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--- electrical_engineering_1:dc_circuit_transients [2023/11/30 00:06]
mexleadmin [Bearbeiten - Panel]
+++ electrical_engineering_1:dc_circuit_transients [2023/12/03 16:53] (aktuell)
mexleadmin [Exercises]
@@ Zeile 474: / Zeile 474: @@
 </WRAP></WRAP></panel>
-#@TaskTitle_HTML@#5.2.2 Charge balance of two capacitors \\ <fs medium>(educational exercise, not part of an exam)</fs>#@TaskText_HTML@#
+#@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:
+  * $R_1 = 1.0 \rm k\Omega$
+  * $R_2 = 2.0 \rm k\Omega$
+  * $R_3 = 3.0 \rm k\Omega$
+  * $C   = 1 \rm \mu F$
+  * $S_1$ 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 $S_1$ will be opened. \\ 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 \approx 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: $U_1 = 5.0 ~\rm V$ 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 $S_1$ is open and $S_2$ is closed.
+{{drawio>electrical_engineering_1:Exercise522sol4.svg}}
+We see that for the time constant, we now need to use $R=R_3 + R_2$.
+\begin{align*}
+\tau_2 &= R\cdot C \\
+       &= (R_3 + R_2) \cdot C \\
+\end{align*}
+#@HiddenEnd_HTML~Solution4,Solution ~@#
+#@HiddenBegin_HTML~Result4,Result~@#
+\begin{align*}
+\tau_2 &= 5~\rm ms \\
+\end{align*}
+#@HiddenEnd_HTML~Result4,Result~@#
+.2 Calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$.
+#@HiddenBegin_HTML~Solution5,Solution~@#
+To calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$, we first have to find out the value of $u_{R2}(t_2 = 10 ~\rm ms)$, when $S_2$ just got closed. \\
+  * Starting from $t_2 = 10 ~\rm ms$, the voltage source $U_2$ charges up the capacitor $C$ further.
+  * Before at $t_1$, when $S_1$ got opened, the value of $u_c$ was: $u_c(t_1) = 4/5 \cdot U_1 = 4 ~\rm V$.
+  * This is also true for $t_2$, since between $t_1$ and $t_2$ the charge on $C$ does not change: $u_c(t_2) = 4 ~\rm V$.
+  * In the first moment after closing $S_2$ at $t_2$, the voltage drop on $R_3 + R_2$ is: $U_{R3+R2} = U_2 - u_c(t_2) = 6 ~\rm V$.
+  * So the voltage divider of $R_3 + R_2$ lead to $ \boldsymbol{u_{R2}(t_2 = 10 ~\rm ms)} =  {{R_2}\over{R_3 + R2}} \cdot U_{R3+R2} = {{2 {~\rm k\Omega}}\over{3 {~\rm k\Omega} + 2 {~\rm k\Omega} }} \cdot 6 ~\rm V =  \boldsymbol{2.4 ~\rm V} $
+We see that the voltage on $R_2$ has to decrease from $2.4 ~\rm V $ to $1/10 \cdot U_2 = 1 ~\rm V$. \\
+To calculate this, there are multiple ways. In the following, one shall be retraced:
+  * We know, that the current $i_C = i_{R2}$ subsides exponentially: $i_{R2}(t) = I_{R2~ 0} \cdot {\rm e}^{-t/\tau}$
+  * So we can rearrange the task to focus on the change in current instead of the voltage.
+  * The exponential decay is true regardless of where it starts.
+So from ${{i_{R2}(t)}\over{I_{R2~ 0}}} =  {\rm e}^{-t/\tau}$ we get
+\begin{align*}
+{{i_{R2}(t_3)}\over{i_{R2}(t_2)}} &=                                 {\rm exp} \left( -{{t_3 - t_2}\over{\tau_2}}       \right) \\
+-{{t_3 - t_2}\over{\tau_2}}       &=                                 {\rm ln } \left( {{i_{R2}(t_3)}\over{i_{R2}(t_2)}} \right) \\
+   t_3                            &= t_2          - \tau_2     \cdot {\rm ln } \left( {{i_{R2}(t_3)}\over{i_{R2}(t_2)}} \right) \\
+   t_3                            &= 10 ~{\rm ms} - 5~{\rm ms} \cdot {\rm ln } \left( {{1 ~\rm V   }\over{2.4 ~\rm V }} \right) \\
+\end{align*}
+#@HiddenEnd_HTML~Solution5,Solution ~@#
+#@HiddenBegin_HTML~Result5,Result~@#
+\begin{align*}
+t_3 &= 14.4~\rm ms \\
+\end{align*}
+#@HiddenEnd_HTML~Result5,Result~@#
+.3 Draw the course of time of the voltage $u_C(t)$ over the capacitor.
+{{drawio>electrical_engineering_1:Exercise522task6.svg}}
+#@HiddenBegin_HTML~Result6,Result~@#
+{{drawio>electrical_engineering_1:Exercise522sol6.svg}}
+#@HiddenEnd_HTML~Result6,Result~@#
+#@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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-{{page>aufgabe_7.2.6_mit_rechnung&nofooter}}