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| Both sides previous revision Previous revision Next revision | Previous revision | ||
| electrical_engineering_and_electronics_2:block08 [2026/04/21 04:01] – mexleadmin | electrical_engineering_and_electronics_2:block08 [2026/05/19 02:12] (current) – mexleadmin | ||
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| Line 25: | Line 25: | ||
| ===== Conceptual overview ===== | ===== Conceptual overview ===== | ||
| - | <callout icon=" | + | |
| - | - ... | + | |
| - | </ | + | |
| ===== Core content ===== | ===== Core content ===== | ||
| Line 116: | Line 114: | ||
| By this, the course from low to high frequencies is easier to see. The following simulation in <imgref imageNo5> | By this, the course from low to high frequencies is easier to see. The following simulation in <imgref imageNo5> | ||
| - | <WRAP centeralign> | ||
| - | < | ||
| - | </ | ||
| - | {{url> | ||
| - | </ | ||
| + | < | ||
| + | </ | ||
| For further consideration, | For further consideration, | ||
| This allows for a generalized representation. This representation is called **normalization**: | This allows for a generalized representation. This representation is called **normalization**: | ||
| - | < | + | < |
| \begin{align*} | \begin{align*} | ||
| \large{\underline{A} | \large{\underline{A} | ||
| = \frac {\omega L} {\sqrt{R^2 + (\omega L)^2}}\cdot {\rm e}^{{\rm j}\left(\frac{\pi}{2} - \arctan \frac{\omega L}{R} \right)}} | = \frac {\omega L} {\sqrt{R^2 + (\omega L)^2}}\cdot {\rm e}^{{\rm j}\left(\frac{\pi}{2} - \arctan \frac{\omega L}{R} \right)}} | ||
| - | | + | |
| + | \end{align*} | ||
| + | </ | ||
| + | <WRAP right> | ||
| + | \begin{align*} | ||
| \large{\underline{A}_{norm} | \large{\underline{A}_{norm} | ||
| = \frac {\omega L / R}{\sqrt{1 | = \frac {\omega L / R}{\sqrt{1 | ||
| Line 136: | Line 135: | ||
| \end{align*} | \end{align*} | ||
| </ | </ | ||
| - | |||
| This equation behaves quite the same as the one considered so far. | This equation behaves quite the same as the one considered so far. | ||
| Line 213: | Line 211: | ||
| The cut-off frequency is again given by $f_{\rm c} = \frac{R}{2 \pi \cdot L}$. | The cut-off frequency is again given by $f_{\rm c} = \frac{R}{2 \pi \cdot L}$. | ||
| - | <WRAP centeralign> | + | < |
| - | < | + | </imgcaption> |
| - | </ | + | |
| - | {{url> | + | |
| - | </WRAP> | + | |
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| Line 244: | Line 239: | ||
| \end{align*} | \end{align*} | ||
| - | <WRAP centeralign> | + | < |
| - | < | + | </imgcaption> |
| - | </ | + | |
| - | {{url> | + | |
| - | </WRAP> | + | |
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| Line 268: | Line 261: | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | <WRAP centeralign> | + | < |
| - | < | + | </imgcaption> |
| - | </ | + | |
| - | {{url> | + | |
| - | </WRAP> | + | |
| ==== RLC - Series Resonant Circuit ==== | ==== RLC - Series Resonant Circuit ==== | ||
| Line 353: | Line 343: | ||
| < | < | ||
| + | |||
| + | ==== RLC - Parallel Resonant Circuit ==== | ||
| + | |||
| + | For parallel circuits the current has to be investigated: | ||
| + | |||
| + | \begin{align*} | ||
| + | \underline{I}_I | ||
| + | & | ||
