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electrical_engineering_2:polyphase_networks [2023/03/22 20:29]
mexleadmin 		electrical_engineering_2:polyphase_networks [2023/09/19 23:52] (aktuell)
mexleadmin
Zeile 1: Zeile 1:
-====== 7. Polyphase Networks and Power in AC Circuits ======+====== 7 Polyphase Networks and Power in AC Circuits ======
  
 emphasizing the importance of power considerations emphasizing the importance of power considerations
Zeile 5: Zeile 5:
   * three-phase four-wire systems   * three-phase four-wire systems
  
-=== 7.0 Recap of complex two-terminal networks ===+===== 7.0 Recap of complex two-terminal networks =====
  
 In the last semester, AC current, AC voltage, and their effects have been considered on a circuit that had simply included an AC voltage source. \\ These circuits can be now understood as. In the last semester, AC current, AC voltage, and their effects have been considered on a circuit that had simply included an AC voltage source. \\ These circuits can be now understood as.
Zeile 23: Zeile 23:
 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) \\ 
 \end{align*} \end{align*}
  
Zeile 34: Zeile 34:
 Out of the last formula we derived the following instantaneous voltage $u(t)$  Out of the last formula we derived the following instantaneous voltage $u(t)$ 
 \begin{align*}  \begin{align*} 
-u(t) &= -\hat{U}  \cdot \sin (\omega t + \varphi_0) \\  +u(t) &= \hat{U}  \cdot \sin (\omega t + \varphi_0) \\  
-     &=  \hat{U}  \cdot \sin (\omega t + \varphi'_0) \\  +     &= \sqrt{2} U\cdot \sin (\omega t + \varphi_0) \\ 
-     &= \sqrt{2} U\cdot \sin (\omega t + \varphi'_0) \\ +
 \end{align*} \end{align*}
  
Zeile 145: Zeile 144:
  
   - Ohmic load: The instantaneous voltage is in phase with the instantaneous current. The instantaneous power is always non-negative. The average power is $P=U^2/R = {{1}\over{2}} \hat{U}^2/R= {{1}\over{2}}(6V)^2/1 ~\rm k\Omega = 18 ~\rm mW$   - Ohmic load: The instantaneous voltage is in phase with the instantaneous current. The instantaneous power is always non-negative. The average power is $P=U^2/R = {{1}\over{2}} \hat{U}^2/R= {{1}\over{2}}(6V)^2/1 ~\rm k\Omega = 18 ~\rm mW$
-  - Inductive load: The voltage is ahead of the current. The phase angle is $+90°$ (which also reflects the $+j$ in the inductive impedance $+j\omega L$). The instantaneous is half positive, half negative; the average power is zero (in the simulation not completely visible). +  - Inductive load: The voltage is ahead of the current. The phase angle is $+90°$ (which also reflects the $+{\rm j}$ in the inductive impedance $+{\rm j}\omega L$). The instantaneous is half positive, half negative; the average power is zero (in the simulation not completely visible). 
-  - Capacitive load: The voltage is lagging the current. The phase angle is $-90°$ (which also reflects the $-j$ in the capacitive impedance ${{1}\over{j\omega C}}$). The instantaneous is again half positive, half negative; the average power is zero (in the simulation not completely visible).+  - Capacitive load: The voltage is lagging the current. The phase angle is $-90°$ (which also reflects the $-{\rm j}$ in the capacitive impedance ${{1}\over{{\rm j}\omega C}}$). The instantaneous is again half positive, half negative; the average power is zero (in the simulation not completely visible).
  
 <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3EZjAZhWALADgwVhZAGyFYBMhKIuEuIGluApgLTIBQATiK2KSKRkjdkfMAHZCUcPDYZSQ5qVyTxk1oSGqphDGwDuwjf0HDexoZH11s5ugE5JAi7PkgUuUROtZwXoTqsbJzc+YMsDFE9JGy1w8AwMWzA7UJNLOSEYr0xE2JAAgHMQ20j+ZSlLABtwFN8VWrCoWDBcOxbuGDhcBIwxMWwxVDbCZChZSEpS0vkfHKksAH0wBcgFgA8sMEhxBdwV2CIlBdYF0lPjloXFlH24G9ZdQUm+achZhPmlhbF1gEMCMCEW4IYHbC5fTpgLCDQh2OHUN52MRYRaQi6PCbcJSSUrMBzxRKaCHrTbbH57TqHPYnM5nVh7ZY3Sn3TDjSiKcq4-FzInLH5rX5bQG7fbLE7LTrI3gU5oXFBstwoHF8aiiD6ZYkbIXk4GEI4086sH4YBZMg4ssQKtDKqhmHl0L78wXbIEysXLZYysHqFyTFBiEB2WgYPWB2jyX2KgP4kN8fERuR+sPgXA+IP8dKkJN2XKpwOJBNZrHdZOsUh2ZOF9lKRI54Tl-MZtgANyxYlSClI7bqUggW3g+SkrGguDEkDEDkihAn2AckhguDYRUU3eCK9C5WcBnXPY5Kj8Vj3th3YSXiscJmtZXnYwiSp77iit7cHhK99P28Bmi8PCfcWST5zLEgRZNE3IHgAxvY+5gReQiiCObRMMwPikNAWB2JAHjyAkIxvGg3CEGw8gBjuWhkbUiQAKorMR47FFopQAXQIA0ZYJHng+97MdRtFrMWPh4sI1CESxULgNAAA6ADOAD2AAWAC2ACWEEyZUsm-AAJmw-HoNwswjNwwbgKhUnScpAB2WkAK4QQALspzaMOpmk6Xp+koXQY7GWJPgoOZEG-AADr8EHKY5zmudpbCyeA8UBhsAKME08CIAc2xSJQMACO4eD+vQ-RYHqoaQvMIABjl9DdPg-QFcV5B8GVmQVU0uU1QV9UlU1zRjCgqGUAAYhAlCYXAKYDqwICDRwjAAI42YwlkQQAnmwQA noborder}} </WRAP> <WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3EZjAZhWALADgwVhZAGyFYBMhKIuEuIGluApgLTIBQATiK2KSKRkjdkfMAHZCUcPDYZSQ5qVyTxk1oSGqphDGwDuwjf0HDexoZH11s5ugE5JAi7PkgUuUROtZwXoTqsbJzc+YMsDFE9JGy1w8AwMWzA7UJNLOSEYr0xE2JAAgHMQ20j+ZSlLABtwFN8VWrCoWDBcOxbuGDhcBIwxMWwxVDbCZChZSEpS0vkfHKksAH0wBcgFgA8sMEhxBdwV2CIlBdYF0lPjloXFlH24G9ZdQUm+achZhPmlhbF1gEMCMCEW4IYHbC5fTpgLCDQh2OHUN52MRYRaQi6PCbcJSSUrMBzxRKaCHrTbbH57TqHPYnM5nVh7ZY3Sn3TDjSiKcq4-FzInLH5rX5bQG7fbLE7LTrI3gU5oXFBstwoHF8aiiD6ZYkbIXk4GEI4086sH4YBZMg4ssQKtDKqhmHl0L78wXbIEysXLZYysHqFyTFBiEB2WgYPWB2jyX2KgP4kN8fERuR+sPgXA+IP8dKkJN2XKpwOJBNZrHdZOsUh2ZOF9lKRI54Tl-MZtgANyxYlSClI7bqUggW3g+SkrGguDEkDEDkihAn2AckhguDYRUU3eCK9C5WcBnXPY5Kj8Vj3th3YSXiscJmtZXnYwiSp77iit7cHhK99P28Bmi8PCfcWST5zLEgRZNE3IHgAxvY+5gReQiiCObRMMwPikNAWB2JAHjyAkIxvGg3CEGw8gBjuWhkbUiQAKorMR47FFopQAXQIA0ZYJHng+97MdRtFrMWPh4sI1CESxULgNAAA6ADOAD2AAWAC2ACWEEyZUsm-AAJmw-HoNwswjNwwbgKhUnScpAB2WkAK4QQALspzaMOpmk6Xp+koXQY7GWJPgoOZEG-AADr8EHKY5zmudpbCyeA8UBhsAKME08CIAc2xSJQMACO4eD+vQ-RYHqoaQvMIABjl9DdPg-QFcV5B8GVmQVU0uU1QV9UlU1zRjCgqGUAAYhAlCYXAKYDqwICDRwjAAI42YwlkQQAnmwQA noborder}} </WRAP>
Zeile 208: Zeile 207:
  
