Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
| electrical_engineering_1:dc_circuit_transients [2021/10/16 14:49] – [Bearbeiten - Panel] slinn | electrical_engineering_1:dc_circuit_transients [2024/11/08 15:21] (aktuell) – mexleadmin | ||
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| Zeile 1: | Zeile 1: | ||
| - | ====== | + | ====== |
| <WRAP onlyprint> | <WRAP onlyprint> | ||
| - | | + | |
| - | - Charge / discharge FET capacitor. | + | |
| + | - Charge/ | ||
| </ | </ | ||
| < | < | ||
| - | < | + | < |
| - | < | + | < |
| - | </ | + | </ |
| - | {{drawio> | + | \\ {{drawio> |
| </ | </ | ||
| + | Here we will shortly introduce the basic idea behind a capacitor. A more detailed analysis will follow in electrical engineering II. \\ \\ | ||
| + | A capacitor consists of two insulated conductors (electrodes) separated by an insulator (cf. <imgref imageNo01 >). \\ | ||
| + | The electrodes serve as " | ||
| - | At the previous chapter the capacitor was already described. It consists of two insulated conductors separated by an insulator (cf. <imgref imageNo01 >). \\ | ||
| - | They serve as energy storage. This is done in the following manner: | ||
| - An external source draws charge carriers from one of the electrodes and carries them to the other electrode. | - An external source draws charge carriers from one of the electrodes and carries them to the other electrode. | ||
| - If the external source is a voltage source with the voltage $U$, a stationary state is reached after a certain time. \\ In this state there is a fixed number of $+Q$ on the positive electrode and $-Q$ on the negative electrode. | - If the external source is a voltage source with the voltage $U$, a stationary state is reached after a certain time. \\ In this state there is a fixed number of $+Q$ on the positive electrode and $-Q$ on the negative electrode. | ||
| - These charges form an electric field in the space between the electrodes. This field stores the supplied energy. | - These charges form an electric field in the space between the electrodes. This field stores the supplied energy. | ||
| - | It is true that the larger the voltage $U$, the more charges $Q$ are stored on the electrode. | + | As larger the voltage $U$, more charges $Q$ are stored on the electrode. This relationship is directly proportional to the proportionality constant $C$: |
| - | This relationship is directly proportional to the proportionality constant $C$: | + | |
| - | \begin{align*} | + | \begin{align*} |
| - | C = {{Q}\over{U}} \quad \text{with: | + | C = {{Q}\over{U}} \quad \text{with: |
| \end{align*} | \end{align*} | ||
| - | + | ||
| But it is not always directly recognizable that a structure contains a capacitor. \\ So the following examples are also capacitors: | But it is not always directly recognizable that a structure contains a capacitor. \\ So the following examples are also capacitors: | ||
| - | * **open switch**: If there is voltage between the two metal parts, charges can also accumulate there. \\ Since the distances are usually large and 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 $100Hz$ and many times that (harmonics). Peak discharge results in the 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 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, | ||
| - | * **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 delayed signal transmission which characterizes the disease patterns. | ||
| - | < | + | * **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 $100~\rm Hz$ 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). |
| - | {{drawio> | + | * **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, |
| - | </ | + | * **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. |
| + | < | ||
| - | 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 charging of the capacitor, besides the voltage source $U_q$ 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 resistor | + | 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 |
| - | 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 | ||
| - | * resistance of $0\Omega$ in closed state (" | + | * resistance of $0~\Omega$ in the closed state (" |
| - | * resistance $\rightarrow \infty$ in open state ("open line") | + | * resistance $\rightarrow \infty$ in the open state ("open line") |
