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electrical_engineering_1:dc_circuit_transients [2023/03/31 12:50] – mexleadminelectrical_engineering_1:dc_circuit_transients [2024/11/08 15:21] (aktuell) – mexleadmin
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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>
  
Zeile 33: Zeile 33:
  
   * **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 $U_s$ 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:
  
   * infinitely fast switching   * infinitely fast switching
Zeile 52: Zeile 52:
 In this chapter also time-varying quantities are considered. These are generally marked with lowercase letters. Examples of time-varying quantities are: In this chapter also time-varying quantities are considered. These are generally marked with lowercase letters. Examples of time-varying quantities are:
  
-  * A **time-varying voltage $u_C(t)$ across a capacitor** or the **voltage $u_s$ of an ac voltage source**  as opposed to a constant voltage $U_s$ across a constant voltage source.+  * A **time-varying voltage $u_C(t)$ across a capacitor** or the **voltage $U_{\rm s}$ of an ac voltage source**  as opposed to a constant voltage $U_{\rm s}$ across a constant voltage source.
   * A **time-varying current $i_L(t)$ across a coil** or **time-varying current $i_C(t)$ across a capacitor**.   * A **time-varying current $i_L(t)$ across a coil** or **time-varying current $i_C(t)$ across a capacitor**.
  
Zeile 75: Zeile 75:
 In the simulation below you can see the circuit mentioned above in a slightly modified form: In the simulation below you can see the circuit mentioned above in a slightly modified form:
  
-  * The capacitance $C$ can be charged via the resistor $R$ if the toggle switch $S$ connects the DC voltage source $U_s$ to the two.+  * The capacitance $C$ can be charged via the resistor $R$ if the toggle switch $S$ connects the DC voltage source $U_{\rm s}$ to the two.
   * But it is also possible to short-circuit the series circuit of $R$ and $C$ via the switch $S$.   * But it is also possible to short-circuit the series circuit of $R$ and $C$ via the switch $S$.
   * Furthermore, the current $i_C$ and the voltage $u_C$ are displayed in the oscilloscope as data points over time and in the circuit as numerical values.   * Furthermore, the current $i_C$ and the voltage $u_C$ are displayed in the oscilloscope as data points over time and in the circuit as numerical values.
Zeile 82: Zeile 82:
 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?
  
Zeile 94: Zeile 94:
 {{youtube>8nyNamrWcyE?start=907&stop=1495}} {{youtube>8nyNamrWcyE?start=907&stop=1495}}
  
-To understand the charging process of a capacitor, an initially uncharged capacitor with capacitance $C$ is to be charged by a DC voltage source $U_s$ via a resistor $R$.+To understand the charging process of a capacitor, an initially uncharged capacitor with capacitance $C$ is to be charged by a DC voltage source $U_{\rm s}$ via a resistor $R$.
  
-  * In order that the voltage $U_s$ acts at a certain time $t_0 = 0 ~s$ the switch $S$ is closed at this time. +  * In order that the voltage $U_{\rm s}$ acts at a certain time $t_0 = 0 ~s$ the switch $S$ is closed at this time. 
-  * Directly after the time $t_0$ the maximum current ("charging current") flows in the circuit. This is only limited by the resistor $R$. The uncharged capacitor has a voltage $u_C(t_0)=0~V$ at that time. The maximum voltage $u_R(t_0)=U_s$ is applied to the resistor. The current is $i_C(t_0)={{U_s}\over{R}}$.+  * Directly after the time $t_0$ the maximum current ("charging current") flows in the circuit. This is only limited by the resistor $R$. The uncharged capacitor has a voltage $u_C(t_0)=0~V$ at that time. The maximum voltage $u_R(t_0)=U_{\rm s}$ is applied to the resistor. The current is $i_C(t_0)={{U_{\rm s}}\over{R}}$.
   * The current causes charge carriers to flow from one electrode to the other. Thus the capacitor is charged and its voltage increases $u_C$.   * The current causes charge carriers to flow from one electrode to the other. Thus the capacitor is charged and its voltage increases $u_C$.
   * Thus the voltage $u_R$ across the resistor is reduced and so is the current $i_R$.   * Thus the voltage $u_R$ across the resistor is reduced and so is the current $i_R$.
   * With the current thus reduced, less charge flows on the capacitor.   * With the current thus reduced, less charge flows on the capacitor.
-  * Ideally, the capacitor is not fully charged to the specified voltage $U_s$ until $t \rightarrow \infty$. It then carries the charge: $q(t \rightarrow \infty) = Q = C \cdot U_s$+  * 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*}
  
