Unterschiede

Hier werden die Unterschiede zwischen zwei Versionen angezeigt.

--- electrical_engineering_and_electronics_1:block07 [2025/09/29 01:13] – angelegt mexleadmin
+++ electrical_engineering_and_electronics_1:block07 [2025/09/29 01:27] (aktuell) – mexleadmin
@@ Zeile 24: / Zeile 24: @@
 ===== Conceptual overview =====
 <callout icon="fa fa-lightbulb-o" color="blue">
-  - Real sources are modeled by an **ideal source** plus **internal resistance** $R_{\rm i}$; the terminal voltage **drops under load**. :contentReference[oaicite:0]{index=0}
+  - Real sources are modeled by an **ideal source** plus **internal resistance** $R_{\rm i}$; the terminal voltage **drops under load**.
-  - **Efficiency** $\eta$ compares *delivered* to *drawn* power. In the simple DC source–load case, $\displaystyle \eta=\frac{R_{\rm L}}{R_{\rm L}+R_{\rm i}}$ (dimensionless). High-efficiency design wants $R_{\rm L}\gg R_{\rm i}$. :contentReference[oaicite:1]{index=1}
+  - **Efficiency** $\eta$ compares *delivered* to *drawn* power. In the simple DC source–load case, $\displaystyle \eta=\frac{R_{\rm L}}{R_{\rm L}+R_{\rm i}}$ (dimensionless). High-efficiency design wants $R_{\rm L}\gg R_{\rm i}$.
-  - **Utilization rate** $\varepsilon$ compares delivered power to the **maximum** available from the ideal source: $\displaystyle \varepsilon=\frac{R_{\rm L}R_{\rm i}}{(R_{\rm L}+R_{\rm i})^2}$. It peaks at $R_{\rm L}=R_{\rm i}$ with $\varepsilon_{\max}=25~\%$. This is the **maximum power transfer** condition. :contentReference[oaicite:2]{index=2}
+  - **Utilization rate** $\varepsilon$ compares delivered power to the **maximum** available from the ideal source: $\displaystyle \varepsilon=\frac{R_{\rm L}R_{\rm i}}{(R_{\rm L}+R_{\rm i})^2}$. It peaks at $R_{\rm L}=R_{\rm i}$ with $\varepsilon_{\max}=25~\%$. This is the **maximum power transfer** condition.
   - Different goals → different $R_{\rm L}$:
-    * **Power engineering**: maximize $\eta$ → $R_{\rm L}\gg R_{\rm i}$. :contentReference[oaicite:3]{index=3}
+    * **Power engineering**: maximize $\eta$ → $R_{\rm L}\gg R_{\rm i}$.
-    * **Communications** (matching, antennas, RF): maximize $P_{\rm L}$ → $R_{\rm L}=R_{\rm i}$, $\eta=50~\%$. :contentReference[oaicite:4]{index=4}
+    * **Communications** (matching, antennas, RF): maximize $P_{\rm L}$ → $R_{\rm L}=R_{\rm i}$, $\eta=50~\%$.
 </callout>
-<callout icon="fa fa-exclamation" color="red" title="Sign convention (recap)">
-We use **conventional current** and the **passive/active sign convention** as introduced earlier. For a **consumer** (passive convention), $P=U\cdot I>0$ means absorption; for a **source** (active convention), $P=U\cdot I>0$ means delivery. See [[simple_circuits#sign_and_arrow-systems]].
-</callout>
 ~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== Core content =====
@@ Zeile 87: / Zeile 83: @@
 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 Efficiency ====
 To understand the lower diagram in <imgref imageNo14 >, the definition equations of the two reference quantities shall be described here again:
@@ Zeile 113: / Zeile 109: @@
             = {{R_{\rm L}}                                 \over {(R_{\rm L}+R_{\rm i})}}\cdot {{R_{\rm i}} \over {(R_{\rm L}+R_{\rm i})}}}
 \end{align*}
+==== The Utilization Rate ====
 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: $\varepsilon = 25~\%$.
