Unterschiede

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--- electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2023/09/19 23:37] – mexleadmin
+++ electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2024/10/31 08:26] (aktuell) – mexleadmin
@@ Zeile 4: / Zeile 4: @@
 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 ideal voltage source concept 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 51: / Zeile 51: @@
 This realization shall now be described with some technical terms:
-  * It is called **open circuit**  when no current is drawn from an active two-terminal network: $I_{\rm LL}=0$. \\ The voltage corresponds to the **open circuit voltage**  $U=U_{\rm OC}$ (German: //Leerlaufspannung//). \\ The open circuit power is $P_{\rm OC}=U_{\rm OC} \cdot I_{\rm OC} = 0$.
+  * It is called **open circuit**  when no current is drawn from an active two-terminal network: $I_{\rm SC}=0$. \\ The voltage corresponds to the **open circuit voltage**  $U=U_{\rm OC}$ (German: //Leerlaufspannung// $U_{\rm LL}$). \\ The open circuit power is $P_{\rm OC}=U_{\rm OC} \cdot I_{\rm OC} = 0$.
-  * The term **short circuit**  is used when the terminals of the two-terminal network are bridged without resistance. The current then flowing is called the **short-circuit current**  $I=I_{\rm SC}$ (German: //Kurzschlussstrom//). \\ The short-circuit voltage is $U_{\rm SC}=0~\rm V$. \\ Also, the short-circuit power is $P_{\rm SC}=U_{\rm SC} \cdot I_{\rm SC} = 0$.
+  * The term **short circuit**  is used when the terminals of the two-terminal network are bridged without resistance. The current then flowing is called the **short-circuit current**  $I=I_{\rm SC}$ (German: //Kurzschlussstrom// $I_{\rm KS}$). \\ The short-circuit voltage is $U_{\rm SC}=0~\rm V$. \\ Also, the short-circuit power is $P_{\rm SC}=U_{\rm SC} \cdot I_{\rm SC} = 0$.
-  * In the region between no-load and short-circuit, the active two-terminal network outputs power to a connected load.
+  * The active two-terminal network outputs power to a connected load in the region between no-load and short-circuit.
-Important: As will be seen in the following, the short-circuit current can cause considerable power loss inside the two-terminal network and thus a lot of waste heat.
+Important: As seen in the following, the short-circuit current can cause considerable power loss inside the two-terminal network and thus a lot of waste heat.
 Not every real two-terminal network is designed for this.
@@ Zeile 87: / Zeile 87: @@
 \begin{align*} U = U_0 - R_{\rm i} \cdot I \end{align*}
-The source voltage $U_0$ of the ideal voltage source is to be measured at the terminals of the two-terminal network if this is unloaded. Then no current flows through the internal resistor $R_i$ and there is no voltage drop there. Therefore: The source voltage is equal to the open circuit voltage $U_0 = U_{\rm LL}$.
+The source voltage $U_0$ of the ideal voltage source will be measured at the terminals of the two-terminal network if this is unloaded. Then no current flows through the internal resistor $R_i$ and there is no voltage drop there. Therefore: The source voltage is equal to the open circuit voltage $U_0 = U_{\rm OC}$.
 \begin{align*} U = U_{\rm OC} - R_{\rm i} \cdot I \end{align*}
@@ 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 ====
-In order to understand the lower diagram in <imgref imageNo14 >, the definition equations of the two reference quantities shall be described here again:
+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):
+The **efficiency**  $\eta$ describes the delivered power (consumer power) concerning the supplied power (power of the ideal source):
 \begin{align*}
 \eta = {{P_{\rm out}}\over{P_{\rm in}}}
@@ Zeile 367: / Zeile 367: @@
 \end{align*}
-The **utilization rate**  $\varepsilon$ describes the delivered power in relation to the maximum possible power of the ideal source.
+Once we want to get the **relative maximum power** out of a system (so maximum power related to the input power) the efficiency should go towards $\eta \rightarrow 100\%$. This situation close to (1.) in <imgref imageNo14>.
+Application:
+  - In __power engineering__ $\eta \rightarrow 100\%$ is often desired: We want the 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 **utilization rate**  $\varepsilon$ describes the delivered power $P_{\rm out}$ concerning the maximum possible power $P_{\rm in, max}$ 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*}
@@ Zeile 378: / Zeile 383: @@
 \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.
+In other applications, the **absolute maximum power** has to be taken from the source, without consideration of the losses via the internal resistance. This corresponds to the situation (2.) in <imgref imageNo14>. For this purpose, the internal resistance of the source and the load 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: $\varepsilon = 25~\%$.
-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: $\varepsilon = 25~\%$
+Application:
+  - In __communications engineering__ the impedance matching of the source (the antenna) and the load (the signal-acquiring microcontroller) uses resistors, capacitors, and inductors.  There, we want to get the maximum power out of an antenna. For this purpose, the internal resistance of the source (e.g., a receiver) and the load (e.g., the downstream evaluation) are matched. An example can be seen in this {{electrical_engineering_1:anp084a_en_-_impedance_matching_for_near_field_com.pdf#page=4|application note for near field communication}}.
+  - Furthermore, also for __photovoltaic cells__ one wants to get the maximum power out. In this case, the concept is often called **{{https://en.wikipedia.org/wiki/Maximum_power_point_tracking|Maximum Power Point Tracking (MPPT)}}  **
 The impedance matching/power matching is also [[https://www.youtube.com/watch?v=BJLlXUD6CsM|here]] explained in a German video.
@@ Zeile 387: / Zeile 393: @@
 ~~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 442: @@
 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 460: @@
 </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)