Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
| electrical_engineering_1:introduction_in_alternating_current_technology [2022/12/06 20:29] – [6.2.4 Comparison of the different Averages] mexleadmin | electrical_engineering_1:introduction_in_alternating_current_technology [2024/12/04 14:43] (aktuell) – [Bearbeiten - Panel] mexleadmin | ||
|---|---|---|---|
| Zeile 1: | Zeile 1: | ||
| - | ====== 6. Introduction | + | ====== 6 Introduction |
| - | Up to now we had analysed | + | Up to now, we had analyzed |
| - | - Often the voltage given by the **power plant is AC**. This is true for example in all power plants which use electric generators. In these, | + | - Often the voltage given by the **power plant is AC**. This is true for example in all power plants which use electric generators. In these, |
| - | - For long-range power transfer the power losses $P_{loss}$ can be reduced by reducing the currents $I$ since $P_{loss}=R\cdot I^2$. Therefore, for constant power transfer the voltage | + | - For long-range power transfer the power losses $P_{\rm loss}$ can be reduced by reducing the currents $I$ since $P_{\rm loss}=R\cdot I^2$. Therefore, for constant power transfer, the voltage |
| - AC signals have **at least one more value** which can be used for understanding the situation of the source or load. This simplifies the power and load management in a complex power network. | - AC signals have **at least one more value** which can be used for understanding the situation of the source or load. This simplifies the power and load management in a complex power network. | ||
| This does not mean that DC power lines are useless or only full of disadvantages: | This does not mean that DC power lines are useless or only full of disadvantages: | ||
| - | * A lot of modern loads need DC voltages, like battery based systems (laptops, electric cars, smartphones). Others can simpliy | + | * A lot of modern loads need DC voltages, like battery-based systems (laptops, electric cars, smartphones). Others can simply |
| * Long-range power transfer with DC voltages show often much lower power losses. | * Long-range power transfer with DC voltages show often much lower power losses. | ||
| - | Besides the applications in power systems AC values are also important in communication engineering. Acoustic and visual signals like sound and images can often be considered as wavelike AC signals. Additionally, | + | Besides the applications in power systems AC values are also important in communication engineering. Acoustic and visual signals like sound and images can often be considered as wavelike AC signals. Additionally, |
| - | In order to understand these systems a bit more, we will start in this chapter with a first introduction | + | To understand these systems a bit more, we will start this chapter with a first introduction |
| < | < | ||
| - | If you have trouble | + | If you have trouble |
| * The question " | * The question " | ||
| * How to calculate with complex numbers (sum, difference, product) can be seen in [[https:// | * How to calculate with complex numbers (sum, difference, product) can be seen in [[https:// | ||
| Zeile 31: | Zeile 31: | ||
| - know which types of time-dependent waveforms there are and be able to assign them | - know which types of time-dependent waveforms there are and be able to assign them | ||
| - Know the relationship between amplitude and peak-to-peak value. | - Know the relationship between amplitude and peak-to-peak value. | ||
| - | - Know the relationship between period, frequency and angular frequency. | + | - Know the relationship between period, frequency, and angular frequency. |
| - Know the difference between zero phase angle and phase shift angle. | - Know the difference between zero phase angle and phase shift angle. | ||
| - Know the direction of the phase shift angle. | - Know the direction of the phase shift angle. | ||
| Zeile 43: | Zeile 43: | ||
| As already used in chapter [[dc_circuit_transients|5.]] for the time-dependent values lowercase letters will be written. | As already used in chapter [[dc_circuit_transients|5.]] for the time-dependent values lowercase letters will be written. | ||
| - | By this time-dependent values any temporal form of the voltage / current curves | + | By these time-dependent values, any temporal form of the voltage/ |
| - | * We distinguish periodic and non periodic signals | + | * We distinguish periodic and non-periodic signals |
| - | * One important family of periodic signals | + | * One important family of periodic signals |
| * Sinusoidal signals can be mixed with DC signals | * Sinusoidal signals can be mixed with DC signals | ||
| Zeile 51: | Zeile 51: | ||
| < | < | ||
| </ | </ | ||
| - | \\ {{drawio> | + | \\ {{drawio> |
| </ | </ | ||
| - | In the following we will investigate mainly pure AC signals. | + | In the following, we will investigate mainly pure AC signals. |
| ==== 6.1.2 Descriptive Values of AC Signals ==== | ==== 6.1.2 Descriptive Values of AC Signals ==== | ||
| Zeile 61: | Zeile 61: | ||
| < | < | ||
| </ | </ | ||
| - | \\ {{drawio> | + | \\ {{drawio> |
| </ | </ | ||
| - | There are some important characteristic values when investigating AC signals (<imgref pic02>). For the singal | + | There are some important characteristic values when investigating AC signals (<imgref pic02>). For the signal |
| - | * The **DC voltage** or DC offset is given by the value $U_{DC}$ of $V_{DC}$ (in German: Gleichanteil). The DC component also defines the average value of an AC signal. | + | * The **DC voltage** or DC offset is given by the value $U_{\rm DC}$ of $V_{\rm DC}$ (in German: Gleichanteil). The DC component also defines the average value of an AC signal. |
| - | * The maximum deviation from the DC value is called **peak voltage** $U_p$ (in German : Spitzespannung). Specifically for sinusidal | + | * The maximum deviation from the DC value is called **peak voltage** $U_\rm p$ (in German : //Spitzespannung//). Specifically for sinusoidal |
| - | * The voltage difference between maximum and minimum deviation is called **peak-to-peak voltage** $U_{pp}$ (in German: Spitze-Spitze-Spannung). \\ Be aware, that in English texts often amplitude is also used for (non sinusidal) $U_{pp}$ | + | * The voltage difference between maximum and minimum deviation is called **peak-to-peak voltage** $U_{\rm pp}$ (in German: |
| - | Additionally, | + | Additionally, |