 \begin{align*}  \begin{align*} 
-\underline{S} &= S         \cdot e^{j\varphi} \\  +\underline{S} &= S         \cdot {\rm e}^{{\rm j}\varphi} \\  
-              &= U \cdot I \cdot e^{j\varphi} +              &= U \cdot I \cdot {\rm e}^{{\rm j}\varphi} 
 \end{align*} \end{align*}
  
Zeile 215: Zeile 214:
  
 \begin{align*}  \begin{align*} 
-\underline{S} &=             U                                       \cdot             I \cdot e^{j(\varphi_U - \varphi_I)} \\  +\underline{S} &=             U                                                   \cdot             I \cdot {\rm e}^{ {\rm j}(\varphi_U - \varphi_I)} \\  
-              &= \underbrace{U \cdot e^{j\varphi_U}}_{\underline{U}} \cdot \underbrace{I \cdot e^{-j\varphi_I}}_{\underline{I}^*} +              &= \underbrace{U \cdot {\rm e}^{{\rm j}\varphi_U}}_{\underline{U}} \cdot \underbrace{I \cdot {\rm e}^{-{\rm j}\varphi_I}}_{\underline{I}^*} 
 \end{align*} \end{align*}
  
Zeile 223: Zeile 222:
 <callout icon="fa fa-exclamation" color="red" title="Notice:"> The apparent power $\underline{S} $ is given by: <callout icon="fa fa-exclamation" color="red" title="Notice:"> The apparent power $\underline{S} $ is given by:
  
-  * $\underline{S} = UI \cdot e^{j\varphi}$ +  * $\underline{S} = UI \cdot {\rm e}^{{\rm j}\varphi}$ 
-  * $\underline{S} = UI \cdot (\cos\varphi + j \sin\varphi)$ +  * $\underline{S} = UI \cdot (\cos\varphi + {\rm j} \sin\varphi)$ 
-  * $\underline{S} = P + jQ$+  * $\underline{S} = P + {\rm j}Q$
   * $\underline{S} = \underline{U} \cdot \underline{I}^*$   * $\underline{S} = \underline{U} \cdot \underline{I}^*$
  
Zeile 327: Zeile 326:
 Thus, the voltage phasors $\underline{U}_1 ... \underline{U}_m$ form a symmetrical star. \\  Thus, the voltage phasors $\underline{U}_1 ... \underline{U}_m$ form a symmetrical star. \\ 
 Example: A 3-phase system is symmetrical for $\varphi = 360°/3 = 120°$ between the voltages of the windings:  Example: A 3-phase system is symmetrical for $\varphi = 360°/3 = 120°$ between the voltages of the windings: 
-$\underline{U}_1 = \sqrt{2} \cdot U \cdot e ^{j(\omega t + 0°)}$,  +$\underline{U}_1 = \sqrt{2} \cdot U \cdot {\rm e} ^{{\rm j}(\omega t + 0°)}$,  
-$\underline{U}_2 = \sqrt{2} \cdot U \cdot e ^{j(\omega t - 120°)}$,  +$\underline{U}_2 = \sqrt{2} \cdot U \cdot {\rm e} ^{{\rm j}(\omega t - 120°)}$,  
-$\underline{U}_3 = \sqrt{2} \cdot U \cdot e ^{j(\omega t - 240°)}$ \\ +$\underline{U}_3 = \sqrt{2} \cdot U \cdot {\rm e} ^{{\rm j}(\omega t - 240°)}$ \\ 
 <WRAP><imgcaption imageNo05 | Visible Representations of the a symmetric and asymmetric System></imgcaption>{{drawio>technicalTermispolySys2.svg}}</WRAP> <WRAP><imgcaption imageNo05 | Visible Representations of the a symmetric and asymmetric System></imgcaption>{{drawio>technicalTermispolySys2.svg}}</WRAP>
 </WRAP> </WRAP>
Zeile 362: Zeile 361:
  
 ==== 7.2.2 Three-Phase System ==== ==== 7.2.2 Three-Phase System ====
 +
 +See also: [[https://de.mathworks.com/videos/series/what-is-3-phase-power.html|MATHWORKS Onramp Video: What is 3-phase power?]]
  
 The most commonly used polyphase system is the three-phase system. The three-phase system has advantages over a DC system or single-phase AC system: The most commonly used polyphase system is the three-phase system. The three-phase system has advantages over a DC system or single-phase AC system:
Zeile 402: Zeile 403:
  
