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electrical_engineering_1:aufgabe_7.2.6_mit_rechnung [2023/03/09 13:36]
mexleadmin
electrical_engineering_1:aufgabe_7.2.6_mit_rechnung [2024/02/08 15:40] (aktuell)
mexleadmin
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-<panel type="info" title="Exercise 5.2.6: Charging and Discharging of RC elements (exam task, ca. 11% of a 60 minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.2.3: Charging and Discharging of RC elements (exam task, ca. 11 % of a 60-minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
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 The circuit shown right is given with the following data: The circuit shown right is given with the following data:
  
-  * $U = 10 ~V$ +  * $U = 10 ~{\rm V}
-  * $I = 4 ~mA$+  * $I = 4 ~{\rm mA}$
   * $R_1 = 100 ~\Omega, R_2 = 80 ~\Omega, R_3 = 50 ~\Omega, R_4 = 10 ~\Omega$   * $R_1 = 100 ~\Omega, R_2 = 80 ~\Omega, R_3 = 50 ~\Omega, R_4 = 10 ~\Omega$
-  * $C = 40 ~nF$+  * $C = 40 ~{\rm nF}$
  
-At first the voltage drop on the capacitor $u_C = 0$ and all switches are open. The switch S1 will be closed at $t = 0$.+At firstthe voltage drop on the capacitor $u_C = 0$and all switches are open. The switch S1 will be closed at $t = 0$.
  
 <button size="xs" type="link" collapse="Loesung_7_2_6_6_Simu">{{icon>eye}} Simulation</button><collapse id="Loesung_7_2_6_6_Simu" collapsed="true">  <button size="xs" type="link" collapse="Loesung_7_2_6_6_Simu">{{icon>eye}} Simulation</button><collapse id="Loesung_7_2_6_6_Simu" collapsed="true"> 
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-The electrical components $R_1$, $R_2$ und $C$ are connected in series with a source $U$. The time constant $\tau$ is therefore: \begin{align*} \tau &= (R_1 + R_2) \cdot C \\ \tau &= 180 ~\Omega \cdot 40 ~nF \end{align*}+The electrical components $R_1$, $R_2$, and $C$ are connected in series with a source $U$.  
 +The time constant $\tau$ is therefore:  
 +\begin{align*}  
 +\tau &= (R_1 + R_2) \cdot C \\  
 +\tau &= 180 ~\Omega \cdot 40 ~{\rm nF}  
 +\end{align*}
  
 </collapse> </collapse>
  
-<button size="xs" type="link" collapse="Loesung_7_2_6_1_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_1_Endergebnis" collapsed="true"> \begin{align*} \tau = 7.2 ~\mu s \end{align*} \\ </collapse>+<button size="xs" type="link" collapse="Loesung_7_2_6_1_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_1_Endergebnis" collapsed="true">  
 +\begin{align*} \tau = 7.2 ~{\rm µs} 
 +\end{align*} \\  
 +</collapse>
  
-2. What is the value of the voltage $u_C(t)$ drop over the capacitor $C$ at $t=10 ~\mu s$?+2. What is the value of the voltage $u_C(t)$ drop over the capacitor $C$ at $t=10 ~{\rm µs}$?
  
 <button size="xs" type="link" collapse="Loesung_7_2_6_2_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_7_2_6_2_Lösungsweg" collapsed="true"> <button size="xs" type="link" collapse="Loesung_7_2_6_2_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_7_2_6_2_Lösungsweg" collapsed="true">
  
 \begin{align*}  \begin{align*} 
-U_C(t) = U \cdot (1 - e^{-t/\tau}) \\  +U_C(t) = U           \cdot (1 - e^{-t/\tau}) \\  
-U_C(t) = 10 ~V \cdot (1 - e^{-10 ~\mu s/7.2 ~\mu s}) +U_C(t) = 10 ~{\rm V\cdot (1 - e^{-10 ~{\rm µs}/7.2 ~{\rm µs}}) 
 \end{align*} \end{align*}
  
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-\begin{align*} U_C(t) = 7.506 ~V \rightarrow 7.5 ~V \end{align*} \\ </collapse>+\begin{align*} U_C(t) = 7.506 ~{\rm V\rightarrow 7.5 ~{\rm V\end{align*} \\ </collapse>
  
-3. What is the value of the energy, when the capacitor is fully charged?+3. What is the value of the stored energy in the capacitor, when it is fully charged?
  
