Unterschiede

Hier werden die Unterschiede zwischen zwei Versionen angezeigt.

Link zu dieser Vergleichsansicht

Beide Seiten der vorigen Revision Vorhergehende Überarbeitung
Nächste Überarbeitung
Vorhergehende Überarbeitung
Nächste Überarbeitung Beide Seiten der Revision
electrical_engineering_1:dc_circuit_transients [2021/10/31 20:51]
tfischer
electrical_engineering_1:dc_circuit_transients [2021/11/22 04:50]
tfischer
Zeile 74: Zeile 74:
 </callout> </callout>
  
-<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BOJyWoVaYEBYDsuBmbSbMSAVmwDZcqRyJyRsCAoAN3G2xACZiuPMNyiiBTSKJjlWAJxAECdYT2KSVospFYBjZnEH7J-SadjxIvKebCsA7oZN8BT7Q6dO1z0-e9+nYLRQrADKfAAc4Y7kyiKmIABmAIYANgDOAKaivL68MeC8URpghcEOxaXFcbn5YFTGAnU+7o31foF02gAezrjgkOi8VATgJczOICGsPXUjYOFW8+rhTDz8IACq0+Dh6ghzCHt0QlEAjtsEuEJYCggQJVZrPABKF228Awr4fGNPIADCrEuxkihkUsSEUlYAAdmNgiiJuFEAlCHEi-F5XL50RovBpunwEMoxgRIPcwKsJgBLC5kgqSAhgOa8EYnAHbPBCb7w5S7cbzdk9bDkdCBSR4RaEcbrACurAA9nxwGIBugYBZaLx9jhyIy8hgLAhaDF9cYlSMqQB9f4AGgAOmkZdagcjrBY6erDWhvUgAUloUkdFSAC5JAB2OgyDsBBCKmnU8BAzwyaSpaVDEajaVeQA 600,400 noborder}} </WRAP>+In the simulation on the below you can see the circuit mentioned above in a slightly modified form:
  
-In the simulation on the right 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 $U_q$ 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 87: Zeile 85:
   - 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 $R=\{ 10\Omega, 100\Omega, 1k\Omega\}$ 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 $R=\{ 10\Omega, 100\Omega, 1k\Omega\}$ and $C=\{ 1\mu F, 10 \mu F\}$. 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?
 +
 +<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BOJyWoVaYEBYDsuBmbSbMSAVmwDZcqRyJyRsD6BTAWjDACgA3cNmwgATMUHCwQqDPFNIMmOR4AnEAQJ0pw4gu0yykHgGNmcCWYViFN2PEgjFd3gHcL10eI9G3Hj7s8bHl8vUL1aKB4AZVEADlj3ci1pGxAAMwBDABsAZzYZEWDRJPARBP0wMsi3CqqKlKKRErAqK3EWoJC9VsDwCKMAD09ccEh0ESpWbkdhMRAoniGWqdjHMFi9WKZZ4QBVRfAN8AQV9Cw6SQScg4JcSSx1BAhKmc8QACUbnpEx9XxRSrMN4AYR4tys8QsGmSkkUPAADsxsOVpEIEh4IJ0kejxAFvEU0RYAvpBqIEFpAQRIM8wNs3gBLG7U0oKAhgKYiViXECgoZ4ST-ZFaI7c3nMchnXAKPBrQhAuYAVx4AHtROBZGN0DB7LQRCccOQ2U0MPYELQksarGrWPSAPrAgA0AB0cgr7WD0U57Mztaa0P6kDyMvCMsZ6QAXDIAO2MbBdoII5QMengHzYOXpOUjMbjOU+QA noborder}} </WRAP>
  
 At the following, this circuit is divided into two separate circuits, which consider only charging and only discharging. At the following, this circuit is divided into two separate circuits, which consider only charging and only discharging.
Zeile 92: Zeile 92:
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-<WRAP> <imgcaption imageNo02 | circuit for viewing the charge curve> </imgcaption> {{drawio>SchaltungEntladekurve2}} </WRAP>+Here a short introduction about the transient behavior of an RC element (starting at 15:07 until 24:55) 
 +{{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 $U_q$ 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 $U_s$ via a resistor $R$.
  
