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electrical_engineering_and_electronics_2:block06 [2026/04/11 12:05] – mexleadminelectrical_engineering_and_electronics_2:block06 [2026/04/21 03:48] (current) – mexleadmin
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-====== Block xx - xxx ======+====== Block 06 - Complex Power ======
  
 ===== Learning objectives ===== ===== Learning objectives =====
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 Thus, the induced voltage $u(t)$ is given by:  Thus, the induced voltage $u(t)$ is given by: 
 \begin{align*}  \begin{align*} 
-u(t) &=               -\frac{{\rm d}                  \Psi}            {{\rm d}t} \\  +u(t) &=                -\frac{{\rm d~}                  \Psi}            {{\rm d}t} \\  
-     &= -N  \cdot       \frac{{\rm d}                  \Phi}            {{\rm d}t} \\  +     &= -N  \cdot       \frac{{\rm d~}                  \Phi}            {{\rm d}t} \\  
-     &= -NBA\cdot       \frac{{\rm d}       \cos \varphi(t)}            {{\rm d}t} \\  +     &= -NBA\cdot       \frac{{\rm d~}       \cos \varphi(t)}            {{\rm d}t} \\  
-     &= -\hat{\Psi}\cdot\frac{{\rm {\rm d}} \cos (\omega t + \varphi_0)}{{\rm d}t} \\ +     &= -\hat{\Psi}\cdot\frac{{\rm {\rm d~}} \cos (\omega t + \varphi_0)}{{\rm d}t} \\ 
      &= \omega \hat{\Psi}           \cdot \sin (\omega t + \varphi_0) \\       &= \omega \hat{\Psi}           \cdot \sin (\omega t + \varphi_0) \\ 
      &= \hat{U}                     \cdot \sin (\omega t + \varphi_0) \\       &= \hat{U}                     \cdot \sin (\omega t + \varphi_0) \\ 
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   - The use of the $\sqrt{2}$ in the definition $\color{blue}{u_R(t)} = \sqrt{2} U \sin(\omega t + \varphi_u)$ leads to the average power as $P_R = {{U^2}\over{R}}$. This formula for the power is exactly like the formula for the power in pure DC situations.   - The use of the $\sqrt{2}$ in the definition $\color{blue}{u_R(t)} = \sqrt{2} U \sin(\omega t + \varphi_u)$ leads to the average power as $P_R = {{U^2}\over{R}}$. This formula for the power is exactly like the formula for the power in pure DC situations.
  
-==== Ideal Inductivity L ====+==== Ideal Capacity C ====
  
-A similar approach is done for the ideal inductivity.  +Also here, we start with the basic definition of the instantaneous voltage
-We again start with the basic definition of the instantaneous voltage+
  
-\begin{align*} \color{blue}{u_{\rm L}(t)} &= \sqrt{2}U \sin(\omega t + \varphi_u) \end{align*}+\begin{align*} \color{blue}{u_C(t)} &= \sqrt{2}U \sin(\omega t + \varphi_u) \end{align*}
  
-With the defining formula for inductivity, we get: +With the defining formula for the capacity, we get: 
 \begin{align*}  \begin{align*} 
-           \color{blue}{u_{\rm L}(t)} &=  L\cdot    {{{\rm d}\color{red} {i_{\rm L}(t)}}\over{{\rm d}t}} \\  +\color{red}{i_C(t)} &= C {{{\rm d}\color{blue}{u_C(t)}}\over{{\rm d}t}} \\  
-\rightarrow \color{red}{i_{\rm L}(t)} &= {{1}\over{L}} \int  \color{blue}{u_{\rm L}(t)}       {\rm d}t \\  +                    &= \sqrt{2} U \omega C \cos(\omega t + \varphi_u) 
-                                &= - \sqrt{2} {{U}\over{\omega L}} \cos(\omega t + \varphi_u) +
 \end{align*} \end{align*}
  
-This leads to an instantaneous power $p_L(t)$ of+This leads to an instantaneous power $p_C(t)$ of
  
 \begin{align*}  \begin{align*} 
-p_L(t) &=   \color{blue}{u(t)}            \cdot \color{red}{i(t)} \\  +p_C(t) &= \color{blue}{u_C(t)} \cdot \color{red}{i_C(t)} \\  
-       &= - 2\cdot {{U^2}\over{\omega L}} \cdot \sin(          \omega t + \varphi_u) \cos(\omega t + \varphi_u) \\  +       &= 2\cdot U^2 \omega C  \cdot \sin(         \omega t + \varphi_u) \cos(\omega t + \varphi_u) \\  
-       &= - {{U^2}\over{\omega L}}        \cdot \sin( 2\cdot (\omega t + \varphi_u)) \\ +       &= +      U^2 \omega C  \cdot \sin( 2\cdot (\omega t + \varphi_u)) \\ 
 \end{align*} \end{align*}
- 
-Again a trigonometric identity ({{https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle_formulae|Double-angle formula}}  "$\sin(2x) = 2 \sin(x)\cos(x)$") was used. 
  