| * The shortest time difference for the signal to repeat is called **period** $T$. | * The shortest time difference for the signal to repeat is called **period** $T$. | ||
| - | * Based on the period $T$ the frequency $f = {{1}\over{T}}$ can be derived. The unit of the frequency is $1 Hz = 1 Hertz$. | + | * Based on the period $T$ the frequency $f = {{1}\over{T}}$ can be derived. The unit of the frequency is $1 ~{\rm Hz} = 1 ~\rm Hertz$. |
| - | * For calculation, | + | * For calculation, |
| - | * Another handy value is the time offset between the start of the sinus wave ($u(t)=0V$ and rising) and $t=0s$. This difference is often written based on an angular difference and is called the **phase | + | * Another handy value is the time offset between the start of the sinus wave ($u(t)=0~\rm V$ and rising) and $t=0 ~\rm s$. This difference is often written based on an angular difference and is called the **phase angle** or **initial phase** $\varphi_U$ (in German: |
| Mathematically, | Mathematically, | ||
| - | $$u(t)=\hat{U}\cdot sin(\omega t + \varphi_U)$$ | + | $$u(t)=\hat{U}\cdot |
| - | $$i(t)=\hat{I}\cdot sin(\omega t + \varphi_I)$$ \\ | + | $$i(t)=\hat{I}\cdot |
| - | Between the AC voltages and currents there is also another important characteristic: | + | Between the AC voltages and currents, there is also another important characteristic: |
| <callout icon=" | <callout icon=" | ||
| < | < | ||
| - | The initial phase $\varphi_0$ has an direction / sign which have to be considered. In the case **a)** in the picture the zero-crossing of the sinusidal | + | The initial phase $\varphi_0$ has a direction/ |
| - | {{drawio> | + | {{drawio> |
| - | Similarly also for the phase difference $\Delta \varphi$ the direction has to be taken into account. In the following image the zero-crossing of the voltage curve is before the zero-crossing of the current. This leads to a positive phase difference $\Delta \varphi$. | + | Similarly also for the phase difference $\Delta \varphi$ the direction has to be taken into account. In the following image, the zero-crossing of the voltage curve is before the zero-crossing of the current. This leads to a positive phase difference $\Delta \varphi$. |
| - | {{drawio> | + | {{drawio> |
| </ | </ | ||
| Zeile 101: | Zeile 101: | ||
| By the end of this section, you will be able to: | By the end of this section, you will be able to: | ||
| - | - calculate the arithmetic mean, the rectified value and the rms value. | + | - calculate the arithmetic mean, the rectified value, and the RMS value. |
| - know these mean values for sinusoidal quantities. | - know these mean values for sinusoidal quantities. | ||
| - | - know the reason for using the rms value. | + | - know the reason for using the RMS value. |
| </ | </ | ||
| - | In order to analyse | + | To analyze |
| - the arithmetic mean $\overline{X}$ | - the arithmetic mean $\overline{X}$ | ||
| - | - the recified | + | - the rectified |
| - | - the rms value $X$ | + | - the RMS value $X$ |
| These shall be discussed in the following. The video " | These shall be discussed in the following. The video " | ||
| Zeile 123: | Zeile 123: | ||
| $$\overline{X}={{1}\over{n}}\cdot \sum_{i=1}^n x_i$$ | $$\overline{X}={{1}\over{n}}\cdot \sum_{i=1}^n x_i$$ | ||
| - | For functions it is given by: | + | For functions, it is given by: |
| - | $$\boxed{\overline{X}={{1}\over{T}}\cdot \int_{t=t_0}^{t_0 + T} x(t) dt}$$ | + | $$\boxed{\overline{X}={{1}\over{T}}\cdot \int_{t=t_0}^{t_0 + T} x(t) {\rm d}t}$$ |
| - | For pure AC signals, the arithmetic mean $\overline{X}=0$, | + | For pure AC signals, the arithmetic mean is $\overline{X}=0$, |
| ==== 6.2.2 The Rectified Value ==== | ==== 6.2.2 The Rectified Value ==== | ||
| - | Since the arithmetic mean of pure AC signals with $\overline{X}=0$ does not really give an insight into the signal, different other (weighted) | + | Since the arithmetic mean of pure AC signals with $\overline{X}=0$ does not really give an insight into the signal, different other (weighted) |
| - | One of them is the rectified value. For this the signal is first recified | + | One of them is the rectified value. For this, the signal is first rectified |
| - | For finite values the rectified value is given by: | + | For finite values, the rectified value is given by: |
| $$\overline{|X|}={{1}\over{n}}\cdot \sum_{i=1}^n |x_i|$$ | $$\overline{|X|}={{1}\over{n}}\cdot \sum_{i=1}^n |x_i|$$ | ||
| - | For functions it is given by: | + | For functions, it is given by: |
| - | $$\boxed{\overline{|X|}={{1}\over{T}}\cdot \int_{t=t_0}^{t_0 + T} |x(t)| | + | $$\boxed{\overline{|X|}={{1}\over{T}}\cdot \int_{t=t_0}^{t_0 + T} |x(t)| |
| < | < | ||
| Zeile 142: | Zeile 142: | ||
| \begin{align*} | \begin{align*} | ||
| - | \overline{|X|} &= {{1}\over{T}}\cdot \int_{t=t_0}^{t_0 + T} |\hat{X}\cdot sin(\omega t + \varphi) | + | \overline{|X|} &= {{1}\over{T}}\cdot \int_{t=t_0}^{t_0 + T} |\hat{X}\cdot |
| \end{align*} | \end{align*} | ||
| Without limiting the generality, we use $\varphi=0$ and $t_0 = 0$ | Without limiting the generality, we use $\varphi=0$ and $t_0 = 0$ | ||
| \begin{align*} | \begin{align*} | ||
| - | \overline{|X|} &= {{1}\over{T}}\cdot \int_{t=0 | + | \overline{|X|} &= {{1}\over{T}}\cdot \int_{t=0 |
| \end{align*} | \end{align*} | ||
| - | Since $sin(\omega t)\geq0$ for $t\in [0,\pi]$, the integral can changed and the absolute value bars can be excluded like the following | + | Since $sin(\omega t)\geq0$ for $t\in [0,\pi]$, the integral can be changed and the absolute value bars can be excluded like the following |
| \begin{align*} | \begin{align*} | ||
| - | \overline{|X|} | + | \overline{|X|} |
| - | | + | &= 2 \cdot {{1}\over{T}}\cdot [-\hat{X}\cdot {{T}\over{2\pi}}\cdot |
| - | | + | &= 2 \cdot {{1}\over{T}}\cdot {{T}\over{2\pi}}\cdot |
| - | | + | &= {{1}\over{\pi}}\cdot \hat{X} \cdot [1+1] \\ |
| - | \boxed{\overline{|X|} = {{2}\over{\pi}}\cdot \hat{X} \approx 0.6366\cdot \hat{X}}\\ | + | \boxed{\overline{|X|} |
| + | = {{2}\over{\pi}}\cdot \hat{X} \approx 0.6366 \cdot \hat{X}}\\ | ||
| \end{align*} | \end{align*} | ||
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| - | Calculate the rectified value of rectangular and triangular signals! Use similar symmetry | + | Calculate the rectified value of rectangular and triangular signals! Use similar symmetry |
| </ | </ | ||
| Zeile 168: | Zeile 169: | ||
| ==== 6.2.3 The RMS Value ==== | ==== 6.2.3 The RMS Value ==== | ||