| * no capacitive effect | * no capacitive effect | ||
| Zeile 55: | Zeile 50: | ||
| < | < | ||
| - | In this chapter also time-varying quantities are considered. These are generally marked with small letters. Examples of time-varying quantities are: | + | In this chapter also time-varying quantities are considered. These are generally marked with lowercase |
| - | * A **time-varying voltage $u_C(t)$ across a capacitor** or the **voltage $u$ of an ac voltage source** as opposed to a constant voltage $U_q$ across a constant voltage source. | + | |
| - | * A **time-varying current $i_L(t)$ across a coil** or **time-varying current $i_L(t)$ across a capacitor**. | + | |
| - | Since the time dependence is already clear from the small letter, these quantities are occasionally not indicated by the trailing $(t)$. So it is $u = u(t)$. | + | * A **time-varying voltage $u_C(t)$ across a capacitor** or the **voltage $U_{\rm s}$ of an ac voltage source** |
| + | * A **time-varying current $i_L(t)$ across a coil** or **time-varying current $i_C(t)$ across a capacitor**. | ||
| + | |||
| + | Since the time dependence is already clear from the lowercase | ||
| </ | </ | ||
| - | ===== 7.1 Time course | + | ===== 5.1 Time Course |
| < | < | ||
| - | === Goals === | + | === Learning Objectives |
| - | After this lesson, you should: | + | By the end of this section, you will be able to: |
| - | + | - know the time constant $\tau$ and in particularly | |
| - | - know the time constant $\tau$ and in particular be able to calculate it. | + | - determine the time characteristic of the currents and voltages at the RC element for a given resistance and capacitance. |
| - | - Be able to determine the time characteristic of the currents and voltages at the RC element for a given resistance and capacitance. | + | |
| - know the continuity conditions of electrical quantities. | - know the continuity conditions of electrical quantities. | ||
| - | - know when (=according to which measure) the capacitor is considered to be fully charged / discharged, i.e. a steady state can be considered to have been reached. | + | - know when (=according to which measure) the capacitor is considered to be fully charged/ |
| </ | </ | ||
| - | < | + | In the simulation below you can see the circuit |
| - | </ | + | |
| - | In the simulation on the right you can see the circuit mentioned above in a slightly modified form: | + | |
| - | | + | * But it is also possible to short-circuit the series |
| - | * But it is also possible to short-circuit the rich 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. | + | * Additionally it is possible to change the capacitance value $C$ and resistance value $R$ with the sliders '' |
| - | * Additionally it is possible to change the capacitance value $C$ and resistance value $R$ with the sliders '' | + | |
| Exercises: | Exercises: | ||
| - | | + | |
| + | | ||
| - 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? | ||
| - | \\ | + | < |
| - | At the following, this circuit is divided into two separate circuits, which consider only charging and only discharging. | + | In the following, this circuit is divided into two separate circuits, which consider only charging and only discharging. |
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | < | + | Here is a short introduction about the transient behavior of an RC element (starting at 15:07 until 24:55) |
| - | < | + | {{youtube>8nyNamrWcyE? |
| - | </ | + | |
| - | {{drawio>SchaltungEntladekurve2}} \\ | + | 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 $R$. |
| - | </ | + | |
| - | 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 $R$. | + | |
| - | | + | * Directly after the time $t_0$ the maximum current (" |
| - | * Directly after the time $t_0$ the maximum current (" | + | |
| * 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 $U_{\rm s}$ until $t \rightarrow \infty$. It then carries the charge: $q(t \rightarrow \infty) = Q = C \cdot U_{\rm s}$ |
| - | The process is now to be summarized in detail in formulas. \\ | + | < |
| - | Linear components are used in the circuit, i.e. the component values | + | |
| - | Then definition equations for the resistor $R$ and the capacitance $C$ are also valid for time-varying or infinitesimal quantities: | + | |
