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electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2023/03/22 19:41]
mexleadmin [Duality of Linear Sources] 		electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2023/12/04 00:22] (aktuell)
mexleadmin
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-====== 3. Linear Sources and two-terminal Networks ======+====== 3 Linear Sources and two-terminal Networks ======
  
 It is known from everyday life that battery voltages drop under heavy loads. This can be seen, for example, when turning the ignition key in winter: The load from the starter motor is sometimes so great that the low beam or radio briefly cuts out.\\ It is known from everyday life that battery voltages drop under heavy loads. This can be seen, for example, when turning the ignition key in winter: The load from the starter motor is sometimes so great that the low beam or radio briefly cuts out.\\
 Another example is a $1.5~\rm V$ battery: If such a battery is short-circuited by a piece of wire, not so much current flows that the piece of wire glows, but noticeably less. Another example is a $1.5~\rm V$ battery: If such a battery is short-circuited by a piece of wire, not so much current flows that the piece of wire glows, but noticeably less.
  
-So it makes sense here to develop the concept of the ideal voltage source further. In addition, we will see that this also opens up a possibility to convert and simplify more complicated circuits.+So it makes sense here to develop the concept of the ideal voltage source further. In addition, we will see that this also opens up the possibility of converting and simplifying more complicated circuits.
  
 <WRAP> <imgcaption imageNo1 | passive two-terminal network> </imgcaption> {{drawio>PassiverZweipol.svg}} </WRAP> <WRAP> <imgcaption imageNo1 | passive two-terminal network> </imgcaption> {{drawio>PassiverZweipol.svg}} </WRAP>
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   - __From linear voltage source to linear current source__: <WRAP>   - __From linear voltage source to linear current source__: <WRAP>
-Given: Source voltage $U_0$, resp. open circuit voltage  $U_{OC}$, internal resistance $R_i$ \\  +Given: Source voltage $U_0$, resp. open circuit voltage  $U_{\rm OC}$, internal resistance $R_\rm i$ \\  
-in question: source current $I_0$, resp. short circuit current $I_{SC}$, internal conductance $G_i$ \\  +in question: source current $I_0$, resp. short circuit current $I_{\rm SC}$, internal conductance $G_\rm i$ \\  
-$\boxed{I_{SC} = {{U_{OC}}\over{R_i}}}$ , $\boxed{G_i = {{1}\over{R_i}}}$ +$\boxed{I_{\rm SC} = {{U_{\rm OC}}\over{R_\rm i}}}$ , $\boxed{G_\rm i = {{1}\over{R_\rm i}}}$ 
 </WRAP> </WRAP>
   - __From linear current source to linear voltage source__: <WRAP>   - __From linear current source to linear voltage source__: <WRAP>
-Given: Source current $I_0$, resp. short-circuit current $I_{SC}$, internal resistance $G_i$ \\  +Given: Source current $I_0$, resp. short-circuit current $I_{\rm SC}$, internal resistance $G_\rm i$ \\  
-in question: source voltage $U_0$, resp. open-circuit voltage $U_{OC}$, internal resistance $R_i$ \\   +in question: source voltage $U_0$, resp. open-circuit voltage $U_{\rm OC}$, internal resistance $R_\rm i$ \\   
-$\boxed{U_{OC} = {{I_{SC}}\over{G_i}}}$ , $\boxed{R_i = {{1}\over{G_i}}}$+$\boxed{U_{\rm OC} = {{I_{\rm SC}}\over{G_\rm i}}}$ , $\boxed{R_\rm i = {{1}\over{G_\rm i}}}$
 </WRAP> </WRAP>
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
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 <WRAP> <imgcaption imageNob7 | Resistance of linear sources> </imgcaption> \\ {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgDOB0YzCsICMZICYaoOyYMxgByoBsAnCZkiAgCw5UCmAtIogFABuIJRIqcP3JJh4QI1CAlFQ4rAE5ceiYUMXKIyGKwDuC3vxV6RrVIh6pU1Q7wsH1265b6qz+sPaXOHt1gEsQOdCsA9TUoeHtggxx8Sw9wOX8YqKSncCRYCJT9SNS3HWjY5WoiQqN8rJ5ix1djUxAqqwa4uwBzL1TzR3x8NLc2htSB7t7WACMQIkQkOFiiOixUePHUcmme6hZeTEW3AA9eElicGjgexGj6pB6AGx8AO3oAQ1kAHQBnAGMAV1lZejuAC7vN4Aex+H3orH2JSQODMeFhJEuiBu9yerze7BB1wBjxa9GBYNkENYQA noborder}} </WRAP> <WRAP> <imgcaption imageNob7 | Resistance of linear sources> </imgcaption> \\ {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgDOB0YzCsICMZICYaoOyYMxgByoBsAnCZkiAgCw5UCmAtIogFABuIJRIqcP3JJh4QI1CAlFQ4rAE5ceiYUMXKIyGKwDuC3vxV6RrVIh6pU1Q7wsH1265b6qz+sPaXOHt1gEsQOdCsA9TUoeHtggxx8Sw9wOX8YqKSncCRYCJT9SNS3HWjY5WoiQqN8rJ5ix1djUxAqqwa4uwBzL1TzR3x8NLc2htSB7t7WACMQIkQkOFiiOixUePHUcmme6hZeTEW3AA9eElicGjgexGj6pB6AGx8AO3oAQ1kAHQBnAGMAV1lZejuAC7vN4Aex+H3orH2JSQODMeFhJEuiBu9yerze7BB1wBjxa9GBYNkENYQA noborder}} </WRAP>
  