-  * In order that the voltage $U_q$ acts at a certain time $t_0 = 0 s$ the switch $S$ is closed at this time. +  * In order that the voltage $U_s$ 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 $u_C(t_0)=0V$ at that time. The maximum voltage $u_R(t_0)=U_q$ is applied to the resistor. The current is $i_C(t_0)={{U_q}\over{R}}$.+  * 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 $u_C(t_0)=0V$ at that time. The maximum voltage $u_R(t_0)=U_s$ is applied to the resistor. The current is $i_C(t_0)={{U_s}\over{R}}$.
   * 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 $U_q$ until $t \rightarrow \infty$. It then carries the charge: $q(t \rightarrow \infty)=Q = C \cdot U_q$+  * Ideally, the capacitor is not fully charged to the specified voltage $U_s$ until $t \rightarrow \infty$. It then carries the charge: $q(t \rightarrow \infty)=Q = C \cdot U_s$ 
 + 
 +<WRAP> <imgcaption imageNo02 | circuit for viewing the charge curve> </imgcaption> {{drawio>SchaltungEntladekurve2}} </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:
Zeile 118: Zeile 121:
  
 \begin{align*}  \begin{align*} 
-U_q =u_R + u_C = R \cdot i_C + u_C \tag{5.1.2} +U_s =u_R + u_C = R \cdot i_C + u_C \tag{5.1.2} 
 \end{align*} \end{align*}
  
Zeile 136: Zeile 139:
  
 \begin{align*}  \begin{align*} 
-U_q &= u_R                               + u_C \\+U_s &= u_R                               + u_C \\
     &= R \cdot C \cdot {{du_C}\over{dt}} + u_C      &= R \cdot C \cdot {{du_C}\over{dt}} + u_C 
 \end{align*} \end{align*}
Zeile 149: Zeile 152:
  
 \begin{align*} \begin{align*}
-U_q &= R \cdot C \cdot {{d}\over{dt}}(\mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C}) + \mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C} \\+U_s &= R \cdot C \cdot {{d}\over{dt}}(\mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C}) + \mathcal{A} \cdot 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 e^{\mathcal{B}\cdot t} + \mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C} \\
-U_q - \mathcal{C} & ( R \cdot C \cdot \mathcal{AB} + \mathcal{A} ) \cdot e^{\mathcal{B}\cdot t} \\+U_s - \mathcal{C} & ( R \cdot C \cdot \mathcal{AB} + \mathcal{A} ) \cdot e^{\mathcal{B}\cdot t} \\
 \end{align*} \end{align*}
  
Zeile 167: Zeile 170:
  
 \begin{align*} \begin{align*}
-u_C(t) = \mathcal{A} \cdot e^{\large{- {{t}\over{R C}} }} + U_q+u_C(t) = \mathcal{A} \cdot e^{\large{- {{t}\over{R C}} }} + U_s
 \end{align*} \end{align*}
  
Zeile 173: Zeile 176:
  
 \begin{align*} \begin{align*}
-0 &= \mathcal{A} \cdot e^{\large{0}} + U_q \\ +0 &= \mathcal{A} \cdot e^{\large{0}} + U_s \\ 
-0 &= \mathcal{A} U_q \\ +0 &= \mathcal{A} U_s \\ 
-\mathcal{A} &= - U_q+\mathcal{A} &= - U_s
 \end{align*} \end{align*}
  
Zeile 182: Zeile 185:
  
 \begin{align*} \begin{align*}
-u_C(t) &= - U_q \cdot e^{\large{- {{t}\over{R C}}}} + U_q +u_C(t) &= - U_s \cdot e^{\large{- {{t}\over{R C}}}} + U_s
 \end{align*} \end{align*}
  