   * **String voltages/currents**  $U_\rm S$, $I_\rm S$ (alternatively: winding voltages/currents, in German: //Strangspannungen/Strangströme//): <WRAP>   * **String voltages/currents**  $U_\rm S$, $I_\rm S$ (alternatively: winding voltages/currents, in German: //Strangspannungen/Strangströme//): <WRAP>
-The string voltages/currents are the values measured on the windings - independent on the winding connection. \\  +The string voltages/currents are the values measured on the windings - independent of the winding connection. \\  
-These voltages are shown in the previous images as $u_\rm U$, $u_\rm V$, $u_\rm W$.+These voltages are shown in the previous images as $u_\rm U$, $u_\rm V$, and $u_\rm W$.
 </WRAP> </WRAP>
   * **Phase voltages/currents**  $U_\rm L$, $I_\rm L$ (alternatively: phase-to-phase voltages/currents, line-to-line voltages/currents, external conductor voltages/currents, in German: //Außenleiterspannungen/Außenleiterströme//): <WRAP>    * **Phase voltages/currents**  $U_\rm L$, $I_\rm L$ (alternatively: phase-to-phase voltages/currents, line-to-line voltages/currents, external conductor voltages/currents, in German: //Außenleiterspannungen/Außenleiterströme//): <WRAP> 
Zeile 479: Zeile 480:
 \underline{I}_2 = {{\underline{U}_{2 \rm N}}\over{\underline{Z}_2^\phantom{O}}} \quad , \quad  \underline{I}_2 = {{\underline{U}_{2 \rm N}}\over{\underline{Z}_2^\phantom{O}}} \quad , \quad 
 \underline{I}_3 = {{\underline{U}_{3 \rm N}}\over{\underline{Z}_3^\phantom{O}}} \\ \end{align*}</WRAP> \underline{I}_3 = {{\underline{U}_{3 \rm N}}\over{\underline{Z}_3^\phantom{O}}} \\ \end{align*}</WRAP>
-  - The **true power** $P_x$ for each string is given by the apparent power $S_x$ of the string times the indivitual phase angle $\varphi_x$ of the string: <WRAP> +  - The **true power** $P_x$ for each string is given by the apparent power $S_x$ of the string times the individual phase angle $\varphi_x$ of the string: <WRAP> 
 \begin{align*} P_x &= S_x \cdot \cos \varphi_x = U_{\rm S} \cdot I_x \cdot \cos \varphi_x \end{align*} \\  \begin{align*} P_x &= S_x \cdot \cos \varphi_x = U_{\rm S} \cdot I_x \cdot \cos \varphi_x \end{align*} \\ 
 Therefore, the resulting true power for the full load is: \\  Therefore, the resulting true power for the full load is: \\ 
Zeile 505: Zeile 506:
 \begin{align*}  \begin{align*} 
 \underline{I}_1 &= {{\underline{U}_{\rm 1N}}\over{\underline{Z}_1}}  \underline{I}_1 &= {{\underline{U}_{\rm 1N}}\over{\underline{Z}_1}} 
-                &= &{{231 ~\rm V}\over{10 ~\Omega + j \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 1 {~\rm mH}}}  +                &= &{{231 ~\rm V}\over{10 ~\Omega + {\rm j} \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 1 {~\rm mH}}}  
-                &= &+23.08 {~\rm A} &- j \cdot 0.72 {~\rm A} &= &23.09 ~{~\rm A} \quad &\angle -1.8° \\ +                &= &+23.08 {~\rm A} &- {\rm j} \cdot 0.72 {~\rm A} &= &23.09 ~{~\rm A} \quad &\angle -1.8° \\ 
 \underline{I}_2 &= {{\underline{U}_{\rm 2N}}\over{\underline{Z}_2}}  \underline{I}_2 &= {{\underline{U}_{\rm 2N}}\over{\underline{Z}_2}} 
-                &= &{{231 {~\rm V} \cdot \left( -{{1}\over{2}}-j{{1}\over{2}}\sqrt{3}\right)}\over{5 ~ \Omega + {{1}\over{j \cdot 2\pi\cdot 50{~\rm Hz} \cdot 100 {~\rm µF}}}}}  +                &= &{{231 {~\rm V} \cdot \left( -{{1}\over{2}}-{\rm j}{{1}\over{2}}\sqrt{3}\right)}\over{5 ~ \Omega + {{1}\over{{\rm j} \cdot 2\pi\cdot 50{~\rm Hz} \cdot 100 {~\rm µF}}}}}  
-                &= &+ 5.58 {~\rm A} &- j \cdot 4.50 {~\rm A} &= & 7.17 {~\rm A} \quad &\angle -38.9° \\ +                &= &+ 5.58 {~\rm A} &- {\rm j} \cdot 4.50 {~\rm A} &= & 7.17 {~\rm A} \quad &\angle -38.9° \\ 
 \underline{I}_3 &= {{\underline{U}_{\rm 3N}}\over{\underline{Z}_3}}  \underline{I}_3 &= {{\underline{U}_{\rm 3N}}\over{\underline{Z}_3}} 
-                &= &{{231{~\rm V}\cdot \left( -{{1}\over{2}}+j{{1}\over{2}}\sqrt{3}\right)}\over{20 ~\Omega}}  +                &= &{{231{~\rm V}\cdot \left( -{{1}\over{2}}+{\rm j}{{1}\over{2}}\sqrt{3}\right)}\over{20 ~\Omega}}  
-                &= &-5.78{~\rm A} &+ j \cdot 10.00{~\rm A} &= &11.55 {~\rm A} \quad &\angle -240.0°  \\ \\ +                &= &-5.78{~\rm A} &+ {\rm j} \cdot 10.00{~\rm A} &= &11.55 {~\rm A} \quad &\angle -240.0°  \\ \\ 
 \underline{I}_{\rm N}  \underline{I}_{\rm N} 
-                &= \underline{I}_1 + \underline{I}_2 + \underline{I}_3 & & & = &+22.88 {~\rm A} &+ j \cdot 4.77 {~\rm A} &= &23.37 {~\rm A} \quad &\angle +11.8° +                &= \underline{I}_1 + \underline{I}_2 + \underline{I}_3 & & & = &+22.88 {~\rm A} &+ {\rm j} \cdot 4.77 {~\rm A} &= &23.37 {~\rm A} \quad &\angle +11.8° 
 \end{align*} </WRAP> \end{align*} </WRAP>
   - The true power is calculated by: <WRAP>    - The true power is calculated by: <WRAP> 
Zeile 540: Zeile 541:
 In the case of a symmetric load, the situation and the formulas get much simpler: In the case of a symmetric load, the situation and the formulas get much simpler:
   - The **phase-voltages** $U_\rm L$ and star-voltages $U_{\rm Y} = U_{\rm S}$ are equal to the asymmetric load: $U_{\rm L} = \sqrt{3}\cdot U_{\rm S}$.   - The **phase-voltages** $U_\rm L$ and star-voltages $U_{\rm Y} = U_{\rm S}$ are equal to the asymmetric load: $U_{\rm L} = \sqrt{3}\cdot U_{\rm S}$.
-  - For equal impedances the absolute value of all **phase currents** $I_x$ are the same: $|\underline{I}_x|= |\underline{I}_{\rm S}| = \left|{{\underline{U}_{\rm S}}\over{\underline{Z}_{\rm S}^\phantom{O}}} \right|$. Since the phase currents have the same absolute value and have the same $\varphi$, they will add up to zero. Therefore there is no current on the neutral line: $I_{\rm N} =0$ +  - For equal impedances the absolute value of all **phase currents** $I_x$ are the same: $|\underline{I}_x|= |\underline{I}_{\rm S}| = \left|{{\underline{U}_{\rm S}}\over{\underline{Z}_{\rm S}^\phantom{O}}} \right|$. \\ Since the phase currents have the same absolute value and have the same $\varphi$, they will add up to zero. Therefore there is no current on the neutral line: $I_{\rm N} =0$ 
-  - The **true power** is three times the true power of a single phase: $P = 3 \cdot U_{\rm S} I_{\rm S} \cdot \cos \varphi$. Based on the line voltages $U_{\rm L}$, the formula is $P = \sqrt{3} \cdot U_{\rm L} I_{\rm S} \cdot \cos \varphi$ +  - The **true power** is three times the true power of a single phase: $P = 3 \cdot U_{\rm S} I_{\rm S} \cdot \cos \varphi$. \\ Based on the line voltages $U_{\rm L}$, the formula is $P = \sqrt{3} \cdot U_{\rm L} I_{\rm S} \cdot \cos \varphi$ 
-  - The **(collective) apparent power** - given the formula above - is: $S_\Sigma = \sqrt{3}\cdot U_{\rm S} \cdot \sqrt{3\cdot I_{\rm S}^2} = 3 \cdot U_{\rm S} I_{\rm S}$. This corresponds to three times the apparent power of a single phase.+  - The **(collective) apparent power** - given the formula above - is: $S_\Sigma = \sqrt{3}\cdot U_{\rm S} \cdot \sqrt{3\cdot I_{\rm S}^2} = 3 \cdot U_{\rm S} I_{\rm S}$. \\ This corresponds to three times the apparent power of a single phase.
   - The **reactive power** leads to:  $Q_\Sigma = \sqrt{S_\Sigma^2 - P^2} = 3 \cdot U_{\rm S} I_{\rm S} \cdot \sin (\varphi)$.   - The **reactive power** leads to:  $Q_\Sigma = \sqrt{S_\Sigma^2 - P^2} = 3 \cdot U_{\rm S} I_{\rm S} \cdot \sin (\varphi)$.
 </callout> </callout>
Zeile 574: Zeile 575:
   * With the switch $S$, the star potential can short-circuited to the neutral potential; so set $\underline{U}_{\rm SN}=0$. This enables a comparison with the previous four-wire three-phase system.   * With the switch $S$, the star potential can short-circuited to the neutral potential; so set $\underline{U}_{\rm SN}=0$. This enables a comparison with the previous four-wire three-phase system.
  