-In the simulation, a measuring current $I_\Omega$ is used to determine the resistance value. ((This concept will also be used in an electrical engineering lab experiment on [[:elektrotechnik_labor:1._widerstaende|resistor]]s in the 3rd semester.)) Let us have a look at the properties of the ohmmeter in the simulation by double-clicking on the ohmmeter. Here, a very large measuring current of $I_\Omega=1~\rm A$ is used. This could lead to high voltages or the destruction of components in real setups. \\+In the simulation, a measuring current $I_\Omega$ is used to determine the resistance value. ((This concept will also be used in an electrical engineering lab experiment on [[elektrotechnik_labor:1_widerstaende|resistor]]s in the 3rd semester.)) Let us have a look at the properties of the ohmmeter in the simulation by double-clicking on the ohmmeter. Here, a very large measuring current of $I_\Omega=1~\rm A$ is used. This could lead to high voltages or the destruction of components in real setups. \\
  \\  \\
 In order to understand why is this nevertheless chosen so high in the simulation, do the following: Set the measuring current for both linear sources to (more realistic) $1~\rm mA$. What do you notice? In order to understand why is this nevertheless chosen so high in the simulation, do the following: Set the measuring current for both linear sources to (more realistic) $1~\rm mA$. What do you notice?
Zeile 224: Zeile 224:
   - In the second step, the linear voltage source $U_4$ formed in 1. with $R_4$ can be connected to the resistor $R_3$. From this again a linear current source can be created. This now has a resistance of $R_5 = R_3+R_4$ and an ideal current source with $I_5 = {{U_4}\over{R_3+R_4}}= {{I_4}\cdot{R_4}\over{R_3+R_4}} $.   - In the second step, the linear voltage source $U_4$ formed in 1. with $R_4$ can be connected to the resistor $R_3$. From this again a linear current source can be created. This now has a resistance of $R_5 = R_3+R_4$ and an ideal current source with $I_5 = {{U_4}\over{R_3+R_4}}= {{I_4}\cdot{R_4}\over{R_3+R_4}} $.
   - The circuit diagram that now emerges is a parallel circuit of ideal current sources and resistors. This can be used to determine the values of the ideal equivalent current source and the equivalent resistance:   - The circuit diagram that now emerges is a parallel circuit of ideal current sources and resistors. This can be used to determine the values of the ideal equivalent current source and the equivalent resistance:
-      - ideal equivalent current source $I_{\rm eq}$: \begin{align*} I_ = I_1 + I_3 + I_5 = I_1 + I_3 + I_4\cdot{{R_4}\over{R_3+R_4}} \end{align*}+      - ideal equivalent current source $I_{\rm eq}$: \begin{align*} I_{\rm eq} = I_1 + I_3 + I_5 = I_1 + I_3 + I_4\cdot{{R_4}\over{R_3+R_4}} \end{align*}
       - Substitute conductance $G_{\rm eq}$: \begin{align*} G_{\rm eq} = \Sigma G_i = {{1}\over{R_1}}+{{1}\over{R_2}}+{{1}\over{R_5}}={{1}\over{R_1}}+{{1}\over{R_2}}+{{1}\over{R_3+R_4}} \end{align*}       - Substitute conductance $G_{\rm eq}$: \begin{align*} G_{\rm eq} = \Sigma G_i = {{1}\over{R_1}}+{{1}\over{R_2}}+{{1}\over{R_5}}={{1}\over{R_1}}+{{1}\over{R_2}}+{{1}\over{R_3+R_4}} \end{align*}
  
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 First, it is necessary to consider how to determine the power. The power meter (or wattmeter) consists of a combined ammeter and voltmeter. First, it is necessary to consider how to determine the power. The power meter (or wattmeter) consists of a combined ammeter and voltmeter.
  
-In <imgref imageNo12 > the wattmeter with the circuit symbol can be seen as a round network with crossed measuring inputs. The circuit also shows one wattmeter each for the (not externally measurable) output power of the ideal source $P_S$ and the input power of the load $P_L$.+In <imgref imageNo12 > the wattmeter with the circuit symbol can be seen as a round network with crossed measuring inputs.  
 +The circuit also shows one wattmeter each for the (not externally measurable) output power of the ideal source $P_\rm S$ and the input power of the load $P_\rm L$.
  
 <WRAP> <imgcaption imageNo12 | Power measurement on linear voltage source> </imgcaption> {{drawio>LeistungsmessungLineareSpannungsquelle.svg}} <WRAP> <WRAP> <imgcaption imageNo12 | Power measurement on linear voltage source> </imgcaption> {{drawio>LeistungsmessungLineareSpannungsquelle.svg}} <WRAP>
Zeile 323: Zeile 324:
 The simulation in <imgref imageNo13 > shows the following: The simulation in <imgref imageNo13 > shows the following:
  
-  * The circuit with linear voltage source ($U_0$ and $R_i$), and a resistive load $R_L$. +  * The circuit with linear voltage source ($U_0$ and $R_\rm i$), and a resistive load $R_\rm L$. 
-  * A simulated wattmeter, where the ammeter is implemented by a measuring resistor $R_S$ (English: shunt) and a voltage measurement for $U_S$. The power is then: $P_L = {{1}\over{R_S}}\cdot U_S \cdot U_L$.+  * A simulated wattmeter, where the ammeter is implemented by a measuring resistor $R_\rm S$ (English: shunt) and a voltage measurement for $U_\rm S$. The power is then: $P_\rm L = {{1}\over{R_\rm S}}\cdot U_\rm S \cdot U_\rm L$.
   * in the oscilloscope section (below).   * in the oscilloscope section (below).
-      * On the left is the power $P_L$ plotted against time in a graph.+      * On the left is the power $P_\rm L$ plotted against time in a graph.
       * On the right is the already-known current-voltage diagram of the current values.       * On the right is the already-known current-voltage diagram of the current values.
-  * The slider load resistance $R_L$, with which the value of the load resistance $R_L$ can be changed.+  * The slider load resistance $R_\rm L$, with which the value of the load resistance $R_\rm L$ can be changed.
  
