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Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
electrical_engineering_1:introduction_in_alternating_current_technology [2021/11/01 19:27] tfischer |
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 | + | ====== 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 understanding the complex numbers please refer to the following videos: | ||
+ | * The question " | ||
+ | * How to calculate with complex numbers (sum, difference, product) can be seen in [[https:// | ||
+ | * The geometric interpretation of the complex multiplication or: "Why is the amount and angle to be added in the multiplication?" | ||
+ | </ | ||
- | ===== 6.1 Classification | + | |
+ | ===== 6.1 Description | ||
< | < | ||
- | ==== Goals ==== | + | === Learning Objectives |
- | After this lesson, you should: | + | By the end of this section, you will be able to: |
- 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 period, frequency, and angular frequency. | ||
+ | - Know the difference between zero phase angle and phase shift angle. | ||
+ | - Know the direction of the phase shift angle. | ||
+ | - know the formula symbols of the above-mentioned quantities. | ||
</ | </ | ||
+ | |||
+ | ==== 6.1.1 Description of Classification of time-dependent Signals ==== | ||
Voltages and currents in the following chapters will be time-dependent values. | Voltages and currents in the following chapters will be time-dependent values. | ||
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 36: | Zeile 51: | ||
< | < | ||
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
- | In the following we will investigate mainly pure AC signals. | + | In the following, we will investigate mainly pure AC signals. |
- | ===== 6.2 Descriptive Values of AC Signals ===== | + | ==== 6.1.2 Descriptive Values of AC Signals ==== |
- | + | ||
- | + | ||
- | < | + | |
- | ==== Goals ==== | + | |
- | + | ||
- | After this lesson, you should: | + | |
- | + | ||
- | - Know the relationship between amplitude and peak-to-peak value. | + | |
- | - Know the relationship between period, frequency and angular frequency. | + | |
- | - Know the difference between zero phase angle and phase shift angle. | + | |
- | - Know the direction of the phase shift angle. | + | |
- | - know the formula symbols of the above-mentioned quantities. | + | |
- | </ | + | |
< | < | ||
< | < | ||
</ | </ | ||
- | \\ {{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> |
</ | </ | ||
</ | </ | ||
- | + | ===== 6.2 Averaging of AC Signals ===== | |
- | ===== 6.3 Averaging of AC Signals ===== | + | |
< | < | ||
- | ==== Goals ==== | + | === Learning Objectives |
- | After this lesson, you should: | + | By the end of this section, you will be able to: |
- | - 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 116: | Zeile 117: | ||
- | ==== The Arithmetic Mean ==== | + | ==== 6.2.1 The Arithmetic Mean ==== |
The arithmetic mean is given by the (equally weighted) averaging of the signed measuring points. \\ | The arithmetic mean is given by the (equally weighted) averaging of the signed measuring points. \\ | ||
Zeile 122: | 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 $\overline{X}=0$, |
- | ==== 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 141: | 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 |
</ | </ | ||
- | ==== 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 values | + | 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 202: | Zeile 204: | ||
\begin{align*} | \begin{align*} | ||
- | X & | + | X & |
- | & | + | & |
- | & | + | & |
- | & | + | & |
& | & | ||
& | & | ||
Zeile 212: | Zeile 214: | ||
This can also be seen in this [[https:// | This can also be seen in this [[https:// | ||
+ | </ | ||
<callout icon=" | <callout icon=" | ||
- | In the following we will often ose $\sqrt{2}X$ instead of $\hat{X}$, e.g.: $u(t)=\hat{X}\cdot sin(\omega t + \varphi_u) \rightarrow | + | In the following |
