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electrical_engineering_and_electronics_2:block08 [2026/05/12 02:49] – mexleadminelectrical_engineering_and_electronics_2:block08 [2026/05/19 02:12] (current) – mexleadmin
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 \end{align*} \end{align*}
  
-Choose the voltage as reference: 
-\begin{align*} 
-\underline{U}=U\angle 0^\circ 
-\end{align*} 
- 
-The branch currents are: 
-\begin{align*} 
-\underline{I}_R &= \frac{U}{R}\angle 0^\circ \\ 
-\underline{I}_L &= \frac{U}{X_L}\angle -90^\circ \\ 
-\underline{I}_C &= \frac{U}{X_C}\angle +90^\circ 
-\end{align*} 
- 
-with the reactances 
-\begin{align*} 
-X_L &= \omega L \\ 
-X_C &= \frac{1}{\omega C} 
-\end{align*} 
- 
-The source current is the phasor sum: 
-\begin{align*} 
-\underline{I} 
-= \underline{I}_R+\underline{I}_L+\underline{I}_C 
-\end{align*} 
 </WRAP> </WRAP>
 #@PathEnd_HTML@# #@PathEnd_HTML@#
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 <WRAP half column> <WRAP half column>
 #@ResultBegin_HTML~601~@# #@ResultBegin_HTML~601~@#
 +{{drawio>electrical_engineering_and_electronics_2:diagram601.svg}}
 \begin{align*} \begin{align*}
 \underline{U}_R \underline{U}_R
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 <WRAP half column> <WRAP half column>
 #@ResultBegin_HTML~602~@# #@ResultBegin_HTML~602~@#
 +
 +[[https://www.falstad.com/afilter/circuitjs.html?cct=$+1+Infinity+0.75+50+5+50%0A%25+2+24389.095776063856%0Ac+400+160+400+224+0+0.000001+0%0Ar+400+272+400+320+0+0.001%0AO+400+272+432+272+0%0Ag+400+320+400+336+0%0A170+240+160+208+160+3+20+1000+5+0.1%0Al+336+160+336+224+0+0.0025+0%0Ar+272+160+272+224+0+10000%0Aw+240+160+272+160+0%0Aw+272+160+336+160+0%0Aw+336+160+400+160+0%0Aw+400+224+336+224+0%0Aw+336+224+272+224+0%0Aw+400+272+400+224+0%0AB+368+240+448+368+2+Box%0Ax+452+259+540+281+0+16+current%5Cnmeasurment%0A|{{:electrical_engineering_and_electronics_2:imageno602.jpg|click to start simulation}}]]  
 +
 +
 \begin{align*} \begin{align*}
 f_0 \approx 3.18~{\rm kHz} f_0 \approx 3.18~{\rm kHz}
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 #@PathBegin_HTML~604~@# #@PathBegin_HTML~604~@#
 <WRAP leftalign> <WRAP leftalign>
 +Choose the voltage as reference:
 +\begin{align*}
 +\underline{U}=U\angle 0^\circ
 +\end{align*}
 +
 +The branch currents are:
 +\begin{align*}
 +\underline{I}_R &= \frac{U}{R}\angle 0^\circ \\
 +\underline{I}_L &= \frac{U}{X_L}\angle -90^\circ \\
 +\underline{I}_C &= \frac{U}{X_C}\angle +90^\circ
 +\end{align*}
 +
 +with the reactances
 +\begin{align*}
 +X_L &= \omega L \\
 +X_C &= \frac{1}{\omega C}
 +\end{align*}
 +
 At $f=f_0$: At $f=f_0$:
 \begin{align*} \begin{align*}
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 \end{align*} \end{align*}
  
-The branch currents are:+we get:
 \begin{align*} \begin{align*}
 \underline{I}_R \underline{I}_R
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 \underline{I}_C \underline{I}_C
 &= 0.24~{\rm A}\angle +90^\circ &= 0.24~{\rm A}\angle +90^\circ
 +\end{align*}
 +
 +
 +The source current is the phasor sum:
 +\begin{align*}
 +\underline{I}
 += \underline{I}_R+\underline{I}_L+\underline{I}_C
 \end{align*} \end{align*}
  
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 \underline{U} &= 12~{\rm V}\angle 0^\circ \\ \underline{U} &= 12~{\rm V}\angle 0^\circ \\
 \underline{I}_R &= 1.2~{\rm mA}\angle 0^\circ \\ \underline{I}_R &= 1.2~{\rm mA}\angle 0^\circ \\
-\underline{I}_L &= 0.24~{\rm A}\angle -90^\circ \\ +\underline{I}_L &= 240~{\rm mA}\angle -90^\circ \\ 
-\underline{I}_C &= 0.24~{\rm A}\angle +90^\circ \\+\underline{I}_C &= 240~{\rm mA}\angle +90^\circ \\
 \underline{I} &= 1.2~{\rm mA}\angle 0^\circ \underline{I} &= 1.2~{\rm mA}\angle 0^\circ
 \end{align*} \end{align*}
 +
 +{{drawio>electrical_engineering_and_electronics_2:diagram604.svg}}
 +
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 </WRAP> </WRAP>
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 The input reactive power is: The input reactive power is:
 \begin{align*} \begin{align*}
-Q = 0+Q = 0 ~\rm var
 \end{align*} \end{align*}
  