@@ Zeile 121: / Zeile 119: @@
 The impedance matching/power matching is also [[https://www.youtube.com/watch?v=BJLlXUD6CsM|here]] explained in a German video.
+~~PAGEBREAK~~ ~~CLEARFIX~~
+<panel type="info" title="Exercise">
+**Given:** $U_0=12.0~\rm V$, $R_{\rm i}=0.50~\Omega$, $R_{\rm L}=5.0~\Omega$.
+**Find:** $U_{\rm L}$, $I_{\rm L}$, $P_{\rm L}$, $\eta$, $\varepsilon$.
+\begin{align*}
+I_{\rm L} &= \frac{12.0~{\rm V}}{0.50~\Omega+5.0~\Omega} = 2.182~{\rm A} \\
+U_{\rm L} &= I_{\rm L} R_{\rm L}= 2.182~{\rm A}\cdot 5.0~\Omega = 10.91~{\rm V} \\
+P_{\rm L} &= U_{\rm L}I_{\rm L} = 10.91~{\rm V}\cdot 2.182~{\rm A}= 23.8~{\rm W} \\
+P_{\rm in,max}&= \frac{U_0^2}{R_{\rm i}}=\frac{(12.0~{\rm V})^2}{0.50~\Omega}=288~{\rm W} \\
+\eta &= \frac{R_{\rm L}}{R_{\rm i}+R_{\rm L}}=\frac{5.0~\Omega}{5.50~\Omega}=0.909=90.9~\% \\
+\varepsilon &= \frac{R_{\rm L}R_{\rm i}}{(R_{\rm L}+R_{\rm i})^2}=\frac{5.0\cdot 0.50}{(5.50)^2}=0.0826=8.26~\%
+\end{align*}
+Interpretation: very **efficient** (small $R_{\rm i}$) but using only **8.26 %** of the source’s ideal maximum capability $U_0^2/R_{\rm i}$—which is fine for power engineering aims.
+</panel>
 ~~PAGEBREAK~~ ~~CLEARFIX~~
@@ Zeile 147: / Zeile 163: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Utilization rate $\varepsilon$ ====
-Define the **best-case input power** of the ideal source as $P_{\rm in,max}=\dfrac{U_0^2}{R_{\rm i}}$. Then
-\begin{align*}
-\boxed{\;\varepsilon \;=\; \frac{P_{\rm L}}{P_{\rm in,max}} \;=\; \frac{R_{\rm L}R_{\rm i}}{(R_{\rm L}+R_{\rm i})^2}\;}
-\end{align*}
-Maximization gives $R_{\rm L}=R_{\rm i}$ and $\varepsilon_{\max}=25~\%$. (At this point $\eta=50~\%$.) :contentReference[oaicite:7]{index=7}
-~~PAGEBREAK~~ ~~CLEARFIX~~
-<panel type="info" title="Exercise">
-**Given:** $U_0=12.0~\rm V$, $R_{\rm i}=0.50~\Omega$, $R_{\rm L}=5.0~\Omega$.
-**Find:** $U_{\rm L}$, $I_{\rm L}$, $P_{\rm L}$, $\eta$, $\varepsilon$.