| - | Often it is important be able to compare AC signals to DC signals by having equivalent values. But what does equivalent mean? \\ | + | Often it is important |
| - | Most importantly, | + | Most importantly, |
| - | How do we come to this values? | + | How do we come to these values? |
| We want to find the voltage $U_{DC}$ and $I_{DC}$ of a DC source, that the output power $P_{DC}$ on a resistor $R$ is similar to the output power $P_{AC}$ of an AC source with the instantaneous values $u(t)$ and $i(t)$. For this, we have to consider the instantaneous power $p(t)$ for a distinct time $t$ and integrate this over one period $T$. | We want to find the voltage $U_{DC}$ and $I_{DC}$ of a DC source, that the output power $P_{DC}$ on a resistor $R$ is similar to the output power $P_{AC}$ of an AC source with the instantaneous values $u(t)$ and $i(t)$. For this, we have to consider the instantaneous power $p(t)$ for a distinct time $t$ and integrate this over one period $T$. | ||
| \begin{align*} | \begin{align*} | ||
| - | | + | P_{\rm DC} |
| - | U_{DC} \cdot I_{DC} &= {{1}\over{T}} \int_{0}^{T} u(t) \cdot i(t) dt \\ | + | U_{DC} \cdot I_{\rm DC} |
| - | | + | |
| - | | + | |
| - | \rightarrow I_{DC} &= \sqrt{{{1}\over{T}} \int_{0}^{T} i^2(t) | + | \rightarrow |
| \end{align*} | \end{align*} | ||
| - | The similar approach can be used on instantaneous voltage $u(t)$. Generally, the RMS value of $X$ is given by | + | A similar approach can be used on instantaneous voltage $u(t)$. Generally, the RMS value of $X$ is given by |
| \begin{align*} | \begin{align*} | ||
| - | \boxed{X_{RMS} = \sqrt{{{1}\over{T}} \int_{0}^{T} x^2(t) | + | \boxed{X_{\rm RMS} = \sqrt{{{1}\over{T}} \int_{0}^{T} x^2(t) |
| \end{align*} | \end{align*} | ||
| What is the meaning of RMS? Simple: | What is the meaning of RMS? Simple: | ||
| - | {{drawio> | + | {{drawio> |
| - | By this abbreviation, | + | By this abbreviation, |
| <callout icon=" | <callout icon=" | ||
| - | * The heat dissipation on an resistor $R$ of an AC current with the rms value of $I_{rms}=1A$ is equal to the heat dissipation of an DC current with $I_{DC}=1A$. | + | * The heat dissipation on a resistor $R$ of an AC current with the RMS value of $I_{\rm RMS} = 1~\rm A$ is equal to the heat dissipation of a DC current with $I_{\rm DC} = 1~\rm A$. |
| - | * To shorten writing formulas, the values of AC signals given with uppercase letters will represent the RMS value in the following: $U = U_{RMS}$, $I = I_{RMS}$. | + | * To shorten writing formulas, the values of AC signals given with uppercase letters will represent the RMS value in the following: $U = U_{\rm RMS}$, $I = I_{\rm RMS}$. |
| - | * It holds for AC signals | + | * It holds for AC signals |
| * The resistance is $R={{U}\over{I}}$ | * The resistance is $R={{U}\over{I}}$ | ||
| * The power dissipation on a resistor is $P=U\cdot I$ | * The power dissipation on a resistor is $P=U\cdot I$ | ||
| Zeile 203: | Zeile 204: | ||
| \begin{align*} | \begin{align*} | ||
| - | X & | + | X & |
| - | & | + | & |
| - | & | + | & |
| - | & | + | & |
| & | & | ||
| & | & | ||
| Zeile 217: | Zeile 218: | ||
| <callout icon=" | <callout icon=" | ||
| In the following chapters, we will often use for a physical value $x(t)$ a dependency on $\sqrt{2}X$ instead of $\hat{X}$. | In the following chapters, we will often use for a physical value $x(t)$ a dependency on $\sqrt{2}X$ instead of $\hat{X}$. | ||
| - | Therefore, the sinusidal | + | Therefore, the sinusoidal |
| </ | </ | ||
| Zeile 223: | Zeile 224: | ||
| <panel type=" | <panel type=" | ||
| - | Calculate the RMS value of rectangular and triangular signals! Use similar symmetry | + | Calculate the RMS value of rectangular and triangular signals! Use similar symmetry |
| + | Compare it to the values shown in <imgref imageNo5> | ||
| </ | </ | ||
| Zeile 229: | Zeile 231: | ||
| ==== 6.2.4 Comparison of the different Averages ==== | ==== 6.2.4 Comparison of the different Averages ==== | ||
| - | The following simulation shows the different values for averaging a rectangular, | + | The following simulation shows the different values for averaging a rectangular, |
| - | Be aware that one has to wait for ont full period | + | Be aware that one has to wait for a full period to see the resulting values on the right outputs of the average generating blocks. |
| < | < | ||
| Zeile 253: | Zeile 255: | ||
| In the chapters [[simple_circuits|2. Simple Circuits]] and [[non-ideal_sources_and_two_terminal_networks|3 Non-ideal Sources and Two-terminal Networks]] we already have seen, that it is possible to reduce complex circuitries down to equivalent resistors (and ideal sources). This we will try to adopt for AC components, too. | In the chapters [[simple_circuits|2. Simple Circuits]] and [[non-ideal_sources_and_two_terminal_networks|3 Non-ideal Sources and Two-terminal Networks]] we already have seen, that it is possible to reduce complex circuitries down to equivalent resistors (and ideal sources). This we will try to adopt for AC components, too. | ||
| - | We want to analyze how relationship between the current through a component and the voltage | + | We want to analyze how the relationship between the current through a component and the voltage |
| ==== 6.3.1 Resistance ==== | ==== 6.3.1 Resistance ==== | ||
| - | We start with Ohms law, which states, that the instantaneous voltage $u(t)$ is proportional to the instantaneous current $i(t)$ by the factor $R$. | + | We start with Ohm' |
| $$u(t) = R \cdot i(t)$$ | $$u(t) = R \cdot i(t)$$ | ||
| - | Then we insert the functions representing the instantaneous signals: $x(t)= \sqrt{2}{X}\cdot sin(\omega t + \varphi_x)$: | + | Then we insert the functions representing the instantaneous signals: $x(t)= \sqrt{2}{X}\cdot |
| - | $$\sqrt{2}{U}\cdot sin(\omega t + \varphi_u) = R \cdot \sqrt{2}{I}\cdot sin(\omega t + \varphi_i)$$ | + | $$\sqrt{2}{U}\cdot |
| Since we know, that $u(t)$ must be proportional to $i(t)$ we conclude that for a resistor $\varphi_u=\varphi_i$! | Since we know, that $u(t)$ must be proportional to $i(t)$ we conclude that for a resistor $\varphi_u=\varphi_i$! | ||
| \begin{align*} | \begin{align*} | ||
| - | R &= {{\sqrt{2}{U}\cdot sin(\omega t + \varphi_i)}\over{\sqrt{2}{I}\cdot sin(\omega t + \varphi_i) }} \\ | + | R &= {{\sqrt{2}{U}\cdot |
| &= {{U}\over{I}} | &= {{U}\over{I}} | ||
| \end{align*} | \end{align*} | ||
| Zeile 273: | Zeile 275: | ||