 Also, this result is interesting: Also, this result is interesting:
  
   - The part $\sin( 2\cdot (\omega t + \varphi_u))$ has an average value of $0$.   - The part $\sin( 2\cdot (\omega t + \varphi_u))$ has an average value of $0$.
-  - Therefore, the average value of $p_L(t)=0$+  - Therefore, also the average value of $p_C(t)=0$
  
-==== Ideal Capacity C ====+==== Ideal Inductivity L ====
  
-Also here, we start with the basic definition of the instantaneous voltage+A similar approach is done for the ideal inductivity.  
 +We again start with the basic definition of the instantaneous voltage
  
-\begin{align*} \color{blue}{u_C(t)} &= \sqrt{2}U \sin(\omega t + \varphi_u) \end{align*}+\begin{align*} \color{blue}{u_{\rm L}(t)} &= \sqrt{2}U \sin(\omega t + \varphi_u) \end{align*}
  
-With the defining formula for the capacity, we get: +With the defining formula for inductivity, we get: 
 \begin{align*}  \begin{align*} 
-\color{red}{i_C(t)} &= C {{{\rm d}\color{blue}{u_C(t)}}\over{{\rm d}t}} \\  +           \color{blue}{u_{\rm L}(t)} &=  L\cdot    {{{\rm d}\color{red} {i_{\rm L}(t)}}\over{{\rm d}t}} \\  
-                    &= \sqrt{2} U \omega C \cos(\omega t + \varphi_u) +\rightarrow \color{red}{i_{\rm L}(t)} &= {{1}\over{L}} \int  \color{blue}{u_{\rm L}(t)}       {\rm d}t \\  
 +                                &= - \sqrt{2} {{U}\over{\omega L}} \cos(\omega t + \varphi_u) 
 \end{align*} \end{align*}
  
-This leads to an instantaneous power $p_C(t)$ of+Since we assume pure AC signals the integration constant has to be 0. 
 +\\ 
 +This formulas lead to an instantaneous power $p_L(t)$ of
  
 \begin{align*}  \begin{align*} 
-p_C(t) &= \color{blue}{u_C(t)} \cdot \color{red}{i_C(t)} \\  +p_L(t) &=   \color{blue}{u(t)}            \cdot \color{red}{i(t)} \\  
-       &= 2\cdot U^2 \omega C  \cdot \sin(         \omega t + \varphi_u) \cos(\omega t + \varphi_u) \\  +       &= - 2\cdot {{U^2}\over{\omega L}} \cdot \sin(          \omega t + \varphi_u) \cos(\omega t + \varphi_u) \\  
-       &= +      U^2 \omega C  \cdot \sin( 2\cdot (\omega t + \varphi_u)) \\ +       &= - {{U^2}\over{\omega L}}        \cdot \sin( 2\cdot (\omega t + \varphi_u)) \\ 
 \end{align*} \end{align*}
 +
 +Again a trigonometric identity ({{https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle_formulae|Double-angle formula}}  "$\sin(2x) = 2 \sin(x)\cos(x)$") was used.
  
 Again this result leads to: Again this result leads to:
  
   - The part $\sin( 2\cdot (\omega t + \varphi_u))$ has an average value of $0$.   - The part $\sin( 2\cdot (\omega t + \varphi_u))$ has an average value of $0$.
-  - Therefore, also the average value of $p_C(t)=0$+  - Therefore, the average value of $p_L(t)=0$
  
   * Instantaneous values of power at $R$, $L$, $C$   * Instantaneous values of power at $R$, $L$, $C$
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 Alternatively, a bad power factor can be compensated with a counteracting complex impedance. This compensating impedance has to provide enough power with the opposite sign to cancel out the unwanted reactive power. The following simulation shows an uncompensated circuit and a circuit with power factor correction. In the ladder, the voltage on the load resistor is the same, but the current provided by the power supply is smaller. Alternatively, a bad power factor can be compensated with a counteracting complex impedance. This compensating impedance has to provide enough power with the opposite sign to cancel out the unwanted reactive power. The following simulation shows an uncompensated circuit and a circuit with power factor correction. In the ladder, the voltage on the load resistor is the same, but the current provided by the power supply is smaller.
  