| - | \begin{align*} | + | 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: |
| - | R = {{u_R(t)}\over{i_R(t)}} = {{du_R}\over{di_R}} = const. \\ | + | |
| - | C = {{q(t)}\over{u_C(t)}} = {{dq}\over{du_C}} = const. | + | \begin{align*} |
| + | R = {{u_R(t)}\over{i_R(t)}} = {{{\rm d}u_R}\over{{\rm d}i_R}} = {\rm const.} \\ | ||
| + | C = {{q(t)} | ||
| \end{align*} | \end{align*} | ||
| - | The following explanations are also well explained in these two videos on [[https:// | + | The following explanations are also well explained in these two videos on [[https:// |
| ==== Charging a capacitor at time t=0 ==== | ==== Charging a capacitor at time t=0 ==== | ||
| Zeile 126: | Zeile 119: | ||
| By considering the loop, the general result is: the voltage of the source is equal to the sum of the two voltages across the resistor and capacitor. | By considering the loop, the general result is: the voltage of the source is equal to the sum of the two voltages across the resistor and capacitor. | ||
| - | \begin{align*} | + | \begin{align*} |
| - | U_q =u_R + u_C = R \cdot i_C + u_C \tag{7.1.2} | + | U_{\rm s} =u_R + u_C = R \cdot i_C + u_C \tag{5.1.2} |
| \end{align*} | \end{align*} | ||
| - | At the first instant $dt$, an infinitesimally small charge " | + | At the first instant ${\rm d}t$, an infinitesimally small charge " |
| - | For this, $(7.1.1)$ gives: | + | |
| - | \begin{align*} | + | \begin{align*} |
| - | i_C = {{dq}\over{dt}} \quad \text{and} \quad dq = C \cdot du_C | + | i_C = {{{\rm d}q}\over{{\rm d}t}} \quad \text{and} \quad {\rm d}q = C \cdot {\rm d}u_C |
| \end{align*} | \end{align*} | ||
| The charging current $i_C$ can be determined from the two formulas: | The charging current $i_C$ can be determined from the two formulas: | ||
| - | \begin{align*} | + | \begin{align*} |
| - | i_C = C \cdot {{du_C}\over{dt}} \tag{7.1.3} | + | i_C = C \cdot {{{\rm d}u_C}\over{{\rm d}t}} \tag{5.1.3} |
| \end{align*} | \end{align*} | ||
| - | Thus $(7.1.2)$ becomes: | + | Thus $(5.1.2)$ becomes: |
| - | \begin{align*} | + | \begin{align*} |
| - | U_q &=u_R + u_C \\ | + | U_{\rm s} &= u_R |
| - | &= R \cdot C \cdot {{du_C}\over{dt}} + u_C | + | &= R \cdot C \cdot {{{\rm d}u_C}\over{{\rm d}t}} + u_C |
| \end{align*} | \end{align*} | ||
| --> here follows some mathematics: | --> here follows some mathematics: | ||
| - | This result represents a 1st order differential equation. | + | This result represents a 1st order differential equation. This should generally be rewritten so that the part that depends (on the variable) is on one side and the rest is on the other. This is already present here. The appropriate approach to such a problem is: |
| - | This should generally be rewritten so that the part that depends (on the variable) is on one side and the rest is on the other. | + | |
| - | This is already present here. The appropriate approach to such a problem is: | + | |
| \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*} | ||
| - | 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_{\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 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} \\ |
| - | U_q - \mathcal{C} &= ( R \cdot C \cdot \mathcal{AB} + \mathcal{A} ) \cdot e^{\mathcal{B}\cdot t} \\ | + | U_{\rm s} - \mathcal{C} & |
| \end{align*} | \end{align*} | ||
| - | This equation must hold for every $t$. This is only possible if the left as well as the right term become equal to 0. \\ Thus: | + | This equation must hold for every $t$. This is only possible if the left, as well as the right term, become equal to 0. \\ Thus: |
| \begin{align*} | \begin{align*} | ||
| - | \mathcal{C} = U_q \ | + | \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 |
| - | R \cdot C \cdot \mathcal{B} &= - 1 \\ | + | R \cdot C \cdot \mathcal{B} &= - 1 \\ |
| - | \mathcal{B} &= - {{1}\over{R C}} \\ | + | \mathcal{B} &= - {{1}\over{R C}} \\ |
| \end{align*} | \end{align*} | ||
| Zeile 179: | Zeile 169: | ||
| \begin{align*} | \begin{align*} | ||
| - | u_C(t) = \mathcal{A} \cdot e^{\large{- {{t}\over{R C}} }} + U_q | + | u_C(t) = \mathcal{A} \cdot {\rm e}^{\large{- {{t}\over{R C}} }} + U_{\rm s} |
| \end{align*} | \end{align*} | ||
| Zeile 185: | Zeile 175: | ||
| \begin{align*} | \begin{align*} | ||
| - | 0 &= \mathcal{A} \cdot e^{\large{0}} + U_q \\ | + | 0 &= \mathcal{A} \cdot {\rm e}^{\large{0}} + U_{\rm s} \\ |