| < | < | ||
| </ | </ | ||
| - | \\ {{drawio> | + | \\ {{drawio> |
| </ | </ | ||
| - | This was not too hard und quite obvious. But, what about the other types of passive two-terminal networks - namely the capacitance and inductance? | + | This was not too hard and quite obvious. But, what about the other types of passive two-terminal networks - namely the capacitance and inductance? |
| ==== 6.3.2 Capacitance ==== | ==== 6.3.2 Capacitance ==== | ||
| - | For the capacitance we have the basis formula: | + | For the capacitance we have the basic formula: |
| $$C={{Q}\over{U}}$$ | $$C={{Q}\over{U}}$$ | ||
| - | This formula | + | This formula |
| $$C={{q(t)}\over{u(t)}}$$ | $$C={{q(t)}\over{u(t)}}$$ | ||
| - | Additionally we know, that the instantaneous current is defined by $i(t)={{dq(t)}\over{dt}}$. | + | Additionally, we know, that the instantaneous current is defined by $i(t)={{{\rm d}q(t)}\over{{\rm d}t}}$. |
| By this we can set up the formula: | By this we can set up the formula: | ||
| \begin{align*} | \begin{align*} | ||
| - | i(t) &= {{dq(t)}\over{dt}} \\ | + | i(t) &= {{{\rm d}q(t)}\over{{\rm d}t}} \\ |
| - | & | + | & |
| \end{align*} | \end{align*} | ||
| Now, we insert the functions representing the instantaneous signals and calculate the derivative: | Now, we insert the functions representing the instantaneous signals and calculate the derivative: | ||
| \begin{align*} | \begin{align*} | ||
| - | | + | |
| - | & | + | |
| - | {I}\cdot sin(\omega t + \varphi_i) & | + | {I}\cdot |
| \end{align*} | \end{align*} | ||
| Zeile 308: | Zeile 310: | ||
| \omega t + \varphi_i &= \omega t + \varphi_u + {{1}\over{2}}\pi \\ | \omega t + \varphi_i &= \omega t + \varphi_u + {{1}\over{2}}\pi \\ | ||
| | | ||
| - | \varphi_u -\varphi_i & | + | \varphi_u -\varphi_i & |
| \end{align*} | \end{align*} | ||
| Zeile 315: | Zeile 317: | ||
| <callout icon=" | <callout icon=" | ||
| - | In order not to mix up the definitions, | + | In order not to mix up the definitions, |
| The impedance is generally defined as | The impedance is generally defined as | ||
| $$Z = {{U}\over{I}}$$ | $$Z = {{U}\over{I}}$$ | ||
| - | Only for pure resistor as a two-terminal network the impedance $Z_R$ is equal to the value of the resistance: $Z_R=R$. | + | Only for a pure resistor as a two-terminal network, the impedance $Z_R$ is equal to the value of the resistance: $Z_R=R$. |
| - | For the pure capacitive as a two-terminal network the impedance $Z_C$ is $Z_C={{1}\over{\omega \cdot C}}$. | + | For the pure capacitive as a two-terminal network, the impedance $Z_C$ is $Z_C={{1}\over{\omega \cdot C}}$. |
| </ | </ | ||
| Zeile 327: | Zeile 329: | ||
| < | < | ||
| </ | </ | ||
| - | \\ {{drawio> | + | \\ {{drawio> |
| </ | </ | ||
| Zeile 333: | Zeile 335: | ||
| < | < | ||
| </ | </ | ||
| - | {{url> | + | {{url> |
| </ | </ | ||
| Zeile 339: | Zeile 341: | ||
| The inductance will here be introduced shortly - the detailed introduction is part of [[electrical_engineering_2: | The inductance will here be introduced shortly - the detailed introduction is part of [[electrical_engineering_2: | ||
| - | For the capacitance $C$ we had the situation, that it reacts to a voltage change ${{d}\over{dt}}u(t)$ with a counteracting current: | + | For the capacitance $C$ we had the situation, that it reacts to a voltage change ${{\rm d}\over{{\rm d}t}}u(t)$ with a counteracting current: |
| - | $$i(t)= C \cdot {{d}\over{dt}}u(t)$$ | + | $$i(t)= C \cdot {{\rm d}\over{{\rm d}t}}u(t)$$ |
| - | This is due to the fact, that the capacity stores charge carriers $q$. It appears that "the capacitance does not like voltage changes and reacts with a compensating current" | + | This is due to the fact, that the capacity stores charge carriers $q$. |
| + | It appears that "the capacitance does not like voltage changes and reacts with a compensating current" | ||
| + | When the voltage on a capacity drops, the capacity supplies a current - when the voltage rises the capacity drains a current. | ||
| - | For an inductance $L$ it is just the other way around: "the inductance does not like current changes and reacts with a compensating voltage drop". Once the current changes the inductance will create a voltage drop that counteracts and continues the current: A current change ${{d}\over{dt}}i(t)$ leads to a voltage drop $u(t)$: | + | For an inductance $L$ it is just the other way around: "the inductance does not like current changes and reacts with a compensating voltage drop". Once the current changes the inductance will create a voltage drop that counteracts and continues the current: A current change ${{\rm d}\over{{\rm d}t}}i(t)$ leads to a voltage drop $u(t)$: |
| - | $$u(t)= L \cdot {{d}\over{dt}}i(t)$$ | + | $$u(t)= L \cdot {{\rm d}\over{{\rm d}t}}i(t)$$ |
| - | The proportionality factor here is $L$, the value of the inductance, and it is measured in $[L] = 1H = 1\; Henry$. | + | The proportionality factor here is $L$, the value of the inductance, and it is measured in $[L] = 1~\rm H = 1~Henry$. |
| We can now again insert the functions representing the instantaneous signals and calculate the derivative: | We can now again insert the functions representing the instantaneous signals and calculate the derivative: | ||
| \begin{align*} | \begin{align*} | ||
| - | | + | |
| - | &= L \cdot \sqrt{2}{I}\cdot \omega \cdot cos(\omega t + \varphi_i) \\ \\ | + | |
| - | | + | |
| \end{align*} | \end{align*} | ||
| Zeile 357: | Zeile 361: | ||
| \begin{align*} | \begin{align*} | ||
| U & | U & | ||
| - | \boxed{Z_L = {{U}\over{I}} = \omega \cdot L} | + | \boxed {Z_L = {{U}\over{I}} = \omega \cdot L} |
| \end{align*} | \end{align*} | ||
| and: | and: | ||
| Zeile 363: | Zeile 367: | ||
| \omega t + \varphi_u &= \omega t + \varphi_i + {{1}\over{2}}\pi \\ | \omega t + \varphi_u &= \omega t + \varphi_i + {{1}\over{2}}\pi \\ | ||
| | | ||
| - | \boxed{\varphi = \varphi_u -\varphi_i = + {{1}\over{2}}\pi } | + | \boxed{\varphi = \varphi_u -\varphi_i = + {{1}\over{2}}\pi } |
| \end{align*} | \end{align*} | ||
| Zeile 369: | Zeile 373: | ||
| < | < | ||
| - | < | + | < |
| </ | </ | ||
| - | \\ {{drawio> | + | \\ {{drawio> |
| </ | </ | ||
| < | < | ||
| - | < | + | < |
| </ | </ | ||
| {{url> | {{url> | ||
| Zeile 391: | Zeile 395: | ||
| </ | </ | ||
| - | One way to memorize the phase shift is given bei the word **CIVIL**: | + | One way to memorize the phase shift is given by the word **CIVIL**: |
