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electrical_engineering_1:dc_circuit_transients [2023/03/31 14:51]
mexleadmin 		electrical_engineering_1:dc_circuit_transients [2023/12/03 16:53] (aktuell)
mexleadmin [Exercises]
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-====== 5. DC Circuit Transients (on RC elements) ======+====== 5 DC Circuit Transients (on RC elements) ======
  
 <WRAP onlyprint> <WRAP onlyprint>
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 <imgcaption imageNo01 | Capacitor in electrical circuit>  <imgcaption imageNo01 | Capacitor in electrical circuit> 
 </imgcaption>  </imgcaption> 
-\\ {{drawio>KondensatorImStromkreis}} \\ +\\ {{drawio>KondensatorImStromkreis.svg}} \\ 
 </WRAP> </WRAP>
  
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   * **Human body**: The human body can likewise pick up charge. The charge thus absorbed forms a capacitor with respect to other objects. This can be charged up to some $kV$. This is a particular problem in electrical laboratories, as the mere touching of components can destroy them.   * **Human body**: The human body can likewise pick up charge. The charge thus absorbed forms a capacitor with respect to other objects. This can be charged up to some $kV$. This is a particular problem in electrical laboratories, as the mere touching of components can destroy them.
   * **Membrane of nerve cells**: Nerve cells also result in a capacitor due to the lipid bilayer (membrane of the nerve cell) and the two cellular fluids with different electrolytes (ions). The nerve cells are surrounded by a thick layer (myelin layer) for faster transmission. This lowers the capacitance and thus increases the successive charging of successive parts of the nerve cell. In diseases such as Creutzfeldt-Jakob or multiple sclerosis, this layer thins out. This leads to the delayed signal transmission which characterizes the disease patterns.   * **Membrane of nerve cells**: Nerve cells also result in a capacitor due to the lipid bilayer (membrane of the nerve cell) and the two cellular fluids with different electrolytes (ions). The nerve cells are surrounded by a thick layer (myelin layer) for faster transmission. This lowers the capacitance and thus increases the successive charging of successive parts of the nerve cell. In diseases such as Creutzfeldt-Jakob or multiple sclerosis, this layer thins out. This leads to the delayed signal transmission which characterizes the disease patterns.
-<WRAP> <imgcaption imageNo02 | Circuit for viewing charge and discharge curve> </imgcaption> \\ {{drawio>SchaltungEntladekurve}} \\ </WRAP>+<WRAP> <imgcaption imageNo02 | Circuit for viewing charge and discharge curve> </imgcaption> \\ {{drawio>SchaltungEntladekurve.svg}} \\ </WRAP>
  
 In the following, the charging process of a capacitor is to be considered in more detail. For this purpose, one has to realize, that during the charging of the capacitor, besides the voltage source $U_{\rm s}$ and the capacitor $C$, there is always a resistance $R$ in the circuit. This is composed of the internal resistance of the (non-ideal) voltage source, the internal resistance of the capacitor, and the parasitic (=interfering) resistance of the line. In practical applications, it is often desired that capacitors charge in a certain time range. For this purpose, another real resistor is inserted into the circuit. The resulting series of resistors and capacitors is called an **RC element**. It resembles a voltage divider in which a resistor has been replaced by a capacitor. \\ To start the charging, an (ideal) switch $S$ is inserted. The circuit to be considered then looks like shown in <imgref imageNo02 >. \\ An ideal switch is characterized by: In the following, the charging process of a capacitor is to be considered in more detail. For this purpose, one has to realize, that during the charging of the capacitor, besides the voltage source $U_{\rm s}$ and the capacitor $C$, there is always a resistance $R$ in the circuit. This is composed of the internal resistance of the (non-ideal) voltage source, the internal resistance of the capacitor, and the parasitic (=interfering) resistance of the line. In practical applications, it is often desired that capacitors charge in a certain time range. For this purpose, another real resistor is inserted into the circuit. The resulting series of resistors and capacitors is called an **RC element**. It resembles a voltage divider in which a resistor has been replaced by a capacitor. \\ To start the charging, an (ideal) switch $S$ is inserted. The circuit to be considered then looks like shown in <imgref imageNo02 >. \\ An ideal switch is characterized by:
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 Exercises: Exercises:
  