Zeile 120: Zeile 120:
  
 \begin{align*}  \begin{align*} 
-U_s =u_R + u_C = R \cdot i_C + u_C \tag{5.1.2} +U_{\rm s} =u_R + u_C = R \cdot i_C + u_C \tag{5.1.2} 
 \end{align*} \end{align*}
  
Zeile 138: Zeile 138:
  
 \begin{align*}  \begin{align*} 
-U_s &= u_R                                           + u_C \\+U_{\rm s} &= u_R                                           + u_C \\
     &= R \cdot C \cdot {{{\rm d}u_C}\over{{\rm d}t}} + u_C      &= R \cdot C \cdot {{{\rm d}u_C}\over{{\rm d}t}} + u_C 
 \end{align*} \end{align*}
Zeile 151: Zeile 151:
  
 \begin{align*} \begin{align*}
-U_s &= R \cdot C \cdot {{\rm d}\over{{\rm d}t}}(\mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C}) + \mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C} \\+U_{\rm s} &= R \cdot C \cdot {{\rm d}\over{{\rm d}t}}(\mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C}) + \mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C} \\
     &= R \cdot C \cdot                                        \mathcal{AB} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C} \\     &= R \cdot C \cdot                                        \mathcal{AB} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C} \\
-U_s - \mathcal{C} &=                                                           ( R \cdot C \cdot \mathcal{AB} + \mathcal{A})\cdot {\rm e}^{\mathcal{B}\cdot t} \\+U_{\rm s} - \mathcal{C} &=                                                           ( R \cdot C \cdot \mathcal{AB} + \mathcal{A})\cdot {\rm e}^{\mathcal{B}\cdot t} \\
 \end{align*} \end{align*}
  
Zeile 159: Zeile 159:
  
 \begin{align*} \begin{align*}
-\mathcal{C} = U_s \\ \\+\mathcal{C} = U_{\rm s} \\ \\
  
 R \cdot C \cdot \mathcal{AB} + \mathcal{A} &= 0  \quad  \quad     | : \mathcal{A} \quad | -1 \\ R \cdot C \cdot \mathcal{AB} + \mathcal{A} &= 0  \quad  \quad     | : \mathcal{A} \quad | -1 \\
Zeile 169: Zeile 169:
  
 \begin{align*} \begin{align*}
-u_C(t) = \mathcal{A} \cdot {\rm e}^{\large{- {{t}\over{R C}} }} + U_s+u_C(t) = \mathcal{A} \cdot {\rm e}^{\large{- {{t}\over{R C}} }} + U_{\rm s}
 \end{align*} \end{align*}
  
Zeile 175: Zeile 175:
  
 \begin{align*} \begin{align*}
-0 &= \mathcal{A} \cdot {\rm e}^{\large{0}} + U_s \\ +0 &= \mathcal{A} \cdot {\rm e}^{\large{0}} + U_{\rm s} \\ 
-0 &= \mathcal{A}  + U_s \\ +0 &= \mathcal{A}  + U_{\rm s} \\ 
-\mathcal{A} &= - U_s+\mathcal{A} &= - U_{\rm s}
 \end{align*} \end{align*}
  