| * **<fc # | * **<fc # | ||
| * **CI<fc # | * **CI<fc # | ||
| </ | </ | ||
| - | For the concept of AC two-terminal networks, we are also able to use the DC methods of network analysis | + | For the concept of AC two-terminal networks, we are also able to use the DC methods of network analysis to solve AC networks. |
| - | ===== 6.4 Complex Values in Electical | + | ===== 6.4 Complex Values in Electrical |
| < | < | ||
| Zeile 423: | Zeile 427: | ||
| ==== 6.4.1 Representation and Interpretation | ==== 6.4.1 Representation and Interpretation | ||
| - | Up to now, we used for the AC signals the formula $x(t)= \sqrt{2} X \cdot sin (\omega t + \varphi_x)$ - which was quite obvious. \\ | + | Up to now, we used for the AC signals the formula $x(t)= \sqrt{2} X \cdot \sin (\omega t + \varphi_x)$ - which was quite obvious. \\ |
| - | However, there is an alternative way to look onto the alternating | + | However, there is an alternative way to look at the alternating |
| - | For this, we look first on a different, but already familiar problem (see <imgref pic06>). | + | For this, we look first at a different, but already |
| - | - A mechanical, linear spring with the characteristic constant $D$ is displaced due a mass $m$ in the Earth' | + | - A mechanical, linear spring with the characteristic constant $D$ is displaced due to a mass $m$ in the Earth' |
| - At the time $t_0=0$ , we deflect this spring a bit more to $X_0 + x(t_0)=X_0 + \hat{X}$ and therefore induce energy into the system. | - At the time $t_0=0$ , we deflect this spring a bit more to $X_0 + x(t_0)=X_0 + \hat{X}$ and therefore induce energy into the system. | ||
| - | - When the mass is released, the mass will spring up and down for $t>0$. The signal can be shown as shadow when the mass is illuminated sideways. \\ For $t>0$, the energy is continuously shifted between | + | - When the mass is released, the mass will spring up and down for $t>0$. The signal can be shown as a shadow when the mass is illuminated sideways. \\ For $t>0$, the energy is continuously shifted between |
| - | - When looking onto the course of time of $x(t)$, the signal will behave as: $x(t)= \hat{X} \cdot sin (\omega t + \varphi_x)$ | + | - When looking onto the course of time of $x(t)$, the signal will behave as: $x(t)= \hat{X} \cdot \sin (\omega t + \varphi_x)$ |
| - | - The movement of the shadow can also be created by a the sideways shadow of a stick on a rotating disc. \\ This means, that a two dimensional rotation is reduced down to a single dimension. | + | - The movement of the shadow can also be created by the sideways shadow of a stick on a rotating disc. \\ This means, that a two-dimensional rotation is reduced down to a single dimension. |
| < | < | ||
| - | < | + | < |
| </ | </ | ||
| - | \\ {{drawio> | + | \\ {{drawio> |
| </ | </ | ||
| - | The transformation of the two dimensional rotation to a one dimensional | + | The transformation of the two-dimensional rotation to a one-dimensional |
| < | < | ||
| - | < | + | < |
| </ | </ | ||
| < | < | ||
| {{url> | {{url> | ||
| - | Klick on the box " | + | Click on the box " |
| </ | </ | ||
| Zeile 455: | Zeile 459: | ||
| - | The two dimensional rotation can be represented with a complex number in Euler' | + | The two-dimensional rotation can be represented with a complex number in Euler' |
| - | $$ \underline{x}(t)=\hat{X}\cdot e^{j(\omega t + \varphi_x)} = \Re(\underline{x}) + j\cdot \Im(\underline{x})$$ | + | It combines the exponential representation with real part $\Re$ and imaginary part $\Im$ of a complex value: |
| + | $$ \underline{x}(t)=\hat{X}\cdot | ||
| - | For the imaginary unit $i$ the letter $j$ is used in electical | + | For the imaginary unit ${\rm i}$ the letter ${\rm j}$ is used in electrical |
| < | < | ||
| - | < | + | < |
| </ | </ | ||
| - | \\ {{drawio> | + | \\ {{drawio> |
| </ | </ | ||
| ==== 6.4.2 Complex Current and Voltage ==== | ==== 6.4.2 Complex Current and Voltage ==== | ||
| - | The concepts of complex numbers shall now be applied to voltages and currents. Up to now we used the following formula to represent alternating voltages: | + | The concepts of complex numbers shall now be applied to voltages and currents. |
| + | Up to now, we used the following formula to represent alternating voltages: | ||
| - | $$u(t)= \sqrt{2} | + | $$u(t)= \sqrt{2} U \cdot \sin (\varphi)$$ |
| - | This is now interpreted as the instantaneuos | + | This is now interpreted as the instantaneous |
| < | < | ||
| - | < | + | < |
| </ | </ | ||
| - | \\ {{drawio> | + | \\ {{drawio> |
| </ | </ | ||
| The parts on the complex plane are then given by: | The parts on the complex plane are then given by: | ||
| - | - The real part $\Re{(\underline{u}(t))} = \sqrt{2}U \cdot cos (\omega t + \varphi_u)$ | + | - The real part $\Re{(\underline{u}(t))} = \sqrt{2}U \cdot \cos (\omega t + \varphi_u)$ |
| - | - The imaginary part $\Im{(\underline{u}(t))} = \sqrt{2}U \cdot sin (\omega t + \varphi_u)$ | + | - The imaginary part $\Im{(\underline{u}(t))} = \sqrt{2}U \cdot \sin (\omega t + \varphi_u)$ |
| - | This is equivalent to the complex phasor $\underline{u}(t)=\sqrt{2}U \cdot e ^{j (\omega t + \varphi_u)}$ | + | This is equivalent to the complex phasor $\underline{u}(t)=\sqrt{2}U \cdot {\rm e} ^{{\rm j} |
| The complex phasor can be separated: | The complex phasor can be separated: | ||
| \begin{align*} | \begin{align*} | ||
| - | \underline{u}(t) & | + | \underline{u}(t) & |
| - | & | + | |
| - | & | + | \cdot {\rm e}^{{\rm j} \omega t} \\ |
| + | | ||
| \end{align*} | \end{align*} | ||
| - | The **fixed phasor** (in German: komplexer Festzeiger) of the voltage is given by $\color{blue}{\underline{U}}= \color{blue}{U | + | The **fixed phasor** (in German: |
| Generally, from now on not only the voltage will be considered as a phasor, but also the current $\underline{I}$ and derived quantities like the impedance $\underline{X}$. \\ | Generally, from now on not only the voltage will be considered as a phasor, but also the current $\underline{I}$ and derived quantities like the impedance $\underline{X}$. \\ | ||
| - | Therefore, the from Mathematics 101 known properties of complex numbers can be applied: | + | Therefore, the known properties of complex numbers |
| * A multiplication with $j$ equals a phase shift of $+90°$ | * A multiplication with $j$ equals a phase shift of $+90°$ | ||
| - | * A multiplication with $-j$ equals a phase shift of $-90°$ | + | * A multiplication with ${{1}\over{j}}$ equals a phase shift of $-90°$ |