-Also here the "path": calculate voltages $\rightarrow$ calculate currents $\rightarrow$ calculate true power $\rightarrow$ calculate apparent and reactive power is the best way to get to all wanted values. +Also here, the "path": calculate voltages $\rightarrow$ calculate currents $\rightarrow$ calculate true power $\rightarrow$ calculate apparent and reactive power is the best way to get to all wanted values. 
-  - **Voltages**: Here, only the phase voltages ($\underline{U}_{12}$, $\underline{U}_{23}$, $\underline{U}_{31}$) are applied by the three-phase net, independently of the load. The star-voltages of the load $\underline{U}_{x \rm S}$ are not given by the network anymore, since the neutral potential is not provided. The network star-voltages and the load star-voltages can be connected in the following way: The calculation of the star-voltage $\underline{U}_{\rm SN}$ is explained after investigating the currents. <WRAP> +  - **Voltages**: Here, only the phase voltages ($\underline{U}_{12}$, $\underline{U}_{23}$, $\underline{U}_{31}$) are applied by the three-phase net, independently of the load. The star-voltages of the load $\underline{U}_{x \rm S}$ are not given by the network anymore since the neutral potential is not provided. The network star-voltages and the load star-voltages can be connected in the following way: The calculation of the star-voltage $\underline{U}_{\rm SN}$ is explained after investigating the currents. <WRAP> 
 \begin{align*}  \begin{align*} 
 \underline{U}_{\rm 1S} &= \underline{U}_{\rm 1N}  - \underline{U}_{\rm SN} \\  \underline{U}_{\rm 1S} &= \underline{U}_{\rm 1N}  - \underline{U}_{\rm SN} \\ 
Zeile 603: Zeile 604:
   - Since the three-wire system has no current out of the network star point, the **apparent power** $\underline{S}_x$ for each string is given by the string voltage and the current through the string $\underline{S}_x = \underline{U}_{x \rm S} \cdot \underline{I}_x^*$. This leads to an overall apparent power $\underline{S}$ of <WRAP>    - Since the three-wire system has no current out of the network star point, the **apparent power** $\underline{S}_x$ for each string is given by the string voltage and the current through the string $\underline{S}_x = \underline{U}_{x \rm S} \cdot \underline{I}_x^*$. This leads to an overall apparent power $\underline{S}$ of <WRAP> 
 \begin{align*}  \begin{align*} 
-\underline{S} &= P + j\cdot Q = \sum_x \underline{S}_x = \sum_x  \left( \underline{U}_{x \rm S} \cdot \underline{I}_x^* \right) \end{align*} \\ +\underline{S} &= P + {\rm j}\cdot Q = \sum_x \underline{S}_x = \sum_x  \left( \underline{U}_{x \rm S} \cdot \underline{I}_x^* \right) \end{align*} \\ 
 In order to simplify the calculation, it would be better to have a formula based on the network star-voltages: \\  In order to simplify the calculation, it would be better to have a formula based on the network star-voltages: \\ 
 \begin{align*}  \begin{align*} 
Zeile 621: Zeile 622:
 \end{align*}  </WRAP> \end{align*}  </WRAP>
   - The abolute **reactive power** $Q$ can be calulated by the apparent power: <WRAP>    - The abolute **reactive power** $Q$ can be calulated by the apparent power: <WRAP> 
-\begin{align*} j\cdot Q &= \underline{S} - P  \end{align*}  \\ +\begin{align*} {\rm j}\cdot Q &= \underline{S} - P  \end{align*}  \\ 
 the **collective reactive power** $Q_\Sigma$ is given by the collective apparent power:  \\  the **collective reactive power** $Q_\Sigma$ is given by the collective apparent power:  \\ 
 \begin{align*} Q &= \sqrt{S_\Sigma^2 - P^2}  \end{align*}  </WRAP> \begin{align*} Q &= \sqrt{S_\Sigma^2 - P^2}  \end{align*}  </WRAP>
Zeile 635: Zeile 636:
 \end{align*} \\  \end{align*} \\ 
 Once investigating the numerator $\sum_x \big( {{1}\over{\underline{Z}_x^\phantom{O}}} \cdot \underline{U}_{x \rm N} \big)$, once can see, that it just equals the sum of the phase currents of the four-wire system. So, the numerator equals the (in the three-wire system: fictive) current on the neutral line. \\  Once investigating the numerator $\sum_x \big( {{1}\over{\underline{Z}_x^\phantom{O}}} \cdot \underline{U}_{x \rm N} \big)$, once can see, that it just equals the sum of the phase currents of the four-wire system. So, the numerator equals the (in the three-wire system: fictive) current on the neutral line. \\ 
-The numerator is therefore: $22.88 {~\rm A} + j \cdot 4.77 {~\rm A}$ (see calculation for the four-wire system). \\ +The numerator is therefore: $22.88 {~\rm A} + {\rm j} \cdot 4.77 {~\rm A}$ (see calculation for the four-wire system). \\ 
 The denominator is: \\  The denominator is: \\ 
 \begin{align*}  \begin{align*} 
-\sum_x  {{1}\over{\underline{Z}_x^\phantom{O}}}  &= {{1}\over{10~\Omega +          j \cdot 2\pi\cdot 50{~\rm Hz} \cdot 1   {~\rm mH} }} +\sum_x  {{1}\over{\underline{Z}_x^\phantom{O}}}  &= {{1}\over{10~\Omega +          {\rm j} \cdot 2\pi\cdot 50{~\rm Hz} \cdot 1   {~\rm mH} }} 
-                                                  + {{1}\over{5~\Omega + {{1}\over{j \cdot 2\pi\cdot 50{~\rm Hz} \cdot 100 {~\rm µF} }} }}+                                                  + {{1}\over{5~\Omega + {{1}\over{{\rm j} \cdot 2\pi\cdot 50{~\rm Hz} \cdot 100 {~\rm µF} }} }}
                                                   + {{1}\over{20~\Omega }} \\ \\                                                    + {{1}\over{20~\Omega }} \\ \\ 
-                                                  &= 0.1547  ~1/\Omega + j \cdot 0.02752 ~1/\Omega  \end{align*} \\ +                                                  &= 0.1547  ~1/\Omega + {\rm j} \cdot 0.02752 ~1/\Omega  \end{align*} \\ 
 The star-voltage $\underline{U}_{\rm SN}$ of the load is:  The star-voltage $\underline{U}_{\rm SN}$ of the load is: 
 \begin{align*}   \begin{align*}  
-\underline{U}_{\rm SN} &= {{22.88 {~\rm A} + j \cdot 4.77 {~\rm A}}\over{0.1547 ~1/\Omega + j \cdot 0.0275 ~1/\Omega}} \\ \\ +\underline{U}_{\rm SN} &= {{22.88 {~\rm A} + {\rm j} \cdot 4.77 {~\rm A}}\over{0.1547 ~1/\Omega + {\rm j} \cdot 0.0275 ~1/\Omega}} \\ \\ 
-                       &= 148.7{~\rm V} + j \cdot 4.41 {~\rm V} \end{align*} \\ +                       &= 148.7   {~\rm V} + {\rm j} \cdot 4.41 {~\rm V} \end{align*} \\ 
 Given this star-voltage $\underline{U}_{\rm SN}$ of the load, the phase currents are: \\  Given this star-voltage $\underline{U}_{\rm SN}$ of the load, the phase currents are: \\ 
 \begin{align*}  \begin{align*} 
 \underline{I}_1 &= {{\underline{U}_{\rm 1N} - \underline{U}_{\rm SN}}\over{\underline{Z}_1^\phantom{O}}}  \underline{I}_1 &= {{\underline{U}_{\rm 1N} - \underline{U}_{\rm SN}}\over{\underline{Z}_1^\phantom{O}}} 
-                & = & {{231{~\rm V} - 148.7{~\rm V} - j \cdot 4.41 {~\rm V}}\over{10~\Omega + j \cdot 2\pi\cdot 50{~\rm Hz} \cdot 1{~\rm mH} }}  +                & = & {{231{~\rm V} - 148.7{~\rm V} - {\rm j} \cdot 4.41 {~\rm V}}\over{10~\Omega + {\rm j} \cdot 2\pi\cdot 50{~\rm Hz} \cdot 1{~\rm mH} }}  
-                & = & +8.21{~\rm A} - j \cdot 0.70{~\rm A} &=&  8.24 {~\rm A} \quad  \angle -4.9° \\ +                & = & +8.21{~\rm A}                 - {\rm j} \cdot 0.70{~\rm A} &=&  8.24 {~\rm A} \quad  \angle -4.9° \\ 
 \underline{I}_2 &= {{\underline{U}_{\rm 2N} - \underline{U}_{\rm SN}}\over{\underline{Z}_2^\phantom{O}}}  \underline{I}_2 &= {{\underline{U}_{\rm 2N} - \underline{U}_{\rm SN}}\over{\underline{Z}_2^\phantom{O}}} 
