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electrical_engineering_1:aufgabe_2.7.8_mit_rechnung [2021/10/07 09:21] – [Bearbeiten - Panel] tfischerelectrical_engineering_1:aufgabe_2.7.8_mit_rechnung [2023/03/19 19:04] (current) – mexleadmin
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-<panel type="info" title="Task 2.7.8: Simplifying Circuits II (written exam task, approx 8% of a 60-minute written exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 2.7.8: Simplifying Circuits II (written exam task, approx 8% of a 60-minute written exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
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 Given is the adjoining circuit with \\ Given is the adjoining circuit with \\
-$R_1=5 \Omega$\\ +$R_1=5 ~\Omega$\\ 
-$R_2=10 \Omega$\\ +$R_2=10 ~\Omega$\\ 
-$R_3=20 \Omega$\\+$R_3=20 ~\Omega$\\
  
-1. determine the equivalent resistance $R_{eq}$ between A and B by summing the resistances.+1. Determine the equivalent resistance $R_{\rm eq}$ between A and B by summing the resistances.
  
 <button size="xs" type="link" collapse="Solution_2_7_8_1_Tips">{{icon>eye}} Tips for solving</button><collapse id="Solution_2_7_8_1_Tips" collapsed="true"> <button size="xs" type="link" collapse="Solution_2_7_8_1_Tips">{{icon>eye}} Tips for solving</button><collapse id="Solution_2_7_8_1_Tips" collapsed="true">
   * How can the circuit be better represented or pulled apart?   * How can the circuit be better represented or pulled apart?
-  * Switches (when used) should be replaced by an open of closed circuit.+  * Switches (when used) should be replaced by an open or closed circuit.
   * Does this result in equal potentials at different nodes that can be cleverly used?   * Does this result in equal potentials at different nodes that can be cleverly used?
 </collapse> </collapse>
  
-<button size="xs" type="link" collapse="Solution_2_7_8_1_Solution_path">{{icon>eye}} Solution path</button><collapse id="Solution_2_7_8_1_Solution_path" collapsed="true">+<button size="xs" type="link" collapse="Solution_2_7_8_1_Solution_path">{{icon>eye}} Solution</button><collapse id="Solution_2_7_8_1_Solution_path" collapsed="true">
  
 First of all, it is a good idea to reshape the circuit so that the actual structure becomes visible. \\ First of all, it is a good idea to reshape the circuit so that the actual structure becomes visible. \\
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 {{elektrotechnik_1:schaltung_klws2020_2_2_1_loesung2.jpg?300}} {{elektrotechnik_1:schaltung_klws2020_2_2_1_loesung2.jpg?300}}
  
-Here it helps to consider the potential of the nodes K1, K2 and K3. For K2, the resistances $R_2 || R_3 || R_2$ must be combined at the top and bottom. Thus, the same resistance values at the top and bottom result. Also at the nodes K1 and K2 the same resistance values at the top and at the bottom result. With the same ratios of the resistances at K1, K2 and K3 respectively, it can be concluded that no current flows across the resistors $R_3$ between K1 and K2 or K2 and K3. Thus, these do not contribute to the total resistance. In such a case, a short circuit or an open line can be freely chosen between the relevant nodes for the calculation. In the following an open line is chosen. Additionally the parallel strings can be reordered. \\+Here it helps to consider the potential of the nodes K1, K2, and K3. For K2, the resistances $R_2 || R_3 || R_2$ must be combined at the top and bottom. Thus, the same resistance values at the top and bottom result. Also at the nodes K1 and K2 the same resistance values at the top and at the bottom result. With the same ratios of the resistances at K1, K2, and K3 respectively, it can be concluded that no current flows across the resistors $R_3$ between K1 and K2 or K2 and K3. Thus, these do not contribute to the total resistance. In such a case, a short circuit or an open line can be freely chosen between the relevant nodes for the calculation. In the following, an open line is chosen. Additionally, the parallel strings can be reordered. \\ 
 + 
 +<WRAP> 
 +<imgcaption BildExercisNr1 | Simulation> 
 +</imgcaption> \\ 
 +{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-Wm22O13e-2BIPc2ppyI2ApqZAS7QmIAZ0iHIgPI+IV8gL3Yp6HrAqRYQGAgImnYmPIDPW4I4-dEgzRnMBy27ABbbsABsAEMBHbVtWwbAAHKtF1rbsYKgzdKj2cg2FAv0bD0TwAGl+FzWhbUkfD2Eoa4S1AkByMfWhpxovlyCQa5uJI5jCCHSQakIfBhFwMS2D40jJD4Qh6Jo25LieK4ICzOTDEeJThBSLTVL1QgdLeH4ziVLS3jeRQgA noborder}} 
 +</WRAP> 
 This results in: This results in:
  
