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--- electrical_engineering_1:dc_circuit_transients [2023/11/29 23:59] – mexleadmin
+++ electrical_engineering_1:dc_circuit_transients [2023/12/02 00:55] – [Exercises] mexleadmin
@@ Zeile 474: / Zeile 474: @@
 </WRAP></WRAP></panel>
-<panel type="info" title="Exercise 5.2.2 Charge balance of two capacitors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+#@TaskTitle_HTML@# Exercise 5.2.2 Capacitor charging/discharging #@TaskText_HTML@#
+The following circuit shows a charging/discharging circuit for a capacitor.
+The values of the components shall be the following:
+  * $R_1 = 1 \rm k\Omega$
+  * $R_2 = 2 \rm k\Omega$
+  * $R_3 = 3 \rm k\Omega$
+  * $C   = 1 \rm \mu F$
+  * $S_1$ and $S_2$ are opened in the beginning (open-circuit)
+{{drawio>electrical_engineering_1:Exercise522setup.svg}}
+. For the first tasks, the switch $S_1$ gets closed at $t=t_0 = 0s$. \\
+.1 What is the value of the time constant $\tau_1$?
+#@HiddenBegin_HTML~Solution1,Solution~@#
+The time constant $\tau$ is generally given as: $\tau= R\cdot C$. \\
+Now, we try to determine which $R$ and $C$ must be used here. \\
+To find this out, we have to look at the circuit when $S_1$ gets closed.
+{{drawio>electrical_engineering_1:Exercise522sol1.svg}}
+We see that for the time constant, we need to use $R=R_1 + R_2$.
+#@HiddenEnd_HTML~Solution1,Solution ~@#
+#@HiddenBegin_HTML~Result1,Result~@#
+\begin{align*}
+\tau_1 &= R\cdot C \\
+       &= (R_1 + R_2) \cdot C \\
+       &= 3~\rm ms \\
+\end{align*}
+#@HiddenEnd_HTML~Result1,Result~@#
+.2 What is the formula for the voltage $u_{R2}$ over the resistor $R_2$? Derive a general formula without using component values!
+#@HiddenBegin_HTML~Solution2,Solution~@#
+To get a general formula, we again take a look at the circuit, but this time with the voltage arrows.
+{{drawio>electrical_engineering_1:Exercise522sol2.svg}}
+We see, that: $U_1 = u_C + u_{R2}$ and there is only one current in the loop: $i = i_C = i_{R2}$\\
+The current is generally given with the exponential function: $i_c = {{U}\over{R}}\cdot e^{-t/\tau}$, with $R$ given here as $R = R_1 + R_2$.
+Therefore, $u_{R2}$ can be written as:
+\begin{align*}
+u_{R2} &= R_2 \cdot i_{R2} \\
+       &= U_1 \cdot {{R_2}\over{R_1 + R_2}} \cdot e^{-t/ \tau}
+\end{align*}
+#@HiddenEnd_HTML~Solution2,Solution ~@#
+#@HiddenBegin_HTML~Result2,Result~@#
+\begin{align*}
+u_{R2} = U_1 \cdot {{R_2}\over{R_1 + R_2}} \cdot e^{t/ \tau}
+\end{align*}
+#@HiddenEnd_HTML~Result2,Result~@#
+. At a distinct time $t_1$, the voltage $u_C$ is charged up to $4/5 \cdot U_1$.
+At this point, the switch $S_2$ will be closed. \\ Calculate $t_1$!
+#@HiddenBegin_HTML~Solution3,Solution~@#
+We can derive $u_{C}$ based on the exponential function: $u_C(t) = U_1 \cdot (1-e^{-t/\tau})$. \\
+Therefore we get $t_1$ by:
+\begin{align*}
+u_C = 4/5 \cdot U_1              &=  U_1 \cdot (1-e^{-t/\tau}) \\
+/5                        &=             1-e^{-t/\tau} \\
+      e^{-t/\tau}                &=             1-4/5 = 1/5\\
+         -t/\tau                 &=             \rm ln (1/5) \\
+          t                      &= -\tau \cdot \rm ln (1/5) \\
+\end{align*}
+#@HiddenEnd_HTML~Solution3,Solution ~@#
+#@HiddenBegin_HTML~Result3,Result~@#
+\begin{align*}
+          t                      &=  3~{\rm ms} \cdot 1.61 = 4.8~\rm ms \\
+\end{align*}
+#@HiddenEnd_HTML~Result3,Result~@#
+. The switch $S_2$ will get closed at the moment $t_2 = 10 ~\rm ms$. The values of the voltage sources are now: $U_1 = 5 ~\rm V$ and $U_2 = 10 ~\rm V$.
+.1 What is the new time constant $\tau_2$?
+#@HiddenBegin_HTML~Solution4,Solution~@#
+Again the time constant $\tau$ is given as: $\tau= R\cdot C$. \\
+Again, we try to determine which $R$ and $C$ must be used here. \\
+To find this out, we have to look at the circuit when both $S_1$ and $S_2$ are closed. \\
+In this case, we can "fold" over the resistor $R_3$ to the left-side of the capacitor $C$.
+{{drawio>electrical_engineering_1:Exercise522sol4.svg}}
+We see that for the time constant, we now need to use $R=R_1 || R_3 + R_2$.
+\begin{align*}
+\tau_2 &= R\cdot C \\
+       &= (R_1 || R_3 + R_2) \cdot C \\
+       &= \left({{\rm  1 ~k\Omega \cdot 3 ~k\Omega }\over{\rm  1 ~k\Omega + 3 ~k\Omega}} + 2 ~{\rm k\Omega}\right) \cdot 1 ~{\rm \mu F} \\
+\end{align*}
+#@HiddenEnd_HTML~Solution4,Solution ~@#
+#@HiddenBegin_HTML~Result4,Result~@#
+\begin{align*}
+\tau_2 &= 2.75~\rm ms \\
+\end{align*}
+#@HiddenEnd_HTML~Result4,Result~@#
+.2 Calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$.
+.3 Draw the course of time of the voltage $u_C(t)$ over the capacitor.
+#@TaskEnd_HTML@#
+{{page>aufgabe_7.2.6_mit_rechnung&nofooter}}
+#@TaskTitle_HTML@#5.2.4 Charge balance of two capacitors \\ <fs medium>(educational exercise, not part of an exam)</fs>#@TaskText_HTML@#
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@@ Zeile 507: / Zeile 631: @@
       - What are the energies and the total energy? \\ How is this understandable with the previous total energy?
-</WRAP></WRAP></panel>
+#@TaskEnd_HTML@#
-{{page>aufgabe_7.2.6_mit_rechnung&nofooter}}