Zeile 189: Zeile 192:
 And this results in:  And this results in: 
 \begin{align*} \begin{align*}
-u_C(t) &U_q \cdot (1 - e^{\large{- {{t}\over{R C}}}})+u_C(t) &U_s \cdot (1 - 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_q}\over{R}} \cdot e^{\large{- {{t}\over{R C}} } }+i_C(t) &= {{U_s}\over{R}} \cdot e^{\large{- {{t}\over{R C}} } }
 \end{align*} \end{align*}
  
Zeile 209: Zeile 212:
  
   * 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_q \cdot (1 - e^{- 1}) = U_q \cdot (1 - {{1}\over{e}}) = U_q \cdot ({{e-1}\over{e}}) = 0.63 \cdot U_q = 63\% \cdot U_q $. \\ So, **the capacitor is charged to $63\%$ after one $\tau$.**  +  * At time $t=\tau$, we get: $u_C(t) = U_s \cdot (1 - e^{- 1}) = U_s \cdot (1 - {{1}\over{e}}) = U_s \cdot ({{e-1}\over{e}}) = 0.63 \cdot U_s = 63\% \cdot U_s $. \\ So, **the capacitor is charged to $63\%$ after one $\tau$.**  
-  * At time $t=2 \cdot \tau$ we get: $u_C(t) = U_q \cdot (1 - e^{- 2}) = 86\% \cdot U_q = (63\% + (1-63\%) \cdot 63\% ) \cdot U_q$. So, **after each additional $\tau$, the uncharged remainder ($1-63\%$) is recharged to $63\%$**. +  * At time $t=2 \cdot \tau$ we get: $u_C(t) = U_s \cdot (1 - e^{- 2}) = 86\% \cdot U_s = (63\% + (1-63\%) \cdot 63\% ) \cdot U_s$. 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 229: Zeile 232:
 The following situation is considered for the discharge: The following situation is considered for the discharge:
  
-  * A capacitor charged to voltage $U_q$ with capacitance $C$ is short-circuited across a resistor $R$ at time $t=t_0$. +  * A capacitor charged to voltage $U_s$ with capacitance $C$ is short-circuited across a resistor $R$ at time $t=t_0$. 
-  * As a result, the full voltage $U_q$ is initially applied to the resistor: $u_R(t_0)=U_q$+  * As a result, the full voltage $U_s$ is initially applied to the resistor: $u_R(t_0)=U_s$
   * 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 256: Zeile 259:
  
 \begin{align*} \begin{align*}
-U_q &= R \cdot C \cdot {{d}\over{dt}}(\mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C}) + \mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C} \\+U_s &= R \cdot C \cdot {{d}\over{dt}}(\mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C}) + \mathcal{A} \cdot 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 e^{\mathcal{B}\cdot t} + \mathcal{A} \cdot e^{\mathcal{B}\cdot t} + \mathcal{C} \\
-U_q - \mathcal{C} & ( R \cdot C \cdot \mathcal{AB} + \mathcal{A} ) \cdot e^{\mathcal{B}\cdot t} \\+U_s - \mathcal{C} & ( R \cdot C \cdot \mathcal{AB} + \mathcal{A} ) \cdot e^{\mathcal{B}\cdot t} \\
 \end{align*} \end{align*}
  
Zeile 264: Zeile 267:
  
 \begin{align*} \begin{align*}
-\mathcal{C} = U_q \\ \\+\mathcal{C} = U_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 274: Zeile 277:
  
 \begin{align*} \begin{align*}
-u_C(t) = \mathcal{A} \cdot e^{\large{- {{t}\over{R C}} }} + U_q+u_C(t) = \mathcal{A} \cdot e^{\large{- {{t}\over{R C}} }} + U_s
 \end{align*} \end{align*}
  