-  - Become familiar with how the capacitor current $i_C$ and capacitor voltage $u_C$ depend on the given capacitance $C$ and resistance $R$. \\ To do this, use for $R=\{ 10~\Omega, 100~\Omega, 1~k\Omega\}$ and $C=\{ 1~\rm µF, 10 ~ µF}$. How fast does the capacitor voltage $u_C$ increase in each case n?+  - Become familiar with how the capacitor current $i_C$ and capacitor voltage $u_C$ depend on the given capacitance $C$ and resistance $R$. \\ To do this, use for $R=\{ 10~\Omega, 100~\Omega, 1~k\Omega\}$ and $C=\{ 1~\rm µF, 10 ~ µF\}$. How fast does the capacitor voltage $u_C$ increase in each case n?
   - Which quantity ($i_C$ or $u_C$) is continuous here? Why must this one be continuous? Why must the other quantity be discontinuous?   - Which quantity ($i_C$ or $u_C$) is continuous here? Why must this one be continuous? Why must the other quantity be discontinuous?
  
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   * Ideally, the capacitor is not fully charged to the specified voltage $U_{\rm s}$ until $t \rightarrow \infty$. It then carries the charge: $q(t \rightarrow \infty) = Q = C \cdot U_{\rm s}$   * Ideally, the capacitor is not fully charged to the specified voltage $U_{\rm s}$ until $t \rightarrow \infty$. It then carries the charge: $q(t \rightarrow \infty) = Q = C \cdot U_{\rm s}$
  
-<WRAP> <imgcaption imageNo02 | circuit for viewing the charge curve> </imgcaption> {{drawio>SchaltungEntladekurve2}} </WRAP>+<WRAP> <imgcaption imageNo02 | circuit for viewing the charge curve> </imgcaption> {{drawio>SchaltungEntladekurve2.svg}}</WRAP>
  
 The process is now to be summarized in detail in formulas. Linear components are used in the circuit, i.e. the component values for the resistor $R$ and the capacitance $C$ are independent of the current or the voltage. Then definition equations for the resistor $R$ and the capacitance $C$ are also valid for time-varying or infinitesimal quantities: The process is now to be summarized in detail in formulas. Linear components are used in the circuit, i.e. the component values for the resistor $R$ and the capacitance $C$ are independent of the current or the voltage. Then definition equations for the resistor $R$ and the capacitance $C$ are also valid for time-varying or infinitesimal quantities:
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 <WRAP> <imgcaption imageNo04 | charging curve>  <WRAP> <imgcaption imageNo04 | charging curve> 
 </imgcaption>  </imgcaption> 
-{{drawio>Ladekurve}} +{{drawio>Ladekurve.svg}} 
 </WRAP> </WRAP>
  
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 <imgcaption imageNo15 | circuit for viewing discharge curve>  <imgcaption imageNo15 | circuit for viewing discharge curve> 
 </imgcaption>  </imgcaption> 
-{{drawio>SchaltungEntladekurve3}} +{{drawio>SchaltungEntladekurve3.svg}} 
 </WRAP> </WRAP>
  
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 <imgcaption imageNo05 | discharge curve>  <imgcaption imageNo05 | discharge curve> 
 </imgcaption>  </imgcaption> 
-{{drawio>Entladekurve}} +{{drawio>Entladekurve.svg}} 
 </WRAP> </WRAP>
  
