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#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Pure Resistor Network Simplification I       \\ <fs medium>(written test, approx. 14 % of a 60-minute written test, SS2023)</fs>
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The circuit below shall be given.

{{drawio>electrical_engineering_1:erlctd760zmvox0tCircuit.svg}}

1. What is the equivalent resistance $R_{\rm eq}$?

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Part of the circuit is shorted. Here the resistors (marked in red) are shorted by the connections marked in blue:
{{drawio>electrical_engineering_1:erlctd760zmvox0tCircuitSolve1.svg}}

The circuit can then be rearranged for better interpretation:
{{drawio>electrical_engineering_1:erlctd760zmvox0tCircuitSolve2.svg}}

Therefore, $R_{\rm eq}$ is given as:
\begin{align*}
R_{\rm eq} &= (2R||2R + R + R)||6R &&+ 6R ||(2R + 2R + 4R||4R) \\
           &= (R + R + R)||6R &&+ 6R ||(2R + 2R + 2R) \\
           &= 3R||6R &&+ 6R ||6R \\
           &= {{3R\cdot 6R}\over{3R+6R}} &&+3R \\
\end{align*}

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\begin{align*}
R_{\rm eq} &= 5R
\end{align*}

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2. Now the input voltage shall be given as $U_{\rm AB} = 60 ~\rm V$. What is the value for $U$ in the circuit?

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The current through the circuit is given as $I_{\rm AB} = U_{\rm AB} \cdot R_{\rm eq} $. \\
This current has to flow in summary through parallel branches. The voltage $U$ in question in the upper right branch given by $(4R||4R)+2R+2R$. Its resistance is just the same as the upper left branch $6R$. \\
Therefore, half of the current flows to the left half to the right side. \\
The voltage $U$ is consequently:
\begin{align*}
U &= {{I_{\rm AB} }\over{ 2\cdot R_{\rm eq} }} \\
  &= {{U_{\rm AB} \cdot 2R }\over{ 2\cdot R_{\rm eq} }} \\
  &= {{60 ~\rm V }\over{ 5 }}
\end{align*}


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\begin{align*}
U &= 12 ~\rm V \\
\end{align*}
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