-For the solution it must still hold that at time $t_0=0$ $u_C(t_0) = U_q$ just holds:+For the solution it must still hold that at time $t_0=0$ $u_C(t_0) = U_s$ just holds:
  
 \begin{align*} \begin{align*}
-0 &= \mathcal{A} \cdot e^{\large{0}} + U_q \\ +0 &= \mathcal{A} \cdot e^{\large{0}} + U_s \\ 
-0 &= \mathcal{A} U_q \\ +0 &= \mathcal{A} U_s \\ 
-\mathcal{A} &= - U_q+\mathcal{A} &= - U_s
 \end{align*} \end{align*}
  
Zeile 288: Zeile 291:
  
 \begin{align*} \begin{align*}
-u_C(t) &= - U_q \cdot e^{\large{- {{t}\over{R C}}}} + U_q +u_C(t) &= - U_s \cdot e^{\large{- {{t}\over{R C}}}} + U_s 
 \end{align*} \end{align*}
  
Zeile 301: Zeile 304:
 And this results in:  And this results in: 
 \begin{align*} \begin{align*}
-u_C(t) &U_q \cdot e^{\large{- {{t}\over{\tau}}}} \quad \text{with} \quad \tau = R C+u_C(t) &U_s \cdot 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_q}\over{R}} \cdot e^{\large{- {{t}\over{R C}} } }+i_C(t) &={{U_s}\over{R}} \cdot e^{\large{- {{t}\over{R C}} } }
 \end{align*} \end{align*}
  
-In <imgref imageNo05 > the two time histories are again shown; this time for the discharge voltage $u_C(t)$ and the discharge current $i_C(t)$ of the capacitor. Since the current now flows out of the capacitor, the sign of $i_C$ is negative.+In <imgref imageNo05 > the two time courses are again shown; now for the discharge voltage $u_C(t)$ and the discharge current $i_C(t)$ of the capacitor. Since the current now flows out of the capacitor, the sign of $i_C$ is negative.
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
Zeile 343: Zeile 346:
 <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}} </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:basics_and_basic_concepts#determination_of_electrical_power_in_the_dc_circuit|Basics and Basic Concepts]], 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|Basics and Basic Concepts]], the power for constant values (DC) is defined as:
  
 \begin{align*} \begin{align*}
Zeile 365: Zeile 368:
 During the charging process During the charging process
 \begin{align*} \begin{align*}
-u_C(t) = U_q\cdot (1 - e^{ - {{t}\over{\tau}} })  \\ +u_C(t) = U_s\cdot (1 - e^{ - {{t}\over{\tau}} })  \\ 
-i_C(t) = {{U_q}\over{R}} \cdot e^{ -{{t}\over{\tau}} } \tag{5.2.2}+i_C(t) = {{U_s}\over{R}} \cdot e^{ -{{t}\over{\tau}} } \tag{5.2.2}
 \end{align*} \end{align*}
  
Zeile 388: Zeile 391:
 \end{align*} \end{align*}
  
-Thus, for a fully discharged capacitor ($U_q=0V$), the energy stored when charging to voltage $U_q$ is $\delta W_C={{1}\over{2}} C \cdot U_q^2$.+Thus, for a fully discharged capacitor ($U_s=0V$), the energy stored when charging to voltage $U_s$ is $\delta W_C={{1}\over{2}} C \cdot U_s^2$.
  
 === Energy consideration on the resistor === === Energy consideration on the resistor ===
Zeile 401: Zeile 404:
  
 \begin{align*} \begin{align*}
-\Delta W_R & R \cdot \int_{0}^t \left( { {U_q}\over{R}} \cdot e^ { -{{t}\over{\tau}}} \right)^2  dt \\ +\Delta W_R & R \cdot \int_{0}^t \left( { {U_s}\over{R}} \cdot e^ { -{{t}\over{\tau}}} \right)^2  dt \\ 
-   & { {U_q^2}\over{R}} \cdot \int_{0}^t  e^ { -{{2 \cdot t}\over{\tau}}}  dt \\ +   & { {U_s^2}\over{R}} \cdot \int_{0}^t  e^ { -{{2 \cdot t}\over{\tau}}}  dt \\ 
-   & { {U_q^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_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 \\ 
-   &  -{{1}\over{2}} \cdot {U_q^2}\cdot{C} \cdot   \left[ e^ { -{{2 \cdot t}\over{\tau}}} \right]_{0}^t \\+   &  -{{1}\over{2}} \cdot {U_s^2}\cdot{C} \cdot   \left[ e^ { -{{2 \cdot t}\over{\tau}}} \right]_{0}^t \\
 \end{align*} \end{align*}
  