-                & = & {{231{~\rm V} \cdot \left( -{{1}\over{2}}-j{{1}\over{2}}\sqrt{3}\right) - 148.7{~\rm V} - j \cdot 4.41 {~\rm V}}\over{5~\Omega + {{1}\over{j \cdot 2\pi\cdot 50{~\rm Hz} \cdot 100{~\rm µF} }}}}  +                & = & {{231{~\rm V} \cdot \left( -{{1}\over{2}}-{\rm j}{{1}\over{2}}\sqrt{3}\right) - 148.7{~\rm V} - {\rm j} \cdot 4.41 {~\rm V}}\over{5~\Omega + {{1}\over{{\rm j} \cdot 2\pi\cdot 50{~\rm Hz} \cdot 100{~\rm µF} }}}}  
-                & = & +5.00{~\rm A} + j \cdot 9.08{~\rm A} & =& 10.36 {~\rm A} \quad \angle -61.2° \\+                & = & +5.00{~\rm A} + {\rm j} \cdot 9.08{~\rm A} & =& 10.36 {~\rm A} \quad \angle -61.2° \\
 \underline{I}_3 &= {{\underline{U}_{\rm 3N} - \underline{U}_{\rm SN}}\over{\underline{Z}_3^\phantom{O}}}  \underline{I}_3 &= {{\underline{U}_{\rm 3N} - \underline{U}_{\rm SN}}\over{\underline{Z}_3^\phantom{O}}} 
-                & = & {{231{~\rm V} \cdot \left( -{{1}\over{2}}+j{{1}\over{2}}\sqrt{3}\right) - 148.7{~\rm V} - j \cdot 4.41 {~\rm V}}\over{20~\Omega }}  +                & = & {{231{~\rm V} \cdot \left( -{{1}\over{2}}+{\rm j}{{1}\over{2}}\sqrt{3}\right) - 148.7{~\rm V} - {\rm j} \cdot 4.41 {~\rm V}}\over{20~\Omega }}  
-                &= & -13.21{~\rm A} + j \cdot 9.78{~\rm A} & =& 16.44{~\rm A} \quad  \angle +143.5° \end{align*} \\ </WRAP>+                &= & -13.21{~\rm A} + {\rm j} \cdot 9.78{~\rm A} & =& 16.44{~\rm A} \quad  \angle +143.5° \end{align*} \\ </WRAP>
   - The true power is calculated by: \\ \begin{align*} P = 231{~\rm V} \cdot \big( 8.24 {~\rm A} \cdot \cos (0° - (-4.9°))+ 10.36{~\rm A} \cdot \cos (-120° - (-61.2°)) + 16.44 {~\rm A} \cdot \cos (-240° - (+143.5°)\big) = 6.62 {~\rm kW} \end{align*}   - The true power is calculated by: \\ \begin{align*} P = 231{~\rm V} \cdot \big( 8.24 {~\rm A} \cdot \cos (0° - (-4.9°))+ 10.36{~\rm A} \cdot \cos (-120° - (-61.2°)) + 16.44 {~\rm A} \cdot \cos (-240° - (+143.5°)\big) = 6.62 {~\rm kW} \end{align*}
   - The apparent power $\underline{S}$ is: <WRAP>    - The apparent power $\underline{S}$ is: <WRAP> 
 \begin{align*}  \begin{align*} 
 \underline{S} &= \underline{U}_{13} \cdot \underline{I}_1^* + \underline{U}_{23} \cdot \underline{I}_2^*  \underline{S} &= \underline{U}_{13} \cdot \underline{I}_1^* + \underline{U}_{23} \cdot \underline{I}_2^* 
-              &=& 400{~\rm V} \cdot (- e^{-j \cdot 7/6 \pi} \cdot (8.21 {~\rm A} + j \cdot 0.70 {~\rm A} )  +  e^{- j \cdot 3/6 \pi} \cdot (5.00{~\rm A} - j \cdot 9.08{~\rm A}) )  +           &=& 400{~\rm V} \cdot (- {\rm e}^{-{\rm j} \cdot 7/6 \pi} \cdot (8.21{~\rm A} + {\rm j}\cdot 0.70 {~\rm A})  + {\rm e}^{-{\rm j} \cdot 3/6 \pi} \cdot (  5.00{~\rm A} -{\rm j} \cdot 9.08{~\rm A}) )  
-              &= 6.62 {~\rm kW} - j \cdot 3.40 {~\rm kVAr} \\  +             &= 6.62 {~\rm kW} - {\rm j} \cdot 3.40 {~\rm kVAr} \\  
-              &= \underline{U}_{12} \cdot \underline{I}_1^* + \underline{U}_{32} \cdot \underline{I}_3^*  +             &= \underline{U}_{12} \cdot \underline{I}_1^* + \underline{U}_{32} \cdot \underline{I}_3^*  
-              &=& 400{~\rm V} \cdot (e^{j \cdot 1/6 \pi} \cdot (8.21{~\rm A} + j \cdot 0.70{~\rm A}) - e^{-j \cdot 3/6 \pi} \cdot (-13.21{~\rm A} -j \cdot 9.78{~\rm A}))  +           &=& 400{~\rm V} \cdot (  {\rm e}^{{\rm j} \cdot 1/6 \pi} \cdot (8.21{~\rm A} +  {\rm j}\cdot 0.70{~\rm A})   - {\rm e}^{-{\rm j} \cdot 3/6 \pi} \cdot (-13.21{~\rm A} -{\rm j} \cdot 9.78{~\rm A}))  
-              &= 6.62 {~\rm kW} - j \cdot 3.40 {~\rm kVAr} \\ +              &= 6.62 {~\rm kW} - {\rm j} \cdot 3.40 {~\rm kVAr} \\ 
               &= \underline{U}_{21} \cdot \underline{I}_2^* + \underline{U}_{31} \cdot \underline{I}_3^*                &= \underline{U}_{21} \cdot \underline{I}_2^* + \underline{U}_{31} \cdot \underline{I}_3^* 
-              &=& 400{~\rm V} \cdot (- e^{j \cdot 1/6 \pi} \cdot (5.00{~\rm A} - j \cdot 9.08{~\rm A}) + e^{- j \cdot 7/6 \pi} \cdot (-13.21{~\rm A} - j \cdot 9.78{~\rm A}))  +           &=& 400{~\rm V} \cdot (- {\rm e}^{{\rm j} \cdot 1/6 \pi} \cdot (5.00{~\rm A} -  {\rm j}\cdot 9.08{~\rm A}) + {\rm e}^{- {\rm j} \cdot 7/6 \pi} \cdot (-13.21{~\rm A} - {\rm j} \cdot 9.78{~\rm A}))  
-              &= 6.62 {~\rm kW} - j \cdot 3.40 {~\rm kVAr} \\ +              &= 6.62 {~\rm kW} - {\rm j} \cdot 3.40 {~\rm kVAr} \\ 
               & = 7.44 {~\rm kVA} \quad \angle -27.2°               & = 7.44 {~\rm kVA} \quad \angle -27.2°
 \end{align*}  \\  \end{align*}  \\ 
Zeile 677: Zeile 678:
           = 8.45 {~\rm kVA} \end{align*}  </WRAP>           = 8.45 {~\rm kVA} \end{align*}  </WRAP>
   - The reactive power is: <WRAP>    - The reactive power is: <WRAP> 
-\begin{align*} Q &= -j \cdot (\underline{S} - P) =  -3.40{~\rm kVAr} \\  +\begin{align*} Q &= -{\rm j} \cdot (\underline{S} - P) =  -3.40{~\rm kVAr} \\  
 \end{align*} \\  \end{align*} \\ 
 The collective reactive power is: \\  The collective reactive power is: \\ 
Zeile 722: Zeile 723:
   - **Voltages**: Here, the string voltages of the load are applied by the three-phase net: <WRAP>    - **Voltages**: Here, the string voltages of the load are applied by the three-phase net: <WRAP> 
 \begin{align*}  \begin{align*} 
-\underline{U}_{12} &=& U_{\rm L} \cdot e^{  j\cdot {{1}\over{6}}} \\  +\underline{U}_{12} &=& U_{\rm L} \cdot {\rm e}^{  {\rm j}\cdot {{1}\over{6}}} \\  
-\underline{U}_{23} &=& U_{\rm L} \cdot e^{- j\cdot {{3}\over{6}}} \\  +\underline{U}_{23} &=& U_{\rm L} \cdot {\rm e}^{- {\rm j}\cdot {{3}\over{6}}} \\  
-\underline{U}_{31} &=& U_{\rm L} \cdot e^{- j\cdot {{7}\over{6}}} +\underline{U}_{31} &=& U_{\rm L} \cdot {\rm e}^{- {\rm j}\cdot {{7}\over{6}}} 
 \end{align*}</WRAP> \end{align*}</WRAP>
   - **Currents**: For the phase currents one can focus on the nodes between the phase lines and the strings. An incomming single-phase current onto a note divides into two string currents: <WRAP>   - **Currents**: For the phase currents one can focus on the nodes between the phase lines and the strings. An incomming single-phase current onto a note divides into two string currents: <WRAP>
Zeile 747: Zeile 748:
 \begin{align*}  \begin{align*} 
 \boxed{ \boxed{
-\underline{S} = P + j \cdot Q +\underline{S} = P + {\rm j} \cdot Q 
               = U_{\rm L}^2 \cdot \left( {{1}\over{\underline{Z}_{12}^* }}                = U_{\rm L}^2 \cdot \left( {{1}\over{\underline{Z}_{12}^* }} 
                                       +  {{1}\over{\underline{Z}_{23}^* }}                                        +  {{1}\over{\underline{Z}_{23}^* }} 
Zeile 753: Zeile 754:
 \end{align*} \\  \end{align*} \\ 
 The **collective apparent power** $S_\Sigma$ here is the same as for the three-wire or four-wire connection. \\  The **collective apparent power** $S_\Sigma$ here is the same as for the three-wire or four-wire connection. \\ 
-In the Delta connection the phase currents $I_x$ have to be calculated since the formula only applies for them:  \\ +In the Delta connection the phase currents $I_x$ have to be calculated since the formula only applies to them:  \\ 
 \begin{align*}  \begin{align*} 
 S_\Sigma &= \sqrt{ {{1}\over{3}} (U_{12}^2 + U_{23}^2 + U_{31}^2) } \cdot \sqrt{\sum_x I_x^2} &= U_{\rm L} \cdot \sqrt{\sum_x I_x^2}   S_\Sigma &= \sqrt{ {{1}\over{3}} (U_{12}^2 + U_{23}^2 + U_{31}^2) } \cdot \sqrt{\sum_x I_x^2} &= U_{\rm L} \cdot \sqrt{\sum_x I_x^2}  