-Now try to vary the value of the load resistance $R_L$ (slider) in the simulation so that the maximum power is achieved. Which resistance value is set?+Now try to vary the value of the load resistance $R_\rm L$ (slider) in the simulation so that the maximum power is achieved. Which resistance value is set?
  
 <WRAP> <imgcaption imageNo13 | power adjustment> </imgcaption> \\ {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0mQrFaB2AzADgJwIGyU1qpDvBPOiIgCyqUCmAtGGAFABuIOVIATFZJ25gq3AQJ6JEYpPBYB3QeBHgwPJaJYAnFWuFDkOdVHBxI83pAp7e-I2e38B1xzemOzCl3wGpUh71DmAQFcroGecEZUodYeIF620dwBZgBGIKhCYMjgmDk8yBBmAB4ZVpC68IhgCPHgFGwA9gA2AC4AhgDmdAA6AM59jQCumgDGdCylEmroiKg8uvh1YBQADo1ydJr9PT0AdgC2dK1bAJZ7nUGWRtmGsSyr8THKSWFFgajIPuhCL5HW7xKZRA6EM6BmmGWFGa5zo7U0kxBah4xBAmGSSyEFAAqoiCLxVCZkmBqOA1ECwERwMhdFRdMhSSsQAAlRHCO7IKxUKzoCBCNSnRF+fzMDKcgncLEsoXocRgQwYEWQ-kgAAy5lQ8Fp3Awvw0pXg1zA2EowlytBV6tKVEwtBWtEN1VBUJAuNKqBNqjUHruEmWaitZTt8rKVmNLoAkiw6eJbNZfHdlAJMDgNX4jJrtYF3cgcs6qKKsC7mo12gATFiNZbGAs4O3QCSMKwN+CMGrGNQ6yu8HIQHL8TCQnjQKSwP0QIqwSDc8G8iQsDB1JyRdxq0tl-rMuh9U59Dp7cabgD66qAA noborder}} </WRAP> <WRAP> <imgcaption imageNo13 | power adjustment> </imgcaption> \\ {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0mQrFaB2AzADgJwIGyU1qpDvBPOiIgCyqUCmAtGGAFABuIOVIATFZJ25gq3AQJ6JEYpPBYB3QeBHgwPJaJYAnFWuFDkOdVHBxI83pAp7e-I2e38B1xzemOzCl3wGpUh71DmAQFcroGecEZUodYeIF620dwBZgBGIKhCYMjgmDk8yBBmAB4ZVpC68IhgCPHgFGwA9gA2AC4AhgDmdAA6AM59jQCumgDGdCylEmroiKg8uvh1YBQADo1ydJr9PT0AdgC2dK1bAJZ7nUGWRtmGsSyr8THKSWFFgajIPuhCL5HW7xKZRA6EM6BmmGWFGa5zo7U0kxBah4xBAmGSSyEFAAqoiCLxVCZkmBqOA1ECwERwMhdFRdMhSSsQAAlRHCO7IKxUKzoCBCNSnRF+fzMDKcgncLEsoXocRgQwYEWQ-kgAAy5lQ8Fp3Awvw0pXg1zA2EowlytBV6tKVEwtBWtEN1VBUJAuNKqBNqjUHruEmWaitZTt8rKVmNLoAkiw6eJbNZfHdlAJMDgNX4jJrtYF3cgcs6qKKsC7mo12gATFiNZbGAs4O3QCSMKwN+CMGrGNQ6yu8HIQHL8TCQnjQKSwP0QIqwSDc8G8iQsDB1JyRdxq0tl-rMuh9U59Dp7cabgD66qAA noborder}} </WRAP>
Zeile 339: Zeile 340:
  
   * Diagram top: current-voltage diagram of a linear voltage source.   * Diagram top: current-voltage diagram of a linear voltage source.
-  * Diagram in the middle: source power $P_S$ and consumer power $P_L$ versus delivered voltage $U_L$. +  * Diagram in the middle: source power $P_\rm S$ and consumer power $P_\rm L$ versus delivered voltage $U_\rm L$. 
-  * Diagram below: Reference quantities over delivered voltage $U_L$.+  * Diagram below: Reference quantities over delivered voltage $U_\rm L$.
  
 The two powers are defined as follows: The two powers are defined as follows:
  
-  * source power: $\, \, \large{ P_S = U_0 \cdot I_L} $ +  * source power: $\, \, \large{ P_\rm S = U_0 \cdot I_\rm L} $ 
-  * consumer power: $\large{ P_L = U_L \cdot I_L} $+  * consumer power: $\large{ P_\rm L = U_\rm L \cdot I_\rm L} $
  
-  - Both power $P_S$ and $P_L$ are equal to 0 without current flow. The source power becomes maximum, at maximum current flow, that is, when the load resistance $R_L=0$. In this case, all the power flows out through the internal resistor. The efficiency drops to $0~\%$. This is the case, for example, with a battery shorted by a wire. +  - Both power $P_\rm S$ and $P_\rm L$ are equal to 0 without current flow. The source power becomes maximum, at maximum current flow, that is when the load resistance $R_\rm L=0$. In this case, all the power flows out through the internal resistor. The efficiency drops to $0~\%$. This is the case, for example, with a battery shorted by a wire. 
-  - If the load resistance becomes just as large as the internal resistance $R_L=R_i$, the result is a voltage divider where the load voltage becomes just half the open circuit voltage: $U_L = {{1}\over{2}}\cdot U_{OC}$. On the other hand, the current is also half the short-circuit current $I_L=I_{SC}$, since the resistance at the ideal voltage source is twice that in the short-circuit case. +  - If the load resistance becomes just as large as the internal resistance $R_\rm L=R_\rm i$, the result is a voltage divider where the load voltage becomes just half the open circuit voltage: $U_\rm L = {{1}\over{2}}\cdot U_{\rm OC}$. On the other hand, the current is also half the short-circuit current $I_\rm L=I_{\rm SC}$, since the resistance at the ideal voltage source is twice that in the short-circuit case. 
-  - If the load resistance becomes high impedance $R_L\rightarrow\infty$, less and less current flows, but more and more voltage drops across the load. Thus, the efficiency increases and approaches $100~\%$ for $R_L\rightarrow\infty$.+  - If the load resistance becomes high impedance $R_{\rm L}\rightarrow\infty$, less and less current flows, but more and more voltage drops across the load. Thus, the efficiency increases and approaches $100~\%$ for $R_{\rm L}\rightarrow\infty$.
  
