electrical_engineering_1:aufgabe_4.5.2_mit_rechnung [MEXLE Wiki]

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A circuit is given with the following parameters\\
$R_1=5 ~\Omega$\\
$U_1=2 ~\rm V$\\
$I_2=1 ~\rm A$\\
$R_3=20 ~\Omega$\\
$U_3=8 ~\rm V$\\
$R_4=10 ~\Omega$

Determine the open circuit voltage between A and B using the principle of superposition.\\

<button size="xs" type="link" collapse="Loesung_4_5_2_1_Tipps">{{icon>eye}} Tips </button><collapse id="Loesung_4_5_2_1_Tipps" collapsed="true">
  * 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?
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<button size="xs" type="link" collapse="Loesung_4_5_2_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_4_5_2_1_Lösungsweg" collapsed="true">

First, the individual circuits must be created, from which the effect of the individual sources between points A and B can be determined.
\\ \\
**(Voltage) source $U_1$**

  * substitute the current source $I_2$ with a short-circuit
  * substitute the voltage source $U_3$ with an open circuit

{{elektrotechnik_1:schaltung_klws2020_2_3_1_q1.jpg?400}}

The components can be moved in order to understand the circuit s bit better.

{{elektrotechnik_1:schaltung_klws2020_2_3_1_q1_1.jpg?300}}

For the open circuit, no current is flowing through any resistor. Therefore, 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

{{elektrotechnik_1:schaltung_klws2020_2_3_1_q2.jpg?400}}

Also here, the components can be shifted for a better understanding:

{{elektrotechnik_1:schaltung_klws2020_2_3_1_q2_1.jpg?300}}

Here, the current source $I_2$ creates a voltage drop $U_{AB_2}$ on the resistor $R_2$ : $U_{\rm AB,2} = - R_1 \cdot I_2$

**(Voltage) source $U_3$**

  * substitute the voltage source $U_1$ with an open circuit
  * substitute the current source $I_2$ with a short-circuit

{{elektrotechnik_1:schaltung_klws2020_2_3_1_q3.jpg?400}}

Again, rearranging the circuit might help for an understanding: 

{{elektrotechnik_1:schaltung_klws2020_2_3_1_q3_1.jpg?300}}

In this case, between the unloaded outputs $\rm A$ and $\rm B$ 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_{\rm AB,3} = \frac{R_4}{R_3 + R_4} \cdot U_3
\end{align*}

\\ \\
**resulting voltage**

\begin{align*}
U_{\rm AB} &=  U_1 - R_1 \cdot I_2 + \frac{R_4}{R_3 + R_4} \cdot U_3   \\
\end{align*}

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<button size="xs" type="link" collapse="Loesung_4_5_2_1_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_4_5_2_1_Endergebnis" collapsed="true">
\begin{align*}
U_{\rm AB} &=  2 ~{\rm V} - 5 ~\Omega \cdot 1 ~{\rm A} + \frac{10 ~\Omega}{20 ~\Omega + 10 ~\Omega} \cdot 8 ~{\rm V} \\ \\
U_{\rm AB} & = 0.333... ~{\rm V} \rightarrow 0.3 ~{\rm V} \\
\end{align*}
 \\
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