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electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2023/12/03 18:20]
mexleadmin 		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 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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 </WRAP></WRAP></WRAP></panel> </WRAP></WRAP></WRAP></panel>
  
-#@TaskTitle_HTML@#3.3.3 Power of two pole components \\ <fs medium>(written test, approx. xxx % of a xxx-minute written test, WS2xxx)</fs>#@TaskText_HTML@#+#@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 $U_{\rm S} = 3.3 ~\rm V$, $R_{\rm i} = 0.1 ~\Omega$). Two heater resistors (both with $R_\rm L = 0.5 ~\Omega$) shall be supplied with two lithium-ion-batteries (both with $U_{\rm S} = 3.3 ~\rm V$, $R_{\rm i} = 0.1 ~\Omega$).
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 #@HiddenBegin_HTML~Solution333_2,Solution~@# #@HiddenBegin_HTML~Solution333_2,Solution~@#
  
-At the maximum utilization rate $\varepsilon= 0.25$ the maximum power $P_{\rm L ~max}$ can be achieved. \\ +At the maximum utilization rate $\varepsilon = 0.25$ the maximum power $P_{\rm L ~max}$ can be achieved. \\ 
-The utilization rates for the different circuits are: +The utilization rate is given as:
-  - Nummerierter Listenpunkt +
- +
 \begin{align*} \begin{align*}
-\varepsilon = {{R_{\rm L}\cdot R_{\rm i}}                  \over {(R_{\rm L}+R_{\rm i})^2}} \\+\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*} \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~@#
  
-{{drawio>electrical_engineering_1:diagram333_2.svg}} 
-#@HiddenEnd_HTML~Solution333_2,Solution ~@# 
  
 3. What is the value of the maximum power $P_{\rm L ~max}$? 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? 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? 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@# #@TaskEnd_HTML@#