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electrical_engineering_1:task_tb6pi8dgh0m2e2pw_with_calculation [2023/02/12 01:52] – angelegt mexleadminelectrical_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 exam_WS2022}} 
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-<panel type="info" > <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
-<fs x-large>**Exercise ~~#~~ : Chargine Capacitors ** \\ (written test, approx. 16% of a 60-minute written test, WS2022) \\ \\ </fs> 
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-The circuit shown in the following is used to control the brightness when turning on a small light bulb.  
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-{{drawio>electrical_engineering_1:tb6pi8dgh0m2e2pwCircuit.svg}} 
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-The circuit contains a voltage source $U=12 V$, a switch $S_1$, a resistor of $R_1=20 \Omega$ and a capacitor of $C=100 \mu F$. \\ 
-The switch $S_2$ to an additional consumer $R_2$ will be considered to be open for the first tasks. At the moment $t_0=0 s$ the switch $S_1$ is closed, the voltage across the capacitor is $u_c (t_0 )=0 V$.  
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-1. First do not consider the light bulb – it is not connected to the RC-circuit. \\ 
-Calculate the point of time $t_1$ when $u_c (t_1 )=0.5\cdot U$.  
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-<button size="xs" type="link" collapse="tb6pi8dgh0m2e2pw_1_path">{{icon>eye}} Solution</button><collapse id="tb6pi8dgh0m2e2pw_1_path" collapsed="true"> 
-{{drawio>electrical_engineering_1:tb6pi8dgh0m2e2pwCircuitSolution1}} \\ 
-The following formula describes the time course of $u_C(t)$ 
-\begin{align*} 
-u_c (t) = U \cdot (1- e^{t_1/\tau}) 
-\end{align*} 
-The resulting circuit can again be transformed: 
-{{drawio>electrical_engineering_1:6tqttque1e2nf2c7CircuitSolution3.svg}} \\ 
-Here, the $U_{24}$ is calculated by $I_{24}$ as the following: 
-\begin{align*} 
-U_{24} &= R_{135} \cdot I_{24} \\ 
-       &= ({{U_2}\over{R_1}} - I_4) \cdot R_1||R_3||R_5 
-\end{align*} 
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-On the right side of the last circuit there is a voltage divider given by $R_{135}$, $R_6$ and $R_7$. \\ 
-Therefore the voltage between $A$ and $B$ is given as: 
-\begin{align*} 
-U_{AB} &= U_{24} \cdot {{R_7}\over{R_6 + R_7 + R_1||R_3||R_5}} \\ 
-       &= ({{U_2}\over{R_1}} - I_4) \cdot {{R_7 \cdot R_1||R_3||R_5}\over{R_6 + R_7 + R_1||R_3||R_5}} \\ 
-\end{align*} 
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-For the internal resistance $R_i$ the ideal voltage source is substituted by its resistance ($=0\Omega$, so a short-circuit): 
-\begin{align*} 
-R_{AB} &= R_7 || ( R_6 + R_1||R_3||R_5) \\ 
-\end{align*} 
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-with $R_1||R_3||R_5 = 5 \Omega || 10 \Omega || 10 \Omega =  5 \Omega ||  5 \Omega = 2.5 \Omega$: 
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-\begin{align*} 
-U_{AB} &= ({{6.0 V}\over{5.0 \Omega}} - 4.2 \Omega) \cdot {{15 \Omega \cdot 2.5 \Omega}\over{7.5 \Omega + 15 \Omega + 2.5 \Omega}} \\ 
-R_{AB} &= 15 \Omega|| ( 7.5 \Omega + 2.5 \Omega) \\ 
-\end{align*} 
-</collapse> 
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-<button size="xs" type="link" collapse="tb6pi8dgh0m2e2pw_1_solution">{{icon>eye}} Final result</button><collapse id="tb6pi8dgh0m2e2pw_1_solution" collapsed="true"> 
-\begin{align*} 
-U_{AB} &= 4.5 V\\ 
-R_{AB} &= 6 \Omega \\ 
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-\end{align*} 
- \\ 
-</collapse> 
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-2. Calculate the overall energy dissipated by $R_1$ while charging the capacitor $0 V$ to $12 V$. 
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-3. 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 $0 V$ at the moment $t_0=0 s$ when the switch $S_1$ is closed.  
-Calculate the voltage $u_c (t_2 )$ across the capacitor at $t_2=1 ms$ after closing the switch. \\ 
-Hint: To solve this, first create an equivalent linear voltage source from $U$, $R_1$ and $R_B$. 
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-4. Explain (without calculation) how the situation in 3. would change once also $S_2$ is closed from the beginning on. 
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