 \end{align*}  </WRAP> \end{align*}  </WRAP>
   - The abolute **reactive power** $Q$ can be calulated by the apparent power: <WRAP>    - The abolute **reactive power** $Q$ can be calulated by the apparent power: <WRAP> 
-\begin{align*} j\cdot Q &= \underline{S} - P  \end{align*}  \\ +\begin{align*} {\rm j}\cdot Q &= \underline{S} - P  \end{align*}  \\ 
 the **collective reactive power** $Q_\Sigma$ is given by the collective apparent power:  \\  the **collective reactive power** $Q_\Sigma$ is given by the collective apparent power:  \\ 
 \begin{align*} Q &= \sqrt{S_\Sigma^2 - P^2}  \end{align*}  </WRAP> \begin{align*} Q &= \sqrt{S_\Sigma^2 - P^2}  \end{align*}  </WRAP>
Zeile 773: Zeile 774:
 The phasors of the string voltages of the network are given as \\  The phasors of the string voltages of the network are given as \\ 
 {{drawio>ThreeWireStringPhaseVoltageFormula.svg}}</WRAP> {{drawio>ThreeWireStringPhaseVoltageFormula.svg}}</WRAP>
-  - Based on the string voltages of the network and the given impedances the string currents $\underline{I}_{12}$, $\underline{I}_{23}$, $\underline{I}_{31}$ of the load can be calculated : <WRAP> +  - Based on the string voltages of the network and the given impedances the string currents $\underline{I}_{12}$, $\underline{I}_{23}$, $\underline{I}_{31}$ of the load can be calculated: <WRAP> 
 \begin{align*}  \begin{align*} 
-\underline{I}_{12} &=& {{ 400 {~\rm V} \cdot\left(+{{1}\over{2}}\sqrt{3}+{{1}\over{2}} \cdot j \right)}\over{ 10~\Omega + j \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 1 {~\rm mH} }}  +\underline{I}_{12} &=& {{ 400 {~\rm V} \cdot\left(+{{1}\over{2}}\sqrt{3}+{{1}\over{2}} \cdot {\rm j} \right)}\over{ 10~\Omega + {\rm j} \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 1 {~\rm mH} }}  
-                   &=& 35.24 {~\rm A} + j \cdot 18.90 {~\rm A} &=& 40 {~\rm A} \quad &\angle 28.2°  \\  +                   &=& 35.24 {~\rm A} + {\rm j} \cdot 18.90 {~\rm A} &=& 40 {~\rm A} \quad &\angle 28.2°  \\  
-\underline{I}_{23} &=& {{400 \cdot j}\over{ 5~\Omega + {{1}\over{j \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 100 {~\rm µF}}} }}   +\underline{I}_{23} &=& {{400 \cdot {\rm j}}\over{ 5~\Omega + {{1}\over{{\rm j} \cdot 2\pi\cdot 50 {~\rm Hz} \cdot 100 {~\rm µF}}} }}   
-                   &=& 12.27 {~\rm A} - j \cdot 1.93 {~\rm A} &=& 12.42 {~\rm A} \quad &\angle -8.9° \\  +                   &=& 12.27 {~\rm A} - {\rm j} \cdot 1.93 {~\rm A} &=& 12.42 {~\rm A} \quad &\angle -8.9° \\  
-\underline{I}_{31} &=& {{400 {~\rm V} \cdot\left(-{{1}\over{2}}\sqrt{3}+{{1}\over{2}} \cdot j \right)}\over{ 20 ~\Omega}}   +\underline{I}_{31} &=& {{400 {~\rm V} \cdot\left(-{{1}\over{2}}\sqrt{3}+{{1}\over{2}} \cdot {\rm j} \right)}\over{ 20 ~\Omega}}   
-                   &=& -17.33 {~\rm A} + j \cdot 10.00 {~\rm A} &=& 20.01 {~\rm A} \quad &\angle 150°  \end{align*} \\ +                   &=& -17.33 {~\rm A} + {\rm j} \cdot 10.00 {~\rm A} &=& 20.01 {~\rm A} \quad &\angle 150°  \end{align*} \\ 
 By these voltages the phase currents $\underline{I}_x$ can be calculated: \\  By these voltages the phase currents $\underline{I}_x$ can be calculated: \\ 
 \begin{align*}  \begin{align*} 
-\underline{I}_{1} &=& ( 35.24 {~\rm A} + j \cdot 18.90 {~\rm A}) - (-17.33 {~\rm A} + j \cdot 10.00 {~\rm A}) &=&  52.57 {~\rm A} + j \cdot 8.90 {~\rm A} +\underline{I}_{1} &=& ( 35.24 {~\rm A} + {\rm j} \cdot 18.90 {~\rm A}) - (-17.33 {~\rm A} + {\rm j} \cdot 10.00 {~\rm A}) &=&  52.57 {~\rm A} + {\rm j} \cdot 8.90 {~\rm A} 
                   &=& 53.32 {~\rm A} \quad &\angle 9.6°  \\                   &=& 53.32 {~\rm A} \quad &\angle 9.6°  \\
-\underline{I}_{2} &=& ( 12.27 {~\rm A} - j \cdot  1.93 {~\rm A}) - ( 35.24 {~\rm A} + j \cdot 18.90 {~\rm A}) &=& -22.98 {~\rm A} - j \cdot 20.83 {~\rm A}  +\underline{I}_{2} &=& ( 12.27 {~\rm A} - {\rm j} \cdot  1.93 {~\rm A}) - ( 35.24 {~\rm A} + {\rm j} \cdot 18.90 {~\rm A}) &=& -22.98 {~\rm A} - {\rm j} \cdot 20.83 {~\rm A}  
                   &=& -31.01 {~\rm A} \quad &\angle -137.8° \\                   &=& -31.01 {~\rm A} \quad &\angle -137.8° \\
-\underline{I}_{3} &=& (-17.33 {~\rm A} + j \cdot 10.00 {~\rm A}) - ( 12.27 {~\rm A} - j \cdot  1.93 {~\rm A}) &=& -29.59 {~\rm A} + j \cdot 11.93 {~\rm A} +\underline{I}_{3} &=& (-17.33 {~\rm A} + {\rm j} \cdot 10.00 {~\rm A}) - ( 12.27 {~\rm A} - {\rm j} \cdot  1.93 {~\rm A}) &=& -29.59 {~\rm A} + {\rm j} \cdot 11.93 {~\rm A} 
                   &=& 31.90 {~\rm A} \quad &\angle 158.0°                     &=& 31.90 {~\rm A} \quad &\angle 158.0°  
 \end{align*} \\</WRAP> \end{align*} \\</WRAP>
Zeile 799: Zeile 800:
 \begin{align*}  \begin{align*} 
 \underline{S} &= \underline{U}_{13} \cdot \underline{I}_1^* + \underline{U}_{23} \cdot \underline{I}_2^*  \underline{S} &= \underline{U}_{13} \cdot \underline{I}_1^* + \underline{U}_{23} \cdot \underline{I}_2^* 
-              &=& 400 {~\rm V} \cdot (- e^{-j \cdot 7/6 \pi} \cdot (52.57  {~\rm A} - j \cdot 8.90  {~\rm A} )  +  e^{- j \cdot 3/6 \pi} \cdot (-22.98 {~\rm A} + j \cdot 20.83 {~\rm A}) )  +              &=& 400 {~\rm V} \cdot (- {\rm e}^{-{\rm j} \cdot 7/6 \pi} \cdot ( 52.57 {~\rm A} - {\rm j} \cdot  8.90 {~\rm A}) +  
-              &= 24.77  {~\rm kW} - j \cdot 4.41  {~\rm kVAr} \\ +                                        {\rm e}^{-{\rm j} \cdot 3/6 \pi} \cdot (-22.98 {~\rm A} + {\rm j} \cdot 20.83 {~\rm A}))  
 +              &= 24.77  {~\rm kW} - {\rm j} \cdot 4.41  {~\rm kVAr} \\ 
               &= \underline{U}_{12} \cdot \underline{I}_1^* + \underline{U}_{32} \cdot \underline{I}_3^*                &= \underline{U}_{12} \cdot \underline{I}_1^* + \underline{U}_{32} \cdot \underline{I}_3^* 
-              &=& 400 {~\rm V} \cdot (e^{j \cdot 1/6 \pi} \cdot (52.57  {~\rm A} - j \cdot 8.90  {~\rm A}) - e^{-j \cdot 3/6 \pi} \cdot (-29.59 {~\rm A} - j \cdot 11.93 {~\rm A}))  +              &=& 400 {~\rm V} \cdot (  {\rm e}^{ {\rm j} \cdot 1/6 \pi} \cdot ( 52.57 {~\rm A} - {\rm j} \cdot  8.90 {~\rm A}) -  
-              &= 24.77  {~\rm kW} - j \cdot 4.41  {~\rm kVAr} \\ +                                        {\rm e}^{-{\rm j} \cdot 3/6 \pi} \cdot (-29.59 {~\rm A} - {\rm j} \cdot 11.93 {~\rm A}))  
 +              &= 24.77  {~\rm kW} - {\rm j} \cdot 4.41  {~\rm kVAr} \\ 
               &= \underline{U}_{21} \cdot \underline{I}_2^* + \underline{U}_{31} \cdot \underline{I}_3^*                &= \underline{U}_{21} \cdot \underline{I}_2^* + \underline{U}_{31} \cdot \underline{I}_3^* 
-              &=& 400 {~\rm V} \cdot (- e^{j \cdot 1/6 \pi} \cdot (-22.98 {~\rm A} + j \cdot 20.83 {~\rm A}) + e^{- j \cdot 7/6 \pi} \cdot (-29.59 {~\rm A} - j \cdot 11.93 {~\rm A}))  +              &=& 400 {~\rm V} \cdot (- {\rm e}^{ {\rm j} \cdot 1/6 \pi} \cdot (-22.98 {~\rm A} + {\rm j} \cdot 20.83 {~\rm A}) +  
-              &= 24.77  {~\rm kW} - j \cdot 4.41  {~\rm kVAr} \\ +                                        {\rm e}^{-{\rm j} \cdot 7/6 \pi} \cdot (-29.59 {~\rm A} - {\rm j} \cdot 11.93 {~\rm A}))  
 +              &= 24.77  {~\rm kW} - {\rm j} \cdot 4.41  {~\rm kVAr} \\ 
               & = 25.16  {~\rm kVA} \quad \angle -10.09°\end{align*}  \\                & = 25.16  {~\rm kVA} \quad \angle -10.09°\end{align*}  \\ 
 The collective apparent power is: \\  The collective apparent power is: \\ 
Zeile 865: Zeile 869:
 \end{align*} \end{align*}
  
