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electrical_engineering_1:task_tb6pi8dgh0m2e2pw_with_calculation [2023/02/12 05:51] – 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_ee1_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"> 
-<panel>{{drawio>electrical_engineering_1:tb6pi8dgh0m2e2pwCircuitSolution1}} \\ 
-So, here only R_1 and C gives the time constant: $\tau = R_1 \cdot C$ 
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-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*} 
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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"><panel> 
-\begin{align*} 
-     t          = 1.39 ms 
-\end{align*}</panel> 
- \\ 
-</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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-<button size="xs" type="link" collapse="tb6pi8dgh0m2e2pw_2_path">{{icon>eye}} Solution</button><collapse id="tb6pi8dgh0m2e2pw_2_path" collapsed="true"> 
-<panel> 
-\begin{align*} 
-\Delta W_R &= \Delta W_C \\ 
-           &= {{1}\over{2}}\cdot C \cdot U^2 \\ 
-           &= {{1}\over{2}}\cdot 100 \mu F \cdot (12 V)^2  
-\end{align*} 
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-</panel></collapse> 
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-<button size="xs" type="link" collapse="tb6pi8dgh0m2e2pw_2_solution">{{icon>eye}} Final result</button><collapse id="tb6pi8dgh0m2e2pw_2_solution" collapsed="true"><panel> 
-\begin{align*} 
-\Delta W_R = 7.2 mWs 
-\end{align*}</panel> 
- \\ 
-</collapse> 
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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 $R_B$. \\ 
-<button size="xs" type="link" collapse="tb6pi8dgh0m2e2pw_3_path">{{icon>eye}} Solution</button><collapse id="tb6pi8dgh0m2e2pw_3_path" collapsed="true"> 
-<panel> 
-{{drawio>electrical_engineering_1:tb6pi8dgh0m2e2pwCircuitSolution2}} 
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-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*} 
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-\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^{1ms/(10 \Omega \cdot 100 \mu F)}) 
-\end{align*} 
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-\begin{align*} 
-u_c (t_2) = 3.79 V 
-\end{align*}</panel> 
- \\ 
-</collapse> 
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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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-<button size="xs" type="link" collapse="tb6pi8dgh0m2e2pw_4_solution">{{icon>eye}} Final result</button><collapse id="tb6pi8dgh0m2e2pw_4_solution" collapsed="true"><panel> 
-It does not change anything. The ideal voltage source $U$ provides the voltage of $U=12V$ 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 $R_2$. 
-</panel></collapse> 
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