| ===== 6.5 Complex Impedance ===== | ===== 6.5 Complex Impedance ===== | ||
| Zeile 507: | Zeile 514: | ||
| By the end of this section, you will be able to: | By the end of this section, you will be able to: | ||
| - draw and read pointer diagrams. | - draw and read pointer diagrams. | ||
| - | - know and apply the complex value formulas of impedance, reactance, resistance. | + | - know and apply the complex value formulas of impedance, reactance, |
| </ | </ | ||
| - | ==== 6.5.1 Introduction | + | ==== 6.5.1 Introduction |
| The complex impedance is " | The complex impedance is " | ||
| Zeile 517: | Zeile 524: | ||
| \begin{align*} | \begin{align*} | ||
| \underline{Z}& | \underline{Z}& | ||
| - | &= \Re{(\underline{Z})} + j \cdot \Im{(\underline{Z})} \\ | + | &= \Re{(\underline{Z})} + {\rm j} \cdot \Im{(\underline{Z})} \\ |
| - | & | + | |
| - | & | + | |
| - | & | + | |
| \end{align*} | \end{align*} | ||
| With | With | ||
| - | * the resistance $R$ (in German: Widerstand) as the pure real part | + | * the resistance $R$ (in German: |
| - | * the reactance $X$ (in German: Blindwiderstand) as the pure imaginary part | + | * the reactance |
| - | * the impedance $Z$ (in German: Scheinwiderstand) as the complex number given by the __complex__ addition of resistance and the reactance as a complex number | + | * the impedance |
| - | The impedance can be transformed from cartesian | + | The impedance can be transformed from Cartesian |
| * $Z=\sqrt{R^2 + X^2}$ | * $Z=\sqrt{R^2 + X^2}$ | ||
| - | * $\varphi = arctan | + | * $\varphi = \arctan |
| The other way around it is possible to transform by: | The other way around it is possible to transform by: | ||
| - | * $R = Z cos \varphi$ | + | * $R = Z \cos \varphi$ |
| - | * $X = Z sin \varphi$ | + | * $X = Z \sin \varphi$ |
| - | ==== 6.5.2 Application on pure Loads ==== | + | y ==== 6.5.2 Application on pure Loads ==== |
| With the complex impedance in mind, the <tabref tab01> can be expanded to: | With the complex impedance in mind, the <tabref tab01> can be expanded to: | ||
| Zeile 542: | Zeile 549: | ||
| ^ Load $\phantom{U\over I}$ ^ ^ integral representation $\phantom{U\over I}$ ^ complex impedance $\underline{Z}={{\underline{U}}\over{\underline{I}}}$ | ^ Load $\phantom{U\over I}$ ^ ^ integral representation $\phantom{U\over I}$ ^ complex impedance $\underline{Z}={{\underline{U}}\over{\underline{I}}}$ | ||
| - | | Resistance | + | | Resistance |
| - | | Capacitance | + | | Capacitance |
| - | | Inductance | + | | Inductance |
| </ | </ | ||
| \\ \\ | \\ \\ | ||
| - | The relationship between $j$ and integral calculus should be clear: | + | The relationship between ${\rm j}$ and integral calculus should be clear: |
| - | - The derivative of a sinusidal | + | - The derivative of a sinusoidal |
| - | - The integral of a sinusidal | + | - The integral of a sinusoidal |
| + | \begin{align*} | ||
| + | \int {\rm e}^{{\rm j}(\omega t + \varphi_x)} | ||
| + | | ||
| + | | ||
| + | \end{align*} | ||
| + | </ | ||
| Once a fixed input voltage is given, the voltage phasor $\underline{U}$, | Once a fixed input voltage is given, the voltage phasor $\underline{U}$, | ||
| Zeile 557: | Zeile 570: | ||
| < | < | ||
| </ | </ | ||
| - | \\ {{drawio> | + | \\ {{drawio> |
| </ | </ | ||
| Zeile 563: | Zeile 576: | ||
| === Simple Networks === | === Simple Networks === | ||
| - | In the capter | + | In the chapter |
| - | These formulas not only apply for ohmic resistors but also for impedances: | + | These formulas not only apply to ohmic resistors but also to impedances: |
| < | < | ||
| < | < | ||
| </ | </ | ||
| - | \\ {{drawio> | + | \\ {{drawio> |
| </ | </ | ||
| - | Similarly, the voltage divider, the current divider, the star-delta-transformation, | + | Similarly, the voltage divider, the current divider, the star-delta transformation, |
| - | This means for example, every linear source can be represented by an output impedance $\underline{Z}_o$ and a ideal voltage source $\underline{U}$. | + | This means for example, every linear source can be represented by an output impedance $\underline{Z}_o$ and an ideal voltage source $\underline{U}$. |
| === More " | === More " | ||
| - | For more complex problems having AC values in circuitries the following approach is beneficial. \\ | + | For more complex problems having AC values in circuitries, the following approach is beneficial. \\ |
| This concept will be used in the next chapter and in circuit design. | This concept will be used in the next chapter and in circuit design. | ||
| Zeile 583: | Zeile 596: | ||
| < | < | ||
| </ | </ | ||
| - | \\ {{drawio> | + | \\ {{drawio> |
| </ | </ | ||
| \\ | \\ | ||
| Zeile 593: | Zeile 606: | ||
| - the absolute value (e.g. the absolute value of the impedance) and the phase | - the absolute value (e.g. the absolute value of the impedance) and the phase | ||
| - | Therefore, instead of the form $\underline{Z}=Z\cdot e^{j\varphi}$ for the phasors often the form $Z\angle{\varphi}$ is used. | + | Therefore, instead of the form $\underline{Z}=Z\cdot |
| </ | </ | ||
| </ | </ | ||
| Zeile 601: | Zeile 614: | ||
| <panel type=" | <panel type=" | ||
| - | A coil has a reactance | + | A coil has a impedance |
| - | - $85 \Omega$ | + | - $85 ~\Omega$ |
| - | - $120 \Omega$ | + | - $120 ~\Omega$ |
| - | - $44 \Omega$ | + | - $44 ~\Omega$ |
| + | |||
| + | <button size=" | ||
| + | When the frequency changes the reactance changes but the inductance is constant. Therefore, the inductance is needed. \\ | ||
| + | It can be calculated by the given reactance for $f_0 = 500 ~\rm Hz$. | ||
| + | \begin{align*} | ||
| + | X_{L0}& | ||
| + | L & | ||
| + | \end{align*} | ||
| + | |||
| + | On the other hand, one can also use the rule of proportion here, and circumvent the calculation of inductance.\\ | ||
| + | It is possible to calculate the reactance at other frequencies with the given reactance. | ||
| + | \begin{align*} | ||
| + | X_L& | ||
| + | f & | ||
| + | & | ||
| + | \end{align*} | ||
| + | |||
| + | With the values given: | ||
| + | \begin{equation*} | ||
| + | f_1 = \frac{85 ~\Omega}{80~\Omega}\cdot500~{\rm Hz}\qquad | ||
| + | f_2 = \frac{120~\Omega}{80~\Omega}\cdot500~{\rm Hz}\qquad | ||
| + | f_3 = \frac{44 ~\Omega}{80~\Omega}\cdot500~{\rm Hz} | ||
| + | \end{equation*} | ||
| + | |||
| + | </ | ||
| + | \begin{equation*} | ||
