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electrical_engineering_and_electronics_2:block08 [2026/05/16 18:47] – mexleadminelectrical_engineering_and_electronics_2:block08 [2026/05/19 02:12] (current) – mexleadmin
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 \begin{align*} \begin{align*}
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- 
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- 
  
 3. How large is the ratio $I_C/I$ at $f=f_0$? 3. How large is the ratio $I_C/I$ at $f=f_0$?
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 X_L &= \omega L \\ X_L &= \omega L \\
 X_C &= \frac{1}{\omega C} 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*} \end{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}}
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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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 \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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 +
 +
 \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}}
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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.
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-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 ($\nu=5$)?
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-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 ($\nu=1$)?
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 f_5=250~{\rm Hz} f_5=250~{\rm Hz}
 \end{align*} \end{align*}
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 +
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