| - | 0 &= \mathcal{A} + U_q \ | + | 0 &= \mathcal{A} |
| - | \mathcal{A} &= - U_q | + | \mathcal{A} &= - U_{\rm s} |
| \end{align*} | \end{align*} | ||
| + | |||
| So the solution is: | So the solution is: | ||
| \begin{align*} | \begin{align*} | ||
| - | u_C(t) &= - U_q \cdot e^{\large{- {{t}\over{R C}}}} + U_q | + | u_C(t) &= - U_{\rm s} \cdot {\rm e}^{\large{- {{t}\over{R C}}}} + U_{\rm s} |
| \end{align*} | \end{align*} | ||
| <-- | <-- | ||
| - | + | And this results in: | |
| - | And this results in: | + | |
| \begin{align*} | \begin{align*} | ||
| - | u_C(t) & | + | u_C(t) & |
| \end{align*} | \end{align*} | ||
| - | And with $(7.1.3)$, $i_C$ 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_{\rm s}}\over{R}} \cdot {\rm e}^{\large{- {{t}\over{R C}} } } |
| \end{align*} | \end{align*} | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | In <imgref imageNo04 >, the two time courses | + | In <imgref imageNo04 >, the two time course diagrams |
| - | < | + | < |
| - | < | + | </ |
| - | </ | + | {{drawio> |
| - | {{drawio> | + | |
| </ | </ | ||
| <callout icon=" | <callout icon=" | ||
| - | | + | |
| - | * 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 $t=2 \cdot \tau$ 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 $t=\tau$, 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_{\rm s} \cdot (1 - {\rm e}^{- 2}) = 86~\% \cdot U_{\rm s} = (63~\% + (100~\% |
| + | * After about $t=5 \cdot \tau$, the result is a capacitor charged to over $99~\%$. In real circuits, | ||
| * The time constant $\tau$ can be determined graphically in several ways: | * The time constant $\tau$ can be determined graphically in several ways: | ||
| - | | + | |
| - | * Plotting the tangent to the (voltage) charge curve at the time of the discharged capacitor. | + | * Plotting the tangent to the (voltage) charge curve at the time of the discharged capacitor. This intersects a horizontal line at the level of the charging voltage at the point $t=\tau$ (see black and light blue lines in <imgref imageNo04> |
| </ | </ | ||
| + | |||
| ==== Discharging a capacitor at time t=0 ==== | ==== Discharging a capacitor at time t=0 ==== | ||
| - | < | + | < |
| - | < | + | < |
| - | </ | + | </ |
| - | {{drawio> | + | {{drawio> |
| </ | </ | ||
| 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 $t=t_0$. | ||
| - | * As a result, the full voltage $U_q$ is initially applied to the resistor: $u_R(t_0)=U_q$ | ||
| - | * 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: $u_C = {{q(t)}{C}}$ | ||
| - | * Ideally, the capacitor is not fully discharged until $t \rightarrow \infty$. | ||
| - | Also this process now is to put into formula in detail. | + | * A capacitor charged to voltage $U_{\rm s}$ with capacitance $C$ is short-circuited across a resistor $R$ at time $t=t_0$. |
| - | By looking at the loop, the general result is: the sum of the two voltages across the resistor and capacitor | + | * 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 discharging charges lower the voltage of the capacitor $u_C$, since: $u_C = {{q(t)}\over{C}}$ | ||
| + | * Ideally, the capacitor is not fully discharged before $t \rightarrow \infty$. | ||
| + | |||
| + | Also, this process now is to put into a formula in detail. By looking at the loop, the general result is: the sum of the two voltages across the resistor and capacitor | ||
| \begin{align*} | \begin{align*} | ||
| Zeile 251: | Zeile 243: | ||
| \end{align*} | \end{align*} | ||
| - | This gives $(7.1.3)$: | + | This gives $(5.1.3)$: |
| \begin{align*} | \begin{align*} | ||
| - | 0 =u_R + u_C = R \cdot C \cdot {{du_C}\over{dt}} + u_C | + | 0 =u_R + u_C = R \cdot C \cdot {{{\rm d}u_C}\over{{\rm d}t}} + u_C |
| \end{align*} | \end{align*} | ||
| --> also here uses some mathematics: | --> also here uses some mathematics: | ||
| - | This result again represents a 1st order differential equation. | + | This result again represents a 1st order differential equation. The appropriate approach to such a problem is: |
| - | The appropriate approach to such a problem is: | + | |
| \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 {{d}\over{dt}}(\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} |
| - | &= 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} |
