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| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
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electrical_engineering_1:introduction_in_alternating_current_technology [2023/03/23 19:14] mexleadmin |
electrical_engineering_1:introduction_in_alternating_current_technology [2023/12/20 09:55] (aktuell) mexleadmin |
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| Zeile 1: | Zeile 1: | ||
| - | ====== 6. Introduction to Alternating Current Technology ====== | + | ====== 6 Introduction to Alternating Current Technology ====== |
| Up to now, we had analyzed DC signals (chapters 1. - 4.) and abrupt voltage changes for (dis)charging capacitors (chapter 5.). In households, we use alternating voltage (AC) instead of a constant voltage (DC). This is due to at least three main facts | Up to now, we had analyzed DC signals (chapters 1. - 4.) and abrupt voltage changes for (dis)charging capacitors (chapter 5.). In households, we use alternating voltage (AC) instead of a constant voltage (DC). This is due to at least three main facts | ||
| - | - 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, the mechanical energy of a rotating system is transformed into electric energy | + | - 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, the mechanical energy of a rotating system is transformed into electric energy |
| - 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 has to be increased. This is much easier done with AC voltages: **AC enables the transformation of a lower voltage to a higher** by the use of alternating magnetic fields in a transformer. | - 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 has to be increased. This is much easier done with AC voltages: **AC enables the transformation of a lower voltage to a higher** by the use of alternating magnetic fields in a transformer. | ||
| - 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. | ||
| Zeile 12: | Zeile 12: | ||
| 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 this chapter with a first introduction to AC systems. | + | To understand these systems a bit more, we will start this chapter with a first introduction to AC systems. |
| < | < | ||
| Zeile 106: | Zeile 106: | ||
| </ | </ | ||
| - | In order to analyze AC signals more, often different types of averages are taken into account. The most important values are: | + | To analyze AC signals more, often different types of averages are taken into account. The most important values are: |
| - the arithmetic mean $\overline{X}$ | - the arithmetic mean $\overline{X}$ | ||
| - the rectified value $\overline{|X|}$ | - the rectified value $\overline{|X|}$ | ||
| Zeile 124: | Zeile 124: | ||
| 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 is $\overline{X}=0$, | For pure AC signals, the arithmetic mean is $\overline{X}=0$, | ||
| Zeile 136: | Zeile 136: | ||
| 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 be 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*} | ||
| </ | </ | ||
| Zeile 175: | Zeile 176: | ||
| \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*} | ||
| A 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: | ||
| Zeile 192: | Zeile 193: | ||
| <callout icon=" | <callout icon=" | ||
| - | * The heat dissipation on a resistor $R$ of an AC current with the RMS value of $I_{RMS} = 1~A$ is equal to the heat dissipation of a DC current with $I_{DC} = 1~A$. | + | * 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 and their RMS values: | * It holds for AC signals and their RMS values: | ||
| * The resistance is $R={{U}\over{I}}$ | * The resistance is $R={{U}\over{I}}$ | ||
| 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 sinusoidal formula of a physical value $x$ will be : $x(t)=\hat{X}\cdot sin(\omega t + \varphi_x) \rightarrow x(t)=\sqrt{2}{X}\cdot sin(\omega t + \varphi_x) $ | + | Therefore, the sinusoidal formula of a physical value $x$ will be : $x(t)=\hat{X}\cdot |
| </ | </ | ||
| Zeile 223: | Zeile 224: | ||
| <panel type=" | <panel type=" | ||
