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Exercise 4.5.3 -Variation: open circuit voltage via superposition (exam task, approx. 12% of a 60-minute exam, WS2020)
A circuit is given with the following parameters
$R_1=5 \Omega$
$U_1=2 V$
$I_2=1 A$
$R_3=20 \Omega$
$U_3=8 V$
$R_4=10 \Omega$
Determine the open circuit voltage between A and B using the principle of superposition.
\begin{align*} U_{AB,1}=\frac{R_4}{R_1+R_4}U_1=\frac{10\Omega}{5\Omega+10\Omega}\cdot2V=1.33V \end{align*} Case 2: For this case is $U_1=0V$ and $U_3=0V$. The voltage is at $R_3$.
\begin{align*} U_{AB,2}=R_3I_2=20\Omega\cdot1A=20V \end{align*} Case 3: For this case is $U_1=0V$ and $I_2=0A$. The voltage comes from the source $U_3$.
\begin{align*} U_{AB,3}=8V \end{align*} Superposition means adding the voltages of all three cases. \begin{align*} U_{AB}=U_{AB,1}+U_{AB,2}+U_{AB,3}=1.33V+20V+8V \end{align*}