Zeile 410: Zeile 413:
  
 \begin{align*} \begin{align*}
-\Delta W_R & -{{1}\over{2}} \cdot {U_q^2}\cdot{C} \cdot   \left[ e^ { -{{2 \cdot t}\over{\tau}}} \right]_{0}^{\infty} \\ +\Delta W_R & -{{1}\over{2}} \cdot {U_s^2}\cdot{C} \cdot   \left[ e^ { -{{2 \cdot t}\over{\tau}}} \right]_{0}^{\infty} \\ 
-   & -{{1}\over{2}} \cdot {U_q^2}\cdot{C} \cdot   \left[ 0 - 1  \right] \\+   & -{{1}\over{2}} \cdot {U_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_q^2}\cdot{C}} \tag{5.2.4}+\boxed{ \Delta W_R  =  {{1}\over{2}} \cdot {U_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 $U_q$ 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: $u_r = U_q - u_C = U_q - {{q}\over{C}}$.+This means that the energy converted at the resistor is independent of the resistance value (for an ideal constant voltage source $U_s$ 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: $u_r = U_s - u_C = U_s - {{q}\over{C}}$.
  
 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 423: Zeile 426:
 === 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_q^2}\cdot{C} $ (see $(5.2.3)$) and at the resistor the energy $W_R= {{1}\over{2}} \cdot {U_q^2}\cdot{C} $ (see $(5.2.4)$) into heat. So, in total, the voltage source injects the following energy:+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. So, in total, the voltage source injects the following energy:
  
 \begin{align*} \begin{align*}
-\Delta W_0 &=\Delta W_R + \Delta W_C =  {U_q^2}\cdot{C} +\Delta W_0 &=\Delta W_R + \Delta W_C =  {U_s^2}\cdot{C} 
 \end{align*} \end{align*}
  
Zeile 432: Zeile 435:
  
 \begin{align*} \begin{align*}
-\Delta W_0 & \int_{0}^{\infty} u_0 \cdot i_0 \cdot dt \quad | \quad u_0 = U_q \text{ is constant because constant voltage source!} \\ +\Delta W_0 & \int_{0}^{\infty} u_0 \cdot i_0 \cdot dt \quad | \quad u_0 = U_s \text{ is constant because constant voltage source!} \\ 
-&U_q \cdot \int_{0}^{\infty} i_C dt \\ +&U_s \cdot \int_{0}^{\infty} i_C dt \\ 
-&U_q \cdot \int_{0}^{\infty} {{dq}\over{dt}} dt \\ +&U_s \cdot \int_{0}^{\infty} {{dq}\over{dt}} dt \\ 
-&U_q \cdot \int_{0}^Q dq = U_q \cdot Q \quad | \quad \text{where } Q= C \cdot U_q \\ +&U_s \cdot \int_{0}^Q dq = U_s \cdot Q \quad | \quad \text{where } Q= C \cdot U_s \\ 
-&U_q^2 \cdot C \\+&U_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. 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 $\Delta W= {{1}\over{2}}$ alone. $U_q^2 \cdot C = {{1}\over{2}} \cdot (5V)^2 \cdot 1 \mu F = 12.5 \mu Ws$ 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. 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 $\Delta W= {{1}\over{2}}$ alone. $U_s^2 \cdot C = {{1}\over{2}} \cdot (5V)^2 \cdot 1 \mu F = 12.5 \mu Ws$ is given.
  
 <WRAP>  <WRAP>