| - | Calculate the RMS value of rectangular and triangular signals! Use similar symmetry simplifications as shown for AC signals. Compare it to the values shown in <imgref imageNo5> | + | Calculate the RMS value of rectangular and triangular signals! Use similar symmetry simplifications as shown for AC signals. |
| + | Compare it to the values shown in <imgref imageNo5> | ||
| </ | </ | ||
| Zeile 230: | Zeile 232: | ||
| 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 a 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 260: | Zeile 262: | ||
| $$u(t) = R \cdot i(t)$$ | $$u(t) = R \cdot i(t)$$ | ||
| - | 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 |
| - | $$\sqrt{2}{U}\cdot sin(\omega t + \varphi_u) = R \cdot \sqrt{2}{I}\cdot sin(\omega t + \varphi_i)$$ | + | $$\sqrt{2}{U}\cdot |
| Since we know, that $u(t)$ must be proportional to $i(t)$ we conclude that for a resistor $\varphi_u=\varphi_i$! | Since we know, that $u(t)$ must be proportional to $i(t)$ we conclude that for a resistor $\varphi_u=\varphi_i$! | ||
| \begin{align*} | \begin{align*} | ||
| - | R &= {{\sqrt{2}{U}\cdot sin(\omega t + \varphi_i)}\over{\sqrt{2}{I}\cdot sin(\omega t + \varphi_i) }} \\ | + | R &= {{\sqrt{2}{U}\cdot |
| &= {{U}\over{I}} | &= {{U}\over{I}} | ||
| \end{align*} | \end{align*} | ||
| Zeile 284: | Zeile 286: | ||
| This formula is also true for the instantaneous values: | This formula is also true for the instantaneous values: | ||
| $$C={{q(t)}\over{u(t)}}$$ | $$C={{q(t)}\over{u(t)}}$$ | ||
| - | Additionally, | + | Additionally, |
| By this we can set up the formula: | By this we can set up the formula: | ||
| \begin{align*} | \begin{align*} | ||
| - | i(t) &= {{dq(t)}\over{dt}} \\ | + | i(t) &= {{{\rm d}q(t)}\over{{\rm d}t}} \\ |
| - | & | + | & |
| \end{align*} | \end{align*} | ||
| Now, we insert the functions representing the instantaneous signals and calculate the derivative: | Now, we insert the functions representing the instantaneous signals and calculate the derivative: | ||
| \begin{align*} | \begin{align*} | ||
| - | | + | |
| - | & | + | |
| - | {I}\cdot sin(\omega t + \varphi_i) & | + | {I}\cdot |
| \end{align*} | \end{align*} | ||
| Zeile 308: | Zeile 310: | ||
| \omega t + \varphi_i &= \omega t + \varphi_u + {{1}\over{2}}\pi \\ | \omega t + \varphi_i &= \omega t + \varphi_u + {{1}\over{2}}\pi \\ | ||
| | | ||
| - | \varphi_u -\varphi_i & | + | \varphi_u -\varphi_i & |
| \end{align*} | \end{align*} | ||
| Zeile 339: | Zeile 341: | ||
| The inductance will here be introduced shortly - the detailed introduction is part of [[electrical_engineering_2: | The inductance will here be introduced shortly - the detailed introduction is part of [[electrical_engineering_2: | ||
| - | For the capacitance $C$ we had the situation, that it reacts to a voltage change ${{d}\over{dt}}u(t)$ with a counteracting current: | + | For the capacitance $C$ we had the situation, that it reacts to a voltage change ${{\rm d}\over{{\rm d}t}}u(t)$ with a counteracting current: |
| - | $$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 396: | Zeile 400: | ||
| </ | </ | ||
| - | 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 Electrical Engineering ===== | ===== 6.4 Complex Values in Electrical Engineering ===== | ||
| 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 at the alternating sinusoidal signals. | However, there is an alternative way to look at the alternating sinusoidal signals. | ||
| - | For this, we look first at 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 to 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 a shadow when the mass is illuminated sideways. \\ For $t>0$, the energy is continuously shifted between potential energy (deflection $x(t)$ around $X_0$) and kinetic energy (${{d}\over{dt}}x(t)$) | + | - 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 potential energy (deflection $x(t)$ around $X_0$) and kinetic energy (${{\rm d}\over{{\rm d}t}}x(t)$) |
| - | - 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 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. | ||
| 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 electrical engineering since the letter $i$ is already taken for currents. | + | For the imaginary unit ${\rm i}$ the letter ${\rm j}$ is used in electrical engineering since the letter ${\rm i}$ is already taken for currents. |