Zeile 184: Zeile 184:
  
 \begin{align*} \begin{align*}
-u_C(t) &= - U_s \cdot {\rm e}^{\large{- {{t}\over{R C}}}} + U_s+u_C(t) &= - U_{\rm s} \cdot {\rm e}^{\large{- {{t}\over{R C}}}} + U_{\rm s}
 \end{align*} \end{align*}
  
Zeile 191: Zeile 191:
 And this results in:  And this results in: 
 \begin{align*} \begin{align*}
-u_C(t) &= U_s \cdot (1 - {\rm e}^{\large{- {{t}\over{R C}}}})+u_C(t) &= U_{\rm s} \cdot (1 - {\rm e}^{\large{- {{t}\over{R C}}}})
 \end{align*} \end{align*}
  
 And with $(5.1.3)$, $i_C$ becomes:  And with $(5.1.3)$, $i_C$ becomes: 
 \begin{align*} \begin{align*}
-i_C(t) &= {{U_s}\over{R}} \cdot {\rm e}^{\large{- {{t}\over{R C}} } }+i_C(t) &= {{U_{\rm s}}\over{R}} \cdot {\rm e}^{\large{- {{t}\over{R C}} } }
 \end{align*} \end{align*}
  
Zeile 205: Zeile 205:
 <WRAP> <imgcaption imageNo04 | charging curve>  <WRAP> <imgcaption imageNo04 | charging curve> 
 </imgcaption>  </imgcaption> 
-{{drawio>Ladekurve}} +{{drawio>Ladekurve.svg}} 
 </WRAP> </WRAP>
  
Zeile 211: Zeile 211:
  
   * There must be a unitless term in the exponent. So $RC$ must also represent a time. This time is called **time constant**  $\tau =R \cdot C$.    * There must be a unitless term in the exponent. So $RC$ must also represent a time. This time is called **time constant**  $\tau =R \cdot C$. 
-  * At time $t=\tau$, we get: $u_C(t) = U_s \cdot (1 - {\rm e}^{- 1}) = U_s \cdot (1 - {{1}\over{\rm e}}) = U_s \cdot ({{{\rm e}-1}\over{\rm e}}) = 0.63 \cdot U_s = 63~\% \cdot U_s $. \\ So, **the capacitor is charged to $63~\%$ after one $\tau$.**  +  * At time $t=\tau$, we get: $u_C(t) = U_{\rm s} \cdot (1 - {\rm e}^{- 1}) = U_{\rm s} \cdot (1 - {{1}\over{\rm e}}) = U_{\rm s} \cdot ({{{\rm e}-1}\over{\rm e}}) = 0.63 \cdot U_{\rm s} = 63~\% \cdot U_{\rm s} $. \\ So, **the capacitor is charged to $63~\%$ after one $\tau$.**  
-  * At time $t=2 \cdot \tau$ we get: $u_C(t) = U_s \cdot (1 - {\rm e}^{- 2}) = 86~\% \cdot U_s = (63~\% + (100~\% - 63~\%) \cdot 63~\% ) \cdot U_s$. So, **after each additional $\tau$, the uncharged remainder ($1-63~\%$) is recharged to $63~\%$**. +  * At time $t=2 \cdot \tau$ we get: $u_C(t) = U_{\rm s} \cdot (1 - {\rm e}^{- 2}) = 86~\% \cdot U_{\rm s} = (63~\% + (100~\% - 63~\%) \cdot 63~\% ) \cdot U_{\rm s}$. So, **after each additional $\tau$, the uncharged remainder ($1-63~\%$) is recharged to $63~\%$**. 
   * After about $t=5 \cdot \tau$, the result is a capacitor charged to over $99~\%$. In real circuits, **a charged capacitor can be assumed after** $5 \cdot \tau$.   * After about $t=5 \cdot \tau$, the result is a capacitor charged to over $99~\%$. In real circuits, **a charged capacitor can be assumed after** $5 \cdot \tau$.
   * The time constant $\tau$ can be determined graphically in several ways:   * The time constant $\tau$ can be determined graphically in several ways:
Zeile 226: Zeile 226:
 <imgcaption imageNo15 | circuit for viewing discharge curve>  <imgcaption imageNo15 | circuit for viewing discharge curve> 
 </imgcaption>  </imgcaption> 
-{{drawio>SchaltungEntladekurve3}} +{{drawio>SchaltungEntladekurve3.svg}} 
 </WRAP> </WRAP>
  