| + | f_1=531.25~{\rm Hz}\qquad f_2=750~{\rm Hz}\qquad f_3=275~{\rm Hz} | ||
| + | \end{equation*} | ||
| + | </ | ||
| </ | </ | ||
| <panel type=" | <panel type=" | ||
| - | A capacitor with $5 \mu F$ is connected to a voltage source which generates $U_\sim = 200 V$. At which frequencies the following | + | A capacitor with $5 ~{\rm µF}$ is connected to a voltage source which generates $U_\sim = 200 ~{\rm V}$. At which frequencies the following |
| - | - $0.5 A$ | + | - $0.5 ~\rm A$ |
| - | - $0.8 A$ | + | - $0.8 ~\rm A$ |
| - | - $1.3 A$ | + | - $1.3 ~\rm A$ |
| </ | </ | ||
| <panel type=" | <panel type=" | ||
| - | A capacitor shall have a capacity of $4.7 \mu F \pm 10\%$. This capacitor shall be used with a AC voltage of $400V$ and $50Hz$. | + | A capacitor shall have a capacity of $4.7 ~{\rm µF} \pm 10~\%$. This capacitor shall be used with an AC voltage of $400~\rm V$ and $50~\rm Hz$. |
| What is the possible current range which could be found on this component? | What is the possible current range which could be found on this component? | ||
| </ | </ | ||
| <panel type=" | <panel type=" | ||
| - | Two ideal AC voltage sources $1$ and $2$ shall generate the RMS voltage drops $U_1 = 100V$ and $U_2 = 120V$. The phase shift between the two sources shall be $+60°$. The phase of source $1$ shall be $\varphi_1=0°$. | + | Two ideal AC voltage sources $1$ and $2$ shall generate the RMS voltage drops $U_1 = 100~\rm V$ and $U_2 = 120~\rm V$. \\ |
| + | The phase shift between the two sources shall be $+60°$. The phase of source $1$ shall be $\varphi_1=0°$. | ||
| The two sources shall be located in series. | The two sources shall be located in series. | ||
| - | - Draw the phasor diagram for the two voltage phasors and the resulting phasor. | + | <WRAP indent> 1. Draw the phasor diagram for the two voltage phasors and the resulting phasor. |
| - | | + | |
| - | - Is the resulting voltage the RMS value or the amplitude? \\ < | + | <WRAP indent>< |
| - | - Draw the phasor diagram for the two voltage phasors and the resulting phasor for the new circuit. | + | The phasor diagram looks roughly like this: |
| - | | + | {{drawio> |
| + | |||
| + | </ | ||
| + | <WRAP indent>< | ||
| + | By the law of cosine, we get: | ||
| + | \begin{align*} | ||
| + | U&= \sqrt{{{U_1 | ||
| + | & | ||
| + | \end{align*} | ||
| + | |||
| + | The angle is by the tangent of the relation of the imaginary part to the real part of the resulting voltage. | ||
| + | \begin{align*} | ||
| + | \varphi& | ||
| + | & | ||
| + | & | ||
| + | & | ||
| + | \end{align*} | ||
| + | |||
| + | </ | ||
| + | \begin{align*} | ||
| + | U & | ||
| + | \varphi& | ||
| + | \end{align*} | ||
| + | |||
| + | </ | ||
| + | |||
| + | <WRAP indent>< | ||
| + | The resulting voltage is the RMS value. | ||
| + | |||
| + | </ | ||
| + | |||
| + | < | ||
| + | <WRAP indent>< | ||
| + | The phasor diagram looks roughly like this. \\ | ||
| + | But have a look at the solution for question 5! | ||
| + | {{drawio> | ||
| + | |||
| + | </ | ||
| + | |||
| + | <WRAP indent>< | ||
| + | By the law of cosine, we get: | ||
| + | \begin{align*} | ||
| + | U&= \sqrt{{{U_1 | ||
| + | & | ||
| + | \end{align*} | ||
| + | The angle is by the tangent of the relation of the imaginary part to the real part of the resulting voltage. | ||
| + | \begin{align*} | ||
| + | \varphi& | ||
| + | & | ||
| + | & | ||
| + | & | ||
| + | & | ||
| + | \end{align*} | ||
| + | The calculated (positive) horizontal and (negative) vertical dimension for the voltage indicates a phasor in the fourth quadrant. Does it seem right? \\ | ||
| + | The phasor diagram which was shown in answer 4. cannot be correct. \\ | ||
| + | With the correct lengths and angles, the real phasor diagram looks like this: | ||
| + | {{drawio> | ||
| + | Here the phasor is in the fourth quadrant with a negative angle. \\ | ||
| + | |||
| + | </ | ||
| + | \begin{align*} | ||
| + | U & | ||
| + | \varphi& | ||
| + | \end{align*} | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | <callout icon=" | ||
| + | Be aware that some of the calculators only provide $\tan^{-1}$ or $\arctan$ and not $\arctan2$! \\ | ||
| + | Therefore, you have always to check whether the solution lies in the correct quadrant. | ||
| + | </ | ||
| </ | </ | ||
| <panel type=" | <panel type=" | ||
| The following plot is visible on an oscilloscope (= plot tool for voltages and current). | The following plot is visible on an oscilloscope (= plot tool for voltages and current). | ||
| - | {{drawio> | + | {{drawio> |
| - | - What is the RMS value of the current | + | - What is the RMS value of the current |
| - | - How would the equivalent circuit look like, when it is build by two series components? | + | - How would the equivalent circuit look like, when it is built by two series components? |
| - Calculate the equivalent component values ($R$, $C$ or $L$) of the series circuit. | - Calculate the equivalent component values ($R$, $C$ or $L$) of the series circuit. | ||
| - | - How would the equivalent circuit look like, when it is build by two parallel components? | + | - How would the equivalent circuit look like, when it is built by two parallel components? |
| - Calculate the equivalent component values ($R$, $C$ or $L$) of the parallel circuit. | - Calculate the equivalent component values ($R$, $C$ or $L$) of the parallel circuit. | ||
| </ | </ | ||
| Zeile 643: | Zeile 756: | ||
| <panel type=" | <panel type=" | ||
| - | The following | + | The following |
| - | {{drawio> | + | {{drawio> |
| - | in the following | + | This circuit is used with different component values, which are given in the following. |
| - | | + | Calculate the RMS value of the missing voltage and the phase shift $\varphi$ between $U$ and $I$. |
| - | - $U_R = ?$, $U_L = 150 V$, $U_C = 110V$, $U=50V$ | + | <WRAP indent> |
| + | |||
| + | <WRAP indent> | ||
| + | <button size=" | ||
| + | The drawing of the voltage pointers is as follows: | ||
| + | The voltage U is determined by the law of Pythagoras | ||
| + | \begin{align*} | ||
| + | U &= \sqrt{{{U_R | ||
| + | | ||
| + | \end{align*} | ||
| + | The phase shift angle is calculated by simple geometry. | ||
| + | \begin{align*} | ||
| + | \tan(\varphi)& | ||
| + | & | ||
| + | \end{align*} | ||
| + | Considering that the angle is in the fourth quadrant we get: | ||
| + | </ | ||
| + | \begin{equation*} | ||
| + | U=\sqrt{2}\cdot 10~{\rm V} = 14.1~{\rm V} \qquad \varphi=-45° | ||