-For the **series circuit**, the impedances add up like: $R_s + j\cdot X_{Ls} = \underline{Z} $, and $R_s = |\underline{Z}| \cos\varphi$ such as $X_{Ls} = |\underline{Z}| \sin\varphi$.+For the **series circuit**, the impedances add up like: $R_s + {\rm j}\cdot X_{Ls} = \underline{Z} $, and $R_s = |\underline{Z}| \cos\varphi$ such as $X_{Ls} = |\underline{Z}| \sin\varphi$.
 Therefore: Therefore:
 \begin{align*}  \begin{align*} 
Zeile 874: Zeile 878:
  
  
-For the **parallel circuit**, the impedances add up like ${{1}\over{R_p}} + {{1}\over{j\cdot X_{Lp}}}= {{1}\over{\underline{Z}}} $. \\+For the **parallel circuit**, the impedances add up like ${{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}}= {{1}\over{\underline{Z}}} $. \\
 The easiest thing is here to use the formulas of $R_s$ and $X_{Ls}$ from before: The easiest thing is here to use the formulas of $R_s$ and $X_{Ls}$ from before:
  
 \begin{align*}  \begin{align*} 
-{{1}\over{R_p}} + {{1}\over{j\cdot X_{Lp}}} &=& {{1}\over{R_s + j\cdot X_{Ls}}} \\ +{{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}} &=& {{1}\over{R_s + {\rm j}\cdot X_{Ls}}} \\ 
-{{1}\over{R_p}} - j {{1}\over{X_{Lp}}}      &=& {{R_s - j\cdot X_{Ls}}\over{R_s^2 + X_{Ls}^2}} \\ +{{1}\over{R_p}} - {\rm j} {{1}\over{X_{Lp}}}      &=&         {{R_s - {\rm j}\cdot X_{Ls}}\over{R_s^2 + X_{Ls}^2}} \\ 
-                                            &=& {{Z \cdot \cos \varphi - j\cdot Z \cdot \sin \varphi }\over{Z^2}}\\ +                                                  &=& {{Z \cdot \cos \varphi - {\rm j}\cdot Z \cdot \sin \varphi }\over{Z^2}} \\ 
-                                            &=& {{\cos \varphi - j \cdot \sin \varphi }\over{Z}}\\+                                                  &=& {        {\cos \varphi - {\rm j} \cdot \sin \varphi }       \over{Z}} \\
 \end{align*} \end{align*}
  