| + | & | ||
| + | \end{align*} | ||
| + | |||
| + | \begin{align*} | ||
| + | \boxed{ {{1}\over{\underline{Z}_{\rm eq}}} = {{1}\over{R}} + {\rm j} \cdot \left( \omega C - {{1}\over{\omega L }} \right) | ||
| + | \end{align*} | ||
| + | |||
| + | Here, the **input current** $\underline{I}_I$ gets minimal when $ f_0 = \frac{1}{2\pi \sqrt{LC}}$. \\ | ||
| + | However, the currents $ |\underline{I}_C| = |\underline{I}_L|$ can get much larger then $\underline{I}_I$. | ||
| + | |||
| + | |||
| + | ==== Simulations to Resonant Circuits ==== | ||
| < | < | ||
| <panel title=" | <panel title=" | ||
| - | <WRAP> | + | <imgcaption imageNo51|> |
| - | </WRAP> </ | + | [[https:// |
| + | </imgcaption> | ||
| + | </ | ||
| <panel title=" | <panel title=" | ||
| - | <WRAP>{{url> | + | <imgcaption imageNo52|> |
| - | </WRAP> </ | + | [[www.falstad.com/ |
| + | </imgcaption> | ||
| + | </ | ||
| <panel title=" | <panel title=" | ||
| - | <WRAP> | + | <imgcaption imageNo53|> |
| - | </WRAP> ~~PAGEBREAK~~ ~~CLEARFIX~~ | + | [[https://www.falstad.com/ |
| + | </imgcaption> | ||
| + | </ | ||
| <panel title=" | <panel title=" | ||
| - | <WRAP> | + | <imgcaption imageNo54|> |
| - | </WRAP> </ | + | [[https://www.falstad.com/ |
| + | </imgcaption> | ||
| + | </ | ||
| <panel title=" | <panel title=" | ||
| - | <WRAP> | + | <imgcaption imageNo55|> |
| - | </WRAP> </ | + | [[https://www.falstad.com/ |
| + | </imgcaption> | ||
| + | </ | ||
| + | </ | ||
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| Line 396: | Line 417: | ||
| [[https:// | [[https:// | ||
| - | Simulation in Time Domain < | + | [[https:// |
| - | Simulation in Frequency Domain < | + | [[https:// |
| [[https:// | [[https:// | ||
| Line 491: | Line 512: | ||
| + | # | ||
| + | A test circuit for an electrical actuator contains a resistor, an inductor and a capacitor connected in parallel. | ||
| + | The circuit is supplied by an AC voltage source. At one specific frequency, the inductor and capacitor exchange energy with each other, so that the source current becomes minimal. | ||
| + | Data: | ||
| + | \begin{align*} | ||
| + | U &= 12~{\rm V} \\ | ||
| + | R &= 10~{\rm k\Omega} \\ | ||
| + | L &= 2.5~{\rm mH} \\ | ||
| + | C &= 1~{\rm \mu F} | ||
| + | \end{align*} | ||
| + | 1. Draw the circuit and add all current and voltage phasors. | ||
| + | <WRAP group> | ||
| + | <WRAP half column rightalign> | ||
| + | # | ||
| + | <WRAP leftalign> | ||
| + | The circuit consists of three parallel branches: | ||
| + | \begin{align*} | ||
| + | R \parallel L \parallel C | ||
| + | \end{align*} | ||
| + | The same voltage is present across all three components: | ||
| + | \begin{align*} | ||
| + | \underline{U}_R | ||
| + | = \underline{U}_L | ||
| + | = \underline{U}_C | ||
| + | = \underline{U} | ||
| + | \end{align*} | ||
| + | </ | ||
| + | # | ||
| + | </ | ||
| + | <WRAP half column> | ||
| + | # | ||
| + | {{drawio> | ||
| + | \begin{align*} | ||
| + | \underline{U}_R | ||
| + | = \underline{U}_L | ||
| + | = \underline{U}_C | ||
| + | = \underline{U} | ||
| + | \end{align*} | ||
| + | |||
| + | \begin{align*} | ||
| + | \underline{I}_R &: \text{ in phase with } \underline{U} \\ | ||
| + | \underline{I}_L &: \text{ lags } \underline{U} \text{ by } 90^\circ \\ | ||
