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Aufgabe 4.5.2: Leerlaufspannung über Superposition (Klausuraufgabe, ca 12% einer 60minütigen Klausur, 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.
- What do the individual circuits look like, by which the effects of the individual sources can be calculated?
Which equivalent resistor must be used to replace a current or voltage source when calculating the individual effects? - Where are the open-circuit voltages applied when looking at the individual components?
(Voltage) source $U_1$
- substitute the current source $I_2$ with a short-circuit
- substitute the voltage source $U_3$ with an open circuit
The components can be moved in order to understand the circuit s bit better.
For the open circuit, no current is flowing through any resistor. Therefor the effect is: $U_{AB,1} = U_1$
(current) source $I_2$
- substitute the voltage source $U_1$ with an open circuit
- substitute the voltage source $U_3$ with an open circuit
Also here, the components can be shifted for a better understanding:
Here, the current source $I_2$ creates a voltage drop $U_{AB_2}$ on the resistor $R_2$ : $U_{AB,2} = - R_1 \cdot I_2$
(Voltage) source $U_3$
- substitute the voltage source $U_1$ with an open circuit
- substitute the current source $I_2$ with a short-circuit
Again, rearranging the circuit might help for an understanding:
In this case, between the unloaded outputs $A$ and $B$ there will be an unloaded voltage divider given by $R_3$ and $R_4$. On $R_1$ there is no voltage drop, since there is no current flow out of the unloaded outputs.
Therefore:
\begin{align*} U_{AB,3} = \frac{R_4}{R_3 + R_4} \cdot U_3 \end{align*}
resulting voltage
\begin{align*} U_{AB} &= U_1 - R_1 \cdot I_2 + \frac{R_4}{R_3 + R_4} \cdot U_3 \\ \end{align*}