
+#@TaskEnd_HTML@#

impedance inductor exam ee2 ss2021