Dies ist eine alte Version des Dokuments!
Exercise 1 : Equivalent Linear Source
(written test, approx. 12% of a 60-minute written test, WS2022)
The circuit in the following has to be simplified.
Calculated the internal resistance $R_i$ and the source voltage $U_s$ of an equivalent linear voltage source on the connectors $A$ and $B$. \begin{align*} R_1=5.0 \Omega, && U_2=6.0 V, && R_3=10 \Omega, \\ I_4=4.2 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!
The linear voltage source of $U_2$ and $R_1$ can be transormed into a current source $I_2={{U_2}\over{R_1}}$ and $R_1$:
Now a lot of 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:
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_{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_{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_{AB} &= ({{6.0 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_{AB} &= 15 \Omega|| ( 7.5 \Omega + 2.5 \Omega) \\ \end{align*}