Zeile 977: Zeile 981:
  
 \begin{align*}  \begin{align*} 
-\underline{I} &= I_R + j \cdot I_L \\ +\underline{I} &= I_R                 + {\rm j} \cdot I_L \\ 
-              &= I \cdot \cos\varphi - j \cdot I \cdot \sin\varphi +              &= I \cdot \cos\varphi - {\rm j} \cdot I \cdot \sin\varphi 
 \end{align*} \end{align*}
  
Zeile 1090: Zeile 1094:
 Q &= \Re (U) \cdot \Im (I) \\  Q &= \Re (U) \cdot \Im (I) \\ 
   &= U \cdot {{U}\over{X}} \\    &= U \cdot {{U}\over{X}} \\ 
-  &= {{U^2}\over{X}} \\ +  &=       {{U^2}\over{X}} \\ 
 \end{align*} \end{align*}
  
Zeile 1148: Zeile 1152:
 \begin{align*}  \begin{align*} 
 \underline{S}_{\rm net} &=& \underline{S}_1               &+& \underline{S}_2 \\ \underline{S}_{\rm net} &=& \underline{S}_1               &+& \underline{S}_2 \\
-                        &=& P_1 + j \cdot Q_1             &+& P_2 + j \cdot Q_2 \\ +                        &=& P_1 + {\rm j} \cdot Q_1       &+& P_2 + {\rm j} \cdot Q_2 \\ 
-                        &=& P_1 + P_2                     &+& j \cdot (Q_1 + Q_2) \\  +                        &=& P_1 + P_2                     &+& {\rm j} \cdot (Q_1 + Q_2) \\  
-                        &=& 2.7 {~\rm kW} + 3.8 {~\rm kW} &+& j \cdot (1.4 {~\rm kVAr} + 3.2 {~\rm kVAr}) \\  +                        &=& 2.7 {~\rm kW} + 3.8 {~\rm kW} &+& {\rm j} \cdot (1.4 {~\rm kVAr} + 3.2 {~\rm kVAr}) \\  
-                        &=& 6.5 {~\rm kW}                 &+& j \cdot 4.6 {~\rm kVAr} \\  +                        &=& 6.5 {~\rm kW}                 &+& {\rm j} \cdot 4.6 {~\rm kVAr} \\  
-                        &=& P_{\rm net}                   &+& j \cdot Q_{\rm net} \\ +                        &=& P_{\rm net}                   &+& {\rm j} \cdot Q_{\rm net} \\ 
 \end{align*} \\ \end{align*} \\
  
 As a complex value in Euler representation: As a complex value in Euler representation:
 \begin{align*}  \begin{align*} 
-\underline{S}_{\rm net} &=& \sqrt{P_{\rm net}^2    +  Q_{\rm net}^2      } &\cdot& e^{j \cdot \arctan ({{Q_{\rm net}}\over{P_{\rm net}}})} \\ +\underline{S}_{\rm net} &=& \sqrt{P_{\rm net}^2    +  Q_{\rm net}^2      } &\cdot& {\rm e}^{{\rm j} \cdot \arctan ({{Q_{\rm net}}\over{P_{\rm net}}})} \\ 
-                            \sqrt{(6.5 {~\rm kW})^2+  (4.6 {~\rm kVAr})^2} &\cdot& e^{j \cdot \arctan ({{4.6        }\over{6.5    }})} \\ +                            \sqrt{(6.5 {~\rm kW})^2+  (4.6 {~\rm kVAr})^2} &\cdot& {\rm e}^{{\rm j} \cdot \arctan ({{4.6        }\over{6.5    }})} \\ 
-                                   8.0 {~\rm kVA}                          &\cdot& e^{j \cdot 35°} \\+                                   8.0 {~\rm kVA}                          &\cdot& {\rm e}^{{\rm j} \cdot 35°} \\
 \end{align*}  \end{align*} 
 </collapse><button size="xs" type="link" collapse="Loesung_7_2_1_2_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_7_2_1_2_Endergebnis" collapsed="true">  </collapse><button size="xs" type="link" collapse="Loesung_7_2_1_2_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_7_2_1_2_Endergebnis" collapsed="true"> 
 \begin{align*}  \begin{align*} 
-\underline{S}_{\rm net} &=& 6.5 {~\rm kW}+ j \cdot 4.6 {~\rm kVAr} \\  +\underline{S}_{\rm net} &=& 6.5 {~\rm kW} + {\rm j} \cdot 4.6 {~\rm kVAr} \\  
-                        &=& 8.0 {~\rm kVA} \cdot e^{j \cdot 35°} \\+                        &=& 8.0 {~\rm kVA} \cdot {\rm e}^{{\rm j} \cdot 35°} \\
 \end{align*}  \end{align*} 
 </collapse>\\ </collapse>\\