| + | \underline{I}_C &: \text{ leads } \underline{U} \text{ by } 90^\circ | ||
| + | \end{align*} | ||
| + | # | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | 2. At which frequency $f_0$ does the input current become minimal? | ||
| + | <WRAP group> | ||
| + | <WRAP half column rightalign> | ||
| + | # | ||
| + | <WRAP leftalign> | ||
| + | The source current becomes minimal when the inductor current and capacitor current have the same magnitude but opposite phase: | ||
| + | \begin{align*} | ||
| + | I_L = I_C | ||
| + | \end{align*} | ||
| + | |||
| + | Since | ||
| + | \begin{align*} | ||
| + | I_L &= \frac{U}{X_L} \\ | ||
| + | I_C &= \frac{U}{X_C} | ||
| + | \end{align*} | ||
| + | |||
| + | this condition becomes | ||
| + | \begin{align*} | ||
| + | X_L = X_C | ||
| + | \end{align*} | ||
| + | |||
| + | With | ||
| + | \begin{align*} | ||
| + | X_L &= \omega_0 L \\ | ||
| + | X_C &= \frac{1}{\omega_0 C} | ||
| + | \end{align*} | ||
| + | |||
| + | we get | ||
| + | \begin{align*} | ||
| + | \omega_0 L &= \frac{1}{\omega_0 C} \\ | ||
| + | \omega_0 &= \frac{1}{\sqrt{LC}} | ||
| + | \end{align*} | ||
| + | |||
| + | Therefore: | ||
| + | \begin{align*} | ||
| + | f_0 &= \frac{1}{2\pi\sqrt{LC}} \\ | ||
| + | &= \frac{1}{2\pi\sqrt{2.5\cdot 10^{-3}~{\rm H}\cdot 1\cdot 10^{-6}~{\rm F}}} \\ | ||
| + | & | ||
| + | \end{align*} | ||
| + | </ | ||
| + | # | ||
| + | </ | ||
| + | <WRAP half column> | ||
| + | # | ||
| + | |||
| + | [[https:// | ||
| + | |||
| + | |||
| + | \begin{align*} | ||
| + | f_0 \approx 3.18~{\rm kHz} | ||
| + | \end{align*} | ||
| + | # | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | 3. How large is the ratio $I_C/I$ at $f=f_0$? | ||
| + | <WRAP group> | ||
| + | <WRAP half column rightalign> | ||
| + | # | ||
| + | <WRAP leftalign> | ||
| + | At resonance: | ||
| + | \begin{align*} | ||
| + | X_L = X_C | ||
| + | \end{align*} | ||
| + | |||
| + | The common reactance is | ||
| + | \begin{align*} | ||
| + | X_0 &= \omega_0 L | ||
| + | = \frac{1}{\omega_0 C} | ||
| + | = \sqrt{\frac{L}{C}} \\ | ||
| + | &= \sqrt{\frac{2.5\cdot 10^{-3}~{\rm H}}{1\cdot 10^{-6}~{\rm F}}} \\ | ||
| + | &= 50~{\rm \Omega} | ||
| + | \end{align*} | ||
| + | |||
| + | The capacitor current is therefore | ||
| + | \begin{align*} | ||
| + | I_C &= \frac{U}{X_0} | ||
| + | = \frac{12~{\rm V}}{50~{\rm \Omega}} | ||
| + | = 0.24~{\rm A} | ||
| + | \end{align*} | ||
| + | |||
| + | At resonance, $I_L$ and $I_C$ cancel each other in the source current. | ||
| + | Therefore, the source current is only the resistor current: | ||
| + | \begin{align*} | ||
| + | I &= I_R | ||
| + | = \frac{U}{R} | ||
| + | = \frac{12~{\rm V}}{10~{\rm k\Omega}} | ||
| + | = 1.2~{\rm mA} | ||
| + | \end{align*} | ||
| + | |||
| + | Thus: | ||
| + | \begin{align*} | ||
| + | \frac{I_C}{I} | ||
| + | = \frac{0.24~{\rm A}}{1.2~{\rm mA}} | ||
| + | = 200 | ||
| + | \end{align*} | ||
| + | </ | ||
| + | # | ||
| + | </ | ||
| + | <WRAP half column> | ||
| + | # | ||
| + | \begin{align*} | ||
| + | I_C &= 0.24~{\rm A} \\ | ||
| + | I &= 1.2~{\rm mA} \\ | ||
| + | \frac{I_C}{I} &= 200 | ||
| + | \end{align*} | ||
| + | # | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | 4. Draw the phasor diagram for $f=f_0$. | ||