 <WRAP> <imgcaption imageNo14 | current-voltage diagram, power-voltage diagram and efficiency-voltage diagram> </imgcaption> {{drawio>StromSpannungsDiagrammMitLeistung.svg}} </WRAP> <WRAP> <imgcaption imageNo14 | current-voltage diagram, power-voltage diagram and efficiency-voltage diagram> </imgcaption> {{drawio>StromSpannungsDiagrammMitLeistung.svg}} </WRAP>
  
-The whole context can be investigated in this [[https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgzCAMB0mQrFaB2MAOAnAgbJDmxJt4BGEeDckAFgngFMBaEkgKADcRGNsuAmeL25Cw8PlAnVIVabOjxWAJxDZqXUeJJ806sRJJxIrAO4gSyEXr6QdjDVCXhr652GeMBvaQaOm30jz41a1s+IIdTElx+cKiYtV8uHnikoU8HZTC1OysbXXE5QzY+LRBkxlUzUuZtCQhrQzgAKgAKAENGACMASlbOxgBjbpMnAPt0bPtE-xSJlKMAczMMcTtNFa442VZrZGXVypINxjRpNQAFAH0AGVZOzb4RDEoSakpmNDIje+wyYRAwtIwF5WAAPB6IRhSSjUIRSXhqEg6AA2AEsAHb0NqKAA6AGcAGoAe2RABc2gt6PiAMpEgCuigG9DBmzAe2YeTsAQMiEROjapPxpIAFlS8bSGUyRuZLJpajkCiMQikogEsg4AA7gNCTPRzQKIiTTdzjE3OIxs6TaWymsZ6L4RUb8QROg0OcGMZB-bTsij8HU0Mw6ACqLMY8HZ2lWpH4yBhZkV4JYk1ijxcZD5IAASiySIR1EizG4C7yEyBUSztGk0Ig+Ph-eJMznwYQ1adwERY3tEeJbuCpJH69Q8-wMKXSn3wBgIFyIFJIa3A4XQy2DOpkGowHE7F6l72WYQZ+ggZy2d2gyAAJKsS0ufIqXWaI1hvPsvNA2xac+F85E4z0XE8QAWSxPEGXoABbeh0VJPE7lZdkx21cA9iMXYJAqBIuHbC5LmpVgljGApNi8BwSlWcpKhOblaiBAFGmadoul6JihkcNZ7yhG09EKOA2FMf4FS4ahuMVSI0nCQT1TQ0o+GQVYRIBNBP1KK0QE6AB6docQ1GBIBIYYlmBT9eGMsxSLQyA9jMriaDgYSdDUehyXgoI-mSYdTJBcE3M2KyaC3Py1EzABhNpkQGOlkQFVEiXRfEiQAM3xABRRLEtRAZUWggYAE8CNSFIpLybYln+AJki2Mj-LkhSdFqrhKlw-ClmSaRKosnZ-PKYt5P4YKQEAMeBLggvhGC6vZKLUPrwwRIaRr4HSlWU+91SEyy9ga5heC24tcMnKiDB0Eh4GyHkl35QU8RFMVriJNoABMJoBST7CCG1ajUYbRvG9DlSEpFJmoAbvsWjVnsEw73BdL6RpIJb0ME4sajVEG4d+-zbKEz1VjAYHA2++HwcR5H7BOFxHPmiCSHGj0I3UNA0iPbBKdKX9-1xHF0RAto8WgXMkLsNBNrrdRsHjNm-wAnEuZ5vnWCJAEH0kKJG2gZ4Nc1rWMCYOISF11ClfECBLxaYNunxc5kSJUkFZoQ1pFBT59N4EhoAEJAyDIAp3a4Cx7fAJAEAkNKMqynLcvxZy2jt4EzAkJ2DDiRAYG9pW-n91T489XgJUZMV2elvErmpfEWgWRR6Ggi28TadEHvxO7HstqXAKua4y8rh7hnQQOADEHR8Oz7OYEBrl50ljFRB6AOuuuG7xLMblYIA|Simulation]].+The whole context can be investigated in this [[https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgzCAMB0mQrFaB2MAOAnAgbJDmxJt4BGEeDckAFgngFMBaEkgKADcRGNsuAmeL25Cw8PlAnVIVabOjxWAJxDZqXUeJJ806sRJJxIrAO4gSyEXr6QdjDVCXhr652GeMBvaQaOm30jz41a1s+IIdTElx+cKiYtV8uHnikoU8HZTC1OysbXXE5QzY+LRBkxlUzUuZtCQhrQzgAKgAKAENGACMASlbOxgBjbpMnAPt0bPtE-xSJlKMAczMMcTtNFa442VZrZGXVypINxjRpNQAFAH0AGVZOzb4RDEoSakpmNDIje+wyYRAwtIwF5WAAPB6IRhSSjUIRSXhqEg6AA2AEsAHb0NqKAA6AGcAGoAe2RABc2gt6PiAMpEgCuigG9DBmzAe2YeTsAQMiEROjapPxpIAFlS8bSGUyRuZLJpajkCiMQikogEsg4AA7gNCTPRzQKIiTTdzjE3OIxs6TaWymsZ6L4RUb8QROg0OcGMZB-bTsij8HU0Mw6ACqLMY8HZ2lWpH4yBhZkV4JYk1ijxcZD5IAASiySIR1EizG4C7yEyBUSztGk0Ig+Ph-eJMznwYQ1adwERY3tEeJbuCpJH69Q8-wMKXSn3wBgIFyIFJIa3A4XQy2DOpkGowHE7F6l72WYQZ+ggZy2d2gyAAJKsS0ufIqXWaI1hvPsvNA2xac+F85E4z0XE8QAWSxPEGXoABbeh0VJPE7lZdkx21cA9iMXYJAqBIuHbC5LmpVgljGApNi8BwSlWcpKhOblaiBAFGmadoul6JihkcNZ7yhG09EKOA2FMf4FS4ahuMVSI0nCQT1TQ0o+GQVYRIBNBP1KK0QE6AB6docQ1GBIBIYYlmBT9eGMsxSLQyA9jMriaDgYSdDUehyXgoI-mSYdTJBcE3M2KyaC3Py1EzABhNpkQGOlkQFVEiXRfEiQAM3xABRRLEtRAZUWggYAE8CNSFIpLybYln+AJki2Mj-LkhSdFqrhKlw-ClmSaRKosnZ-PKYt5LvNRADHgPhGC6vZKLUPrMMDIadKVZT73VITLL2BrmF4Vbi1wycqIMHQSHgbIeSXflBTxEUxWuIk2gAE1GgFJPsIIxmBabhru5UhKRSZqGCkAZo1O7BJ29wmr+khZvQwTixqIjppIEbIcUoSprsF7BvBgHIeh+w1vUNGwZGj0I3UNA0iPbBHLLX9-1xHF0RAto8WgSt7LsDAyD4MBJlQPcQGpgCcTphmmdYIkAQfSQokbaBnlluX5YwJg4hIJXUPF8QIEvFpg26fFzmRIlSVFmhDWkUFPn03gSGgAQkDIDmkFWCwTfAJAEAkNKMqynLcvxZy2mNl6HXNgw4kQGAHafT0reIv5nYlRkxX5wCrmpfEWgWRR6Gg3W8TadFrvxS6br1v8BbxK5rnTrPruGdBXYAMQdHw7NZshrkZ0ljFRa6ALO-PC7xLMblYIA|Simulation with a resistor]] or [[https://www.falstad.com/circuit/circuitjs.html?ctz=CQAgzCAMB0lw7ADgIwCYCsVrOWMyAWRA9MSANnMVXInUREwLoFMBaHAKADcQ2BOcnwxCBosOlRRpBSI2lyY6TgCcQ5AnwlS0DNtunI4kTgHcQyeOMkhUkPQZNqNctqkRyXw5ELlGT5mB2wqiadnqooVBmFuSukZo+IZoBfILJaaIi0WoJWjbh+VKKsHDInKhoIOlsGhZVHO7SEHbGcABUABQAhmwARgCUXX1sAMYDMUGuBmDERdGBwW5Rs5rLKZwA5hb8Uvo6u3xJihWQ8Dt7dciHbB4gmgAKAPoAMpx9RzRa-PwWBL8cFDRD7kZCZWzBMC+TgAD0+mDYsl+BFEsiEiQYABsAJYAOxY3RUAB0AM4ANQA9piAC7dTYsUkAZQpAFcVKMWLCjmBzhx7FpXEYmBYGN1qaTqQALBkk5lsjkxSzWHRNfQ2VKFdaxeJREwAB3AczVUlWGQgfkmSxmVsgGx5cncDhsU3mYIdlvi8BNSzyJjhbHgYLciF56AB7k0GJAAFUuWx0Lz3Ht0EHUPBkRZilycGsgokvvswVGAErZshaZAMXB7fDCqrYrl4PbuTBgAhrJP3EUgUtwttZO4GNzwc75kBvOGyRP8BgEXDCfh1qQT8D8CD6SAQWQIshyKOxvtGLTwTT4cSBrtVFdkdezORkBwjy8MACSnHtWhrNjqxos0lQca4LyuD3noaCjt2DwUqYLDEiSACyBIkmyLAALYsLi1Iku83K8ouhrgOcJh2ERfB1K4dyPE8jJbNI+x0cc0SVHsNQ-ncjQMPethtB0PT9EMfHjKon7zIiTrFFgxjlOYYiiUQ8ypEkWqyVqxFVGmezyY6RxVA6IB9AA9D0RJ6jAkDIBM2xQmBQjWbECinOcdlifccB8PJmgsLSOGhEG6RzrZ0Jwr5RxnPc+ChZG3YAMLdJiowspiYrYhSuKkhSABmpIAKIZRl2KjNiGGjAAnrRKl5BV-InNssmuOkjHEWFGnuQwLW1FFzw0ds6RyA1vhMWFNRBCAXoiZogBjwKgbCOdUoh1GNHVdlNJkxNpv6VY4s3tUk7UjVRK4-kYVboGsQrPiAYoStKpIvBS3QACbbVEG2yFoULLdN238r+la5u2n2rSR4JLY0rh1JNyBA2FKkjWD40gBNyAzcDLm-gGNYA5D0O8nDQ5KXg6KI8jcYJloiBZLeVCXlIUEwcSRK4oh3QktA2b4foIYQgCUIZlUdOwUSjPM6znAZbYiARJL4A+tLI3INAmBmQBxYS1LbVvcGnF-sEMAtAo2Bhsgp38MwkC7JwFK2OoKQgDCKCQDYBDQD8rtu+7-DsEkyBe6RxSK-+4CSZAEkvp00YDKSDyYhS1KW-ciTSPbRgPUICuhB7meexwae+zrWBuowfvSLl+WFcVJWkl53TxzoScO6n8gwGCUj+0rmYG+3erQYLJKpddMox4974MBAABihf+K5bkcOOLPUqY2IPbBJK0riD2ksWrycEAA|this one with a variable load]].
  