| < | < | ||
| Zeile 468: | Zeile 473: | ||
| ==== 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 instantaneous value of a complex vector $\underline{u}(t)$, | This is now interpreted as the instantaneous value of a complex vector $\underline{u}(t)$, | ||
| Zeile 477: | Zeile 483: | ||
| < | < | ||
| </ | </ | ||
| - | \\ {{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*} | ||
| Zeile 496: | Zeile 503: | ||
| 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 $j\omega$ equals a phase shift of $+90°$ |
| - | * A multiplication with $-j$ equals a phase shift of $-90°$ | + | * A multiplication with ${{1}\over{j\omega}}$ equals a phase shift of $-90°$ |
| ===== 6.5 Complex Impedance ===== | ===== 6.5 Complex Impedance ===== | ||
| 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 to polar coordinates by: | The impedance can be transformed from Cartesian to polar coordinates by: | ||
| * $Z=\sqrt{R^2 + X^2}$ | * $Z=\sqrt{R^2 + X^2}$ | ||
| - | * $\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 ==== | + | value - and therefore a phasor - can simply |
| 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 sinusoidal value - and therefore a phasor - can simply be written as " | + | - The derivative of a sinusoidal value - and therefore a phasor - can simply be written as " |
| - | - The integral of a sinusoidal value - and therefore a phasor - can simply be written as " | + | - The integral of a sinusoidal value - and therefore a phasor - can simply be written as " |
| + | \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 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 of $80\Omega$ at a frequency of $500 ~Hz$. At which frequencies the impedance will have the following values? | + | A coil has a reactance of $80\Omega$ at a frequency of $500 ~\rm Hz$. At which frequencies the impedance will have the following values? |
| - | - $85 ~\Omega$ | + | - $85 ~\Omega$ |
| - $120 ~\Omega$ | - $120 ~\Omega$ | ||
| - | - $44 ~\Omega$ | + | - $44 ~\Omega$ |
| <button size=" | <button size=" | ||
| - | When the frequency changes the reactance changes but the inductance is constant. Therefore the inductance is needed. \\ | + | 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 $f_0 = 500 ~Hz$. | + | It can be calculated by the given reactance for $f_0 = 500 ~\rm Hz$. |
| \begin{align*} | \begin{align*} | ||
| X_{L0}& | X_{L0}& | ||
| Zeile 615: | Zeile 628: | ||
| On the other hand, one can also use the rule of proportion here, and circumvent the calculation of inductance.\\ | On the other hand, one can also use the rule of proportion here, and circumvent the calculation of inductance.\\ | ||
| - | With the given reactance it is possible to calculate the reactance at other frequencies. | + | It is possible to calculate the reactance at other frequencies |
| \begin{align*} | \begin{align*} | ||
| X_L& | X_L& | ||
| Zeile 624: | Zeile 637: | ||
| With the values given: | With the values given: | ||
| \begin{equation*} | \begin{equation*} | ||
| - | f_1 = \frac{85 ~\Omega}{80~\Omega}\cdot500~Hz\qquad | + | f_1 = \frac{85 ~\Omega}{80~\Omega}\cdot500~{\rm Hz}\qquad |
| - | f_2 = \frac{120~\Omega}{80~\Omega}\cdot500~Hz\qquad | + | f_2 = \frac{120~\Omega}{80~\Omega}\cdot500~{\rm Hz}\qquad |
| - | f_3 = \frac{44 ~\Omega}{80~\Omega}\cdot500~Hz | + | f_3 = \frac{44 ~\Omega}{80~\Omega}\cdot500~{\rm Hz} |
| \end{equation*} | \end{equation*} | ||
| </ | </ | ||
| \begin{equation*} | \begin{equation*} | ||
| - | f_1=531.25~Hz\qquad f_2=750~Hz\qquad f_3=275~Hz | + | f_1=531.25~{\rm Hz}\qquad f_2=750~{\rm Hz}\qquad f_3=275~{\rm Hz} |
| \end{equation*} | \end{equation*} | ||
| </ | </ | ||
| Zeile 637: | Zeile 650: | ||
| <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 currencies can be measured? | + | A capacitor with $5 ~{\rm µF}$ is connected to a voltage source which generates $U_\sim = 200 ~{\rm V}$. At which frequencies the following currencies can be measured? |
| - | - $0.5 ~A$ | + | - $0.5 ~\rm A$ |
| - | - $0.8 ~A$ | + | - $0.8 ~\rm A$ |
| - | - $1.3 ~A$ | + | - $1.3 ~\rm A$ |
| </ | </ | ||
| <panel type=" | <panel type=" | ||
