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ee2:task_ezrkjzifcegttcpc_with_calculation [2024/07/04 10:48] – angelegt mexleadminee2:task_ezrkjzifcegttcpc_with_calculation [2024/07/04 11:34] (current) – mexleadmin
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-{{tag>resonance resonant_circuit RMS exam_ee2_SS2021}}{{include_n>1280}}+{{tag>Multiphase_systems RMS power exam_ee2_SS2021}}{{include_n>1280}}
    
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Resonant Circuit \\+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~ Multiphase systems \\
 <fs medium>(written test, approx. 4 % of a 120-minute written test, SS2021)</fs> #@TaskText_HTML@# <fs medium>(written test, approx. 4 % of a 120-minute written test, SS2021)</fs> #@TaskText_HTML@#
  
-Multiphase systems (10 points)  
 A symmetrical three-phase generator in a delta connection shall be considered in the following. \\ A symmetrical three-phase generator in a delta connection shall be considered in the following. \\
 A voltage with the RMS value $U_{\rm RMS} = 110~\rm V$ is applied between the terminals of each winding. \\ A voltage with the RMS value $U_{\rm RMS} = 110~\rm V$ is applied between the terminals of each winding. \\
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 Since the given voltage of $U_{\rm RMS} = 110~\rm V$ is applied between the terminals of each winding, this is also the string voltage $U_\rm S$. \\ Since the given voltage of $U_{\rm RMS} = 110~\rm V$ is applied between the terminals of each winding, this is also the string voltage $U_\rm S$. \\
 For delta configuration, the phase voltage $U_\rm L$ is equal to the string voltage $U_\rm S$. For delta configuration, the phase voltage $U_\rm L$ is equal to the string voltage $U_\rm S$.
 +
 +{{drawio>ee2:EzrkJzIFCEgTtCpC_solution2.svg}}
 +
 #@HiddenEnd_HTML~EzrkJzIFCEgTtCpC_21,Path~@# #@HiddenEnd_HTML~EzrkJzIFCEgTtCpC_21,Path~@#
  
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 #@HiddenBegin_HTML~EzrkJzIFCEgTtCpC_31,Path~@# #@HiddenBegin_HTML~EzrkJzIFCEgTtCpC_31,Path~@#
 +Since the given current of $I_{\rm RMS} = 5~\rm A$ is running through each winding, this is also the string current $I_\rm S$. \\
 +For the phase current $I_\rm L$, one has to consider that at each node the sum of all the notes must be zero: $\sum_i I_i =0$. \\
 +By this (and showing in the example in the image below), One can see, that $I_\rm L= \sqrt{3} \cdot I_{\rm RMS} = \sqrt{3}  \cdot 5~\rm A$
  
 +{{drawio>ee2:EzrkJzIFCEgTtCpC_solution3.svg}}
  
 #@HiddenEnd_HTML~EzrkJzIFCEgTtCpC_31,Path~@# #@HiddenEnd_HTML~EzrkJzIFCEgTtCpC_31,Path~@#
  
 #@HiddenBegin_HTML~EzrkJzIFCEgTtCpC_32,Result~@# #@HiddenBegin_HTML~EzrkJzIFCEgTtCpC_32,Result~@#
-$f_0 = 205.5 \rm Hz$+  * $I_{\rm S} = 5~\rm A$ 
 +  * $I_{\rm L} =  8.66~\rm A$
 #@HiddenEnd_HTML~EzrkJzIFCEgTtCpC_32,Result~@# #@HiddenEnd_HTML~EzrkJzIFCEgTtCpC_32,Result~@#
  
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 #@HiddenBegin_HTML~EzrkJzIFCEgTtCpC_41,Path~@# #@HiddenBegin_HTML~EzrkJzIFCEgTtCpC_41,Path~@#
-The resonant frequency $f_0$ is given as  
-\begin{align*} 
-f_0 = {{1}\over{ 2\pi \sqrt{LC} }} 
-\end{align*} 
  
-With the values:+The active power $P$ is given by: $P = 3 \cdot U \cdot I \cdot \sin(\varphi)$
 \begin{align*} \begin{align*}
-f_0 &= {{1}\over{ 2\pi \sqrt{20 \cdot 10^{-3} ~\rm H \cdot 30 \cdot 10^{-6} ~\rm F} }} \\ + P &= 3 \cdot  110{~\rm V} \cdot 5{~\rm A} \cdot \sin(25°) \\ 
-    &= 205.4681... \rm Hz+   &= 1'610.888... ~\rm W
 \end{align*} \end{align*}
  
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 #@HiddenBegin_HTML~EzrkJzIFCEgTtCpC_42,Result~@# #@HiddenBegin_HTML~EzrkJzIFCEgTtCpC_42,Result~@#
-$f_0 = 205.5 \rm Hz$+$W = 1.61 ~\rm kW$
 #@HiddenEnd_HTML~EzrkJzIFCEgTtCpC_42,Result~@# #@HiddenEnd_HTML~EzrkJzIFCEgTtCpC_42,Result~@#
  
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#