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 \begin{align*} \begin{align*}
 P &= 14.4~{\rm mW} \\ P &= 14.4~{\rm mW} \\
-Q &= 0 \\+Q &= 0 ~{\rm var}\\
 S &= 14.4~{\rm mVA} S &= 14.4~{\rm mVA}
 \end{align*} \end{align*}
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 \underline{I} \underline{I}
 &= 1.2~{\rm mA} - j(0.48~{\rm A}-0.12~{\rm A}) \\ &= 1.2~{\rm mA} - j(0.48~{\rm A}-0.12~{\rm A}) \\
-&= 0.0012 - j0.36~{\rm A}+&= 0.0012 - j \cdot 0.36~{\rm A}
 \end{align*} \end{align*}
  
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 \begin{align*} \begin{align*}
 I &\approx 0.360~{\rm A} \\ I &\approx 0.360~{\rm A} \\
-\varphi_I &\approx -89.8^\circ+\varphi_I &= -89.8 ... ^\circ
 \end{align*} \end{align*}
  
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 <WRAP half column> <WRAP half column>
 #@ResultBegin_HTML~606~@# #@ResultBegin_HTML~606~@#
 +
 +
 \begin{align*} \begin{align*}
 \underline{U} &= 12~{\rm V}\angle 0^\circ \\ \underline{U} &= 12~{\rm V}\angle 0^\circ \\
 \underline{I}_R &= 1.2~{\rm mA}\angle 0^\circ \\ \underline{I}_R &= 1.2~{\rm mA}\angle 0^\circ \\
-\underline{I}_L &= 0.48~{\rm A}\angle -90^\circ \\ +\underline{I}_L &= 480~{\rm mA}\angle -90^\circ \\ 
-\underline{I}_C &= 0.12~{\rm A}\angle +90^\circ \\ +\underline{I}_C &= 120~{\rm mA}\angle +90^\circ \\ 
-\underline{I} &\approx 0.360~{\rm A}\angle -89.8^\circ+\underline{I} &\approx 360~{\rm mA}\angle -90^\circ
 \end{align*} \end{align*}
 +
 +{{drawio>electrical_engineering_and_electronics_2:diagram606.svg}}
 +
 +
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 </WRAP> </WRAP>
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 #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Harmonic Trap for the 5th Grid Harmonic                    #@TaskText_HTML@# #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Harmonic Trap for the 5th Grid Harmonic                    #@TaskText_HTML@#
 +
 +<WRAP right>{{drawio>electrical_engineering_and_electronics_2:diagram630.svg}}</WRAP>
 +
  
 Many power electronic converters generate harmonics of the grid frequency. Many power electronic converters generate harmonics of the grid frequency.
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 \end{align*} \end{align*}
  
-1. Give the complex input impedance of the harmonic trap.+1. Give the formula of the complex input impedance of the harmonic trap.
 <WRAP group> <WRAP group>
 <WRAP half column rightalign> <WRAP half column rightalign>
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 </WRAP> </WRAP>
  
-2. How large must $L_{\rm trap}$ be so that the 5th harmonic is suppressed?+2. How large must $L_{\rm trap}$ be so that the 5th harmonic is suppressed ($\nu=5$)?
 <WRAP group> <WRAP group>
 <WRAP half column rightalign> <WRAP half column rightalign>
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 </WRAP> </WRAP>
  
-3. How large is the complex input impedance for the fundamental frequency $\nu=1$?+3. How large is the complex input impedance for the fundamental frequency ($\nu=1$)?
 <WRAP group> <WRAP group>
 <WRAP half column rightalign> <WRAP half column rightalign>
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 Thus: Thus:
 \begin{align*} \begin{align*}
-\underline{Z}_{\rm SK,1}+\underline{Z}_{\rm trap,1}
 &= j(X_L-X_C) \\ &= j(X_L-X_C) \\
 &= j(84.9~{\rm \Omega}-2122~{\rm \Omega}) \\ &= j(84.9~{\rm \Omega}-2122~{\rm \Omega}) \\
-&= -j2037~{\rm \Omega}+&= -j \cdot 2037~{\rm \Omega}
 \end{align*} \end{align*}
 </WRAP> </WRAP>
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 #@ResultBegin_HTML~633~@# #@ResultBegin_HTML~633~@#
 \begin{align*} \begin{align*}
-\underline{Z}_{\rm SK,1} +\underline{Z}_{\rm trap,1} 
-= -j2037~{\rm \Omega}+= -j \cdot 2037~{\rm \Omega}
 \end{align*} \end{align*}
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
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 At the fundamental frequency, the harmonic trap has the impedance: At the fundamental frequency, the harmonic trap has the impedance:
 \begin{align*} \begin{align*}
-\underline{Z}_{\rm SK,1} = -j2037~{\rm \Omega}+\underline{Z}_{\rm trap,1} = -j \cdot 2037~{\rm \Omega}
 \end{align*} \end{align*}
  
 The current magnitude is: The current magnitude is:
 \begin{align*} \begin{align*}
-I &= \frac{U}{|Z_{\rm SK,1}|} \\+I &= \frac{U}{|Z_{\rm trap,1}|} \\
   &= \frac{230~{\rm V}}{2037~{\rm \Omega}} \\   &= \frac{230~{\rm V}}{2037~{\rm \Omega}} \\
   &= 0.113~{\rm A}   &= 0.113~{\rm A}
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 </WRAP> </WRAP>
  
-5. Draw the magnitude of the input impedance $Z$ as a function of the frequency $\omega$.+5. Draw the magnitude of the input impedance $Z$ as a function of the frequency $\omega$. (not an exam-style question) 
 <WRAP group> <WRAP group>
 <WRAP half column rightalign> <WRAP half column rightalign>
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 f_5=250~{\rm Hz} f_5=250~{\rm Hz}
 \end{align*} \end{align*}
 +
 +{{drawio>electrical_engineering_and_electronics_2:diagram635.svg}}
 +
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 </WRAP> </WRAP>