- | + | 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 | + | Calculate the RMS value of rectangular and triangular signals! Use similar symmetry |
+ | Compare it to the values shown in <imgref imageNo5> | ||
</ | </ | ||
- | ==== Comparison of the different | + | ==== 6.2.4 Comparison of the different |
+ | |||
+ | The following simulation shows the different values for averaging a rectangular, | ||
+ | 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 236: | Zeile 241: | ||
- | ===== 6.4 AC Two-Poles ===== | + | |
+ | ===== 6.3 AC Two-Terminal Networks | ||
< | < | ||
- | ==== Goals ==== | + | === Learning Objectives |
- | After this lesson, you should: | + | By the end of this section, you will be able to: |
- know that real, lossy components are described by equivalent circuits of ideal components. | - know that real, lossy components are described by equivalent circuits of ideal components. | ||
Zeile 247: | Zeile 253: | ||
</ | </ | ||
- | In the chapters [[simple_circuits|2. Simple Circuits]] and [[non-ideal_sources_and_two_pole_networks|3 Non-ideal Sources and Two Pole 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 |
- | We want to analyze | + | We want to analyze |
- | ==== 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 |
- | $$\sqrt{2}{U}\cdot sin(\omega t + \varphi_u) = R \cdot \sqrt{2}{I}\cdot sin(\omega t + \varphi_i)$$ | + | $$\sqrt{2}{U}\cdot |
Since we know, that $u(t)$ must be proportional to $i(t)$ we conclude that for a resistor $\varphi_u=\varphi_i$! | Since we know, that $u(t)$ must be proportional to $i(t)$ we conclude that for a resistor $\varphi_u=\varphi_i$! | ||
\begin{align*} | \begin{align*} | ||
- | R &= {{\sqrt{2}{I}\cdot sin(\omega t + \varphi_i)}\over{\sqrt{2}{U}\cdot sin(\omega t + \varphi_i) }} \\ | + | R &= {{\sqrt{2}{U}\cdot |
&= {{U}\over{I}} | &= {{U}\over{I}} | ||
\end{align*} | \end{align*} | ||
Zeile 269: | Zeile 275: | ||
< | < | ||
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
- | This was not too hard und quite obvious. But, what about the other types of passive two-poles - namely the capacitance and inductance? | + | This was not too hard and quite obvious. But, what about the other types of passive two-terminal networks |
- | ==== Capacitance ==== | + | ==== 6.3.2 Capacitance ==== |
- | For the capacitance we have the principle | + | For the capacitance we have the basic formula: |
$$C={{Q}\over{U}}$$ | $$C={{Q}\over{U}}$$ | ||
- | This formula | + | This formula |
$$C={{q(t)}\over{u(t)}}$$ | $$C={{q(t)}\over{u(t)}}$$ | ||
- | Additionally we know, that the instantaneous current is defined by $i(t)={{dq(t)}\over{dt}}$. | + | Additionally, we know, that the instantaneous current is defined by $i(t)={{{\rm d}q(t)}\over{{\rm d}t}}$. |
By this we can set up the formula: | By this we can set up the formula: | ||
\begin{align*} | \begin{align*} | ||
- | i(t) &= {{dq(t)}\over{dt}} \\ | + | i(t) &= {{{\rm d}q(t)}\over{{\rm d}t}} \\ |
- | & | + | & |
\end{align*} | \end{align*} | ||
Now, we insert the functions representing the instantaneous signals and calculate the derivative: | Now, we insert the functions representing the instantaneous signals and calculate the derivative: | ||
\begin{align*} | \begin{align*} | ||
- | | + | |
- | & | + | |
- | {I}\cdot sin(\omega t + \varphi_i) & | + | {I}\cdot |
\end{align*} | \end{align*} | ||
- | Equating coefficients in $(6.4.1)$ leads to: | + | Equating coefficients in $(6.3.1)$ leads to: |
\begin{align*} | \begin{align*} | ||
I & | I & | ||
Zeile 304: | Zeile 310: | ||
\omega t + \varphi_i &= \omega t + \varphi_u + {{1}\over{2}}\pi \\ | \omega t + \varphi_i &= \omega t + \varphi_u + {{1}\over{2}}\pi \\ | ||
| | ||
- | \varphi_u -\varphi_i & | + | \varphi_u -\varphi_i & |
\end{align*} | \end{align*} | ||
Zeile 311: | Zeile 317: | ||
<callout icon=" | <callout icon=" | ||
- | In order not to mix up the definitions, | + | In order not to mix up the definitions, |
- | The apparent | + | The impedance is generally defined as |