| - | 0 - \mathcal{C} &= ( R \cdot C \cdot \mathcal{AB} + \mathcal{A} ) \cdot e^{\mathcal{B}\cdot t} \\ | + | - \mathcal{C} & |
| \end{align*} | \end{align*} | ||
| - | This equation must hold for every $t$. This is only possible if the left as well as the right term become equal to 0. \\ Thus: | + | This equation must hold for every $t$. This is only possible if the left, as well as the right term, become equal to 0. Thus: |
| \begin{align*} | \begin{align*} | ||
| - | \mathcal{C} = 0 \\ \ | + | \mathcal{C} = 0 \\ \\ |
| - | 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 |
| - | R \cdot C \cdot \mathcal{B} &= - 1 \\ | + | R \cdot C \cdot \mathcal{B} &= - 1 \\ |
| - | \mathcal{B} &= - {{1}\over{R C}} \\ | + | \mathcal{B} &= - {{1}\over{R C}} \\ |
| \end{align*} | \end{align*} | ||
| Zeile 285: | Zeile 276: | ||
| \begin{align*} | \begin{align*} | ||
| - | u_C(t) = \mathcal{A} \cdot e^{\large{- {{t}\over{R C}} }} | + | 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 $t_0=0$ $u_C(t_0) = U_q$ 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_q &= \mathcal{A} \cdot e^{\large{0}} | + | U_{\rm s} &= \mathcal{A} \cdot {\rm e}^{\large{0}} |
| - | U_q & | + | \mathcal{A} |
| - | \mathcal{A} & | + | \end{align*} |
| + | |||
| + | Therefore, the result is: | ||
| + | |||
| + | \begin{align*} | ||
| + | u_C(t) | ||
| \end{align*} | \end{align*} | ||
| <-- | <-- | ||
| - | < | + | < |
| - | < | + | < |
| - | </ | + | </ |
| - | {{drawio> | + | {{drawio> |
| </ | </ | ||
| - | + | And this results in: | |
| - | And this results in: | + | |
| \begin{align*} | \begin{align*} | ||
| - | u_C(t) & | + | u_C(t) & |
| \end{align*} | \end{align*} | ||
| - | And with $(7.1.3)$, $i_C$ 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_{\rm s}}\over{R}} \cdot {\rm e}^{\large{- {{t}\over{R C}} } } |
| \end{align*} | \end{align*} | ||
| - | In <imgref imageNo05 > the two time histories | + | In <imgref imageNo05 > the two time course diagrams |
| - | Since the current now flows out of the capacitor, the sign of $i_C$ is negative. | + | |
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| Zeile 322: | Zeile 316: | ||
| ==== Periodic switching operations ==== | ==== Periodic switching operations ==== | ||
| - | < | + | < |
| - | </ | + | |
| - | In the simulation on the right, a periodic switching operation can be seen. The capacitor is periodically charged and discharged via the switch. | + | In the simulation on the right, a periodic switching operation can be seen. The capacitor is periodically charged and discharged via the switch. Three sliders are given in the simulation to change the resistance $R$ (< |
| - | Three sliders are given in the simulation to change the resistance $R$ (%%Resistance R%%), the capacity $C$ (%%Capacity C%%) and the frequency $f$ (%%Frequency f%%). \\ | + | |
| - | In the simulation below, the voltage $u_C$ across the capacitor is shown in green and the current $i_C$ is shown in yellow. | + | |
| Exercises: | Exercises: | ||
| - | | + | |
| - | - Now increase the capacitance to $C=10 \mu F$ using the corresponding slider. What is the change for $u_C$ and $i_C$? | + | |
| - | - Now increase the resistance to $R= 1 k\Omega$ using the corresponding slider. What is the change for $u_C$ and $i_C$? | + | - Now increase the capacitance to $C=10 ~{\rm µF}$ using the corresponding slider. What is the change for $u_C$ and $i_C$? |
| + | - Now increase the resistance to $R= 1 ~\rm k\Omega$ using the corresponding slider. What is the change for $u_C$ and $i_C$? | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | ===== 7.2 Energy | + | ===== 5.2 Energy |
| < | < | ||
| - | === Goals === | + | === Learning Objectives |
| - | After this lesson, you should: | + | By the end of this section, you will be able to: |
| + | - calculate the energy content in a capacitor. | ||
| + | - calculate the change in energy of a capacitor resulting from a change in voltage between the capacitor terminals. | ||
| + | - calculate (initial) current, (final) voltage, and charge when balancing the charge of several capacitors (also via resistors). | ||
| - | - Be able to calculate the energy content in a capacitor. | ||
| - | - Be able to calculate the change in energy of a capacitor resulting from a change in voltage between the capacitor terminals. | ||