| + | \end{equation*} | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | </ | ||
| + | <WRAP indent> | ||
| + | |||
| + | <button size=" | ||
| + | The drawing of the voltage pointers is as follows: {{drawio> | ||
| + | The voltage $U_R$ is determined by the law of Pythagoras | ||
| + | \begin{align*} | ||
| + | U_R& | ||
| + | & | ||
| + | \end{align*} | ||
| + | The phase shift angle is calculated by simple geometry. | ||
| + | \begin{align*} | ||
| + | \tan(\varphi)& | ||
| + | & | ||
| + | \end{align*} | ||
| + | Considering that the angle is in the fourth quadrant we get: | ||
| + | </ | ||
| + | |||
| + | <button size=" | ||
| + | \begin{equation*} | ||
| + | U_R= 30~{\rm V}\qquad \varphi=53.13° | ||
| + | \end{equation*} | ||
| + | </ | ||
| + | </ | ||
| + | </ | ||
| </ | </ | ||
| <panel type=" | <panel type=" | ||
| - | The following | + | The following |
| - | {{drawio> | + | {{drawio> |
| - | in the following some of the numbers are given. Calculate the RMS value of the missing | + | in the following, some of the numbers are given. |
| - | - $I_R = 3A$, $I_L = 1A$, $I_C = 5A$, $I=?$ | + | Calculate the RMS value of the missing |
| - | - $I_R = ?$, $I_L = 1.2A$, $I_C = 0.4A$, $I=1A$ | + | - $I_R = 3~\rm A$, $I_L = 1 ~\rm A$, $I_C = 5 ~\rm A$, $I=?$ |
| + | - $I_R = ?$, | ||
| </ | </ | ||
| <panel type=" | <panel type=" | ||
| - | The following two currents with similar | + | The following two currents with similar |
| - | * $i_1(t) = \sqrt{2} \cdot 2 A \cdot cos (\omega t + 20°)$ | + | * $i_1(t) = \sqrt{2} \cdot 2 ~A \cdot \cos (\omega t + 20°)$ |
| - | * $i_2(t) = \sqrt{2} \cdot 5 A \cdot cos (\omega t + 110°)$ | + | * $i_2(t) = \sqrt{2} \cdot 5 ~A \cdot \cos (\omega t + 110°)$ |
| </ | </ | ||
| <panel type=" | <panel type=" | ||
| - | Two complex impedances $\underline{Z}_1$ and $\underline{Z}_2$ are invesigated. The resulting impedance for a series circuit is $60\Omega$. The resulting impedance for a prallel | + | Two complex impedances $\underline{Z}_1$ and $\underline{Z}_2$ are investigated. |
| + | The resulting impedance for a series circuit is | ||
| + | The resulting impedance for a parallel | ||
| What are the values for $\underline{Z}_1$ and $\underline{Z}_2$? | What are the values for $\underline{Z}_1$ and $\underline{Z}_2$? | ||
| + | |||
| + | # | ||
| + | It's a good start to write down all definitions of the given values: | ||
| + | * the given values for the series circuit ($\square_\rm s$) and the parallel circuit ($\square_\rm p$) are: \begin{align*} R_\rm s = 60 ~\Omega , \quad X_\rm s = 0 ~\Omega \\ R_\rm p = 25 ~\Omega , \quad X_\rm p = 0 ~\Omega \\ \end{align*} | ||
| + | * the series circuit and the parallel circuit results into: \begin{align*} | ||
| + | * the unknown values of the two impedances are: \begin{align*} \underline{Z}_1 = R_1 + {\rm j}\cdot X_1 \tag{3} \\ \underline{Z}_2 = R_2 + {\rm j}\cdot X_2 \tag{4} \\ \end{align*} | ||
| + | |||
| + | Based on $(1)$,$(3)$ and $(4)$: | ||
| + | \begin{align*} | ||
| + | R_\rm s & | ||
| + | &= R_1 + {\rm j}\cdot X_1 &&+ R_2 + {\rm j}\cdot X_2 \\ | ||
| + | \rightarrow 0 & | ||
| + | \end{align*} | ||
| + | Real value and imaginary value must be zero: | ||
| + | \begin{align*} | ||
| + | R_1 &= R_{\rm s} - R_2 \tag{5} \\ | ||
| + | X_1 &= - X_2 \tag{6} | ||
| + | \end{align*} | ||
| + | |||
| + | Based on $(2)$ with $R_\rm s = \underline{Z}_1 + \underline{Z}_2$ | ||
| + | \begin{align*} | ||
| + | R_{\rm p} &= {{\underline{Z}_1 \cdot \underline{Z}_2}\over{\underline{Z}_1 + \underline{Z}_2}} \\ | ||
| + | & | ||
| + | R_{\rm p} \cdot R_{\rm s} & | ||
| + | & | ||
| + | & | ||
| + | \end{align*} | ||
| + | |||
| + | Substituting $R_1$ and $X_1$ based on $(5)$ and $(6)$: | ||
| + | \begin{align*} | ||
| + | R_{\rm p} \cdot R_{\rm s} & | ||
| + | \rightarrow 0 & | ||
| + | \end{align*} | ||
| + | |||
| + | Again real value and imaginary value must be zero: | ||
| + | \begin{align*} | ||
| + | 0 & | ||
| + | & | ||
| + | \rightarrow | ||
| + | |||
| + | 0 &= R_{\rm s} R_2 - R_2^2 + X_2^2 - R_{\rm p} \cdot R_{\rm s} \\ | ||
| + | &= R_{\rm s} ({{1}\over{2}} R_{\rm s}) - ({{1}\over{2}} R_{\rm s})^2 - X_2^2 - R_{\rm p} \cdot R_{\rm s} \\ | ||
| + | &= {{1}\over{4}} R_{\rm s}^2 + X_2^2 - R_{\rm p} \cdot R_{\rm s} \\ | ||
| + | \rightarrow | ||
| + | |||
| + | \end{align*} | ||
| + | |||
| + | The concluding result is: | ||
| + | \begin{align*} | ||
| + | (5)+(7): \quad R_1 &= {{1}\over{2}} R_{\rm s} \\ | ||
| + | (7): \quad R_2 &= {{1}\over{2}} R_{\rm s} \\ | ||
| + | (6)+(8) | ||
| + | (8): \quad X_2 &= \pm \sqrt{R_{\rm p} \cdot R_{\rm s} - {{1}\over{4}} R_{\rm s}^2 } | ||
| + | \end{align*} | ||
| + | |||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | R_1 &= 30~\Omega \\ | ||
| + | R_2 &= 30~\Omega \\ | ||
| + | X_1 &= \mp \sqrt{600}~\Omega \approx \mp 24.5~\Omega \\ | ||
| + | X_2 &= \pm \sqrt{600}~\Omega \approx \pm 24.5~\Omega \\ | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| </ | </ | ||
| <panel type=" | <panel type=" | ||
| - | A real coil has both ohmic and inductance behavior. At DC voltage the resistance is measured as $9 \Omega$. With an AC voltage of $5V$ at $50Hz$ a current of $0.5A$ is measured. | + | A real coil has both ohmic and inductance behavior. |
| + | At DC voltage the resistance is measured as $9 ~\Omega$. | ||
| + | With an AC voltage of $5~\rm V$ at $50~\rm Hz$ a current of $0.5~\rm A$ is measured. | ||
| What is the value of the inductance $L$? | What is the value of the inductance $L$? | ||
| Zeile 684: | Zeile 915: | ||
| <panel type=" | <panel type=" | ||
| - | A real coil has both ohmic and inductance behavior. This coil has at $100Hz$ a impedance of $1.5k\Omega$ and a resistance $1k\Omega$. | + | A real coil has both ohmic and inductance behavior. |
| + | This coil has at $100~\rm Hz$ an impedance of $1.5~\rm k\Omega$ and a resistance $1~\rm k\Omega$. | ||
| What is the value of the reactance and inductance? | What is the value of the reactance and inductance? | ||
| Zeile 690: | Zeile 922: | ||
| <panel type=" | <panel type=" | ||
| - | An ideal capacitor is in series with a resistor $R=1k\Omega$. The capacitor shows a similar voltage drop like the resistor for $100Hz$. | + | An ideal capacitor is in series with a resistor $R=1~\rm k\Omega$. |
| + | The capacitor shows a similar voltage drop to the resistor for $100~\rm Hz$. | ||
| What is the value of the capacitance? | What is the value of the capacitance? | ||