{{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 $l=3 ~\rm dm$. \\
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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!

1. Draw the equivalent magnetic circuit that fully represents the setup. \\ Name all the necessary magnetic resistances, fluxes, and voltages.

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  * 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.

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2. Calculate the magnetic resistance $R_\rm m$.

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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*}

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#@HiddenBegin_HTML~n1kwu944m7jac3tf_22,Result~@#
$R_{\rm m} = 0.884 \cdot 10^{6}  \rm {{1}\over{H}} $
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3. Calculate the resulting magnetic flux in the core.

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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*}

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#@HiddenBegin_HTML~n1kwu944m7jac3tf_32,Result~@#
$\Phi = 67.8 ~\rm \mu Vs$
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