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--- electrical_engineering_1:aufgabe_7.2.6_mit_rechnung [2021/10/24 00:57] – tfischer
+++ electrical_engineering_1:aufgabe_7.2.6_mit_rechnung [2024/11/20 15:21] (aktuell) – mexleadmin
@@ Zeile 1: / Zeile 1: @@
-<panel type="info" title="Excercise 7.2.6: 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%>
 <WRAP right> {{:elektrotechnik_1:schaltung_klws2020_3_2_1.jpg?400|schaltung_klws2020_3_2_1.jpg}}</WRAP>
-The adjacent circuit with the following data is given:
+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 capacitor is empty and all switches are open. The switsch S1 will be closed at t=0.
+At first, the voltage drop on the capacitor $u_C = 0$, and all switches are open. The switch S1 will be closed at $t = 0$.
+<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">
+<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjOB0AMt-CwFMC0B2E1IGYAsBOaPAJnwFY1cA2MADlzCpDIjJF22dTDACgA3EMWK4Q2MsSEjwaJtEzt5YeSsyQyvAM5TRYWUNq0Zc8CABmAQwA2mpLwBOBo3qbZ6xhcugOx7l79FiMhNabwBjAKFgyPFJeVxUIxVeAHcYiR0xDO9tN119XFCPJXNrWx9CpQKi2IUybzS8rMlK5sxU9hqMilds3gBLZn1iNEketpUYaFwfcZGx-TBiJPAGoaYlowIN5fbtDklN9nwdlYhLGzs0g-BdpqO1+7u-fW8Ae3ZTJWJsaDip6AQLBwIFCT6cACuAH0AMK8IA noborder}} </WRAP>
+</collapse>
 . Determine the time constant $\tau$ for this charging process.
@@ Zeile 24: / Zeile 30: @@
 <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>
-<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 = 7.2 ~{\rm µs}
+\end{align*} \\
+</collapse>
-. How high is the voltage at the capacitor $C$ when $t=10 µs$?
+. 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">
-\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}/7.2 ~{\rm µs}})
+\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 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">
-. How high is the energy when the capacitor is fully charged?
+\begin{align*} U_C(t) = 7.506 ~{\rm V} \rightarrow 7.5 ~{\rm V} \end{align*} \\ </collapse>
+. What is the value of the stored energy in the capacitor, when it is fully charged?
 <button size="xs" type="link" collapse="Loesung_7_2_6_3_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_7_2_6_3_Lösungsweg" collapsed="true">
-\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>
-<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>
-. Determine the new time constant when the capacitor is fully charged. The switch S1 will be opened whereas the switch S2 will be closed simultaneously.
+. Determine the new time constant when the switch $S_1$ will be opened and the switch $S_3$ will be closed simultaneously.
 <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>
-<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 = 5.2 ~{\rm µs}
+\end{align*} \\ </collapse>
-. When the capacitor is empty all switches will be opend. The switch S4 will be closed at $t = 1μs$. \\ How high is the voltage at the capacitor C?
+. When the capacitor is completely discharged, 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">
-  * 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.
@@ Zeile 71: / Zeile 101: @@
 <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 case, the 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>
-<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>
 </WRAP></WRAP></panel>