{{tag>dc_network_analysis pure_resistor_network_simplification delta_wye_transformation exam_ee1_WS2022}}
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#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Equivalent linear Source       \\ <fs medium>(written test, approx. 14 % of a 60-minute written test, WS2022)</fs>
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The circuit in the following has to be simplified. 

{{drawio>electrical_engineering_1:6tqttque1e2nf2c7Circuit.svg}}

Calculated the internal resistance $R_\rm i$ and the source voltage $U_\rm s$ of an equivalent linear voltage source on the connectors $\rm A$ and $\rm B$.
\begin{align*}
 R_1=5.0 ~\Omega, && U_2=6.0 ~\rm{V}, && R_3= 10 ~\Omega, \\
 I_4=4.2 ~\rm{A}, && R_5=10 ~\Omega , && R_6=7.5 ~\Omega, \\
 R_7=15  ~\Omega  &&                 &&
\end{align*}
Use equivalent sources in order to simplify the circuit!

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The best thing is to re-think the wiring like rubber bands and adjust them:
{{drawio>electrical_engineering_1:6tqttque1e2nf2c7CircuitSolution1.svg}} \\

The linear voltage source of $U_2$ and $R_1$ can be transformed into a current source $I_2={{U_2}\over{R_1}}$ and $R_1$:
{{drawio>electrical_engineering_1:6tqttque1e2nf2c7CircuitSolution2.svg}} \\

Now a lot of them can be combined. The resistors $R_1$, $R_3$, $R_5$ are in parallel, like also $I_2$ and $I_4$:
\begin{align*}
R_{135} &= R_1||R_3||R_5\\
I_{24} &= I_2 - I_4 = {{U_2}\over{R_1}} - I_4 
\end{align*}
The resulting circuit can again be transformed:
{{drawio>electrical_engineering_1:6tqttque1e2nf2c7CircuitSolution3.svg}} \\
Here, the $U_{24}$ is calculated by $I_{24}$ as the following:
\begin{align*}
U_{24} &= R_{135} \cdot I_{24} \\
       &= ({{U_2}\over{R_1}} - I_4) \cdot R_1||R_3||R_5
\end{align*}

On the right side of the last circuit, there is a voltage divider given by $R_{135}$, $R_6$, and $R_7$. \\
Therefore the voltage between $A$ and $B$ is given as:
\begin{align*}
U_{\rm AB} &= U_{24} \cdot {{R_7}\over{R_6 + R_7 + R_1||R_3||R_5}} \\
           &= ({{U_2}\over{R_1}} - I_4) \cdot {{R_7 \cdot R_1||R_3||R_5}\over{R_6 + R_7 + R_1||R_3||R_5}} \\
\end{align*}

For the internal resistance $R_i$ the ideal voltage source is substituted by its resistance ($=0\Omega$, so a short-circuit):
\begin{align*}
R_{\rm AB} &= R_7 || ( R_6 + R_1||R_3||R_5) \\
\end{align*}

with $R_1||R_3||R_5 = 5 ~\Omega || 10 ~\Omega || 10 ~\Omega =  5 ~\Omega ||  5 ~\Omega = 2.5 ~\Omega$:

\begin{align*}
U_{\rm AB} &= ({{6.0 ~\rm{V}}\over{5.0 ~\Omega}} - 4.2 ~\Omega) \cdot {{15 ~\Omega \cdot 2.5 ~\Omega}\over{7.5 ~\Omega + 15 ~\Omega + 2.5 ~\Omega}} \\
R_{\rm AB} &= 15 ~\Omega|| ( 7.5 ~\Omega + 2.5 ~\Omega) \\
\end{align*}

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\begin{align*}
U_{\rm s} &= U_{\rm AB} &&= 4.5 ~\rm{V}\\
R_{\rm i} &= R_{\rm AB} &&= 6   ~\Omega 
\end{align*}
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