 ==== The Characteristics: Efficiency and Utilization Rate ==== ==== The Characteristics: Efficiency and Utilization Rate ====
Zeile 359: Zeile 360:
 In order to understand the lower diagram in <imgref imageNo14 >, the definition equations of the two reference quantities shall be described here again: In order to understand the lower diagram in <imgref imageNo14 >, the definition equations of the two reference quantities shall be described here again:
  
-The **efficiency**  $\eta$ describes the delivered power (consumer power) in relation to the supplied power (power of the ideal source): \begin{align*}\eta = {{P_{out}}\over{P_{in}}} = {{R_L\cdot I_L^2}\over{(R_L+R_i) \cdot I_L^2}} \quad \rightarrow \quad \boxed{ \eta = {{R_L}\over{R_L+R_i}} } \end{align*}+The **efficiency**  $\eta$ describes the delivered power (consumer power) in relation to the supplied power (power of the ideal source):  
 +\begin{align*} 
 +\eta = {{P_{\rm out}}\over{P_{\rm in}}}  
 +     = {{R_{\rm L}\cdot I_{\rm L}^2}\over{(R_{\rm L}+R_{\rm i}) \cdot I_{\rm L}^2}} \quad \rightarrow \quad \boxed{ \eta  
 +     = {{R_{\rm L}}                 \over {R_{\rm L}+R_{\rm i}}} }  
 +\end{align*}
  
-The **utilization rate**  $\varepsilon$ describes the delivered power in relation to the maximum possible power of the ideal source. Here, the currently supplied power is not assumed (as in the case of efficiency), but the best possible power of the ideal source, i.e. in the short-circuit case: \begin{align*}\varepsilon = {{P_{out}}\over{P_{in,max}}} = {{R_L\cdot I_L^2}\over{{U_0^2}\over{R_i}}} = {{R_L\cdot R_i \cdot I_L^2} \over {U_0^2}} = {{R_L\cdot R_i \cdot \left({{U_0}\over{R_L+R_i}}\right)^2} \over {U_0^2}} \quad \rightarrow \quad \boxed{\varepsilon = {{R_L\cdot R_i } \over {(R_L+R_i)^2}} = {{R_L} \over {(R_L+R_i)}}\cdot {{R_i} \over {(R_L+R_i)}}} \end{align*}+The **utilization rate**  $\varepsilon$ describes the delivered power in relation to the maximum possible power of the ideal source.  
 +Here, the currently supplied power is not assumed (as in the case of efficiency), but the best possible power of the ideal source, i.e. in the short-circuit case:  
 +\begin{align*} 
 +\varepsilon = {{P_{\rm out}}\over{P_{\rm in, max}}}  
 +            = {{R_{\rm L}                \cdot I_{\rm L}^2}\over{{U_0^2}\over{R_i}}}  
 +            = {{R_{\rm L}\cdot R_{\rm i} \cdot I_{\rm L}^2}\over {U_0^2}}  
 +            = {{R_{\rm L}\cdot R_{\rm i} \cdot \left({{U_0}\over  {R_{\rm L}+R_{\rm i}}}\right)^2} \over {U_0^2}} \quad \rightarrow \quad \boxed{\varepsilon  
 +            = {{R_{\rm L}\cdot R_{\rm i}}                  \over {(R_{\rm L}+R_{\rm i})^2}}  
 +            = {{R_{\rm L}}                                 \over {(R_{\rm L}+R_{\rm i})}}\cdot {{R_{\rm i}} \over {(R_{\rm L}+R_{\rm i})}}}  
 +\end{align*}
  
-In __power engineering__  a situation close to (1.) in <imgref imageNo14 > is desired: maximum power output with the lowest losses at the internal resistance of the source. Thus, the internal resistance of the source should be low compared to the load $R_L \gg R_i $. The efficiency should go towards $\eta \rightarrow 100\%$.+In __power engineering__  a situation close to (1.) in <imgref imageNo14 > is desired: maximum power output with the lowest losses at the internal resistance of the source.  
 +Thus, the internal resistance of the source should be low compared to the load $R_{\rm L} \gg R_{\rm i} $. The efficiency should go towards $\eta \rightarrow 100\%$.
  
 In __communications engineering__, one situation is different and corresponds to the situation (2.): The maximum power is to be taken from the source, without consideration of the losses via the internal resistance. For this purpose, the internal resistance of the source (e.g., a receiver) and the load (e.g., the downstream evaluation) are matched. This case is called **{{https://en.wikipedia.org/wiki/Impedance_matching|impedance matching}}  ** (the impedance is up to for DC circuits equal to the resistance). The utilization rate here becomes maximum: $\varepsilon = 25~\%$ In __communications engineering__, one situation is different and corresponds to the situation (2.): The maximum power is to be taken from the source, without consideration of the losses via the internal resistance. For this purpose, the internal resistance of the source (e.g., a receiver) and the load (e.g., the downstream evaluation) are matched. This case is called **{{https://en.wikipedia.org/wiki/Impedance_matching|impedance matching}}  ** (the impedance is up to for DC circuits equal to the resistance). The utilization rate here becomes maximum: $\varepsilon = 25~\%$
Zeile 371: Zeile 387:
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-<panel type="info" title="Exercise 3.3.1 Simplification by Northon / Thevenin theorem"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <WRAP> 
  