-<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3Z5OfATFaYwGZtgCwAc+ArNpAGwWGoXYjkgkj70kCmAtFgFAA2ITgE4K4fPkFExEyBjioA7KM4w4JQktpDIQ-KlL4F6SDwBOkwiH2zOUgjJBDCJHgHdBIq-htTrUN16yfuQSfiYA5uBg6H5YMSSisiYAblEx3mnSUODgSCCJgmDQJAqQCiLYtOVEIonFZgyQoRkhWbJOLgKx4o0S9tmqkIrKg+qaFNq6+iSGxjypnKiQln6Ly205iPD52dzFpeV0VbqEtRgukWsrGVdWCQMBt-3Cov0m7i+BgkvXSTwAxl8wEJ0rJ+jFoEouEImPtCBR1EIwGRsAk8HJIGAGn4cRlgcZwFxogFcWDol93plwXieiZvPRbrZLNxXgUwAB9DmQdlLIgUTnc1TREjs7g8zns7A8enfdZSFmZWSECXcgAezkxIpFgwoqBFYtQ4u5yuw7MGhrpkHocQsvXABX5XJ5TXhArNsGFoo5hv5UplNqkrRtshFTvVJE17vgXI9qGVBqj8HwQhTqbT6f53JFpvNZp4qu+5M4JBhi2YxdEfUsAFUAHb-AD2AFsAA7sWsAZwAhgAXdgAEwAOh2AMIAS1M-wArmOe-ncsySwwi-cqyBx5OZz3h65ZwALYcABQAYsORw3TKZ2P8e2OG7WeNhLLJjxAkh7wB-uCAAJK1-tTjeXb1uwPBAA noborder}} </WRAP>+<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=DwYwlgTgBAZgvAIgIwKgFwM6IAwDpsEFKEmmEBMqYIOuSSAzA0gCwAcLArA9gGy9tyvBqhAAjRN1QAHCQhYioANwiTUAW0ySApgFp6CAHwAoKFGAAbKAA9EugJy8orFlF3tnLFqngJsqC1psFChqHCo7PAIAdhZeeyRObAY2JN5otmjOBAB6EzNgaFsEdzYochZsNw8XH3CoVQQk-zzTcwB3GztHcsrqsor-WHDWgs7iwd6qnldBur9c-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-jSXfLse2-Rze2KS4PDxYCZNAuTJwU7JoOUpcEOCALNJCLcMOw3C9N4hjiMs1lgnYvwksk+wMsyrLsp+KiRifMzGOgxLgnIKD1HIliuPyny7IKBzPyOaw+x2XROHsNwHzcPZ5muDBmWQKC0G0RAAFUpy7KjpwwN5hszKAAGFIBATCwDQd1mrciU2o65gxJ6m4+oGpAhpGhAlpw1a0DuNa-igAAFAAxRbOwgCBtBANAwE7Kd3REwUnQAodtgjN1ZPkyClNg3x4Mk4INPXNYoFiuj4qMqNiBSgK8pogqryI29EpCCqgmx7ijXq197ME5zP1OaInA8ySQZA3iIdIwLodUkLpPC7Tkbxgyk3R2chhJjmcZ4-T4sJ2hicq-AyZq2yqfqmnSTeTMACtJ1nCNtAGoYIPOt5pDecA0DeCbewACkegBKDQFLjBwNGKBg0rIUgWHjJiNQW03zbWq2YtaPjwAgEwgA noborder}} </WRAP>
  