$$Z = {{U}\over{I}}$$ | $$Z = {{U}\over{I}}$$ | ||
- | Only for pure resistor two pole the apparent | + | Only for a pure resistor |
+ | |||
+ | For the pure capacitive as a two-terminal network, the impedance $Z_C$ is $Z_C={{1}\over{\omega \cdot C}}$. | ||
</ | </ | ||
Zeile 321: | Zeile 329: | ||
< | < | ||
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
Zeile 327: | Zeile 335: | ||
< | < | ||
</ | </ | ||
- | {{url> | + | {{url> |
</ | </ | ||
- | ==== Inductance ==== | + | ==== 6.3.3 Inductance ==== |
The inductance will here be introduced shortly - the detailed introduction is part of [[electrical_engineering_2: | The inductance will here be introduced shortly - the detailed introduction is part of [[electrical_engineering_2: | ||
- | For the capacitance $C$ we had the situation, that it reacts to a voltage change ${{d}\over{dt}}u(t)$ with a counteracting current: | + | For the capacitance $C$ we had the situation, that it reacts to a voltage change ${{\rm d}\over{{\rm d}t}}u(t)$ with a counteracting current: |
- | $$i(t)= C \cdot {{d}\over{dt}}u(t)$$ | + | $$i(t)= C \cdot {{\rm d}\over{{\rm d}t}}u(t)$$ |
- | This is due to the fact, that the capacity stores charge carriers $q$. It appears that "the capacitance does not like voltage changes and reacts with a compensating current" | + | This is due to the fact, that the capacity stores charge carriers $q$. |
+ | It appears that "the capacitance does not like voltage changes and reacts with a compensating current" | ||
+ | When the voltage on a capacity drops, the capacity | ||
- | For an inductance $L$ it is just the other way around: "the inductance does not like current changes and reacts with a compensating voltage drop". Once the current changes | + | For an inductance $L$ it is just the other way around: "the inductance does not like current changes and reacts with a compensating voltage drop". Once the current changes |
- | $$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 the value of the inductance | + | 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*} | ||
- | Equating coefficients in $(6.4.2)$ leads to: | + | Equating coefficients in $(6.3.2)$ leads to: |
\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 357: | 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 363: | Zeile 373: | ||
< | < | ||
- | < | + | < |
</ | </ | ||
- | \\ {{drawio> | + | \\ {{drawio> |
</ | </ | ||
< | < | ||
- | < | + | < |
</ | </ | ||
{{url> | {{url> | ||
Zeile 377: | Zeile 387: | ||
Remember the formulas for the different pure loads: | Remember the formulas for the different pure loads: | ||
- | ^Load ^ | + | < |
- | |Resistance | + | |
- | |Capacitance |$C$| $Z_C = {{1}\over{\omega \cdot C}} $ | $\varphi_C = - {{1}\over{2}}\pi $ | | + | |
- | |Impedance | + | |
+ | ^ Load | ||
+ | | Resistance | ||
+ | | Capacitance | ||
+ | | Inductance | ||
+ | </ | ||
- | 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 # | ||
</ | </ | ||
- | By the concept of AC two-poles we are also able to use the DC methods of network analysis | + | For the concept of AC two-terminal networks, |
- | ===== 6.5 AC Impedance | + | ===== 6.4 Complex Values in Electrical Engineering |
< | < | ||
- | ==== Goals ==== | + | === Learning Objectives |
- | After this lesson, you should: | + | By the end of this section, you will be able to: |
- know how sine variables can be symbolized by a vector. | - know how sine variables can be symbolized by a vector. | ||
- know which parameters can determine a sinusoidal quantity. | - know which parameters can determine a sinusoidal quantity. | ||
- | - be able to graphically derive a pointer diagram for several existing sine variables. | + | - graphically derive a pointer diagram for several existing sine variables. |
- | - be able to plot the phase shift on the vector and time plots. | + | - plot the phase shift on the vector and time plots. |
- | - Be able to add sinusoidal quantities in vector and time representation. | + | - add sinusoidal quantities in vector and time representation. |