-Simplify the following circuits by the Northon theorem to a linear current source (circuits marked with NT) or by Thevenin theorem to a linear voltage source (marked with TT). +==== Exercises ====
  
-<imgcaption BildNr3_0 | Simplification by Northon / Thevenin theorem> </imgcaption> {{drawio>BildNr3_0.svg}} </WRAP>+<panel type="info" title="Exercise 3.3.1 Simplification by Norton / Thevenin theorem"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <WRAP> 
 + 
 +Simplify the following circuits by the Norton theorem to a linear current source (circuits marked with NT) or by Thevenin theorem to a linear voltage source (marked with TT).  
 + 
 +<imgcaption BildNr3_0 | Simplification by Norton / Thevenin theorem> </imgcaption> {{drawio>BildNr3_0.svg}} </WRAP>
  
 <button size="xs" type="link" collapse="Loesung_3_3_1_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_3_3_1_1_Lösungsweg" collapsed="true">  <button size="xs" type="link" collapse="Loesung_3_3_1_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_3_3_1_1_Lösungsweg" collapsed="true"> 
Zeile 381: Zeile 400:
 Shutting down all sources leads to  Shutting down all sources leads to 
 \begin{equation*}  \begin{equation*} 
-R_{i}= 8~\Omega \end{equation*} +R_{\rm i}= 8~\Omega \end{equation*} 
  
 Next, we figure out the current in the short circuit.  Next, we figure out the current in the short circuit. 
Zeile 393: Zeile 412:
 To substitute the circuit in $b)$ first we determine the inner resistance.  To substitute the circuit in $b)$ first we determine the inner resistance. 
 Shutting down all sources leads to  Shutting down all sources leads to 
-\begin{equation*} R_{i}= 4 ~\Omega \end{equation*} +\begin{equation*} R_{\rm i}= 4 ~\Omega \end{equation*} 
  
 Next, we figure out the voltage at the open circuit.  Next, we figure out the voltage at the open circuit. 
 Thus we know the given current flows through the ideal current source as well as the resistor.  Thus we know the given current flows through the ideal current source as well as the resistor. 
 The voltage drop on the resistor is  The voltage drop on the resistor is 
-\begin{equation*} R_{i}= -4~\Omega \cdot 2~A \end{equation*} +\begin{equation*} R_{\rm i}= -4~\Omega \cdot 2~A \end{equation*} 
  
 The voltage at the open circuit is  The voltage at the open circuit is 
-\begin{equation*} U_S= 2~V + 1~V + U_R \end{equation*}+\begin{equation*} U_{\rm S}= 2~V + 1~V + U_R \end{equation*}
  
 </collapse> <button size="xs" type="link" collapse="Loesung_3_3_1_2_Lösungsweg">{{icon>eye}} Final result</button><collapse id="Loesung_3_3_1_2_Lösungsweg" collapsed="true">  </collapse> <button size="xs" type="link" collapse="Loesung_3_3_1_2_Lösungsweg">{{icon>eye}} Final result</button><collapse id="Loesung_3_3_1_2_Lösungsweg" collapsed="true"> 
Zeile 412: Zeile 431:
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
 +<wrap anchor #exercise_3_3_2 /> 
 <panel type="info" title="Exercise 3.3.2 Internal resistances and Efficieny"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <WRAP> <panel type="info" title="Exercise 3.3.2 Internal resistances and Efficieny"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <WRAP>
  
-For the company „HHN Mechatronics & Robotics“ you shall analyze a competitor product: a simple drilling machine. This contains a battery pack, some electronics, and a motor. For this consideration, the battery pack can be treated as a linear voltage source with $U_s = ~11 V$ and internal resistance of $R_i = 0.1 ~\Omega$. The used motor shall be considered as an ohmic resistance $R_m = 1 ~\Omega$. +For the company „HHN Mechatronics & Robotics“ you shall analyze a competitor product: a simple drilling machine. This contains a battery pack, some electronics, and a motor. For this consideration, the battery pack can be treated as a linear voltage source with $U_{\rm s} = ~11 V$ and internal resistance of $R_{\rm i} = 0.1 ~\Omega$. The used motor shall be considered as an ohmic resistance $R_{\rm m} = 1 ~\Omega$. 
  
-The drill has two speed modes:+The drill has two speed-modes:
   - max power: here, the motor is directly connected to the battery.   - max power: here, the motor is directly connected to the battery.
-  - reduced power: in this case, a shunt resistor $R_s = 1 ~\Omega$ is connected in series to the motor.+  - reduced power: in this case, a shunt resistor $R_{\rm s} = 1 ~\Omega$ is connected in series to the motor.
  
 {{drawio>sketchDrillingMachine.svg}} {{drawio>sketchDrillingMachine.svg}}
Zeile 427: Zeile 447:
   - What are the efficiencies for both modes?   - What are the efficiencies for both modes?
   - Which value should the shunt resistor $R_s$ have, when the reduced power should be exactly half of the maximum power?   - Which value should the shunt resistor $R_s$ have, when the reduced power should be exactly half of the maximum power?
-  - Your company uses the reduced power mode instead of the shunt resistor $R_s$ multiple diodes in series $D$, which generates a constant voltage drop of $U_D = 2.8 ~V$. \\ What are the input and output power, such as the efficiency in this case?+  - Your company uses the reduced power mode instead of the shunt resistor $R_{\rm s}$ multiple diodes in series $D$, which generates a constant voltage drop of $U_D = 2.8 ~V$. \\ What are the input and output power, such as the efficiency in this case?
  
