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<WRAP mexle-task-title>
~~#@ee1_taskctr~~ Pure Resistor Network Simplification        \\ <fs medium>(written test, approx. 12 % of a 60-minute written test, SS2023)</fs>

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The circuit below shall be given.

The values in the circuit are
  * $R_1 = 60 ~\Omega$
  * $R_2 = 40 ~\Omega$
  * $R_3 = 40 ~\Omega$
  * $R_4 = 100 ~\Omega$
  * $U_{\rm AB} = 10 ~\rm V$

{{drawio>electrical_engineering_1:cgeyprm6oboukcciCircuit.svg}}

1. Calculate the voltage at node $\rm K$, when switch $\rm S$ is open. \\ It might be beneficial to redraw the circuit first.


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Rearranging the circuit one can get:
{{drawio>electrical_engineering_1:cgeyprm6oboukcciCircuitSolve1.svg}}

Once the switch $\rm S$ is opened, the upper part is a parallel circuit.
Therefore, $R_{\rm eq}$ is given as:
\begin{align*}
R_{\rm eq} &= (R_1+R_2)||(R_1+R_2)+R_4 \\
           &= {{1}\over{2}}\cdot(R_1+R_2)+R_4 \\
           &= {{1}\over{2}}\cdot(60~\Omega + 40~\Omega) + 100~\Omega \\
\end{align*}


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\begin{align*}
R_{\rm eq} &= 150~\Omega
\end{align*}


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2. Calculate the voltage at node $\rm K$, when switch $\rm S$ is closed. 


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The voltage divider for node $\rm K$ has the same proportionality as the voltage divider for node $\rm K'$. Therefore, the potential of $\rm K$ is the same as for $\rm K'$. There will be no current flow through $R_3$. The resistance does not create a voltage drop and therefore does not interfere with the circuit.

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The equivalent resistance is similar to the circuit with opened switch.
\begin{align*}
R_{\rm eq} &= 150~\Omega
\end{align*}

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