- | - know and be able to apply the impedance of components. | + | - know and apply the impedance of components. |
- know the frequency dependence of the impedance of the components. In particular, you should know the effect of the ideal components at very high and very low frequencies and be able to apply it for plausibility checks. | - know the frequency dependence of the impedance of the components. In particular, you should know the effect of the ideal components at very high and very low frequencies and be able to apply it for plausibility checks. | ||
</ | </ | ||
+ | |||
+ | |||
+ | The following two videos explain the basic terms of the complex AC calculus: Impedance, Reactance, Resistance | ||
+ | |||
+ | {{youtube> | ||
+ | |||
+ | {{youtube> | ||
+ | |||
+ | |||
+ | ==== 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. \\ | ||
+ | |||
+ | However, there is an alternative way to look at the alternating sinusoidal signals. | ||
+ | For this, we look first at a different, but already a familiar problem (see <imgref pic06> | ||
+ | - 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. | ||
+ | - 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)$ | ||
+ | - 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> | ||
+ | </ | ||
+ | |||
+ | The transformation of the two-dimensional rotation to a one-dimensional sinusoidal signal is also shown in <imgref BildNr00> | ||
< | < | ||
- | < | + | < |
</ | </ | ||
< | < | ||
- | {{url> | + | {{url> |
+ | Click on the box " | ||
</ | </ | ||
Zeile 415: | Zeile 456: | ||
<button type=" | <button type=" | ||
</ | </ | ||
- | |||
</ | </ | ||
- | ===== 6.6 Insertion: Complex Numbers ===== | ||
- | < | + | The two-dimensional rotation can be represented with a complex number in Euler' |
- | ==== Goals ==== | + | It combines the exponential representation with real part $\Re$ and imaginary part $\Im$ of a complex value: |
+ | $$ \underline{x}(t)=\hat{X}\cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)} | ||
- | After this lesson you should be able to: | + | 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. |
- | - be able to convert complex-valued numbers from polar to Cartesian coordinates and vice versa. | + | < |
- | - know how to add, subtract, multiply, and divide two complex-valued numbers using a formula and complex plane. | + | < |
- | - know what multiplication by / division by j means graphically. | + | </imgcaption> |
- | - know how to find the absolute value of a complex number. | + | \\ {{drawio> |
- | </callout> | + | </WRAP> |
- | ==== Video ==== | + | ==== 6.4.2 Complex Current and Voltage |
- | What's the point of complex numbers? | + | The concepts |
+ | Up to now, we used the following formula to represent alternating voltages: | ||
- | {{youtube> | + | $$u(t)= \sqrt{2} U \cdot \sin (\varphi)$$ |
- | Calculating with complex | + | This is now interpreted as the instantaneous value of a complex |
- | {{youtube>OzU_1EwaF9Y}} | + | < |
+ | < | ||
+ | </ | ||
+ | \\ {{drawio>voltagephasorcomplexplane.svg}} \\ | ||
+ | </ | ||
- | Geometric interpretation of the complex | + | The parts on the complex |
- | Or: Why is the amount and angle to be added in the multiplication? | + | - The real part $\Re{(\underline{u}(t))} = \sqrt{2}U \cdot \cos (\omega t + \varphi_u)$ |
+ | - The imaginary part $\Im{(\underline{u}(t))} = \sqrt{2}U \cdot \sin (\omega t + \varphi_u)$ | ||
- | {{youtube> | + | This is equivalent to the complex phasor $\underline{u}(t)=\sqrt{2}U \cdot {\rm e} ^{{\rm j} |
- | {{youtube> | + | |
+ | The complex phasor can be separated: | ||
+ | \begin{align*} | ||
+ | \underline{u}(t) & | ||
+ | & | ||
+ | \cdot {\rm e}^{{\rm j} \omega t} \\ | ||
+ | & | ||
+ | \end{align*} | ||
+ | |||
+ | The **fixed phasor** (in German: //komplexer Festzeiger// | ||
+ | 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 known properties of complex numbers from Mathematics 101 can be applied: | ||