| - | - Be able to calculate (initial) current, (final) voltage and charge when balancing the charge of several capacitors (also via resistors). | ||
| - | |||
| </ | </ | ||
| - | < | + | < |
| - | < | + | |
| - | </ | + | |
| - | {{drawio> | + | |
| - | </ | + | |
| - | Now the capacitor as energy storage is to be looked at more closely. This derivation is also explained in [[https:// | + | Now the capacitor as energy storage is to be looked at more closely. This derivation is also explained in [[https:// |
| - | According to the chapter [[Basics and Basic Concepts#Determination_of_electrical_power_in_the_DC_circuit|Basics | + | |
| \begin{align*} | \begin{align*} | ||
| - | P={{delta W}\over{delta t}} = U \cdot I | + | P={{\Delta |
| \end{align*} | \end{align*} | ||
| Zeile 366: | Zeile 352: | ||
| \begin{align*} | \begin{align*} | ||
| - | p={{dw}\over{dt}} = u \cdot i | + | p={{{\rm d}w}\over{{\rm d}t}} = u \cdot i |
| \end{align*} | \end{align*} | ||
| Zeile 374: | Zeile 360: | ||
| \begin{align*} | \begin{align*} | ||
| - | \delta W_C = \int_{t_0}^{t_1} | + | \Delta W_C = \int_{t_0}^{t_1} |
| + | = \int_{0}^t u | ||
| + | = \int_{0}^t u_C \cdot i_C {\rm d}t \tag{5.2.1} | ||
| \end{align*} | \end{align*} | ||
| 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_{\rm s} \cdot (1 - {\rm e}^{ -{{t}\over{\tau}} }) \\ |
| - | i_C(t) = {{U_q}\over{R}} \cdot e^{ -{{t}\over{\tau}} } \tag{7.2.2} | + | i_C(t) = {{U_{\rm s}}\over{R}} \cdot |
| \end{align*} | \end{align*} | ||
| Zeile 387: | Zeile 374: | ||
| \begin{align*} | \begin{align*} | ||
| - | C = {{q(t)}\over{u_C(t)}} \quad & | + | C = {{q(t)}\over{u_C(t)}} |
| - | i_C(t) = {{d q(t)}\over{dt}} \quad & | + | i_C(t) = {{{\rm d} q(t)}\over{{\rm d}t}} \quad & |
| \end{align*} | \end{align*} | ||
| - | Thus, the stored energy from formula $(7.2.1)$: | + | Thus, the stored energy from formula $(5.2.1)$: |
| \begin{align*} | \begin{align*} | ||
| - | \delta W_C &= \int_{0}^t u_C(t) \cdot C \cdot {{d u_C(t)}\over{dt}} dt \quad & | \text{ substitution of integration variable: } t \rightarrow u_C \\ | + | \Delta W_C &= \int_{0}^t |
| - | &= \int_{U_0}^{U_1} u_C(t) \cdot C \cdot du_C \quad & | \text{ Since the capacity is constant, it can be written | + | |
| - | &= C \cdot \int_{U_0}^{U_1} u_C \, d u_C \\ | + | |
| - | &= C \cdot \left[{{1}\over{2}} u_C^2 \right] _{U_0}^{U_1} \\ | + | |
| \end{align*} | \end{align*} | ||
| - | |||
| \begin{align*} | \begin{align*} | ||
| - | \boxed{\delta W_C= {{1}\over{2}} C \cdot (U_1^2-U_0^2)} \tag{7.2.3} | + | \boxed{\Delta |
| \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 ($U_{\rm s}=0~{\rm V}$), the energy stored when charging to voltage $U_{\rm s}$ is $\Delta W_C={{1}\over{2}} C \cdot U_{\rm s}^2$. |
| - | === Energy | + | === Energy |
| - | The converted energy can also be determined for the resistance: | + | The converted energy can also be determined for the resistor: |
| \begin{align*} | \begin{align*} | ||
| - | \delta W_R = \int_{0}^t u_R \cdot i_R dt = \int_{0}^t R \cdot i_R \cdot i_R dt = R \cdot \int_{0}^t i_R^2 dt | + | \Delta W_R = \int_{0}^t u_R \cdot i_R {\rm d}t |
| + | = \int_{0}^t | ||
| + | = R \cdot \int_{0}^t i_R^2 {\rm d}t | ||
| \end{align*} | \end{align*} | ||
| - | Since the current through the capacitor $i_C$ is equal to that through the resistor $i_R$, it follows via $(7.2.2)$: | + | Since the current through the capacitor $i_C$ is equal to that through the resistor $i_R$, it follows via $(5.2.2)$: |
| \begin{align*} | \begin{align*} | ||
| - | \Delta W_R & | + | \Delta W_R & |
| - | | + | |
| - | | + | |
| - | | + | |
| \end{align*} | \end{align*} | ||
| Zeile 426: | Zeile 414: | ||
| \begin{align*} | \begin{align*} | ||
| - | \Delta W_R & | + | \Delta W_R & |
| - | | + | |
| \end{align*} | \end{align*} | ||
| \begin{align*} | \begin{align*} | ||
| - | \boxed{ \Delta W_R = {{1}\over{2}} \cdot {U_q^2}\cdot{C}} \tag{7.2.4} | + | \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 $U_q$ and given capacitor $C$)! At first, this doesn' | + | 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' |
| - | Graphically, | + | |
| - | In real applications, | + | In real applications, |
| - | === Consideration of total energy | + | === Consideration of total energy |
| - | In the previous considerations, | + | In the previous considerations, |
