{{tag>Multiphase_systems RMS power exam_ee2_SS2021}}{{include_n>1280}}
 

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<WRAP mexle-task-title>
~~#@ee1_taskctr~~ Multiphase systems \\
<fs medium>(written test, approx. 4 % of a 120-minute written test, SS2021)</fs> 
</WRAP>



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. \\
Through each of the windings, there is a current with an RMS value $I_{\rm RMS} = 5 ~\rm A$ and a phase shift of $\varphi = +25°$ compared to the voltage. 

a) Draw the circuit diagram.
 

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{{drawio>ee2:EzrkJzIFCEgTtCpC_solution1.svg}}

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b) Specify the RMS value of the phase voltage $U_\rm L$ and the string voltage $U_\rm S$. 


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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 $U_\rm S$. \\
For delta configuration, the phase voltage $U_\rm L$ is equal to the string voltage $U_\rm S$.

{{drawio>ee2:EzrkJzIFCEgTtCpC_solution2.svg}}


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  * $U_{\rm S} = 110~\rm V$
  * $U_{\rm L} = 110~\rm V$

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c) Specify the RMS value of the phase current $I_\rm L$ and the string current $I_\rm S$. 


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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}}


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  * $I_{\rm S} = 5~\rm A$
  * $I_{\rm L} =  8.66~\rm A$

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d) Determine the active power. 


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The active power $P$ is given by: $P = 3 \cdot U \cdot I \cdot \sin(\varphi)$
\begin{align*}
 P &= 3 \cdot  110{~\rm V} \cdot 5{~\rm A} \cdot \sin(25°) \\
   &= 1'610.888... ~\rm W
\end{align*}


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$W = 1.61 ~\rm kW$

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