| - | A capacitor shall have a capacity of $4.7 ~\mu F \pm 10~\%$. This capacitor shall be used with an AC voltage of $400V$ and $50~Hz$. | + | 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 $U_1 = 100~V$ and $U_2 = 120~V$. \\ | + | 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 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. | ||
| Zeile 657: | Zeile 670: | ||
| <WRAP indent>< | <WRAP indent>< | ||
| The phasor diagram looks roughly like this: | The phasor diagram looks roughly like this: | ||
| - | {{drawio> | + | {{drawio> |
| </ | </ | ||
| Zeile 669: | Zeile 682: | ||
| The angle is by the tangent of the relation of the imaginary part to the real part of the resulting voltage. | The angle is by the tangent of the relation of the imaginary part to the real part of the resulting voltage. | ||
| \begin{align*} | \begin{align*} | ||
| - | \varphi& | + | \varphi& |
| - | & | + | & |
| - | & | + | & |
| - | & | + | & |
| \end{align*} | \end{align*} | ||
| </ | </ | ||
| \begin{align*} | \begin{align*} | ||
| - | U & | + | U & |
| \varphi& | \varphi& | ||
| \end{align*} | \end{align*} | ||
| Zeile 692: | Zeile 705: | ||
| The phasor diagram looks roughly like this. \\ | The phasor diagram looks roughly like this. \\ | ||
| But have a look at the solution for question 5! | But have a look at the solution for question 5! | ||
| - | {{drawio> | + | {{drawio> |
| </ | </ | ||
| Zeile 704: | Zeile 717: | ||
| The angle is by the tangent of the relation of the imaginary part to the real part of the resulting voltage. | The angle is by the tangent of the relation of the imaginary part to the real part of the resulting voltage. | ||
| \begin{align*} | \begin{align*} | ||
| - | \varphi& | + | \varphi& |
| - | & | + | & |
| - | & | + | & |
| - | & | + | & |
| - | & | + | & |
| \end{align*} | \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 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. \\ | The phasor diagram which was shown in answer 4. cannot be correct. \\ | ||
| - | With the correct lengths and angles, the real phasor diagram | + | With the correct lengths and angles, the real phasor diagram looks like this: |
| - | {{drawio> | + | {{drawio> |
| Here the phasor is in the fourth quadrant with a negative angle. \\ | Here the phasor is in the fourth quadrant with a negative angle. \\ | ||
| </ | </ | ||
| \begin{align*} | \begin{align*} | ||
| - | U & | + | U & |
| \varphi& | \varphi& | ||
| \end{align*} | \end{align*} | ||
| Zeile 725: | Zeile 738: | ||
| <callout icon=" | <callout icon=" | ||
| - | Be aware that some of the calculators only provide $tan^{-1}$ or $arctan$ and not $arctan2$! \\ | + | 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 into the correct quadrant. | + | Therefore, you have always to check whether the solution lies in the correct quadrant. |
| </ | </ | ||
| </ | </ | ||
| Zeile 734: | Zeile 747: | ||
| {{drawio> | {{drawio> | ||
| - | - What is the RMS value of the current and the voltage? What is the frequency $f$ and the phase $\varphi$? Does the component under test behave ohmic, capacitive or inductive? | + | - What is the RMS value of the current and the voltage? What is the frequency $f$ and the phase $\varphi$? Does the component under test behave ohmic, capacitive, or inductive? |
| - How would the equivalent circuit look like, when it is built 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. | ||
| Zeile 748: | Zeile 761: | ||
| This circuit is used with different component values, which are given 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 $I$. | Calculate the RMS value of the missing voltage and the phase shift $\varphi$ between $U$ and $I$. | ||
| - | <WRAP indent> | + | <WRAP indent> |
| <WRAP indent> | <WRAP indent> | ||
| <button size=" | <button size=" | ||
| - | The drawing of the voltage pointers is as follows: {{drawio> | + | The drawing of the voltage pointers is as follows: |
| The voltage U is determined by the law of Pythagoras | The voltage U is determined by the law of Pythagoras | ||
| \begin{align*} | \begin{align*} | ||
| - | U &= \sqrt{{{U_R}^2}+{({U_L}-{U_C}})^2}\\ | + | U &= \sqrt{{{U_R |
| - | & | + | &= \sqrt{(10~{\rm V})^2+ {({10~{\rm V}}-{20~{\rm V}}})^2} |
| \end{align*} | \end{align*} | ||