 Another explanation of the power factor can be seen here: Another explanation of the power factor can be seen here:
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
 +
 +#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Mains Input of a Control Cabinet: Resistor-Capacitor Parallel Branch                    #@TaskText_HTML@#
 +
 +<WRAP right>{{drawio>electrical_engineering_and_electronics_2:EMIcheckExerciseV01.svg}}</WRAP>
 +
 +At the input of an industrial control cabinet, an RC parallel branch is connected to the AC mains.
 +The resistor represents a damping and discharge path, while the capacitor models an EMI suppression capacitor.
 +For thermal design and power-quality assessment, the effective currents and powers of this branch shall be determined.
 +
 +Data:
 +\begin{align*}
 +u_1 &= \hat U_1 \cos(\omega t) \\
 +\hat U_1 &= 325~{\rm V} \\
 +f &= 50~{\rm Hz} \\
 +R &= 220~{\rm \Omega} \\
 +C &= 4.7~{\rm \mu F}
 +\end{align*}
 +
 +1. Determine the RMS values of the voltage $U_1$ and of the currents $I$, $I_R$, and $I_C$.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~501~@#
 +<WRAP leftalign>
 +First convert the voltage amplitude into the RMS value:
 +\begin{align*}
 +U_1 &= \frac{\hat U_1}{\sqrt{2}}
 +    = \frac{325~{\rm V}}{\sqrt{2}}
 +    = 229.8~{\rm V}
 +\end{align*}
 +
 +The angular frequency is
 +\begin{align*}
 +\omega &= 2\pi f
 +       = 2\pi \cdot 50~{\rm s^{-1}}
 +       = 314.16~{\rm s^{-1}}
 +\end{align*}
 +
 +The resistor current is
 +\begin{align*}
 +I_R &= \frac{U_1}{R}
 +    = \frac{229.8~{\rm V}}{220~{\rm \Omega}}
 +    = 1.045~{\rm A}
 +\end{align*}
 +
 +The capacitor current is
 +\begin{align*}
 +I_C &= U_1 \omega C \\
 +    &= 229.8~{\rm V}\cdot 314.16~{\rm s^{-1}}\cdot 4.7\cdot 10^{-6}~{\rm F} \\
 +    &= 0.339~{\rm A}
 +\end{align*}
 +
 +Because $I_R$ and $I_C$ are perpendicular in the phasor diagram, the total current is
 +\begin{align*}
 +I &= \sqrt{I_R^2+I_C^2} \\
 +  &= \sqrt{(1.045~{\rm A})^2 + (0.339~{\rm A})^2} \\
 +  &= 1.098~{\rm A}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~501~@#
 +\begin{align*}
 +U_1 &= 229.8~{\rm V} \\
 +I_R &= 1.045~{\rm A} \\
 +I_C &= 0.339~{\rm A} \\
 +I   &= 1.098~{\rm A}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +2. What active, reactive, and apparent power does the circuit absorb?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~502~@#
 +<WRAP leftalign>
 +The resistor absorbs only active power:
 +\begin{align*}
 +P &= \frac{U_1^2}{R}
 +   = \frac{(229.8~{\rm V})^2}{220~{\rm \Omega}} \\
 +  &= 240.1~{\rm W}
 +\end{align*}
 +
 +The capacitor absorbs only reactive power. For a capacitor, the reactive power is negative:
 +\begin{align*}
 +Q &= -U_1^2 \omega C \\
 +  &= -(229.8~{\rm V})^2 \cdot 314.16~{\rm s^{-1}} \cdot 4.7\cdot 10^{-6}~{\rm F} \\
 +  &= -78.0~{\rm var}
 +\end{align*}
 +
 +The apparent power is
 +\begin{align*}
 +S &= U_1 I \\
 +  &= 229.8~{\rm V}\cdot 1.098~{\rm A} \\
 +  &= 252.4~{\rm VA}
 +\end{align*}
 +
 +Check with the power triangle:
 +\begin{align*}
 +S &= \sqrt{P^2+Q^2}
 +   = \sqrt{(240.1~{\rm W})^2+(-78.0~{\rm var})^2} \\
 +  &= 252.4~{\rm VA}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~502~@#
 +\begin{align*}
 +P &= 240.1~{\rm W} \\
 +Q &= -78.0~{\rm var} \\
 +S &= 252.4~{\rm VA}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +3. Determine the maximum and minimum value of the instantaneous power.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~503~@#
 +<WRAP leftalign>
 +For a sinusoidal AC circuit, the instantaneous power can be written as
 +\begin{align*}
 +p(t) = P + S\cos(2\omega t + \varphi)
 +\end{align*}
 +where $P$ is the active power, $S$ is the apparent power, and $\varphi$ is the phase angle.
 +
 +Thus the oscillating part has the amplitude $S$, so
 +\begin{align*}
 +p_{\rm max} &= P + S \\
 +p_{\rm min} &= P - S
 +\end{align*}
 +
 +Insert the values:
 +\begin{align*}
 +p_{\rm max} &= 240.1~{\rm W} + 252.4~{\rm W}
 +            = 492.5~{\rm W} \\
 +p_{\rm min} &= 240.1~{\rm W} - 252.4~{\rm W}
 +            = -12.3~{\rm W}
 +\end{align*}
 +
 +The negative minimum value means that for a short time interval, reactive energy is fed back from the capacitor to the source.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~503~@#
 +\begin{align*}
 +p_{\rm max} &= 492.5~{\rm W} \\
 +p_{\rm min} &= -12.3~{\rm W}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +#@TaskEnd_HTML@#