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 </callout> </callout>
  
-<WRAP> <imgcaption imageNo02 | circuit for viewing the charge curve> </imgcaption> {{drawio>SchaltungEntladekurve2}} </WRAP>+<WRAP> <imgcaption imageNo02 | circuit for viewing the charge curve> </imgcaption> {{drawio>SchaltungEntladekurve2.svg}} </WRAP>
  
 Now the capacitor as energy storage is to be looked at more closely. This derivation is also explained in [[https://www.youtube.com/watch?v=PTyB6_Kt_5A&ab_channel=TheOrganicChemistryTutor|this youtube video]]. For this, we consider again the circuit in <imgref imageNo02 > an. According to the chapter [[:electrical_engineering_1:preparation_properties_proportions#power_and_efficiency|Preparation, Properties, and Proportions]], the power for constant values (DC) is defined as: Now the capacitor as energy storage is to be looked at more closely. This derivation is also explained in [[https://www.youtube.com/watch?v=PTyB6_Kt_5A&ab_channel=TheOrganicChemistryTutor|this youtube video]]. For this, we consider again the circuit in <imgref imageNo02 > an. According to the chapter [[:electrical_engineering_1:preparation_properties_proportions#power_and_efficiency|Preparation, Properties, and Proportions]], the power for constant values (DC) is defined as:
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 \end{align*} \end{align*}
  
-Thus, for a fully discharged capacitor ($U_{\rm s}=0~{\rm V}$), the energy stored when charging to voltage $U_{\rm s}$ is $\delta W_C={{1}\over{2}} C \cdot U_{\rm s}^2$.+Thus, for a fully discharged capacitor ($U_{\rm s}=0~{\rm V}$), the energy stored when charging to voltage $U_{\rm s}$ is $\Delta W_C={{1}\over{2}} C \cdot U_{\rm s}^2$.
  
 === Energy Consideration on the Resistor === === Energy Consideration on the Resistor ===
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 <imgcaption imageNo06 | Current, voltage, and energy during charging and discharging>  <imgcaption imageNo06 | Current, voltage, and energy during charging and discharging> 
 </imgcaption>  </imgcaption> 
-{{drawio>LadenStromSpannungEnergie}} +{{drawio>LadenStromSpannungEnergie.svg}} 
 </WRAP> </WRAP>
  
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Exercise 5.2.2 Further capacitor charging/discharging practice Exercise "> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+#@TaskTitle_HTML@# Exercise 5.2.2 Capacitor charging/discharging #@TaskText_HTML@#
  
-{{youtube>a-gPuw6JsxQ}}+The following circuit shows a charging/discharging circuit for a capacitor.
  
-</WRAP></WRAP></panel>+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)
  
-<panel type="info" title="Exercise 5.2.3 Further practice charging the capacitor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+{{drawio>electrical_engineering_1:Exercise522setup.svg}}
  
-{{youtube>L0S_Aw8pBto}}+1. For the first tasks, the switch $S_1$ gets closed at $t=t_0 = 0s$. \\
  
-</WRAP></WRAP></panel>+1.1 What is the value of the time constant $\tau_1$?
  
-<panel type="info" title="Exercise 5.2.4 Charge balance of two capacitors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+#@HiddenBegin_HTML~Solution1,Solution~@#
  
-{{youtube>EMdpkDoMXXE}}+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.
  
-</WRAP></WRAP></panel>+{{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~@# 
 + 
 +1.2 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~@# 
 + 
 +2. 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}) \\ 
 +      4/5                        &=             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~@# 
 + 
 +3. 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 $U_2 = 10 ~\rm V$. 
 + 
 +3.1 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~@# 
 + 
 +3.2 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~@# 
 + 
 +3.3 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@#
  
-<panel type="info" title="Exercise 5.2.5 Charge balance of two capacitors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
  
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Zeile 525: Zeile 668:
       - What are the energies and the total energy? \\ How is this understandable with the previous total energy?       - What are the energies and the total energy? \\ How is this understandable with the previous total energy?
  
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