-\begin{align*}
-I_{\rm L} &= \frac{12.0~{\rm V}}{0.50~\Omega+5.0~\Omega} = 2.182~{\rm A} \\
-U_{\rm L} &= I_{\rm L} R_{\rm L}= 2.182~{\rm A}\cdot 5.0~\Omega = 10.91~{\rm V} \\
-P_{\rm L} &= U_{\rm L}I_{\rm L} = 10.91~{\rm V}\cdot 2.182~{\rm A}= 23.8~{\rm W} \\
-P_{\rm in,max}&= \frac{U_0^2}{R_{\rm i}}=\frac{(12.0~{\rm V})^2}{0.50~\Omega}=288~{\rm W} \\
-\eta &= \frac{R_{\rm L}}{R_{\rm i}+R_{\rm L}}=\frac{5.0~\Omega}{5.50~\Omega}=0.909=90.9~\% \\
-\varepsilon &= \frac{R_{\rm L}R_{\rm i}}{(R_{\rm L}+R_{\rm i})^2}=\frac{5.0\cdot 0.50}{(5.50)^2}=0.0826=8.26~\%
-\end{align*}
-Interpretation: very **efficient** (small $R_{\rm i}$) but using only **8.26 %** of the source’s ideal maximum capability $U_0^2/R_{\rm i}$—which is fine for power engineering aims.
-</panel>
-~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== Exercises =====
@@ Zeile 180: / Zeile 169: @@
   - (a) Choose $R_{\rm L}=9.0~\Omega$. Compute $I_{\rm L}$, $U_{\rm L}$, $P_{\rm L}$, $\eta$, $\varepsilon$.
   - (b) Choose $R_{\rm L}=1.0~\Omega$. Repeat. Which choice maximizes $P_{\rm L}$? Which yields higher $\eta$?
-**Strategy:** use the boxed formulas in this block; for (b) note $R_{\rm L}=R_{\rm i} \Rightarrow \eta=50~\%$. :contentReference[oaicite:12]{index=12}
+**Strategy:** use the boxed formulas in this block; for (b) note $R_{\rm L}=R_{\rm i} \Rightarrow \eta=50~\%$.
 </WRAP></WRAP></panel>
@@ Zeile 193: / Zeile 182: @@
 <imgcaption BildNr3_0 | Simplification by Norton / Thevenin theorem> </imgcaption> {{drawio>BildNr3_0.svg}}
 </WRAP>
-Tip: Short ideal voltage sources and open ideal current sources to determine the internal resistance. :contentReference[oaicite:13]{index=13}
+Tip: Short ideal voltage sources and open ideal current sources to determine the internal resistance.
 </WRAP></WRAP></panel>
-<panel type="info" title="Exercise 7.4 — Battery monitor shunt: loss vs. measurement accuracy"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<wrap anchor #exercise_3_3_2 />
-Design the shunt $\rm R_S$ for the battery monitor (target range $\pm 0.20~\rm V$ at the ADC). Compute measurable current range, minimum current step, shunt loss at maximum current, and the **efficiency penalty** due to the shunt for a given $R_{\rm L}$.
+<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>
-<WRAP>
-<imgcaption BildNr29 | Sketch of the setup>
-</imgcaption>
-{{drawio>SkizzeBatteriemonitor.svg}}
-</WRAP>
-(Use the prompts inside the panel text on the source page for guidance.) :contentReference[oaicite:14]{index=14}
-</WRAP></WRAP></panel>
-~~PAGEBREAK~~ ~~CLEARFIX~~
+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 $R_{\rm m} = 1 ~\Omega$.
-===== Applications & remarks =====
+The drill has two speed-modes:
-  * **Communications / RF**: power matching networks (R, L, C) to get maximum signal power; see {{https://www.youtube.com/watch?v=BJLlXUD6CsM|here}} and {{electrical_engineering_1:anp084a_en_-_impedance_matching_for_near_field_com.pdf#page=4|application note for near field communication}}. :contentReference[oaicite:15]{index=15}
+  - max power: here, the motor is directly connected to the battery.
-  * **Photovoltaics**: operation near **maximum power point** (MPPT) optimizes $P_{\rm L}$ rather than $\eta$ of the source alone. :contentReference[oaicite:16]{index=16}
+  - reduced power: in this case, a shunt resistor $R_{\rm s} = 1 ~\Omega$ is connected in series to the motor.
+{{drawio>sketchDrillingMachine.svg}}
+Tasks:
+  - Calculate the input and output power 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?