| + | <WRAP group> | ||
| + | <WRAP half column rightalign> | ||
| + | # | ||
| + | <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 $f=f_0$: | ||
| + | \begin{align*} | ||
| + | X_L = X_C = 50~{\rm \Omega} | ||
| + | \end{align*} | ||
| + | |||
| + | we get: | ||
| + | \begin{align*} | ||
| + | \underline{I}_R | ||
| + | &= 1.2~{\rm mA}\angle 0^\circ \\ | ||
| + | \underline{I}_L | ||
| + | &= 0.24~{\rm A}\angle -90^\circ \\ | ||
| + | \underline{I}_C | ||
| + | &= 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*} | ||
| + | |||
| + | Since | ||
| + | \begin{align*} | ||
| + | \underline{I}_L + \underline{I}_C = 0 | ||
| + | \end{align*} | ||
| + | |||
| + | the total current is: | ||
| + | \begin{align*} | ||
| + | \underline{I} = \underline{I}_R | ||
| + | = 1.2~{\rm mA}\angle 0^\circ | ||
| + | \end{align*} | ||
| + | |||
| + | So the voltage and total current are in phase. | ||
| + | </ | ||
| + | # | ||
| + | </ | ||
| + | <WRAP half column> | ||
| + | # | ||
| + | \begin{align*} | ||
| + | \underline{U} &= 12~{\rm V}\angle 0^\circ \\ | ||
| + | \underline{I}_R &= 1.2~{\rm mA}\angle 0^\circ \\ | ||
| + | \underline{I}_L &= 240~{\rm mA}\angle -90^\circ \\ | ||
| + | \underline{I}_C &= 240~{\rm mA}\angle +90^\circ \\ | ||
| + | \underline{I} &= 1.2~{\rm mA}\angle 0^\circ | ||
| + | \end{align*} | ||
| + | |||
| + | {{drawio> | ||
| + | |||
| + | # | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | 5. How large are the active, reactive and apparent power at the input of the circuit? | ||
| + | <WRAP group> | ||
| + | <WRAP half column rightalign> | ||
| + | # | ||
| + | <WRAP leftalign> | ||
| + | At resonance, the inductive and capacitive reactive currents cancel at the input. | ||
| + | Therefore the input behaves like a purely resistive load. | ||
| + | |||
| + | The active power is: | ||
| + | \begin{align*} | ||
| + | P &= \frac{U^2}{R} \\ | ||
| + | &= \frac{(12~{\rm V})^2}{10~{\rm k\Omega}} \\ | ||
| + | &= 14.4~{\rm mW} | ||
| + | \end{align*} | ||
| + | |||
| + | The input reactive power is: | ||
| + | \begin{align*} | ||
| + | Q = 0 ~\rm var | ||
| + | \end{align*} | ||
| + | |||
| + | The source current is: | ||
| + | \begin{align*} | ||
| + | I = 1.2~{\rm mA} | ||
| + | \end{align*} | ||
| + | |||
| + | Thus the apparent power is: | ||
| + | \begin{align*} | ||
| + | S &= UI \\ | ||
| + | &= 12~{\rm V}\cdot 1.2~{\rm mA} \\ | ||
| + | &= 14.4~{\rm mVA} | ||
| + | \end{align*} | ||
| + | </ | ||
| + | # | ||
| + | </ | ||
| + | <WRAP half column> | ||
| + | # | ||
| + | \begin{align*} | ||
| + | P &= 14.4~{\rm mW} \\ | ||
| + | Q &= 0 ~{\rm var}\\ | ||
| + | S &= 14.4~{\rm mVA} | ||
| + | \end{align*} | ||
| + | # | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | 6. Draw the phasor diagram for $f=f_0/2$. | ||
| + | <WRAP group> | ||
| + | <WRAP half column rightalign> | ||
| + | # | ||
| + | <WRAP leftalign> | ||
| + | At half the resonance frequency: | ||
| + | \begin{align*} | ||
| + | f &= \frac{f_0}{2} | ||
| + | \end{align*} | ||
| + | |||
| + | The inductive reactance is halved: | ||
| + | \begin{align*} | ||
| + | X_L &= 25~{\rm \Omega} | ||
| + | \end{align*} | ||
| + | |||
| + | The capacitive reactance is doubled: | ||
