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electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2023/11/11 16:00]
mexleadmin [Bearbeiten - Panel]
electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2023/12/04 00:22] (aktuell)
mexleadmin
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 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 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>
Zeile 354: Zeile 354:
 <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 386: Zeile 386:
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
 +
 +
 +==== Exercises ====
  
 <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> <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>
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 $R_{\rm m} = 1 ~\Omega$.  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 $R_{\rm s} = 1 ~\Omega$ 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.
Zeile 451: 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@#