| The phase shift angle is calculated by simple geometry. | The phase shift angle is calculated by simple geometry. | ||
| \begin{align*} | \begin{align*} | ||
| - | \tan(\varphi)& | + | \tan(\varphi)& |
| - | & | + | & |
| \end{align*} | \end{align*} | ||
| Considering that the angle is in the fourth quadrant we get: | Considering that the angle is in the fourth quadrant we get: | ||
| </ | </ | ||
| \begin{equation*} | \begin{equation*} | ||
| - | U=\sqrt{2}\cdot 10~V = 14.1~V \qquad \varphi=-45° | + | U=\sqrt{2}\cdot 10~{\rm V} = 14.1~{\rm V} \qquad \varphi=-45° |
| \end{equation*} | \end{equation*} | ||
| </ | </ | ||
| </ | </ | ||
| - | </ | + | </ |
| <WRAP indent> | <WRAP indent> | ||
| <button size=" | <button size=" | ||
| - | The drawing of the voltage pointers is as follows: {{drawio> | + | The drawing of the voltage pointers is as follows: {{drawio> |
| The voltage $U_R$ is determined by the law of Pythagoras | The voltage $U_R$ is determined by the law of Pythagoras | ||
| \begin{align*} | \begin{align*} | ||
| - | U_R& | + | U_R& |
| - | & | + | & |
| \end{align*} | \end{align*} | ||
| The phase shift angle is calculated by simple geometry. | The phase shift angle is calculated by simple geometry. | ||
| \begin{align*} | \begin{align*} | ||
| - | \tan(\varphi)& | + | \tan(\varphi)& |
| - | & | + | & |
| \end{align*} | \end{align*} | ||
| Considering that the angle is in the fourth quadrant we get: | Considering that the angle is in the fourth quadrant we get: | ||
| Zeile 791: | Zeile 804: | ||
| <button size=" | <button size=" | ||
| \begin{equation*} | \begin{equation*} | ||
| - | U_R= 30~V\qquad \varphi=53.13° | + | U_R= 30~{\rm V}\qquad \varphi=53.13° |
| \end{equation*} | \end{equation*} | ||
| </ | </ | ||
| Zeile 804: | Zeile 817: | ||
| in the following, some of the numbers are given. | in the following, some of the numbers are given. | ||
| - | Calculate the RMS value of the missing | + | Calculate the RMS value of the missing |
| - | - $I_R = 3~A$, $I_L = 1~A$, $I_C = 5~A$, $I=?$ | + | - $I_R = 3~\rm A$, $I_L = 1 ~\rm A$, $I_C = 5 ~\rm A$, $I=?$ |
| - | - $I_R = ?$, $I_L = 1.2~A$, $I_C = 0.4~A$, $I=1~A$ | + | - $I_R = ?$, |
| </ | </ | ||
| <panel type=" | <panel type=" | ||
| - | The following two currents with similar frequencies, | + | The following two currents with similar frequencies, |
| - | * $i_1(t) = \sqrt{2} \cdot 2 ~A \cdot cos (\omega t + 20°)$ | + | * $i_1(t) = \sqrt{2} \cdot 2 ~A \cdot \cos (\omega t + 20°)$ |
| - | * $i_2(t) = \sqrt{2} \cdot 5 ~A \cdot cos (\omega t + 110°)$ | + | * $i_2(t) = \sqrt{2} \cdot 5 ~A \cdot \cos (\omega t + 110°)$ |
| </ | </ | ||
| Zeile 819: | Zeile 832: | ||
| <panel type=" | <panel type=" | ||
| Two complex impedances $\underline{Z}_1$ and $\underline{Z}_2$ are investigated. | Two complex impedances $\underline{Z}_1$ and $\underline{Z}_2$ are investigated. | ||
| - | The resulting impedance for a series circuit is $60~\Omega$. | + | The resulting impedance for a series circuit is |
| - | The resulting impedance for a parallel circuit is $25~\Omega$. | + | The resulting impedance for a parallel circuit is $25~\Omega + \rm j \cdot 0 ~\Omega $. |
| 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*} | ||
| + | # | ||
| + | |||
| </ | </ | ||
| Zeile 829: | Zeile 908: | ||
| A real coil has both ohmic and inductance behavior. | A real coil has both ohmic and inductance behavior. | ||
| At DC voltage the resistance is measured as $9 ~\Omega$. | At DC voltage the resistance is measured as $9 ~\Omega$. | ||
| - | With an AC voltage of $5~V$ at $50~Hz$ a current of $0.5~A$ is measured. | + | 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 837: | Zeile 916: | ||
| <panel type=" | <panel type=" | ||
| A real coil has both ohmic and inductance behavior. | A real coil has both ohmic and inductance behavior. | ||
| - | This coil has at $100~Hz$ an impedance of $1.5~k\Omega$ and a resistance $1~k\Omega$. | + | 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 843: | Zeile 922: | ||
| <panel type=" | <panel type=" | ||
| - | An ideal capacitor is in series with a resistor $R=1~k\Omega$. | + | 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 $100~Hz$. | + | 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? | ||