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electrical_engineering_1:aufgabe_7.2.6_mit_rechnung [2023/02/11 23:08]
mexleadmin
electrical_engineering_1:aufgabe_7.2.6_mit_rechnung [2023/11/30 00:07]
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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 <button size="xs" type="link" collapse="Loesung_7_2_6_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_7_2_6_1_Lösungsweg" collapsed="true"> <button size="xs" type="link" collapse="Loesung_7_2_6_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_7_2_6_1_Lösungsweg" collapsed="true">
  
-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 µs \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.~{\rm µs
 +\end{align*} \\  
 +</collapse>
  
-2. What is the value of the voltage $u_C(t)$ drop over the capacitor $C$ at $t=10 µs$?+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*} U_C(t) = U \cdot (1 - e^{-t/\tau}) \\ U_C(t) = 10 V \cdot (1 - e^{-10 µs/7,2 µs}) \end{align*}+\begin{align*}  
 +U_C(t) = U           \cdot (1 - e^{-t/\tau}) \\  
 +U_C(t) = 10 ~{\rm V\cdot (1 - e^{-10 ~{\rm µs}/7.~{\rm µs}})  
 +\end{align*}
  
 </collapse> </collapse>
  
-<button size="xs" type="link" collapse="Loesung_7_2_6_2_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_2_Endergebnis" collapsed="true"> \begin{align*} U_C(t) = 7,506 V -> 7,5 V \end{align*} \\ </collapse>+<button size="xs" type="link" collapse="Loesung_7_2_6_2_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_2_Endergebnis" collapsed="true">  
 + 
 +\begin{align*} U_C(t) = 7.506 ~{\rm V} \rightarrow 7.~{\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 energy, when the capacitor is fully charged?
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-\begin{align*} W_C &= \frac{1}{2}CU^2 \\ &= \frac{1}{2} \cdot 40nF \cdot (10V)^2 \end{align*}+\begin{align*}  
 +W_C &= \frac{1}{2} C U^2 \\  
 +    &= \frac{1}{2} \cdot 40~{\rm nF} \cdot (10~{\rm V})^2  
 +\end{align*}
  
 </collapse> </collapse>
  
-<button size="xs" type="link" collapse="Loesung_7_2_6_3_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_3_Endergebnis" collapsed="true"> \begin{align*} W_C = 2 µJ \end{align*} \\ </collapse>+<button size="xs" type="link" collapse="Loesung_7_2_6_3_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_3_Endergebnis" collapsed="true">  
 +\begin{align*} W_C = 2 ~{\rm µJ\end{align*} \\  
 +</collapse>
  
 4. Determine the new time constant when the switch $S_1$ will be opened and the switch $S_3$ will be closed simultaneously.  4. Determine the new time constant when the switch $S_1$ will be opened and the switch $S_3$ will be closed simultaneously. 
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 <button size="xs" type="link" collapse="Loesung_7_2_6_4_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_7_2_6_4_Lösungsweg" collapsed="true"> <button size="xs" type="link" collapse="Loesung_7_2_6_4_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_7_2_6_4_Lösungsweg" collapsed="true">
  
-The capacitor $C$ discharges by the series connected resistors $R_2$ und $R_3$. \begin{align*} \tau &= (R_2 + R_3) \cdot C \\ \tau &= 130 \Omega \cdot 40 nF \end{align*}+The capacitor $C$ discharges by the series connected resistors $R_2$ und $R_3$.  
 +\begin{align*}  
 +\tau &= (R_2 + R_3) \cdot C \\  
 +     &= 130 ~\Omega \cdot 40 ~{\rm nF}  
 +\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 µs \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.~{\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 $t = 1μs$?+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 $t = 1 ~ {\rm µs}$?
  
 <button size="xs" type="link" collapse="Loesung_7_2_6_5_Tipps">{{icon>eye}} Tips</button><collapse id="Loesung_7_2_6_5_Tipps" collapsed="true"> <button size="xs" type="link" collapse="Loesung_7_2_6_5_Tipps">{{icon>eye}} Tips</button><collapse id="Loesung_7_2_6_5_Tipps" collapsed="true">
  
-  * 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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 <button size="xs" type="link" collapse="Loesung_7_2_6_5_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_7_2_6_5_Lösungsweg" collapsed="true"> <button size="xs" type="link" collapse="Loesung_7_2_6_5_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_7_2_6_5_Lösungsweg" collapsed="true">
  
-The voltage $U_C$ is in general: $U_C = \frac{Q}{C}$. In this case the constant current I results in $Q = \int I dt = I \cdot t$ \begin{align*} U_C(t) &= \frac{Q}{C} \\ U_C(t) &= \frac{I \cdot t}{C} \\ U_C(1μs) &= \frac{4mA \cdot 1μs}{40nF} = \frac{4 \cdot 10^{-3}A \cdot 1\cdot 10^{-6}s}{40\cdot 10^{-9}F} \\ \end{align*}+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*}  
 +U_C(t)   &= \frac{Q}{C} \\  
 +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*}
  
 </collapse> </collapse>
  
-<button size="xs" type="link" collapse="Loesung_7_2_6_5_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_5_Endergebnis" collapsed="true"> \begin{align*} U_C(1μs) &1V \\ \end{align*} \\ </collapse>+<button size="xs" type="link" collapse="Loesung_7_2_6_5_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_7_2_6_5_Endergebnis" collapsed="true">  
 +\begin{align*}  
 +U_C(1~{\rm µs}) &1~{\rm V} \\  
 +\end{align*} \\  
 +</collapse>
  
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