 +
 +
 +
 +
 +#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  AC Solenoid Branch of a Valve Driver: Parallel Resistor-Inductor Circuit                    #@TaskText_HTML@#
 +
 +<WRAP right>{{drawio>electrical_engineering_and_electronics_2:ACcontrolunitV01.svg}}</WRAP>
 +
 +An industrial valve driver contains an AC solenoid branch that can be modeled by a resistor in parallel with an inductor.
 +The resistor represents electrical losses, while the inductive branch represents the magnetizing behavior of the actuator.
 +For thermal design and grid-side power assessment, the RMS currents and powers of the branch shall be determined.
 +
 +Data:
 +\begin{align*}
 +u_1 &= \hat U_1 \cos(\omega t) \\
 +\hat U_1 &= 170~{\rm V} \\
 +f &= 60~{\rm Hz} \\
 +R &= 220~{\rm \Omega} \\
 +L &= 325~{\rm mH}
 +\end{align*}
 +
 +1. Determine the RMS values of the voltage $U_1$ and of the currents $I$, $I_R$, and $I_L$.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~511~@#
 +<WRAP leftalign>
 +First convert the voltage amplitude into the RMS value:
 +\begin{align*}
 +U_1 &= \frac{\hat U_1}{\sqrt{2}}
 +    = \frac{170~{\rm V}}{\sqrt{2}}
 +    = 120.2~{\rm V}
 +\end{align*}
 +
 +The angular frequency is
 +\begin{align*}
 +\omega &= 2\pi f
 +       = 2\pi \cdot 60~{\rm s^{-1}}
 +       = 377.0~{\rm s^{-1}}
 +\end{align*}
 +
 +The inductive reactance is
 +\begin{align*}
 +\omega L &= 377.0~{\rm s^{-1}} \cdot 0.325~{\rm H}
 +         = 122.5~{\rm \Omega}
 +\end{align*}
 +
 +The resistor current is
 +\begin{align*}
 +I_R &= \frac{U_1}{R}
 +    = \frac{120.2~{\rm V}}{220~{\rm \Omega}}
 +    = 0.546~{\rm A}
 +\end{align*}
 +
 +The inductor current is
 +\begin{align*}
 +I_L &= \frac{U_1}{\omega L}
 +    = \frac{120.2~{\rm V}}{122.5~{\rm \Omega}}
 +    = 0.981~{\rm A}
 +\end{align*}
 +
 +Because $I_R$ and $I_L$ are perpendicular in the phasor diagram, the total current is
 +\begin{align*}
 +I &= \sqrt{I_R^2+I_L^2} \\
 +  &= \sqrt{(0.546~{\rm A})^2 + (0.981~{\rm A})^2} \\
 +  &= 1.123~{\rm A}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~511~@#
 +\begin{align*}
 +U_1 &= 120.2~{\rm V} \\
 +I_R &= 0.546~{\rm A} \\
 +I_L &= 0.981~{\rm A} \\
 +I   &= 1.123~{\rm A}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +2. What active, reactive, and apparent power does the circuit absorb?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~512~@#
 +<WRAP leftalign>
 +The resistor absorbs only active power:
 +\begin{align*}
 +P &= \frac{U_1^2}{R}
 +   = \frac{(120.2~{\rm V})^2}{220~{\rm \Omega}} \\
 +  &= 65.7~{\rm W}
 +\end{align*}
 +
 +The inductor absorbs only reactive power. For an inductor, reactive power is positive:
 +\begin{align*}
 +Q &= \frac{U_1^2}{\omega L}
 +   = \frac{(120.2~{\rm V})^2}{122.5~{\rm \Omega}} \\
 +  &= 117.9~{\rm var}
 +\end{align*}
 +
 +The apparent power is
 +\begin{align*}
 +S &= U_1 I \\
 +  &= 120.2~{\rm V}\cdot 1.123~{\rm A} \\
 +  &= 135.0~{\rm VA}
 +\end{align*}
 +
 +Check with the power triangle:
 +\begin{align*}
 +S &= \sqrt{P^2+Q^2}
 +   = \sqrt{(65.7~{\rm W})^2+(117.9~{\rm var})^2} \\
 +  &= 135.0~{\rm VA}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~512~@#
 +\begin{align*}
 +P &= 65.7~{\rm W} \\
 +Q &= 117.9~{\rm var} \\
 +S &= 135.0~{\rm VA}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +3. Determine the maximum and minimum value of the instantaneous power.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~513~@#
 +<WRAP leftalign>
 +For a sinusoidal AC circuit, the instantaneous power can be written as
 +\begin{align*}
 +p(t) = P + S\cos(2\omega t + \varphi_p)
 +\end{align*}
 +
 +Thus the oscillating part has the amplitude $S$, so
 +\begin{align*}
 +p_{\rm max} &= P + S \\
 +p_{\rm min} &= P - S
 +\end{align*}
 +
 +Insert the values:
 +\begin{align*}
 +p_{\rm max} &= 65.7~{\rm W} + 135.0~{\rm W}
 +            = 200.7~{\rm W} \\
 +p_{\rm min} &= 65.7~{\rm W} - 135.0~{\rm W}
 +            = -69.3~{\rm W}
 +\end{align*}
 +
 +The negative minimum value means that magnetic energy stored in the inductor is temporarily fed back to the source.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~513~@#
 +\begin{align*}
 +p_{\rm max} &= 200.7~{\rm W} \\
 +p_{\rm min} &= -69.3~{\rm W}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +#@TaskEnd_HTML@#
 +
 +
 +