| + | \begin{align*} | ||
| + | X_C &= 100~{\rm \Omega} | ||
| + | \end{align*} | ||
| + | |||
| + | The branch currents are: | ||
| + | \begin{align*} | ||
| + | \underline{I}_R | ||
| + | &= \frac{12~{\rm V}}{10~{\rm k\Omega}}\angle 0^\circ | ||
| + | = 1.2~{\rm mA}\angle 0^\circ \\ | ||
| + | \underline{I}_L | ||
| + | &= \frac{12~{\rm V}}{25~{\rm \Omega}}\angle -90^\circ | ||
| + | = 0.48~{\rm A}\angle -90^\circ \\ | ||
| + | \underline{I}_C | ||
| + | &= \frac{12~{\rm V}}{100~{\rm \Omega}}\angle +90^\circ | ||
| + | = 0.12~{\rm A}\angle +90^\circ | ||
| + | \end{align*} | ||
| + | |||
| + | The total current is: | ||
| + | \begin{align*} | ||
| + | \underline{I} | ||
| + | &= 1.2~{\rm mA} - j(0.48~{\rm A}-0.12~{\rm A}) \\ | ||
| + | &= 0.0012 - j \cdot 0.36~{\rm A} | ||
| + | \end{align*} | ||
| + | |||
| + | Therefore: | ||
| + | \begin{align*} | ||
| + | I & | ||
| + | \varphi_I &= -89.8 ... ^\circ | ||
| + | \end{align*} | ||
| + | |||
| + | The circuit behaves almost like an inductive load at $f=f_0/2$. | ||
| + | </ | ||
| + | # | ||
| + | </ | ||
| + | <WRAP half column> | ||
| + | # | ||
| + | |||
| + | |||
| + | \begin{align*} | ||
| + | \underline{U} &= 12~{\rm V}\angle 0^\circ \\ | ||
| + | \underline{I}_R &= 1.2~{\rm mA}\angle 0^\circ \\ | ||
| + | \underline{I}_L &= 480~{\rm mA}\angle -90^\circ \\ | ||
| + | \underline{I}_C &= 120~{\rm mA}\angle +90^\circ \\ | ||
| + | \underline{I} & | ||
| + | \end{align*} | ||
| + | |||
| + | {{drawio> | ||
| + | |||
| + | |||
| + | # | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | # | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | # | ||
| + | |||
| + | <WRAP right> | ||
| + | |||
| + | |||
| + | Many power electronic converters generate harmonics of the grid frequency. | ||
| + | Typical harmonic frequencies are | ||
| + | \begin{align*} | ||
| + | f_\nu = \nu \cdot f_{\rm grid} | ||
| + | \end{align*} | ||
| + | with | ||
| + | \begin{align*} | ||
| + | \nu = 5, 7, 11, 13, \ldots | ||
| + | \end{align*} | ||
| + | |||
| + | To suppress a selected harmonic, a series resonant circuit is connected as a harmonic trap. | ||
| + | At its resonance frequency, the trap has a very small impedance and therefore offers the selected harmonic current a low-impedance path. | ||
| + | |||
| + | Data: | ||
| + | \begin{align*} | ||
| + | C_{\rm trap} &= 1.5~{\rm \mu F} \\ | ||
| + | U &= 230~{\rm V} \\ | ||
| + | f_{\rm grid} &= 50~{\rm Hz} | ||
| + | \end{align*} | ||
| + | |||
| + | 1. Give the formula of the complex input impedance of the harmonic trap. | ||
| + | <WRAP group> | ||
| + | <WRAP half column rightalign> | ||
| + | # | ||
| + | <WRAP leftalign> | ||
| + | The harmonic trap is a series connection of an inductor and a capacitor: | ||
| + | \begin{align*} | ||
| + | \underline{Z}_{\rm trap} | ||
| + | = \underline{Z}_{L_{\rm trap}}+\underline{Z}_{C_{\rm trap}} | ||
| + | \end{align*} | ||
| + | |||
| + | With | ||
| + | \begin{align*} | ||
| + | \underline{Z}_{L_{\rm trap}} &= j\omega L_{\rm trap} \\ | ||
| + | \underline{Z}_{C_{\rm trap}} &= \frac{1}{j\omega C_{\rm trap}} | ||
| + | = -j\frac{1}{\omega C_{\rm trap}} | ||
| + | \end{align*} | ||
| + | |||
| + | the input impedance becomes: | ||
| + | \begin{align*} | ||
