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#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Pure Resistor Network Simplification  \\ <fs medium>(written test, approx. 13 % of a 60-minute written test, WS2022)</fs> #@TaskText_HTML@#    

The following circuit with $R_1=200 ~\Omega$, $R_2=R_3=100 ~\Omega$ and the switch $S$ is given. \\
{{drawio>electrical_engineering_1:70jjg4yzwctqarsqCircuit.svg}}

1. The switch shall now be open. Calculate the equivalent resistance $R_{\rm eq}$ between $\rm A$ and $\rm B$.

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{{drawio>electrical_engineering_1:70jjg4yzwctqarsqCircuitSolution1.svg}}
The equivalent resistor is given by a parallel configuration of resistors in series:
\begin{align*}
R_{\rm eq} &= (R_2 + R_1 + R_1)||(R_2 + R_2)\\
R_{\rm eq} &= (100 ~\Omega + 200 ~\Omega + 200 ~\Omega )&&||(100 ~\Omega + 100 ~\Omega ) &&\\
R_{\rm eq} &= (500 ~\Omega )&&||(200 ~\Omega )&&\\
R_{\rm eq} &= {{500 ~\Omega \cdot 200 ~\Omega }\over{500 ~\Omega + 200 ~\Omega}}&&\\
\end{align*}

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#@ResultBegin_HTML~1~@#
\begin{align*}
R_{\rm eq} &= 142.8 ~\Omega \\
\end{align*}
 \\
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2. The switch shall now be closed. Calculate the equivalent resistance $R_{\rm eq}$ between $\rm A$ and $\rm B$.

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Now a wye-delta transformation is necessary.
{{drawio>electrical_engineering_1:70jjg4yzwctqarsqCircuitSolution2.svg}}
Since $R_2=R_3$ and based on the equations for the transformation, the transformed $R_Y$ is given as:
\begin{align*}
R_{Y} &= {{R_2 \cdot R_2}\over{R_2 + R_2 + R_2}} \\
      &= {{(100 ~\Omega)^2}\over{3 \cdot 100 ~\Omega}} \\
      &= {{1}\over{3}} \cdot 100 ~\Omega = 33.33 ~\Omega
\end{align*}

The equivalent resistor is given by a parallel configuration of resistors in series:
\begin{align*}
R_{\rm eq} &= R_Y + (R_Y + R_1 + R_1)||(R_Y + R_2)\\
R_{\rm eq} &= 33.33 ~\Omega + (33.33 ~\Omega + 400 ~\Omega)||(33.33 ~\Omega + 100 ~\Omega)\\
\end{align*}

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#@ResultBegin_HTML~2~@#
\begin{align*}
R_{\rm eq} &= 135.3 ~\Omega \\
\end{align*}

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