 +#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  AC Filter and Sensor Front-End: Power Flow in a Composite Reactive Network                    #@TaskText_HTML@#
 +
 +<WRAP right>{{drawio>electrical_engineering_and_electronics_2:ACsensorFrontendV01.svg}}</WRAP>
 +
 +An industrial AC sensor front-end contains a compensation capacitor, a series inductor, and an output branch made of a resistor and another inductor.
 +For thermal design and reactive-power assessment, the active, reactive, and apparent powers at the input and at each component shall be determined.
 +
 +Data:
 +\begin{align*}
 +\omega C &= 0.01~{\rm S} \\
 +\omega L_1 &= 50~{\rm \Omega} \\
 +\omega L_2 &= 200~{\rm \Omega} \\
 +R &= 100~{\rm \Omega} \\
 +\hat U_1 &= 325~{\rm V}
 +\end{align*}
 +
 +1. Calculate the active, reactive, and apparent power at the input and at the individual components of the circuit.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~521~@#
 +<WRAP leftalign>
 +For power calculations, we use RMS values. Therefore, the source voltage is
 +\begin{align*}
 +U_1 &= \frac{\hat U_1}{\sqrt{2}}
 +    = \frac{325~{\rm V}}{\sqrt{2}}
 +    = 229.8~{\rm V}
 +\end{align*}
 +
 +First determine the impedances of the individual elements:
 +\begin{align*}
 +\underline{Z}_C &= \frac{1}{j\omega C}
 += -j\frac{1}{\omega C}
 += -j100~{\rm \Omega} \\
 +\underline{Z}_{L_1} &= j\omega L_1
 += j50~{\rm \Omega} \\
 +\underline{Z}_{L_2} &= j\omega L_2
 += j200~{\rm \Omega}
 +\end{align*}
 +
 +Now the parallel branch:
 +\begin{align*}
 +\underline{Z}_{R\parallel L_2}
 +&= \frac{R\underline{Z}_{L_2}}{R+\underline{Z}_{L_2}} \\
 +&= \frac{100\cdot j200}{100+j200}
 += 80+j40~{\rm \Omega}
 +\end{align*}
 +
 +Thus the total input impedance is
 +\begin{align*}
 +\underline{Z}_{\rm in}
 +&= \underline{Z}_C + \underline{Z}_{L_1} + \underline{Z}_{R\parallel L_2} \\
 +&= (-j100) + (j50) + (80+j40) \\
 +&= 80-j10~{\rm \Omega}
 +\end{align*}
 +
 +The input current is therefore
 +\begin{align*}
 +\underline{I}
 +&= \frac{\underline{U}_1}{\underline{Z}_{\rm in}}
 += \frac{229.8~{\rm V}}{80-j10~{\rm \Omega}} \\
 +&= 2.828 + j0.354~{\rm A}
 +\end{align*}
 +
 +Its magnitude is
 +\begin{align*}
 +I = 2.85~{\rm A}
 +\end{align*}
 +
 +The complex input power is
 +\begin{align*}
 +\underline{S}_{\rm in}
 +&= \underline{U}_1 \underline{I}^{\,*} \\
 +&= 229.8 \cdot (2.828-j0.354)~{\rm VA} \\
 +&= 650 - j81.25~{\rm VA}
 +\end{align*}
 +
 +Hence
 +\begin{align*}
 +P_{\rm in} &= 650~{\rm W} \\
 +Q_{\rm in} &= -81.25~{\rm var} \\
 +S_{\rm in} &= \sqrt{P_{\rm in}^2+Q_{\rm in}^2}
 += 655.1~{\rm VA}
 +\end{align*}
 +
 +Now calculate the powers of the individual components.
 +
 +For the capacitor:
 +\begin{align*}
 +\underline{S}_C
 +&= \underline{U}_C \underline{I}^{\,*}
 += -j812.5~{\rm VA}
 +\end{align*}
 +So
 +\begin{align*}
 +P_C &= 0~{\rm W} \\
 +Q_C &= -812.5~{\rm var} \\
 +S_C &= 812.5~{\rm VA}
 +\end{align*}
 +
 +For the inductor $L_1$:
 +\begin{align*}
 +\underline{S}_{L_1}
 +&= \underline{U}_{L_1}\underline{I}^{\,*}
 += j406.25~{\rm VA}
 +\end{align*}
 +Thus
 +\begin{align*}
 +P_{L_1} &= 0~{\rm W} \\
 +Q_{L_1} &= 406.25~{\rm var} \\
 +S_{L_1} &= 406.25~{\rm VA}
 +\end{align*}
 +
 +The voltage across the parallel branch is
 +\begin{align*}
 +\underline{U}_{R\parallel L_2}
 +&= \underline{I}\,\underline{Z}_{R\parallel L_2} \\
 +&= (2.828+j0.354)(80+j40) \\
 +&= 212.13 + j141.42~{\rm V}
 +\end{align*}
 +
 +The resistor current is
 +\begin{align*}
 +\underline{I}_R
 += \frac{\underline{U}_{R\parallel L_2}}{R}
 += 2.121 + j1.414~{\rm A}
 +\end{align*}
 +
 +The inductor current is
 +\begin{align*}
 +\underline{I}_{L_2}
 += \frac{\underline{U}_{R\parallel L_2}}{j200}
 += 0.707 - j1.061~{\rm A}
 +\end{align*}
 +
 +For the resistor:
 +\begin{align*}
 +\underline{S}_R
 +&= \underline{U}_{R\parallel L_2}\underline{I}_R^{\,*}
 += 650 + j0~{\rm VA}
 +\end{align*}
 +So
 +\begin{align*}
 +P_R &= 650~{\rm W} \\
 +Q_R &= 0~{\rm var} \\
 +S_R &= 650~{\rm VA}
 +\end{align*}
 +
 +For the inductor $L_2$:
 +\begin{align*}
 +\underline{S}_{L_2}
 +&= \underline{U}_{R\parallel L_2}\underline{I}_{L_2}^{\,*}
 += j325~{\rm VA}
 +\end{align*}
 +Thus
 +\begin{align*}
 +P_{L_2} &= 0~{\rm W} \\
 +Q_{L_2} &= 325~{\rm var} \\
 +S_{L_2} &= 325~{\rm VA}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~521~@#
 +\begin{align*}
 +P_{\rm in} &= 650~{\rm W} \\
 +Q_{\rm in} &= -81.25~{\rm var} \\
 +S_{\rm in} &= 655.1~{\rm VA}
 +\end{align*}
 +