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 \begin{align*} \begin{align*}
-R_{ges} &= \left( \left( 2 \cdot R_2 \right) || \left( 2 \cdot R_2 \right) \right) \quad && || \quad \left( \left( 2 \cdot R_3 \right) || \left( 2 \cdot R_3 \right) || \left( 2 \cdot R_3 \right) \right)  \\ +R_{\rm eq} &= \left( \left( 2 \cdot R_2 \right) || \left( 2 \cdot R_2 \right) \right) && || \; \left( \left( 2 \cdot R_3 \right) || \left( 2 \cdot R_3 \right) || \left( 2 \cdot R_3 \right) \right)  \\ 
-R_{ges} &= R_2 \quad && || \quad \left( R_3 || \left( 2 \cdot R_3 \right) \right)  \\ +R_{\rm eq} &= R_2 && || \;\left( R_3 || \left( 2 \cdot R_3 \right) \right)  \\ 
-R_{ges} &= R_2 \quad && || \quad \frac{R_3 \cdot 2 R_3}{R_3 + 2 R_3}   \\ +R_{\rm eq} &= R_2 && || \;\frac{R_3 \cdot 2 R_3}{R_3 + 2 R_3}   \\ 
-R_{ges} &= R_2 \quad && || \quad \frac{2}{3}\cdot R_3 \ +R_{\rm eq} &= R_2 && || \;\frac{2}{3}\cdot R_3 \\ 
-R_{ges} &= \frac{R_2 \cdot \frac{2}{3}\cdot R_3}{R_2 + \frac{2}{3}\cdot R_3} = \frac{R_2 \cdot R_3}{\frac{3}{2}\cdot R_2 + R_3}   \\ \\+R_{\rm eq} &=  \frac{R_2 \cdot \frac{2}{3}\cdot R_3}{R_2 + \frac{2}{3}\cdot R_3} \\ 
 +R_{\rm eq} &=  \frac{R_2 \cdot R_3}{\frac{3}{2}\cdot R_2 + R_3} 
 +   \\ \\
 \end{align*} \end{align*}
  
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 <button size="xs" type="link" collapse="Solution_2_7_8_1_Final_result">{{icon>eye}} Final result</button><collapse id="Solution_2_7_8_1_Final_result" collapsed="true"> <button size="xs" type="link" collapse="Solution_2_7_8_1_Final_result">{{icon>eye}} Final result</button><collapse id="Solution_2_7_8_1_Final_result" collapsed="true">
 \begin{align*} \begin{align*}
-R_{ges} &= \frac{10 \Omega \cdot 20 \Omega}{\frac{3}{2}\cdot 10 \Omega + 20 \Omega} = 5.7143 \Omega -> 5.7 \Omega \.+R_{\rm eq} &= \frac{10 ~\Omega \cdot 20 ~\Omega}{\frac{3}{2}\cdot 10 ~\Omega + 20 ~\Omega} = 5.7143 ~\Omega \rightarrow 5.7 ~\Omega 
 \end{align*} \end{align*}
  \\  \\
 </collapse> </collapse>
  
-2. now let the voltage from A to B be: $U_{AB}=U_0= 20 V$. What is the current $I$?+2. Now let the voltage from A to B be: $U_{AB}=U_0= 20 ~\rm V$. What is the current $I$?
  
-<button size="xs" type="link" collapse="Solution_2_7_8_2_Solution_Path">{{icon>eye}} Solution path</button><collapse id="Solution_2_7_8_2_Solution_Path" collapsed="true">+<button size="xs" type="link" collapse="Solution_2_7_8_2_Solution_Path">{{icon>eye}} Solution</button><collapse id="Solution_2_7_8_2_Solution_Path" collapsed="true">
 The partial current $I$ is obtained directly from the voltage $U_0$: The partial current $I$ is obtained directly from the voltage $U_0$:
 \begin{align*} \begin{align*}
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 <button size="xs" type="link" collapse="Solution_2_7_8_2_Final_result">{{icon>eye}} Final result</button><collapse id="Solution_2_7_8_2_Final_result" collapsed="true"> <button size="xs" type="link" collapse="Solution_2_7_8_2_Final_result">{{icon>eye}} Final result</button><collapse id="Solution_2_7_8_2_Final_result" collapsed="true">
 \begin{align*} \begin{align*}
-I =\frac{20V}{2 \cdot 20 \Omega} = 0.5 A+I =\frac{20~V}{2 \cdot 20 ~\Omega} = 0.5 ~\rm A
 \end{align*} \end{align*}
  \\  \\