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electrical_engineering_1:introduction_in_alternating_current_technology [2023/12/16 00:36] – [Bearbeiten - Panel] mexleadminelectrical_engineering_1:introduction_in_alternating_current_technology [2024/12/04 14:43] (aktuell) – [Bearbeiten - Panel] mexleadmin
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 Up to now, we used the following formula to represent alternating voltages: Up to now, we used the following formula to represent alternating voltages:
  
-$$u(t)= \sqrt{2} \hat{U} \cdot \sin (\varphi)$$+$$u(t)= \sqrt{2} U \cdot \sin (\varphi)$$
  
 This is now interpreted as the instantaneous value of a complex vector $\underline{u}(t)$, which rotates given by the time-dependent angle $\varphi = \omega t + \varphi_u$. This is now interpreted as the instantaneous value of a complex vector $\underline{u}(t)$, which rotates given by the time-dependent angle $\varphi = \omega t + \varphi_u$.
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 Therefore, the known properties of complex numbers from Mathematics 101 can be applied: Therefore, the known properties of complex numbers from Mathematics 101 can be applied:
   * A multiplication with $j$ equals a phase shift of $+90°$   * A multiplication with $j$ equals a phase shift of $+90°$
-  * A multiplication with $-j$ equals a phase shift of $-90°$+  * A multiplication with ${{1}\over{j}}$ equals a phase shift of $-90°$
  
 ===== 6.5 Complex Impedance ===== ===== 6.5 Complex Impedance =====
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   * $X = Z \sin \varphi$   * $X = Z \sin \varphi$
  
-value - and therefore a phasor - can simply ==== 6.5.2 Application on pure Loads ====+y ==== 6.5.2 Application on pure Loads ====
  
 With the complex impedance in mind, the <tabref tab01> can be expanded to:  With the complex impedance in mind, the <tabref tab01> can be expanded to: 
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 \\ \\ \\ \\
 The relationship between ${\rm j}$ and integral calculus should be clear:  The relationship between ${\rm j}$ and integral calculus should be clear: 
-  - The derivative of a sinusoidal value - and therefore a phasor - can simply be written as "$\cdot {\rm j}$", \\ which also means a phase shift of $+90°$: \\ \begin{align*}{{\rm d}\over{{\rm d}t}} {\rm e}^{{\rm j}(\omega t + \varphi_x)} = {\rm j} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}\end{align*} +  - The derivative of a sinusoidal value - and therefore a phasor - can simply be written as "$\cdot {\rm j}\omega$", \\ which also means a phase shift of $+90°$: \\ \begin{align*}{{\rm d}\over{{\rm d}t}} {\rm e}^{{\rm j}(\omega t + \varphi_x)} = {\rm j} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}\end{align*} 
-  - The integral of a sinusoidal value - and therefore a phasor - can simply be written as "$\cdot (-{\rm j})$", \\ which also means a phase shift of $-90°$.((in general, here the integration constant must be considered. This is however often neglectable since only AC values (without a DC value) are considered.)) <WRAP> +  - The integral of a sinusoidal value - and therefore a phasor - can simply be written as "$\cdot (-{{1}\over{ {\rm j}\omega}})$", \\ which also means a phase shift of $-90°$.((in general, here the integration constant must be considered. This is however often neglectable since only AC values (without a DC value) are considered.)) <WRAP> 
 \begin{align*} \begin{align*}
                      \int {\rm e}^{{\rm j}(\omega t + \varphi_x)}                       \int {\rm e}^{{\rm j}(\omega t + \varphi_x)} 
-  = {{1}\over{\rm j}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}  +  = {{1}\over{\rm j\omega}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}  
-  =         - {\rm j} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}+  = -{{\rm j}\over{\omega}} \cdot {\rm e}^{{\rm j}(\omega t + \varphi_x)}
 \end{align*} \end{align*}
 </WRAP> </WRAP>
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 <panel type="info" title="Exercise 6.3.1 Impedance of single Components I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 6.3.1 Impedance of single Components I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-A coil has a reactance of $80\Omega$ at a frequency of $500 ~\rm Hz$. At which frequencies the impedance will have the following values?+A coil has a impedance of $80~\Omega$ at a frequency of $500 ~\rm Hz$. At which frequencies the impedance will have the following values?
   - $85  ~\Omega$   - $85  ~\Omega$
   - $120 ~\Omega$   - $120 ~\Omega$
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 <panel type="info" title="Exercise 6.3.2 Impedance of single Components II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 6.3.2 Impedance of single Components II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-A capacitor with $5 ~{\rm µF}$ is connected to a voltage source which generates $U_\sim = 200 ~{\rm V}$. At which frequencies the following currencies can be measured?+A capacitor with $5 ~{\rm µF}$ is connected to a voltage source which generates $U_\sim = 200 ~{\rm V}$. At which frequencies the following currents can be measured?
   - $0.5 ~\rm A$   - $0.5 ~\rm A$
   - $0.8 ~\rm A$   - $0.8 ~\rm A$
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 <panel type="info" title="Exercise 6.5.5 Complex Calculation I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 6.5.5 Complex Calculation I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-The following two currents with similar frequencies, but different phases have to be added. Use complex calulation!+The following two currents with similar frequencies, but different phases have to be added. Use complex calculation!
   * $i_1(t) = \sqrt{2} \cdot 2 ~A \cdot \cos (\omega t + 20°)$   * $i_1(t) = \sqrt{2} \cdot 2 ~A \cdot \cos (\omega t + 20°)$
   * $i_2(t) = \sqrt{2} \cdot 5 ~A \cdot \cos (\omega t + 110°)$   * $i_2(t) = \sqrt{2} \cdot 5 ~A \cdot \cos (\omega t + 110°)$