 +\begin{align*}
 +P_C &= 0~{\rm W}, & Q_C &= -812.5~{\rm var}, & S_C &= 812.5~{\rm VA} \\
 +P_{L_1} &= 0~{\rm W}, & Q_{L_1} &= 406.25~{\rm var}, & S_{L_1} &= 406.25~{\rm VA} \\
 +P_R &= 650~{\rm W}, & Q_R &= 0~{\rm var}, & S_R &= 650~{\rm VA} \\
 +P_{L_2} &= 0~{\rm W}, & Q_{L_2} &= 325~{\rm var}, & S_{L_2} &= 325~{\rm VA}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +2. Verify the validity of the following reactive-power balance:
 +\[
 +Q = Q_C + Q_{L_1} + Q_{L_2}
 +\]
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~522~@#
 +<WRAP leftalign>
 +Insert the reactive powers found above:
 +\begin{align*}
 +Q_C &= -812.5~{\rm var} \\
 +Q_{L_1} &= 406.25~{\rm var} \\
 +Q_{L_2} &= 325~{\rm var}
 +\end{align*}
 +
 +Then
 +\begin{align*}
 +Q_C + Q_{L_1} + Q_{L_2}
 +&= -812.5 + 406.25 + 325 \\
 +&= -81.25~{\rm var}
 +\end{align*}
 +
 +But the input reactive power is
 +\begin{align*}
 +Q_{\rm in} = -81.25~{\rm var}
 +\end{align*}
 +
 +Therefore,
 +\begin{align*}
 +Q_{\rm in} = Q_C + Q_{L_1} + Q_{L_2}
 +\end{align*}
 +
 +So the reactive-power balance is fulfilled.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~522~@#
 +\begin{align*}
 +Q_C + Q_{L_1} + Q_{L_2}
 += -812.5 + 406.25 + 325
 += -81.25~{\rm var}
 +\end{align*}
 +
 +\begin{align*}
 +Q_{\rm in} = -81.25~{\rm var}
 +\end{align*}
 +
 +\begin{align*}
 +Q_{\rm in} = Q_C + Q_{L_1} + Q_{L_2}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +#@TaskEnd_HTML@#
 +
 +
 +
 +
 +
 +#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Real Current-Sense Choke: Series Model of an Industrial Coil                    #@TaskText_HTML@#
 +
 +A small current-sense choke in an industrial electronics module is used at low frequency for filtering and current shaping.
 +In practice, the coil is not ideal: besides its inductance, it also has a winding resistance.
 +Therefore, the real coil is modeled as a series connection of an inductance and an ohmic resistance.
 +
 +Data:
 +\begin{align*}
 +L_{\rm sp} &= 2.5~{\rm mH} \\
 +R_{\rm sp} &= 100~{\rm m\Omega} \\
 +I_{\rm sp} &= 0.5~{\rm A} \\
 +f &= 50~{\rm Hz}
 +\end{align*}
 +
 +1. Draw the circuit and place all current and voltage phasors in the phasor diagram.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~531~@#
 +<WRAP leftalign>
 +The real coil is modeled as a series connection of
 +\begin{align*}
 +R_{\rm sp} \text{ and } L_{\rm sp}
 +\end{align*}
 +
 +Because this is a series circuit, the same current flows through both elements:
 +\begin{align*}
 +I_R = I_L = I_{\rm sp} = 0.5~{\rm A}
 +\end{align*}
 +
 +Choose the current as the reference phasor:
 +\begin{align*}
 +\underline{I}_{\rm sp} = 0.5~{\rm A}\angle 0^\circ
 +\end{align*}
 +
 +Then:
 +\begin{align*}
 +\underline{U}_R &\text{ is in phase with } \underline{I}_{\rm sp} \\
 +\underline{U}_L &\text{ leads } \underline{I}_{\rm sp} \text{ by } 90^\circ \\
 +\underline{U}_{\rm sp} &= \underline{U}_R + \underline{U}_L
 +\end{align*}
 +
 +So in the phasor diagram:
 +\begin{align*}
 +\underline{I}_{\rm sp} &: \text{horizontal to the right} \\
 +\underline{U}_R &: \text{same direction as } \underline{I}_{\rm sp} \\
 +\underline{U}_L &: \text{vertical upward} \\
 +\underline{U}_{\rm sp} &: \text{diagonal sum of } \underline{U}_R \text{ and } \underline{U}_L
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~531~@#
 +\begin{align*}
 +\underline{I}_{\rm sp} &= 0.5~{\rm A}\angle 0^\circ \\
 +\underline{U}_R &\parallel \underline{I}_{\rm sp} \\
 +\underline{U}_L &\text{ leads by }90^\circ \\
 +\underline{U}_{\rm sp} &= \underline{U}_R+\underline{U}_L
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +2. Calculate the complex input impedance.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~532~@#
 +<WRAP leftalign>
 +First calculate the angular frequency:
 +\begin{align*}
 +\omega &= 2\pi f
 += 2\pi \cdot 50~{\rm s^{-1}}
 += 314.16~{\rm s^{-1}}
 +\end{align*}
 +
 +Then the inductive reactance is
 +\begin{align*}
 +X_L = \omega L_{\rm sp}
 += 314.16~{\rm s^{-1}} \cdot 2.5\cdot 10^{-3}~{\rm H}
 += 0.785~{\rm \Omega}
 +\end{align*}
 +
 +The complex input impedance of the real coil is
 +\begin{align*}
 +\underline{Z}_{\rm sp}
 +&= R_{\rm sp} + jX_L \\
 +&= 0.100 + j0.785~{\rm \Omega}
 +\end{align*}
 +
 +Its magnitude is
 +\begin{align*}
 +\left|\underline{Z}_{\rm sp}\right|
 +&= \sqrt{0.100^2+0.785^2} \\
 +&= 0.792~{\rm \Omega}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~532~@#