+ | * A multiplication with $j\omega$ equals a phase shift of $+90°$ | ||
+ | * A multiplication with ${{1}\over{j\omega}}$ equals a phase shift of $-90°$ | ||
- | ===== 6.7 Complex | + | ===== 6.5 Complex Impedance ===== |
< | < | ||
- | ==== Goals ==== | + | === Learning Objectives |
- | After this lesson | + | By the end of this section, |
- | + | - draw and read pointer diagrams. | |
- | - Be able to draw and read pointer diagrams. | + | - know and apply the complex value formulas of impedance, reactance, |
- | - Know and be able to apply the complex value formulas of impedance, reactance, resistance. | + | |
</ | </ | ||
- | ==== Video ==== | + | ==== 6.5.1 Introduction to Complex Impedance |
- | Pointer diagrams; | + | The complex |
+ | Now the complex impedance is: | ||
- | {{youtube> | + | \begin{align*} |
+ | \underline{Z}& | ||
+ | & | ||
+ | & | ||
+ | & | ||
+ | & | ||
+ | \end{align*} | ||
- | Complex alternating current calculation - basic terms: Impedance, Reactance, Resistance | + | With |
+ | * the resistance $R$ (in German: // | ||
+ | * the reactance | ||
+ | * the impedance | ||
- | {{youtube> | + | The impedance can be transformed from Cartesian to polar coordinates by: |
+ | * $Z=\sqrt{R^2 + X^2}$ | ||
+ | * $\varphi = \arctan | ||
+ | The other way around it is possible to transform by: | ||
+ | * $R = Z \cos \varphi$ | ||
+ | * $X = Z \sin \varphi$ | ||
- | Capacitor | + | value - and therefore a phasor - can simply ==== 6.5.2 Application on pure Loads ==== |
- | {{youtube>LM2G3cunKp4}} | + | With the complex impedance in mind, the <tabref tab01> can be expanded to: |
- | detailed explanation of impedances | + | < |
- | {{youtube> | + | ^ Load $\phantom{U\over I}$ ^ ^ integral representation $\phantom{U\over I}$ ^ complex impedance $\underline{Z}={{\underline{U}}\over{\underline{I}}}$ |
+ | | Resistance | ||
+ | | Capacitance | ||
+ | | Inductance | ||
+ | </ | ||
+ | \\ \\ | ||
+ | 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 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)} | ||
+ | = {{1}\over{\rm j\omega}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)} | ||
+ | = -{{\rm j}\over{\omega}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)} | ||
+ | \end{align*} | ||
+ | </ | ||
- | ===== 6.8 Power in AC ===== | + | Once a fixed input voltage is given, the voltage phasor $\underline{U}$, |
- | <callout> | + | <WRAP> |
- | ==== Goals ==== | + | < |
+ | </ | ||
+ | \\ {{drawio> | ||
+ | </ | ||
- | After this lesson you should: | + | ==== 6.5.3 Application on Impedance Networks ==== |
+ | === Simple Networks === | ||
- | - Know the formula | + | In the chapter [[: |
+ | These formulas not only apply to ohmic resistors but also to impedances: | ||
+ | |||
+ | < | ||
+ | < | ||
+ | </ | ||
+ | \\ {{drawio> | ||
+ | </ | ||
+ | |||
+ | Similarly, | ||
+ | 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 " | ||
+ | |||
+ | For more complex problems having AC values | ||
+ | This concept will be used in the next chapter and in circuit design. | ||
+ | |||
+ | < | ||
+ | < | ||
+ | </ | ||
+ | \\ {{drawio> | ||
+ | </ | ||
+ | \\ | ||
+ | |||
+ | <callout icon=" | ||
+ | < | ||
+ | For a complex number are always two values are needed. These are either | ||
+ | - the real part (e.g. the resistance) and the imaginary part (e.g. the reactance), or | ||
+ | - the absolute value (e.g. the absolute value of the impedance) and the phase | ||
+ | |||
+ | Therefore, instead of the form $\underline{Z}=Z\cdot {\rm e}^{{\rm j}\varphi}$ for the phasors often the form $Z\angle{\varphi}$ is used. | ||
+ | </ | ||
</ | </ | ||
- | ==== Video ==== | ||
- | detailed explanation of power factor in alternating current technology | + | ===== Exercises ===== |
- | {{youtube>Tv_7XWf96gg}} | + | <panel type=" |
+ | 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? | ||
+ | - $85 ~\Omega$ | ||
+ | - $120 ~\Omega$ | ||
+ | - $44 ~\Omega$ | ||
