{{fa>pencil?32}} {{elektrotechnik_1:schaltung_klws2020_2_3_1.jpg?400}} 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.\\ {{icon>eye}} Tips * 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? {{icon>eye}} Solution 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*} {{icon>eye}} Final value \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*} \\