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--- electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2023/07/17 15:09]
mexleadmin [Bearbeiten - Panel]
+++ electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2023/12/04 00:22] (aktuell)
mexleadmin
@@ Zeile 1: / Zeile 1: @@
-====== 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.\\
 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>
@@ Zeile 196: / Zeile 196: @@
 <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?
@@ 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}} $.
   - 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*}
@@ Zeile 354: / Zeile 354: @@
 <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 ====
@@ Zeile 387: / Zeile 387: @@
 ~~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">
@@ Zeile 433: / Zeile 436: @@
 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.
   - reduced power: in this case, a shunt resistor $R_{\rm s} = 1 ~\Omega$ is connected in series to the motor.
@@ Zeile 451: / Zeile 454: @@
 </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 $U_{\rm S} = 3.3 ~\rm V$, $R_{\rm i} = 0.1 ~\Omega$).
+. 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 ~@#
+. 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~@#
+. 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~@#
+. 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~@#
+. 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@#
-<panel type="info" title="Exercise 3.3.n Simplification by Northon / Thevenin theorem"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <WRAP>
+<panel type="info" title="Exercise 3.3.n Simplification by Norton / Thevenin theorem"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <WRAP>
 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)