 +\begin{align*}
 +\underline{Z}_{\rm sp} &= 0.100 + j0.785~{\rm \Omega} \\
 +\left|\underline{Z}_{\rm sp}\right| &= 0.792~{\rm \Omega}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +3. How large is the voltage across the coil, and what is the phase-shift angle?
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~533~@#
 +<WRAP leftalign>
 +The voltage across the winding resistance is
 +\begin{align*}
 +U_R &= I_{\rm sp} R_{\rm sp}
 += 0.5~{\rm A}\cdot 0.100~{\rm \Omega}
 += 0.050~{\rm V}
 +\end{align*}
 +
 +The voltage across the inductance is
 +\begin{align*}
 +U_L &= I_{\rm sp} X_L
 += 0.5~{\rm A}\cdot 0.785~{\rm \Omega}
 += 0.3927~{\rm V}
 +\end{align*}
 +
 +The total coil voltage is the vector sum:
 +\begin{align*}
 +U_{\rm sp}
 +&= I_{\rm sp}\left|\underline{Z}_{\rm sp}\right| \\
 +&= 0.5~{\rm A}\cdot 0.792~{\rm \Omega}
 += 0.396~{\rm V}
 +\end{align*}
 +
 +The phase-shift angle between coil voltage and current is
 +\begin{align*}
 +\varphi &= \arctan\left(\frac{X_L}{R_{\rm sp}}\right)
 += \arctan\left(\frac{0.785}{0.100}\right)
 += 82.74^\circ
 +\end{align*}
 +
 +So the coil voltage leads the current by about $82.7^\circ$.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~533~@#
 +\begin{align*}
 +U_R &= 0.050~{\rm V} \\
 +U_L &= 0.3927~{\rm V} \\
 +U_{\rm sp} &= 0.396~{\rm V} \\
 +\varphi &= 82.74^\circ
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +4. Calculate the active, reactive, and apparent power absorbed by the coil, and determine the power factor.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~534~@#
 +<WRAP leftalign>
 +The active power is dissipated only in the resistance:
 +\begin{align*}
 +P &= I_{\rm sp}^2 R_{\rm sp} \\
 +  &= (0.5~{\rm A})^2 \cdot 0.100~{\rm \Omega} \\
 +  &= 0.0250~{\rm W}
 +\end{align*}
 +
 +The reactive power is taken by the inductance:
 +\begin{align*}
 +Q &= I_{\rm sp}^2 X_L \\
 +  &= (0.5~{\rm A})^2 \cdot 0.785~{\rm \Omega} \\
 +  &= 0.196~{\rm var}
 +\end{align*}
 +
 +The apparent power is
 +\begin{align*}
 +S &= U_{\rm sp} I_{\rm sp} \\
 +  &= 0.396~{\rm V}\cdot 0.5~{\rm A} \\
 +  &= 0.198~{\rm VA}
 +\end{align*}
 +
 +The power factor is
 +\begin{align*}
 +\lambda = \cos\varphi = \frac{P}{S}
 += \frac{0.0250}{0.198}
 += 0.126
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~534~@#
 +\begin{align*}
 +P &= 0.0250~{\rm W} \\
 +Q &= 0.196~{\rm var} \\
 +S &= 0.198~{\rm VA} \\
 +\lambda &= \cos\varphi = 0.126
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +5. Determine the loss factor and the loss angle.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~535~@#
 +<WRAP leftalign>
 +For a real inductor, the loss factor is
 +\begin{align*}
 +d = \frac{P}{Q} = \frac{R_{\rm sp}}{X_L}
 +\end{align*}
 +
 +Thus,
 +\begin{align*}
 +d &= \frac{0.100}{0.785}
 += 0.127
 +\end{align*}
 +
 +The loss angle $\delta$ is related to the loss factor by
 +\begin{align*}
 +\tan\delta = d
 +\end{align*}
 +
 +Hence,
 +\begin{align*}
 +\delta &= \arctan(d)
 += \arctan(0.127)
 += 7.26^\circ
 +\end{align*}
 +
 +As a check:
 +\begin{align*}
 +\delta = 90^\circ - \varphi
 += 90^\circ - 82.74^\circ
 += 7.26^\circ
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~535~@#
 +\begin{align*}
 +d &= 0.127 \\
 +\delta &= 7.26^\circ
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +6. Draw the power triangle and indicate the loss angle.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~536~@#
 +<WRAP leftalign>
 +In the power triangle:
 +\begin{align*}
 +P &= 0.0250~{\rm W} \qquad \text{horizontal axis} \\
 +Q &= 0.196~{\rm var} \qquad \text{vertical axis upward} \\
 +S &= 0.198~{\rm VA} \qquad \text{hypotenuse}
 +\end{align*}
 +
 +The angle between $S$ and the horizontal axis is
 +\begin{align*}
 +\varphi = 82.74^\circ
 +\end{align*}
 +
 +The loss angle is the complementary angle:
 +\begin{align*}
 +\delta = 7.26^\circ
 +\end{align*}
 +
 +So in the sketch, mark $\delta$ between the vertical reactive-power axis and the apparent-power vector.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~536~@#
 +\begin{align*}
 +P &= 0.0250~{\rm W} \\
 +Q &= 0.196~{\rm var} \\
 +S &= 0.198~{\rm VA} \\
 +\varphi &= 82.74^\circ \\
 +\delta &= 7.26^\circ
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +#@TaskEnd_HTML@#
 ===== Embedded resources ===== ===== Embedded resources =====
 <WRAP column half> <WRAP column half>