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--- electrical_engineering_1:task_tb6pi8dgh0m2e2pw_with_calculation [2023/03/10 17:29] – mexleadmin
+++ electrical_engineering_1:task_tb6pi8dgh0m2e2pw_with_calculation [Unbekanntes Datum] (aktuell) – gelöscht - Externe Bearbeitung (Unbekanntes Datum) 127.0.0.1
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-{{tag>Charging_Capacitors dc_network_analysis pure_resistor_network_simplification delta_wye_transformation exam_ee1_WS2022}}
-<panel type="info" > <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-<fs x-large>**Exercise ~~#@ee1_taskctr.#~~ : Chargine Capacitors ** \\ (written test, approx. 16 % of a 60-minute written test, WS2022) \\ \\ </fs>
-The circuit shown in the following is used to control the brightness when turning on a small light bulb.
-{{drawio>electrical_engineering_1:tb6pi8dgh0m2e2pwCircuit.svg}}
-The circuit contains a voltage source $U=12 ~\rm{V}$, a switch $S_1$, a resistor of $R_1=20 ~\Omega$ and a capacitor of $C=100 ~\rm{µF}$. \\
-The switch $S_2$ to an additional consumer $R_2$ will be considered to be open for the first tasks. At the moment $t_0=0 ~\rm{s}$ the switch $S_1$ is closed, the voltage across the capacitor is $u_c (t_0 )=0 ~\rm{V}$.
-. First do not consider the light bulb – it is not connected to the RC-circuit. \\
-Calculate the point of time $t_1$ when $u_c (t_1)=0.5\cdot U$.
-<button size="xs" type="link" collapse="tb6pi8dgh0m2e2pw_1_path">{{icon>eye}} Solution</button><collapse id="tb6pi8dgh0m2e2pw_1_path" collapsed="true">
-<callout type="tip" icon="true">{{drawio>electrical_engineering_1:tb6pi8dgh0m2e2pwCircuitSolution1}} \\
-So, here only R_1 and C gives the time constant: $\tau = R_1 \cdot C$
-The following formula describes the time course of $u_C(t)$ which has to be $u_c (t_1)=0.5\cdot U$:
-\begin{align*}
-u_c (t) = U \cdot (1- e^{t/\tau}) = 0.5\cdot U
-\end{align*}
-It has to be rearranged to $t$
-\begin{align*}
-(1- e^{t/\tau}) &= 0.5 \\
-     e^{t/\tau} &= 0.5 \\
-     t/\tau     &= ln(0.5) \\
-     t          &= \tau \cdot ln(0.5) \\
-     t          &= R_1 \cdot C \cdot ln(0.5)
-\end{align*}
-</callout></collapse>
-<button size="xs" type="link" collapse="tb6pi8dgh0m2e2pw_1_solution">{{icon>eye}} Final result</button><collapse id="tb6pi8dgh0m2e2pw_1_solution" collapsed="true"><callout type="tip" icon="true">
-\begin{align*}
-     t          = 1.39 ~\rm{ms}
-\end{align*}</callout>
- \\
-</collapse>
-. Calculate the overall energy dissipated by $R_1$ while charging the capacitor $0 ~\rm{V}$ to $12 ~\rm{V}$.
-<button size="xs" type="link" collapse="tb6pi8dgh0m2e2pw_2_path">{{icon>eye}} Solution</button><collapse id="tb6pi8dgh0m2e2pw_2_path" collapsed="true">
-<callout type="tip" icon="true">
-\begin{align*}
-\Delta W_R &= \Delta W_C \\
-           &= {{1}\over{2}}\cdot C \cdot U^2 \\
-           &= {{1}\over{2}}\cdot 100 ~\rm{µF} \cdot (12 ~\rm{V})^2
-\end{align*}
-</callout></collapse>
-<button size="xs" type="link" collapse="tb6pi8dgh0m2e2pw_2_solution">{{icon>eye}} Final result</button><collapse id="tb6pi8dgh0m2e2pw_2_solution" collapsed="true"><callout type="tip" icon="true">
-\begin{align*}
-\Delta W_R = 7.2 ~\rm{mWs}
-\end{align*}</callout>
- \\
-</collapse>
-. Now, consider the light bulb as a resistor of $R_B=20 ~\Omega$, and ignore again the left side ($S_2$ is open).
-The voltage across the capacitor is again $0 ~\rm{V}$ at the moment $t_0=0 ~\rm{s}$ when the switch $S_1$ is closed.
-Calculate the voltage $u_c (t_2)$ across the capacitor at $t_2=1 ~\rm{ms}$ after closing the switch. \\
-Hint: To solve this, first create an equivalent linear voltage source from $U$, $R_1$ and $R_B$. \\
-<button size="xs" type="link" collapse="tb6pi8dgh0m2e2pw_3_path">{{icon>eye}} Solution</button><collapse id="tb6pi8dgh0m2e2pw_3_path" collapsed="true">
-<callout type="tip" icon="true">
-{{drawio>electrical_engineering_1:tb6pi8dgh0m2e2pwCircuitSolution2}}
-An equivalent linear voltage source can be given with $U$, $R_1$ and $R_B$ as seen in yellow. \\
-Therefore, the voltage of the equivalent linear voltage source is: $U_s = U \cdot {{R_B}\over{R_1 + R_B}} = 1/2 \cdot U$
-The internal resistance is given by substituting the ideal voltage source with its resistance ($=0 ~\Omega$, short-circuit).
-\begin{align*}
-R_i &= R_1 || R_B \\
-    &= 10 ~\Omega
-\end{align*}
-\begin{align*}
-u_c (t_2) &= U_s \cdot (1- e^{t_2/(R_i\cdot C)}) \\
-          &= {{1}\over{2}} \cdot U \cdot (1- e^{1~\rm{ms}/(10 ~\Omega \cdot 100 ~\rm{µF})})
-\end{align*}
-</callout></collapse>
-<button size="xs" type="link" collapse="tb6pi8dgh0m2e2pw_3_solution">{{icon>eye}} Final result</button><collapse id="tb6pi8dgh0m2e2pw_3_solution" collapsed="true"><callout type="tip" icon="true">
-\begin{align*}
-u_c (t_2) = 3.79 ~\rm{V}
-\end{align*}</callout>
- \\
-</collapse>
-. Explain (without calculation) how the situation in 3. would change once also $S_2$ is closed from the beginning on.
-<button size="xs" type="link" collapse="tb6pi8dgh0m2e2pw_4_solution">{{icon>eye}} Final result</button><collapse id="tb6pi8dgh0m2e2pw_4_solution" collapsed="true"><callout type="tip" icon="true">
-It does not change anything. The ideal voltage source $U$ provides the voltage of $U=12~\rm{V}$ independent of this resistor. \\
-On an alternative view, one can try to create an equivalent linear voltage source again. Then, the internal resistance is given by substituting the ideal voltage source is again short-circuiting $R_2$.
-</callout></collapse>
-</WRAP></WRAP></panel>

charging capacitors dc network analysis pure resistor network simplification delta wye transformation exam ee1 ws2022