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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} \cdot 2~V = 1.33~V \end{align*} Case 2: For this case is $U_1 = 0~V$ and $U_3 = 0~V$. The voltage is at $R_3$.
\begin{align*} U_{AB,2} = R_3 I_2 = 20~\Omega \cdot 1~A = 20~V \end{align*} Case 3: For this case is $U_1 = 0~V$ and $I_2 = 0~A$. The voltage comes from the source $U_3$.
\begin{align*} U_{AB,3} = 8~V \end{align*} Superposition means adding the voltages of all three cases. \begin{align*} U_{AB} = U_{AB,1} + U_{AB,2} + U_{AB,3} = 1.33~V + 20~V + 8~V \end{align*}