| + | \underline{Z}_{\rm trap} | ||
| + | &= j\omega L_{\rm trap} - j\frac{1}{\omega C_{\rm trap}} \\ | ||
| + | &= j\left(\omega L_{\rm trap}-\frac{1}{\omega C_{\rm trap}}\right) | ||
| + | \end{align*} | ||
| + | </ | ||
| + | # | ||
| + | </ | ||
| + | <WRAP half column> | ||
| + | # | ||
| + | \begin{align*} | ||
| + | \underline{Z}_{\rm trap} | ||
| + | = j\left(\omega L_{\rm trap}-\frac{1}{\omega C_{\rm trap}}\right) | ||
| + | \end{align*} | ||
| + | # | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | 2. How large must $L_{\rm trap}$ be so that the 5th harmonic is suppressed ($\nu=5$)? | ||
| + | <WRAP group> | ||
| + | <WRAP half column rightalign> | ||
| + | # | ||
| + | <WRAP leftalign> | ||
| + | The 5th harmonic has the frequency: | ||
| + | \begin{align*} | ||
| + | f_5 &= 5 f_{\rm grid} | ||
| + | = 5\cdot 50~{\rm Hz} | ||
| + | = 250~{\rm Hz} | ||
| + | \end{align*} | ||
| + | |||
| + | The angular frequency is: | ||
| + | \begin{align*} | ||
| + | \omega_5 &= 2\pi f_5 | ||
| + | = 2\pi\cdot 250~{\rm s^{-1}} | ||
| + | = 1570.8~{\rm s^{-1}} | ||
| + | \end{align*} | ||
| + | |||
| + | For a series resonant circuit, the impedance becomes zero when the two reactances have equal magnitude: | ||
| + | \begin{align*} | ||
| + | X_L = X_C | ||
| + | \end{align*} | ||
| + | |||
| + | So: | ||
| + | \begin{align*} | ||
| + | \omega_5 L_{\rm trap} | ||
| + | = \frac{1}{\omega_5 C_{\rm trap}} | ||
| + | \end{align*} | ||
| + | |||
| + | Therefore: | ||
| + | \begin{align*} | ||
| + | L_{\rm trap} | ||
| + | &= \frac{1}{\omega_5^2 C_{\rm trap}} \\ | ||
| + | &= \frac{1}{(1570.8~{\rm s^{-1}})^2\cdot 1.5\cdot 10^{-6}~{\rm F}} \\ | ||
| + | &= 0.270~{\rm H} | ||
| + | \end{align*} | ||
| + | </ | ||
| + | # | ||
| + | </ | ||
| + | <WRAP half column> | ||
| + | # | ||
| + | \begin{align*} | ||
| + | L_{\rm trap} \approx 0.270~{\rm H} | ||
| + | = 270~{\rm mH} | ||
| + | \end{align*} | ||
| + | # | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | 3. How large is the complex input impedance for the fundamental frequency ($\nu=1$)? | ||
| + | <WRAP group> | ||
| + | <WRAP half column rightalign> | ||
| + | # | ||
| + | <WRAP leftalign> | ||
| + | For the fundamental frequency: | ||
| + | \begin{align*} | ||
| + | f_1 &= f_{\rm grid} = 50~{\rm Hz} \\ | ||
| + | \omega_1 &= 2\pi f_1 = 314.16~{\rm s^{-1}} | ||
| + | \end{align*} | ||
| + | |||
| + | The inductive reactance is: | ||
| + | \begin{align*} | ||
| + | X_L &= \omega_1 L_{\rm trap} \\ | ||
| + | &= 314.16~{\rm s^{-1}}\cdot 0.270~{\rm H} \\ | ||
| + | &= 84.9~{\rm \Omega} | ||
| + | \end{align*} | ||
| + | |||
| + | The capacitive reactance is: | ||
| + | \begin{align*} | ||
| + | X_C &= \frac{1}{\omega_1 C_{\rm trap}} \\ | ||
| + | &= \frac{1}{314.16~{\rm s^{-1}}\cdot 1.5\cdot 10^{-6}~{\rm F}} \\ | ||
| + | &= 2122~{\rm \Omega} | ||
| + | \end{align*} | ||
| + | |||
| + | Thus: | ||
| + | \begin{align*} | ||
| + | \underline{Z}_{\rm trap,1} | ||
| + | &= j(X_L-X_C) \\ | ||
| + | &= j(84.9~{\rm \Omega}-2122~{\rm \Omega}) \\ | ||
| + | &= -j \cdot 2037~{\rm \Omega} | ||
| + | \end{align*} | ||
| + | </ | ||
| + | # | ||
| + | </ | ||
| + | <WRAP half column> | ||
| + | # | ||
| + | \begin{align*} | ||
