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electrical_engineering_1:dc_circuit_transients [2023/03/31 14:48]
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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   * **open switch**: If there is a voltage between the two metal parts, charges can also accumulate there. \\ Since the distances are usually large and the air is used as the dielectric, the capacitance of the capacitor formed in this way is very small.   * **open switch**: If there is a voltage between the two metal parts, charges can also accumulate there. \\ Since the distances are usually large and the air is used as the dielectric, the capacitance of the capacitor formed in this way is very small.
-  * **Overhead line**: An overhead line also represents a capacitor against the ground potential of the earth. The charging and discharging by the alternating current leads to the fact that polarizable molecules can align themselves. For example, the water drops near the line are rolled through the field and hum with $100~Hz$ and many times that (harmonics). Peak discharge results in a high-frequency crackle.+  * **Overhead line**: An overhead line also represents a capacitor against the ground potential of the earth. The charging and discharging by the alternating current leads to the fact that polarizable molecules can align themselves. For example, the water drops near the line are rolled through the field and hum with $100~\rm Hz$ and many times that (harmonics). Peak discharge results in a high-frequency crackle.
   * **Conductor trace**: A trace on a PCB can also be a capacitor against a nearby ground plane. This can be a problem for digital signals (see the charge and discharge curves below).   * **Conductor trace**: A trace on a PCB can also be a capacitor against a nearby ground plane. This can be a problem for digital signals (see the charge and discharge curves below).
   * **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 $C=\{ 1~\mu F, 10 ~\mu F\}$. 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:
  
 \begin{align*}  \begin{align*} 
-R = {{u_R(t)}\over{i_R(t)}} = {{{\rm d}u_R}\over{{\rm d}i_R}} = const. \\ +R = {{u_R(t)}\over{i_R(t)}} = {{{\rm d}u_R}\over{{\rm d}i_R}} = {\rm const.} \\ 
-C = {{q(t)}  \over{u_C(t)}} = {{{\rm d}q}  \over{{\rm d}u_C}} = const. \tag{5.1.1} +C = {{q(t)}  \over{u_C(t)}} = {{{\rm d}q}  \over{{\rm d}u_C}} = {\rm const.} \tag{5.1.1} 
 \end{align*} \end{align*}
  
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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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