 The following situation is considered for the discharge: The following situation is considered for the discharge:
  
-  * A capacitor charged to voltage $U_s$ with capacitance $C$ is short-circuited across a resistor $R$ at time $t=t_0$. +  * A capacitor charged to voltage $U_{\rm s}$ with capacitance $C$ is short-circuited across a resistor $R$ at time $t=t_0$. 
-  * As a result, the full voltage $U_s$ is initially applied to the resistor: $u_R(t_0)=U_s$+  * As a result, the full voltage $U_{\rm s}$ is initially applied to the resistor: $u_R(t_0)=U_{\rm s}$
   * The initial discharge current is thus defined by the resistance: $i_C ={{u_R}\over{R}}$   * The initial discharge current is thus defined by the resistance: $i_C ={{u_R}\over{R}}$
   * The discharging charges lower the voltage of the capacitor $u_C$, since: $u_C = {{q(t)}\over{C}}$   * The discharging charges lower the voltage of the capacitor $u_C$, since: $u_C = {{q(t)}\over{C}}$
Zeile 254: Zeile 254:
  
 \begin{align*} \begin{align*}
-u_C(t) = \mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C}+u_C(t) = \mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C}
 \end{align*} \end{align*}
  
 \begin{align*} \begin{align*}
-0 &= R \cdot C \cdot {{\rm d}\over{{\rm d}t}}(\mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C}) + \mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C} \\ +0 &= R \cdot C \cdot {{\rm d}\over{{\rm d}t}}(\mathcal{A} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C}) + \mathcal{A}  \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C} \\ 
-  &= R \cdot C \cdot                                        \mathcal{AB} \cdot e^{\mathcal{B}\cdot t} + \mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C} \\ +  &= R \cdot C \cdot                                        \mathcal{AB} \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{A}  \cdot {\rm e}^{\mathcal{B}\cdot t} + \mathcal{C} \\ 
- - \mathcal{C} &=                                                    ( R \cdot C \cdot \mathcal{AB} + \mathcal{A} ) \cdot e^{\mathcal{B}\cdot t} \\+ - \mathcal{C} &=                                                            ( R \cdot C \cdot \mathcal{AB} + \mathcal{A}) \cdot {\rm e}^{\mathcal{B}\cdot t} \\
 \end{align*} \end{align*}
  
Zeile 276: Zeile 276:
  
 \begin{align*} \begin{align*}
-u_C(t) = \mathcal{A} \cdot e^{\large{- {{t}\over{R C}} }} + 0+u_C(t) = \mathcal{A} \cdot {\rm e}^{\large{- {{t}\over{R C}} }} + 0
 \end{align*} \end{align*}
  
-For the solution it must still hold that at time $t_0=0$ $u_C(t_0) = U_s$ just holds:+For the solution it must still hold that at time $t_0=0$ $u_C(t_0) = U_{\rm s}$ just holds:
  
 \begin{align*} \begin{align*}
-U_s &= \mathcal{A} \cdot e^{\large{0}}  \\ +U_{\rm s} &= \mathcal{A} \cdot {\rm e}^{\large{0}}  \\ 
-\mathcal{A} &= U_s+\mathcal{A} &= U_{\rm s}
 \end{align*} \end{align*}
  
Zeile 289: Zeile 289:
  
 \begin{align*} \begin{align*}
-u_C(t) &= U_s \cdot e^{\large{- {{t}\over{R C}}}}  +u_C(t) &= U_{\rm s} \cdot {\rm e}^{\large{- {{t}\over{R C}}}}  
 \end{align*} \end{align*}
  
