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+{{tag>self:magnetic_circuit exam_ee2_SS2024}}{{include_n>1080}}
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Magnetic Circuit \\
+<fs medium>(written test, approx. 9 % of a 120-minute written test, SS2024)</fs> #@TaskText_HTML@#
+A toroidal core (ferrite, $μ_{\rm r}=900$) has a cross-sectional area of $A=300 ~\rm mm^2$ and an average circumference of $l=3 ~\rm dm$. \\
+{{drawio>ee2:n1kwu944m7jac3tf_question1.svg}}
+On the core, there are three coils with:
+  * Coil 1: $N_1 = 1200$, $I_1=100 ~\rm mA$
+  * Coil 2: $N_2 = 33  $, $I_2=  3 ~\rm A$
+  * Coil 3: $N_3 = 270 $, $I_3=0.3 ~\rm A$
+Refer to the drawing for the direction of the windings, current, and flux!
+. Draw the equivalent magnetic circuit that fully represents the setup. \\ Name all the necessary magnetic resistances, fluxes, and voltages.
+#@HiddenBegin_HTML~n1kwu944m7jac3tf_12,Result~@#
+  * Since the material, and diameter of the core is constant, one can directly simplify the magnetic resistor into a single $R_\rm m$.
+  * For the orientation of the magnetic voltages $\theta_1$, $\theta_2$, and $\theta_3$, the orientation of the coils and the direction of the current has to be taken into account by the right-hand rule.
+  * There is only one flux $\Phi$
+  * The magnetic voltages are antiparallel to the flux for sources and parallel for the load.
+{{drawio>ee2:n1kwu944m7jac3tf_answer1.svg}}
+#@HiddenEnd_HTML~n1kwu944m7jac3tf_12,Result~@#
+. Calculate the magnetic resistance $R_\rm m$.
+#@HiddenBegin_HTML~n1kwu944m7jac3tf_21,Path~@#
+The formula of the magnetic resistance is:
+\begin{align*}
+R_{\rm m} &= {{1}\over{\mu_0 \mu_{\rm r}}} {{l}\over{A}} \\
+          &= {{1}\over{4\pi \cdot 10^{-7} {\rm {{Vs}\over{Am}}} \cdot 900}} {{3 \cdot 10^{-1} ~\rm m}\over{300 \cdot 10^{-6} ~\rm m^2}} \\
+          &= 0.88419... ~ \cdot 10^{6}  \rm {{1}\over{H}} \\
+\end{align*}
+#@HiddenEnd_HTML~n1kwu944m7jac3tf_21,Path~@#
+#@HiddenBegin_HTML~n1kwu944m7jac3tf_22,Result~@#
+$R_{\rm m} = 0.884 \cdot 10^{6}  \rm {{1}\over{H}} $
+#@HiddenEnd_HTML~n1kwu944m7jac3tf_22,Result~@#
+. Calculate the resulting magnetic flux in the core.
+#@HiddenBegin_HTML~n1kwu944m7jac3tf_31,Path~@#
+First we have to calculate the resulting magnetic voltage based on the sources:
+\begin{align*}
+- \theta_{\rm R} + \theta_1 + \theta_2 - \theta_3 &= 0 \\
+\theta_{\rm R} &= \theta_1 + \theta_2 - \theta_3 \\
+               &= I_1 \cdot N_1    + I_2 \cdot N_2 - I_3 \cdot N_3 \\
+      \rm      &= 1200 \cdot 0.1~A + 33 \cdot 3~A  - 270 \cdot 0.3~A \\
+               &= -60~A
+\end{align*}
+To get the flux $\Phi$, the Hopkinson's Law can be applied - similar to the Ohm's Law:
+\begin{align*}
+\Phi &= {{\theta_{\rm R} }\over {R_{\rm m}}} \\
+     &= {{-60~\rm A }\over { 0.884 \cdot 10^{6}  \rm {{1}\over{H}} }} \\
+     &= 67.8 ... \cdot 10^{-6} { \rm A \cdot H} \\
+     &= 67.8 ... ~\rm \mu Wb \\
+     &= 67.8 ... ~\rm \mu Vs \\
+\end{align*}
+#@HiddenEnd_HTML~n1kwu944m7jac3tf_31,Path~@#
+#@HiddenBegin_HTML~n1kwu944m7jac3tf_32,Result~@#
+$\Phi = 67.8 ~\rm \mu Vs$
+#@HiddenEnd_HTML~n1kwu944m7jac3tf_32,Result~@#
+#@TaskEnd_HTML@#

magnetic circuit exam ee2 ss2024