- | [[https://www.geogebra.org/ | + | <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*} | ||
- | ===== 6.9 Generation of AC ===== | + | 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*} | ||
- | <callout> | + | </collapse><button size=" |
- | ==== Goals ==== | + | \begin{equation*} |
+ | f_1=531.25~{\rm Hz}\qquad f_2=750~{\rm Hz}\qquad f_3=275~{\rm Hz} | ||
+ | \end{equation*} | ||
+ | </ | ||
+ | </ | ||
- | After this lesson, you should: | + | <panel type=" |
+ | 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? | ||
+ | - $0.5 ~\rm A$ | ||
+ | - $0.8 ~\rm A$ | ||
+ | - $1.3 ~\rm A$ | ||
+ | </ | ||
+ | <panel type=" | ||
+ | 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? | ||
+ | </ | ||
+ | |||
+ | <panel type=" | ||
+ | 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. | ||
+ | |||
+ | <WRAP indent> 1. Draw the phasor diagram for the two voltage phasors and the resulting phasor. | ||
+ | |||
+ | <WRAP indent>< | ||
+ | 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> 4. Draw the phasor diagram for the two voltage phasors and the resulting phasor for the new circuit. | ||
+ | <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=" | ||
+ | The following plot is visible on an oscilloscope (= plot tool for voltages and current). | ||
+ | {{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? | ||
+ | - 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. | ||
+ | - 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. | ||
+ | </ | ||
- | ===== 6.10 Exercises ===== | ||
- | ==== Video ==== | + | <panel type=" |
+ | The following circuit shall be given. \\ | ||
+ | {{drawio> | ||
- | Parallel connection | + | 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 $I$. | ||
+ | <WRAP indent> | ||
- | {{youtube>u6lE4gIIfBw}} | + | <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 | ||
+ | &= \sqrt{(10~{\rm V})^2+ {({10~{\rm V}}-{20~{\rm V}}})^2} | ||
+ | \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*} | ||
+ | </ | ||
+ | </ | ||
- | more complex exam task: complex circuit I | + | </ |
+ | <WRAP indent> | ||
- | {{youtube>RPK0wkyLyMY}} | + | <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: | ||
+ | </ | ||
- | more complex exam task: complex circuit II | + | <button size=" |
+ | \begin{equation*} | ||
+ | U_R= 30~{\rm V}\qquad \varphi=53.13° | ||
+ | \end{equation*} | ||
+ | </ | ||
+ | </ | ||
+ | </ | ||
+ | </ | ||
- | {{youtube> | ||
- | Examination task: complex | + | <panel type=" |
+ | The following | ||
+ | {{drawio> | ||
- | {{youtube>8MMzeeHNjIw}} | + | in the following, some of the numbers are given. |
+ | Calculate the RMS value of the missing currents and the phase shift $\varphi$ between $U$ and $I$. | ||
+ | - $I_R = 3~\rm A$, $I_L = 1 ~\rm A$, $I_C = 5 ~\rm A$, $I=?$ | ||
+ | - $I_R = ?$, $I_L = 1.2~\rm A$, $I_C = 0.4~\rm A$, $I=1~\rm A$ | ||
+ | </ | ||
+ | <panel type=" | ||
+ | The following two currents with similar frequencies, | ||
+ | * $i_1(t) = \sqrt{2} \cdot 2 ~A \cdot \cos (\omega t + 20°)$ | ||
+ | * $i_2(t) = \sqrt{2} \cdot 5 ~A \cdot \cos (\omega t + 110°)$ | ||
+ | </ | ||
- | {{youtube> | + | <panel type=" |
- | {{youtube> | + | 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 circuit is $25~\Omega + \rm j \cdot 0 ~\Omega $. | ||
+ | |||
+ | 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=" | ||
+ | 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 $L$? | ||
+ | </ | ||
+ | |||
+ | |||
+ | <panel type=" | ||
+ | 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? | ||
+ | </ | ||
+ | |||
+ | <panel type=" | ||
+ | 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? | ||
+ | </ | ||
+ | |||
+ | <panel type=" | ||
+ | {{youtube> | ||
+ | </ | ||
+ | |||
+ | <panel type=" | ||
+ | {{youtube> | ||
+ | </ | ||
+ | |||
+ | <panel type=" | ||
+ | {{youtube> | ||
+ | </ | ||
+ | |||
+ | <panel type=" | ||
+ | {{youtube> | ||
+ | </ | ||
+ | |||
+ | <panel type=" | ||
+ | {{youtube> | ||
+ | </ | ||
+ | |||
+ | <panel type=" | ||
+ | {{youtube> | ||
+ | </ | ||