Zeile 297: Zeile 297:
 <imgcaption imageNo05 | discharge curve>  <imgcaption imageNo05 | discharge curve> 
 </imgcaption>  </imgcaption> 
-{{drawio>Entladekurve}} +{{drawio>Entladekurve.svg}} 
 </WRAP> </WRAP>
  
 And this results in:  And this results in: 
 \begin{align*} \begin{align*}
-u_C(t) &= U_s \cdot e^{\large{- {{t}\over{\tau}}}} \quad \text{with} \quad \tau = R C+u_C(t) &= U_{\rm s} \cdot {\rm e}^{\large{- {{t}\over{\tau}}}} \quad \text{with} \quad \tau = R C
 \end{align*} \end{align*}
  
 And with $(5.1.3)$, $i_C$ becomes:  And with $(5.1.3)$, $i_C$ becomes: 
 \begin{align*} \begin{align*}
-i_C(t) &=- {{U_s}\over{R}} \cdot e^{\large{- {{t}\over{R C}} } }+i_C(t) &=- {{U_{\rm s}}\over{R}} \cdot {\rm e}^{\large{- {{t}\over{R C}} } }
 \end{align*} \end{align*}
  
Zeile 341: Zeile 341:
 </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:
Zeile 367: Zeile 367:
 During the charging process During the charging process
 \begin{align*} \begin{align*}
-u_C(t) =   U_s           \cdot (1 - e^{ - {{t}\over{\tau}} })  \\ +u_C(t) =   U_{\rm s}           \cdot (1 - {\rm e}^{ -{{t}\over{\tau}} })  \\ 
-i_C(t) = {{U_s}\over{R}} \cdot       e^{ -{{t}\over{\tau}} } \tag{5.2.2}+i_C(t) = {{U_{\rm s}}\over{R}} \cdot      {\rm e}^{ -{{t}\over{\tau}} } \tag{5.2.2}
 \end{align*} \end{align*}
  
Zeile 390: Zeile 390:
 \end{align*} \end{align*}
  
-Thus, for a fully discharged capacitor ($U_s=0~{\rm V}$), the energy stored when charging to voltage $U_s$ is $\delta W_C={{1}\over{2}} C \cdot U_s^2$.+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 ===
Zeile 405: Zeile 405:
  
 \begin{align*} \begin{align*}
-\Delta W_R &=                   R \cdot \int_{0}^t \left( { {U_s}\over{R}} \cdot e^ { -{{t}\over{\tau}}} \right)^2  {\rm d}t \\ +\Delta W_R &=                   R \cdot \int_{0}^t \left( { {U_{\rm s}}\over{R}} \cdot {\rm e}^ { -{{t}\over{\tau}}} \right)^2  {\rm d}t \\ 
-           &=  { {U_s^2}\over{R}} \cdot \int_{0}^t                               e^ { -{{2 \cdot t}\over{\tau}}}    {\rm d}t \\ +           &=  { {U_{\rm s}^2}\over{R}} \cdot \int_{0}^t                               {\rm e}^ { -{{2 \cdot t}\over{\tau}}}    {\rm d}t \\ 
-           &=  { {U_s^2}\over{R}} \cdot  \left[ -{{\tau }\over{2}} \cdot         e^ { -{{2 \cdot t}\over{\tau}}} \right]_{0}^t \quad & | \text{with } \tau = R \cdot C \\ +           &=  { {U_{\rm s}^2}\over{R}} \cdot  \left[ -{{\tau }\over{2}} \cdot         {\rm e}^ { -{{2 \cdot t}\over{\tau}}} \right]_{0}^t \quad & | \text{with } \tau = R \cdot C \\ 
-           &=  -{{1}\over{2}}     \cdot {           U_s^2}\cdot{C} \cdot  \left[ e^ { -{{2 \cdot t}\over{\tau}}} \right]_{0}^t \\+           &=  -{{1}\over{2}}     \cdot {           U_{\rm s}^2}\cdot{C} \cdot  \left[ {\rm e}^ { -{{2 \cdot t}\over{\tau}}} \right]_{0}^t \\
 \end{align*} \end{align*}
  