| + | 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_{\rm s}^2}\cdot{C} |
| \end{align*} | \end{align*} | ||
| - | This also follows via $(7.2.1)$: | + | This also follows via $(5.2.1)$: |
| \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 & |
| - | & | + | |
| - | & | + | |
| - | & | + | |
| - | & | + | |
| + | | ||
| \end{align*} | \end{align*} | ||
| - | This means that only half of the energy emitted by the source is stored in the capacitor! Again, | + | This means that only half of the energy emitted by the source is stored in the capacitor! Again, |
| - | + | < | |
| - | < | + | < |
| - | < | + | </ |
| - | </ | + | {{drawio> |
| - | {{drawio> | + | |
| </ | </ | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | This can also be tested in the following simulation. In addition to the RC element shown so far, a power meter and an integrator are also drawn in here. It is possible to display the instantaneous power and the stored energy. Via the slider | + | This can also be tested in the following simulation. In addition to the RC element shown so far, a power meter and an integrator are also drawn here. It is possible to display the instantaneous power and the stored energy. Via the slider |
| * left: Current $u_C$ and voltage $i_C$ at the capacitor. | * left: Current $u_C$ and voltage $i_C$ at the capacitor. | ||
| * middle: Instantaneous power $p_C = u_C \cdot i_C$ of the capacitor. | * middle: Instantaneous power $p_C = u_C \cdot i_C$ of the capacitor. | ||
| - | * right: stored energy $w_C = \int u_C \cdot i_C \; dt$ of the capacitor | + | * right: stored energy $w_C = \int u_C \cdot i_C \; {\rm d}t$ of the capacitor |
| - | + | ||
| - | < | + | |
| - | </ | + | |
| + | < | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | ==== Exercises==== | + | ==== Exercises ==== |
| - | <panel type=" | + | <panel type=" |
| + | <WRAP group>< | ||
| {{youtube> | {{youtube> | ||
| Zeile 486: | Zeile 474: | ||
| </ | </ | ||
| + | # | ||
| + | The following circuit shows a charging/ | ||
| - | <panel type=" | + | 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) | ||
| - | {{youtube>a-gPuw6JsxQ}} | + | {{drawio>electrical_engineering_1: |
| - | </ | + | 1. For the first tasks, the switch $S_1$ gets closed at $t=t_0 = 0s$. \\ |
| + | 1.1 What is the value of the time constant $\tau_1$? | ||
| + | # | ||
| + | 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. | ||
| + | {{drawio> | ||
| - | <panel type=" | + | We see that for the time constant, we need to use $R=R_1 + R_2$. |
| - | {{youtube> | + | # |
| - | </ | + | # |
| + | \begin{align*} | ||
| + | \tau_1 &= R\cdot C \\ | ||
| + | & | ||
| + | & | ||
| + | \end{align*} | ||
| + | # | ||
| + | 1.2 What is the formula for the voltage $u_{R2}$ over the resistor $R_2$? Derive a general formula without using component values! | ||
| + | # | ||
| + | To get a general formula, we again look at the circuit, but this time with the voltage arrows. | ||
| - | <panel type=" | + | {{drawio>electrical_engineering_1: |
| - | {{youtube> | + | 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/ | ||
| + | Therefore, $u_{R2}$ can be written as: | ||
| - | </WRAP></ | + | \begin{align*} |
| + | u_{R2} &= R_2 \cdot i_{R2} \\ | ||
| + | & | ||
| + | \end{align*} | ||
| + | # | ||
| + | # | ||
| + | \begin{align*} | ||
| + | u_{R2} = U_1 \cdot {{R_2}\over{R_1 + R_2}} \cdot e^{t/ \tau} | ||
| + | \end{align*} | ||
| + | # | ||
| + | 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 $t_1$! | ||
| + | # | ||
| - | <panel type=" | + | We can derive $u_{C}$ based on the exponential function: $u_C(t) |
| + | Therefore, we get $t_1$ by: | ||
| - | < | + | \begin{align*} |
| + | u_C = 4/5 \cdot U_1 & | ||
| + | 4/5 & | ||
| + | e^{-t/ | ||
| + | | ||
| + | t &= -\tau \cdot \rm ln (1/5) \\ | ||
| + | \end{align*} | ||
| + | |||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | t & | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | 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 $\tau_2$? | ||
| + | |||
| + | # | ||
| + | |||
| + | 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> | ||
| + | |||
| + | We see that for the time constant, we now need to use $R=R_3 + R_2$. | ||
| + | |||
| + | \begin{align*} | ||
| + | \tau_2 &= R\cdot C \\ | ||
| + | & | ||
| + | \end{align*} | ||