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 \begin{align*}  \begin{align*} 
 W_C &= \frac{1}{2} C U^2 \\  W_C &= \frac{1}{2} C U^2 \\ 
-    &= \frac{1}{2} \cdot 40~nF \cdot (10~V)^2 +    &= \frac{1}{2} \cdot 40~{\rm nF\cdot (10~{\rm V})^2 
 \end{align*} \end{align*}
  
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-\begin{align*} W_C = 2 ~\mu J \end{align*} \\ +\begin{align*} W_C = 2 ~{\rm µJ} \end{align*} \\ 
 </collapse> </collapse>
  
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 \begin{align*}  \begin{align*} 
 \tau &= (R_2 + R_3) \cdot C \\  \tau &= (R_2 + R_3) \cdot C \\ 
-     &= 130 ~\Omega \cdot 40 ~nF +     &= 130 ~\Omega \cdot 40 ~{\rm nF
 \end{align*} \end{align*}
  
 </collapse> </collapse>
  
-<button size="xs" type="link" collapse="Loesung_7_2_6_4_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_4_Endergebnis" collapsed="true"> \begin{align*} \tau = 5.2 ~\mu s \end{align*} \\ </collapse>+<button size="xs" type="link" collapse="Loesung_7_2_6_4_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_4_Endergebnis" collapsed="true">  
 +\begin{align*} \tau = 5.2 ~{\rm µs} 
 +\end{align*} \\ </collapse>
  
-5. When the capacitor is empty all switches will be opened. The switch $S_4$ will be closed at $t= 0$. \\ What is the voltage $u_C$ at the capacitor C after $t = 1 ~\mu s$?+5. When the capacitor is empty all switches will be opened. The switch $S_4$ will be closed at $t= 0$. \\ What is the voltage $u_C$ at the capacitor C after $t = 1 ~ {\rm µs}$?
  
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-  * Through the current source there is a continuous flow of elctric charge into the capacitor.+  * Through the current source there is a continuous flow of electric charge into the capacitor.
   * The resistors passed by the current on the way to the capacitor are irrelevant. They only increase the voltage of an ideal current source to guarantee the current.   * The resistors passed by the current on the way to the capacitor are irrelevant. They only increase the voltage of an ideal current source to guarantee the current.
  
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-The voltage $U_C$ is in general: $U_C = \frac{Q}{C}$. In this case the constant current I results in $Q = \int I dt = I \cdot t$ +The voltage $U_C$ is in general: $U_C = \frac{Q}{C}$. In this casethe constant current I results in $Q = \int I {\rm d}t = I \cdot t$ 
 \begin{align*}  \begin{align*} 
-U_C(t)   &= \frac{Q}{C} \\ U_C(t) &= \frac{I \cdot t}{C} \\  +U_C(t)   &= \frac{Q}{C} \\  
-U_C(1μs) &= \frac{4~mA \cdot 1~\mu s}{40~nF} = \frac{4 \cdot 10^{-3}~A \cdot 1\cdot 10^{-6}~s}{40\cdot 10^{-9}~F} \\ +U_C(t)   &= \frac{I \cdot t}{C} \\  
 +U_C(1μs) &= \frac{4~{\rm mA\cdot 1~{\rm µs}}{40~{\rm nF}}  
 +          = \frac{4          \cdot 10^{-3}~{\rm A\cdot 1\cdot 10^{-6}~{\rm s}}{40\cdot 10^{-9}~{\rm F}} \\ 
 \end{align*} \end{align*}
  
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 \begin{align*}  \begin{align*} 
-U_C(1~\mu s) &= 1~V \\ +U_C(1~{\rm µs}) &= 1~{\rm V\\ 
 \end{align*} \\  \end{align*} \\ 
 </collapse> </collapse>
  
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