| + | \underline{Z}_{\rm trap,1} | ||
| + | = -j \cdot 2037~{\rm \Omega} | ||
| + | \end{align*} | ||
| + | # | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | 4. What reactive power is absorbed by the harmonic trap at $f=f_{\rm grid}$? | ||
| + | <WRAP group> | ||
| + | <WRAP half column rightalign> | ||
| + | # | ||
| + | <WRAP leftalign> | ||
| + | At the fundamental frequency, the harmonic trap has the impedance: | ||
| + | \begin{align*} | ||
| + | \underline{Z}_{\rm trap,1} = -j \cdot 2037~{\rm \Omega} | ||
| + | \end{align*} | ||
| + | |||
| + | The current magnitude is: | ||
| + | \begin{align*} | ||
| + | I &= \frac{U}{|Z_{\rm trap,1}|} \\ | ||
| + | &= \frac{230~{\rm V}}{2037~{\rm \Omega}} \\ | ||
| + | &= 0.113~{\rm A} | ||
| + | \end{align*} | ||
| + | |||
| + | Since the impedance is capacitive, the reactive power is negative: | ||
| + | \begin{align*} | ||
| + | Q &= -UI \\ | ||
| + | &= -230~{\rm V}\cdot 0.113~{\rm A} \\ | ||
| + | &= -26.0~{\rm var} | ||
| + | \end{align*} | ||
| + | |||
| + | Equivalently: | ||
| + | \begin{align*} | ||
| + | Q &= \frac{U^2}{X_L-X_C} \\ | ||
| + | &= \frac{(230~{\rm V})^2}{84.9~{\rm \Omega}-2122~{\rm \Omega}} \\ | ||
| + | &= -26.0~{\rm var} | ||
| + | \end{align*} | ||
| + | </ | ||
| + | # | ||
| + | </ | ||
| + | <WRAP half column> | ||
| + | # | ||
| + | \begin{align*} | ||
| + | I &= 0.113~{\rm A} \\ | ||
| + | Q &= -26.0~{\rm var} | ||
| + | \end{align*} | ||
| + | # | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | 5. Draw the magnitude of the input impedance $Z$ as a function of the frequency $\omega$. (not an exam-style question) | ||
| + | <WRAP group> | ||
| + | <WRAP half column rightalign> | ||
| + | # | ||
| + | <WRAP leftalign> | ||
| + | The input impedance is: | ||
| + | \begin{align*} | ||
| + | \underline{Z}_{\rm trap} | ||
| + | = j\left(\omega L_{\rm trap}-\frac{1}{\omega C_{\rm trap}}\right) | ||
| + | \end{align*} | ||
| + | |||
| + | Therefore, the magnitude is: | ||
| + | \begin{align*} | ||
| + | |Z_{\rm trap}| | ||
| + | = | ||
| + | \left| | ||
| + | \omega L_{\rm trap}-\frac{1}{\omega C_{\rm trap}} | ||
| + | \right| | ||
| + | \end{align*} | ||
| + | |||
| + | For the sketch: | ||
| + | \begin{align*} | ||
| + | |Z_{\rm trap}| & | ||
| + | \qquad \text{for very small } \omega \\ | ||
| + | |Z_{\rm trap}| &= 0 | ||
| + | \qquad \text{at } \omega=\omega_5 \\ | ||
| + | |Z_{\rm trap}| & | ||
| + | \qquad \text{for very large } \omega | ||
| + | \end{align*} | ||
| + | |||
| + | The minimum occurs at: | ||
| + | \begin{align*} | ||
| + | \omega_5 = 1570.8~{\rm s^{-1}} | ||
| + | \end{align*} | ||
| + | |||
| + | or: | ||
| + | \begin{align*} | ||
| + | f_5 = 250~{\rm Hz} | ||
| + | \end{align*} | ||
| + | </ | ||
| + | # | ||
| + | </ | ||
| + | <WRAP half column> | ||
| + | # | ||
| + | \begin{align*} | ||
| + | |Z_{\rm trap}| | ||
| + | = | ||
| + | \left| | ||
| + | \omega L_{\rm trap}-\frac{1}{\omega C_{\rm trap}} | ||
| + | \right| | ||
| + | \end{align*} | ||
| + | |||
| + | \begin{align*} | ||
| + | |Z_{\rm trap}|_{\rm min}=0 | ||
| + | \quad \text{at} \quad | ||
| + | f_5=250~{\rm Hz} | ||
| + | \end{align*} | ||
| + | |||
| + | {{drawio> | ||
| + | |||
| + | # | ||
| + | </ | ||
| + | </ | ||
| + | # | ||