| + | |||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | \tau_2 &= 5~\rm ms \\ | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | 3.2 Calculate the moment $t_3$ when $u_{R2}$ is smaller than $1/10 \cdot U_2$. | ||
| + | |||
| + | # | ||
| + | |||
| + | 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: | ||
| + | * 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/ | ||
| + | \begin{align*} | ||
| + | {{i_{R2}(t_3)}\over{i_{R2}(t_2)}} & | ||
| + | -{{t_3 - t_2}\over{\tau_2}} | ||
| + | | ||
| + | | ||
| + | \end{align*} | ||
| + | |||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | t_3 &= 14.4~\rm ms \\ | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | 3.3 Draw the course of time of the voltage $u_C(t)$ over the capacitor. | ||
| + | |||
| + | {{drawio> | ||
| + | |||
| + | |||
| + | # | ||
| + | {{drawio> | ||
| + | # | ||
| + | |||
| + | # | ||
| + | |||
| + | {{page> | ||
| + | |||
| + | # | ||
| + | |||
| + | |||
| + | < | ||
| - | On the right you see a simulation containing | + | In the simulation, |
| - the capacitances $C_1$ and $C_2$ are shorted, or | - the capacitances $C_1$ and $C_2$ are shorted, or | ||
| - the capacitors $C_1$ and $C_2$ are connected via resistor $R$. | - the capacitors $C_1$ and $C_2$ are connected via resistor $R$. | ||
| - | On the right side of the simulation there are some additional " | + | On the right side of the simulation, there are some additional " |
| - | In the following, the charging and discharging of a capacitor | + | In the following, the charging and discharging of a capacitor |
| - | ~~PAGEBREAK~~ ~~CLEARFIX~~ Under the electrical structure, the following quantities are shown over time: | + | ~~PAGEBREAK~~ ~~CLEARFIX~~ |
| + | |||
| + | Under the electrical structure, the following quantities are shown over time: | ||
| ^Voltage $u_1(C_1)$ of the first capacitor^Voltage $u_2(C_2)$ of the second capacitor^Stored energy $w_1(C_1)$^Stored energy $w_2(C_2)$^Total energy $\sum w$| | ^Voltage $u_1(C_1)$ of the first capacitor^Voltage $u_2(C_2)$ of the second capacitor^Stored energy $w_1(C_1)$^Stored energy $w_2(C_2)$^Total energy $\sum w$| | ||
| - | |Initially charged to $10V$|Initially neutrally charged ($0V$)|Initially holds: \\ $w_1(C_1)= {1 \over 2} \cdot C \cdot U^2 = {1 \over 2} \cdot 10\mu F \cdot (10V)^2 = 500\mu W$ \\ In the oscilloscope, | + | |Initially charged to $10~{\rm V}$|Initially neutrally charged ($0~{\rm V}$)|Initially holds: \\ $w_1(C_1)= {1 \over 2} \cdot C \cdot U^2 = {1 \over 2} \cdot 10~{\rm µF} \cdot (10~{\rm V})^2 = 500~{\rm µW}$ \\ In the oscilloscope, |
| - | The capacitor $C_1$ has thus initially stored the full energy and via a closing of the switch $S_2$ one would expect a balancing of the voltages and an equal distribution of the energy $w_1 + w_2 = 500\mu W$. | + | The capacitor $C_1$ has thus initially stored the full energy and via closing of the switch, $S_2$ one would expect a balancing of the voltages and an equal distribution of the energy $w_1 + w_2 = 500~\rm µW$. |
| - Close the switch $S_2$ (the toggle switch $S_1$ should point to the switch $S_2$). What do you find? | - Close the switch $S_2$ (the toggle switch $S_1$ should point to the switch $S_2$). What do you find? | ||
| - What do the voltages $u_1$ and $u_2$ do? | - What do the voltages $u_1$ and $u_2$ do? | ||
| - | - What 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? |
| - Open $S_2$ - the changeover switch $S_1$ should not be changed. What do you find? | - Open $S_2$ - the changeover switch $S_1$ should not be changed. What do you find? | ||
| - What do the voltages $u_1$ and $u_2$ do? | - What do the voltages $u_1$ and $u_2$ do? | ||
| - | - What 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? |
| - | - Repeat 1. and 2. several times. Can anything be deduced regarding the distribution of the energy? | + | - Repeat 1. and 2. several times. Can anything be deduced regarding the distribution of energy? |
| - Change the switch $S_2$ to the resistor. What do you observe? | - Change the switch $S_2$ to the resistor. What do you observe? | ||
| - What do the voltages $u_1$ and $u_2$ do? | - What do the voltages $u_1$ and $u_2$ do? | ||
| - | - What 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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