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electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2023/12/03 20:18]
mexleadmin 		electrical_engineering_1:non-ideal_sources_and_two_terminal_networks [2023/12/04 00:22] (aktuell)
mexleadmin
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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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 The utilization rate is given as: The utilization rate is given as:
 \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*}
  
-The utilization rates for the different circuits are: +As near the resulting equivalent internal resistance approaches the resulting equivalent load resistance, as higher the utilization rate $\varepsilon$ will be.\\ 
-  - \begin{align*} \varepsilon_1 = {{R_{\rm L} \cdot R_{\rm i} }   \over {R_{\rm L} + R_{\rm i} )^2}} = {{0.5 \Omega \cdot 0.1  \Omega }   \over {(0.5 \Omega + 0.1  \Omega )^2}} = 13.9\% \end{align*} +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. 
-  - \begin{align*} \varepsilon_2 = {{2 R_{\rm L} \cdot 2 R_{\rm i} }   \over {2 R_{\rm L} + 2 R_{\rm i} )^2}} = {{1.0 \Omega \cdot 0.2  \Omega }   \over {(1.0 \Omega + 0.2  \Omega )^2}} = 13.9\% \end{align*} +#@HiddenEnd_HTML~Solution333_2,Solution ~@#
-  - \begin{align*} \varepsilon_3 = {{ {{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}} = 24.9\% \end{align*} +
-  - \begin{align*} \varepsilon_4 = {{2 R_{\rm L} \cdot {{1}\over{2}} R_{\rm i} }   \over {2 R_{\rm L} + 2 R_{\rm i} )^2}} ={{1.0 \Omega \cdot 0.05 \Omega }   \over {(1.0 \Omega + 0.05 \Omega )^2}} = 4.5\% \end{align*} +
-  - \begin{align*} \varepsilon_5 = {{{{1}\over{2}} R_{\rm L} \cdot {{1}\over{2}} R_{\rm i} }   \over {{{1}\over{2}} R_{\rm L} + {{1}\over{2}} R_{\rm i} )^2}} ={{0.5 \Omega \cdot 0.05 \Omega }   \over {(0.5 \Omega + 0.05 \Omega )^2}} = 8.3\% \end{align*}+
  
 +#@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*}
  
-{{drawio>electrical_engineering_1:diagram333_2.svg}} +Therefore, the maximum power is: 
-#@HiddenEnd_HTML~Solution333_2,Solution ~@#+\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*}
  
-3. What is the value of the maximum power $P_{\rm L ~max}$?+#@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@#