+  - 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:
+{{url>http://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&ctz=CQAgjCAMB0l5AWAnC1b0DYqwExgZDgUgOxgkkIKVYDMkIArCArUwKYC0YYAUAG4gkWHCRxCsYHAA4ocguAgNl0RrwBOE8DPBJxU2Sr6acjEWJCnJeucZC0cDcnUfbD4ASzjgSWAk51lL0todxU1AGVdfR0rNzlxADMAQwAbAGd2OVpeAHdLM3AqAutxSDyS+IcAwwq4nnF-RTKKpoavGqgK6p8XJxty-J6wG3qB1u9nDuiu-LabYfH84UsLONEWgA9LCBGSe1lHBgRLE4BVXm3aQ8h9xgY8Y9OQdMv4m3xmJGYTohAAJTe+CwKCKSHAhBY4FkAEs3jJxEQIDhQUQTr8ToDtijDghZLQpAUICcwLIALZvVhODDHajgDBsDEgC7bKiyMAYfYIGk+J6kkAAETetAQ+gZ9kY4I5fmeWIlUrI9kV5HRz1e2PwQmR9CEqr+EV4QA noborder}}
+</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@#
+~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== Summary =====
 <callout>
   - Real sources: $U_{\rm L}=U_0 \dfrac{R_{\rm L}}{R_{\rm i}+R_{\rm L}}$, $I_{\rm L}=\dfrac{U_0}{R_{\rm i}+R_{\rm L}}$; $P_{\rm L}=\dfrac{U_0^2 R_{\rm L}}{(R_{\rm i}+R_{\rm L})^2}$.
-  - **Efficiency**: $\displaystyle \eta=\dfrac{R_{\rm L}}{R_{\rm i}+R_{\rm L}}$; maximize by $R_{\rm L}\gg R_{\rm i}$ (power engineering). :contentReference[oaicite:17]{index=17}
+  - **Efficiency**: $\displaystyle \eta=\dfrac{R_{\rm L}}{R_{\rm i}+R_{\rm L}}$; maximize by $R_{\rm L}\gg R_{\rm i}$ (power engineering).
-  - **Utilization rate**: $\displaystyle \varepsilon=\dfrac{R_{\rm L}R_{\rm i}}{(R_{\rm L}+R_{\rm i})^2}$; peak $\varepsilon_{\max}=25~\%$ at $R_{\rm L}=R_{\rm i}$ (maximum power transfer; $\eta=50~\%$). :contentReference[oaicite:18]{index=18}
+  - **Utilization rate**: $\displaystyle \varepsilon=\dfrac{R_{\rm L}R_{\rm i}}{(R_{\rm L}+R_{\rm i})^2}$; peak $\varepsilon_{\max}=25~\%$ at $R_{\rm L}=R_{\rm i}$ (maximum power transfer; $\eta=50~\%$).
-  - **Chain efficiencies** multiply: $\eta_{\rm total}=\prod \eta_i$. :contentReference[oaicite:19]{index=19}
+  - **Chain efficiencies** multiply: $\eta_{\rm total}=\prod \eta_i$.
-  - Thevenin/Norton help to **separate** source figures ($U_0$, $R_{\rm i}$) from the load and to reuse the same formulas. :contentReference[oaicite:20]{index=20}
+  - Thevenin/Norton help to **separate** source figures ($U_0$, $R_{\rm i}$) from the load and to reuse the same formulas.
   - **Max efficiency $\eta$**: $R_{\rm L}\rightarrow\infty$ (relative to $R_{\rm i}$) → small current, small loss.
   - **Max delivered power $P_{\rm L}$**: $R_{\rm L}=R_{\rm i}$ (impedance matching). See also {{https://en.wikipedia.org/wiki/Impedance_matching|impedance matching}} and {{https://en.wikipedia.org/wiki/Maximum_power_point_tracking|Maximum Power Point Tracking (MPPT)}} for PV systems.
 </callout>
-</callout>
 ===== Common pitfalls checklist =====