Zeile 414: Zeile 414:
  
 \begin{align*} \begin{align*}
-\Delta W_R &=  -{{1}\over{2}} \cdot {U_s^2}\cdot{C} \cdot   \left[ e^ { -{{2 \cdot t}\over{\tau}}} \right]_{0}^{\infty} \\ +\Delta W_R &=  -{{1}\over{2}} \cdot {U_{\rm s}^2}\cdot{C} \cdot   \left[ {\rm e}^ { -{{2 \cdot t}\over{\tau}}} \right]_{0}^{\infty} \\ 
-           &=  -{{1}\over{2}} \cdot {U_s^2}\cdot{C} \cdot   \left[ 0 - 1  \right] \\+           &=  -{{1}\over{2}} \cdot {U_{\rm s}^2}\cdot{C} \cdot   \left[ 0 - 1  \right] \\
 \end{align*} \end{align*}
 \begin{align*} \begin{align*}
-\boxed{ \Delta W_R  =  {{1}\over{2}} \cdot {U_s^2}\cdot{C}} \tag{5.2.4}+\boxed{ \Delta W_R  =  {{1}\over{2}} \cdot {U_{\rm s}^2}\cdot{C}} \tag{5.2.4}
 \end{align*} \end{align*}
  
-This means that the energy converted at the resistor is independent of the resistance value (for an ideal constant voltage source $U_s$ and given capacitor $C$)! At first, this doesn't really sound comprehensible. No matter if there is a very large resistor $R_1$ or a tiny small resistor $R_2$: The same waste heat is always produced. Graphically, this apparent contradiction can be resolved like this: A higher resistor $R_2$ slows down the small charge packets $\Delta q_1$, $\Delta q_2$, … $\Delta q_n$ more strongly. But a considered single charge packet $\Delta q_k$ will nevertheless pass the same voltage across the resistor $R_1$ or $R_2$ since this is given only by the accumulated packets in the capacitor: $u_r = U_s - u_C = U_s - {{q}\over{C}}$.+This means that the energy converted at the resistor is independent of the resistance value (for an ideal constant voltage source $U_{\rm s}$ and given capacitor $C$)! At first, this doesn't really sound comprehensible. No matter if there is a very large resistor $R_1$ or a tiny small resistor $R_2$: The same waste heat is always produced. Graphically, this apparent contradiction can be resolved like this: A higher resistor $R_2$ slows down the small charge packets $\Delta q_1$, $\Delta q_2$, … $\Delta q_n$ more strongly. But a considered single charge packet $\Delta q_k$ will nevertheless pass the same voltage across the resistor $R_1$ or $R_2$ since this is given only by the accumulated packets in the capacitor: $u_r = U_{\rm s} - u_C = U_{\rm s} - {{q}\over{C}}$.
  
 In real applications, as mentioned in previous chapters, ideal voltage sources are not possible. Thus, without a real resistor, the waste heat is dissipated proportionally to the internal resistance of the source and the internal resistance of the capacitor. The internal resistance of the capacitor depends on the frequency but is usually smaller than the internal resistance of the source. In real applications, as mentioned in previous chapters, ideal voltage sources are not possible. Thus, without a real resistor, the waste heat is dissipated proportionally to the internal resistance of the source and the internal resistance of the capacitor. The internal resistance of the capacitor depends on the frequency but is usually smaller than the internal resistance of the source.
Zeile 427: Zeile 427:
 === Consideration of total energy turnover === === Consideration of total energy turnover ===
  
