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electrical_engineering_1:task_tb6pi8dgh0m2e2pw_with_calculation [2023/04/02 00:36] – 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 dc_network_analysis pure_resistor_network_simplification delta_wye_transformation exam_ee1_WS2022}} 
-{{include_n>5000}} 
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-#@TaskTitle_HTML@#~~#@ee1_taskctr#~~  Charging Capacitors  \\ <fs medium>(written test, approx. 16 % of a 60-minute written test, WS2022)</fs> #@TaskText_HTML@#     
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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 $U=12 ~\rm{V}$, 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 $u_c (t_0 )=0 ~\rm{V}$.  
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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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-#@PathBegin_HTML~1~@# 
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-{{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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-#@PathEnd_HTML~1~@# 
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-#@ResultBegin_HTML~1~@# 
-\begin{align*} 
-     t          = 1.39 ~\rm{ms} 
-\end{align*} 
-#@ResultEnd_HTML~1~@# 
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-2. Calculate the overall energy dissipated by $R_1$ while charging the capacitor $0 ~\rm{V}$ to $12 ~\rm{V}$. 
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-#@PathBegin_HTML~2~@# 
-\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*} 
-#@PathEnd_HTML~2~@# 
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-#@ResultBegin_HTML~2~@# 
-\begin{align*} 
-\Delta W_R = 7.2 ~\rm{mWs} 
-\end{align*} 
-#@ResultEnd_HTML~2~@# 
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-3. Now, consider the light bulb as a resistor of $R_\rm B=20 ~\Omega$, 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 $R_\rm B$. \\ 
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-#@PathBegin_HTML~3~@# 
-{{drawio>electrical_engineering_1:tb6pi8dgh0m2e2pwCircuitSolution2}} 
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-An equivalent linear voltage source can be given with $U$, $R_1$, and $R_\rm B$ as seen in yellow. \\ 
-Therefore, the voltage of the equivalent linear voltage source is: $U_s = U \cdot {{R_\rm B}\over{R_1 + R_\rm B}} = 1/2 \cdot U$ 
-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_\rm 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^{1~\rm{ms}/(10 ~\Omega \cdot 100 ~\rm{µF})}) 
-\end{align*} 
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-#@PathEnd_HTML~3~@# 
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-#@ResultBegin_HTML~3~@# 
-\begin{align*} 
-u_c (t_2) = 3.79 ~\rm{V} 
-\end{align*} 
-#@ResultEnd_HTML~3~@# 
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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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-#@ResultBegin_HTML~4~@# 
-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 $R_2$. 
-#@ResultEnd_HTML~4~@# 
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-#@TaskEnd_HTML@# 
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