 You can check your results for the currents, voltages, and powers with the following simulation: You can check your results for the currents, voltages, and powers with the following simulation:
Zeile 434: Zeile 454:
  
 </WRAP></WRAP></WRAP></panel> </WRAP></WRAP></WRAP></panel>
 +
 +#@TaskTitle_HTML@#3.3.3 Power of two pole components #@TaskText_HTML@#
 +
 +Two heater resistors (both with $R_\rm L = 0.5 ~\Omega$) shall be supplied with two lithium-ion-batteries (both with $U_{\rm S} = 3.3 ~\rm V$, $R_{\rm i} = 0.1 ~\Omega$).
 +
 +1. What are the possible ways to connect these components?
 +
 +#@HiddenBegin_HTML~Solution333_1,Solution~@#
 +{{drawio>electrical_engineering_1:diagram333_1.svg}}
 +#@HiddenEnd_HTML~Solution333_1,Solution ~@#
 +
 +2. Which circuit can provide the maximum power $P_{\rm L ~max}$ at the loads?
 +
 +#@HiddenBegin_HTML~Solution333_2,Solution~@#
 +
 +At the maximum utilization rate $\varepsilon = 0.25$ the maximum power $P_{\rm L ~max}$ can be achieved. \\
 +The utilization rate is given as:
 +\begin{align*}
 +\varepsilon &= {{P_{\rm out}}\over{P_{\rm in, max}}}  \\
 +            &= {{R_{\rm L}\cdot R_{\rm i}}                  \over {(R_{\rm L}+R_{\rm i})^2}} \\
 +\end{align*}
 +
 +As near the resulting equivalent internal resistance approaches the resulting equivalent load resistance, as higher the utilization rate $\varepsilon$ will be.\\
 +Therefore, a series configuration of the batteries ($2 R_{\rm i} = 0.2~\Omega$) and a parallel configuration of the load (${{1}\over{2}} R_{\rm L}= 0.25~\Omega$) will have the highest output.
 +#@HiddenEnd_HTML~Solution333_2,Solution ~@#
 +
 +#@HiddenBegin_HTML~Result333_2,Result~@#
 +The following configuration has the maximum output power.
 +
 +{{drawio>electrical_engineering_1:diagram333_3.svg}}
 +#@HiddenEnd_HTML~Result333_2,Result~@#
 +
 +
 +3. What is the value of the maximum power $P_{\rm L ~max}$?
 +
 +#@HiddenBegin_HTML~Solution333_3,Solution~@#
 +The maximum utilization rate is:
 +\begin{align*}
 +\varepsilon &= {{{{1}\over{2}} R_{\rm L} \cdot 2 R_{\rm i} }   \over { ({{1}\over{2}} R_{\rm L} + 2 R_{\rm i} )^2}} \\
 +            &= { {0.25 ~\Omega           \cdot 0.2 ~\Omega }   \over { (           0.25 ~\Omega + 0.2 ~\Omega )^2}} \\
 +            &= 24.6~\%
 +\end{align*}
 +
 +Therefore, the maximum power is:
 +\begin{align*}
 +            \varepsilon  &= {{P_{\rm out}}\over{P_{\rm in, max}}} \\
 +\rightarrow P_{\rm out}  &= \varepsilon   \cdot P_{\rm in, max} \\
 +                         &= \varepsilon   \cdot {{U_s^2}\over{R_{\rm i}}} \\
 +                         &= 24.6~\%       \cdot {{(3.3~\rm V)^2}\over{0.1~\Omega}} \\
 +\end{align*}
 +
 +#@HiddenEnd_HTML~Solution333_3,Solution~@#
 +
 +#@HiddenBegin_HTML~Result333_3,Result~@#
 +\begin{align*}
 +P_{\rm out}  = 26.8 W
 +\end{align*}
 +#@HiddenEnd_HTML~Result333_3,Result~@#
 +
 +4. Which circuit has the highest efficiency?
 +
 +#@HiddenBegin_HTML~Solution333_4,Solution~@#
 +The highest efficiency $\eta$ is given when the output power compared to the input power is minimal. \\
 +A parallel configuration of the batteries (${{1}\over{2}} R_{\rm i} = 0.05~\Omega$) and a series configuration of the load ($2 R_{\rm L}= 1.0~\Omega$) will have the highest efficiency.
 +#@HiddenEnd_HTML~Solution333_4,Solution~@#
 +
 +#@HiddenBegin_HTML~Result333_4,Result~@#
 +{{drawio>electrical_engineering_1:diagram333_4.svg}}
 +#@HiddenEnd_HTML~Result333_4,Result~@#
 +
 +5. What is the value of the highest efficiency?
 +
 +#@HiddenBegin_HTML~Solution333_5,Solution~@#
 +The efficiency $\eta$ is given as:
 +\begin{align*}
 +\eta  &= { {2 R_{\rm L} }\over{ 2 R_{\rm L}+ {{1}\over{2}} R_{\rm i} }} \\
 +      &= { { 1.0~\Omega }\over{ 1.0~\Omega + 0.05~\Omega }}
 +\end{align*}
 +
 +#@HiddenEnd_HTML~Solution333_5,Solution~@#
 +
 +#@HiddenBegin_HTML~Result333_5,Result~@#
 +\begin{align*}
 +\eta  = 95.2~\%
 +\end{align*}
 +#@HiddenEnd_HTML~Result333_5,Result~@#
 +\\ \\
 +#@HiddenBegin_HTML~Details333,Detailed Comparison~@#
 +{{drawio>electrical_engineering_1:diagram333_2.svg}}
 +
 +#@HiddenEnd_HTML~Details333,Detailed Comparison~@#
 +
 +
 +#@TaskEnd_HTML@#
  
  
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 Further German exercises can be found in ILIAS (see [[https://ilias.hs-heilbronn.de/goto.php?target=file_488031_download&client_id=iliashhn|here]], page 13 to 15) Further German exercises can be found in ILIAS (see [[https://ilias.hs-heilbronn.de/goto.php?target=file_488031_download&client_id=iliashhn|here]], page 13 to 15)