-In the previous considerations, the energy conversion during the complete charging process was also considered. It was found that the capacitor stores the energy $W_C= {{1}\over{2}} \cdot {U_s^2}\cdot{C} $ (see $(5.2.3)$) and at the resistor the energy $W_R= {{1}\over{2}} \cdot {U_s^2}\cdot{C} $ (see $(5.2.4)$) into heat. +In the previous considerations, the energy conversion during the complete charging process was also considered. It was found that the capacitor stores the energy $W_C= {{1}\over{2}} \cdot {U_{\rm s}^2}\cdot{C} $ (see $(5.2.3)$) and at the resistor the energy $W_R= {{1}\over{2}} \cdot {U_{\rm s}^2}\cdot{C} $ (see $(5.2.4)$) into heat. 
 So, in total, the voltage source injects the following energy: So, in total, the voltage source injects the following energy:
  
 \begin{align*} \begin{align*}
-\Delta W_0 &=\Delta W_R + \Delta W_C =  {U_s^2}\cdot{C} +\Delta W_0 &=\Delta W_R + \Delta W_C =  {U_{\rm s}^2}\cdot{C} 
 \end{align*} \end{align*}
  
Zeile 437: Zeile 437:
  
 \begin{align*} \begin{align*}
-\Delta W_0 &=           \int_{0}^{\infty}         u_0 \cdot i_0 \cdot {\rm d}t \quad | \quad u_0 = U_s \text{ is constant because constant voltage source!} \\ +\Delta W_0 &=           \int_{0}^{\infty}         u_0 \cdot i_0 \cdot {\rm d}t \quad | \quad u_0 = U_{\rm s} \text{ is constant because constant voltage source!} \\ 
-           &= U_s \cdot \int_{0}^{\infty}                         i_C {\rm d}t \\ +           &= U_{\rm s} \cdot \int_{0}^{\infty}                         i_C {\rm d}t \\ 
-           &= U_s \cdot \int_{0}^{\infty} {{{\rm d}q}\over{{\rm d}t}} {\rm d}t \\ +           &= U_{\rm s} \cdot \int_{0}^{\infty} {{{\rm d}q}\over{{\rm d}t}} {\rm d}t \\ 
-           &= U_s \cdot \int_{0}^Q {\rm d}q  +           &= U_{\rm s} \cdot \int_{0}^Q {\rm d}q  
-            = U_s \cdot Q \quad | \quad \text{where } Q= C \cdot U_s \\ +            = U_{\rm s} \cdot Q \quad | \quad \text{where } Q= C \cdot U_{\rm s} \\ 
-           &= U_s^2 \cdot C \\+           &= U_{\rm s}^2 \cdot C \\
 \end{align*} \end{align*}
  
-This means that only half of the energy emitted by the source is stored in the capacitor! Again, This doesn't really sound comprehensible at first. And again, it helps to look at small packets of charge that have to be transferred from the ideal source to the capacitor. <imgref imageNo06 > shows current and voltage waveforms across the capacitor and the stored energy for different resistance values. There, too, it can be seen that the maximum stored energy (dashed line in the figure at right) is given by $\Delta W= {{1}\over{2}}  {U_s^2}\cdot{C} $ alone. $U_s^2 \cdot C = {{1}\over{2}} \cdot (5~{\rm V})^2 \cdot 1 ~{\rm µF} = 12.5 ~{\rm µWs}$ is given.+This means that only half of the energy emitted by the source is stored in the capacitor! Again, This doesn't really sound comprehensible at first. And again, it helps to look at small packets of charge that have to be transferred from the ideal source to the capacitor. <imgref imageNo06 > shows current and voltage waveforms across the capacitor and the stored energy for different resistance values. There, too, it can be seen that the maximum stored energy (dashed line in the figure at right) is given by $\Delta W= {{1}\over{2}}  {U_{\rm s}^2}\cdot{C} $ alone. $U_{\rm s}^2 \cdot C = {{1}\over{2}} \cdot (5~{\rm V})^2 \cdot 1 ~{\rm µF} = 12.5 ~{\rm µWs}$ is given.
  
 <WRAP>  <WRAP> 
 <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>
  
Zeile 474: Zeile 474:
 </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@# 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 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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