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-====== 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 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 201: Zeile 200:
 Similarly, the currents and voltages can be separated into active, reactive, and apparent values. </callout> Similarly, the currents and voltages can be separated into active, reactive, and apparent values. </callout>
  
-Based on the given formulas the three types of power are connected with each other. Since the apparent power is given by $S=U\cdot I$, the active power $P = U\cdot I \cdot \sin \varphi = S \cdot \sin \varphi $ and the reactive power $Q = S \cdot \cos \varphi $, the relationship can be shown in a triangle (see <imgref imageNo02>).+Based on the given formulas the three types of power are connected with each other. Since the apparent power is given by $S=U\cdot I$, the active power $P = U\cdot I \cdot \cos \varphi = S \cdot \cos \varphi $ and the reactive power $Q = S \cdot \sin \varphi $, the relationship can be shown in a triangle (see <imgref imageNo02>).
  
 <WRAP> <imgcaption imageNo02 | Power Triangle of active, reactive and apparent power></imgcaption> {{drawio>powertriangle.svg}} </WRAP> <WRAP> <imgcaption imageNo02 | Power Triangle of active, reactive and apparent power></imgcaption> {{drawio>powertriangle.svg}} </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 845: Zeile 849:
 <panel type="info" title="Exercise 7.1.1 Power and Power Factor I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 7.1.1 Power and Power Factor I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-A passive component is fed by a sinusoidal AC voltage with the RMS value $U=230~\rm V$ and $f=50.0~\rm Hz$. The RMS current on this component is $I=5.00~\rm A$ with a phase angle of $\varphi=60°$.+A passive component is fed by a sinusoidal AC voltage with the RMS value $U=230~\rm V$ and $f=50.0~\rm Hz$. The RMS current on this component is $I=5.00~\rm A$ with a phase angle of $\varphi=+60°$.
  
-  - Draw the equivalent circuits based on a series and on a parallel circuit. +1. Draw the equivalent circuits based on a series and on a parallel circuit. \\
-  - Calculate the equivalent components for both circuits. +
-  - Calculate the real power, the reactive power, and the apparent power based on the equivalent components for both circuits from 2. . +
-  - Check the solutions from 3. via direct calculation based on the input in the task above. +
- +
-<button size="xs" type="link" collapse="Loesung_7_1_1_1_Rechnung">{{icon>eye}} Result for 1.</button><collapse id="Loesung_7_1_1_1_Rechnung" collapsed="true"> +
  
 +#@HiddenBegin_HTML~71111,Result~@#
 {{drawio>electrical_engineering_2:Sol711EquivCirc.svg}} {{drawio>electrical_engineering_2:Sol711EquivCirc.svg}}
 +#@HiddenEnd_HTML~71111,Result~@#
  
-</collapse>+2. Calculate the equivalent components for both circuits. \\
  
-<button size="xs" type="link" collapse="Loesung_7_1_1_2_Rechnung">{{icon>eye}} Result for 2.</button><collapse id="Loesung_7_1_1_2_Rechnung" collapsed="true"> +#@HiddenBegin_HTML~71112,Solution~@#
  
 The apparent impedance is: The apparent impedance is:
Zeile 865: Zeile 866:
 \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 873: Zeile 874:
 \end{align*} \end{align*}
  
 +\\ \\
 +For the **parallel circuit**, the impedances add up like ${{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}}= {{1}\over{\underline{Z}}} $ with $\underline{Z} = {{U}\over{I}}\cdot e^{j\cdot \varphi}$. \\
  
-For the **parallel circuit**, the impedances add up like ${{1}\over{R_p}} + {{1}\over{j\cdot X_{Lp}}}= {{1}\over{\underline{Z}}} $. \\ +There are multiple ways to solve this problem. Two ways shall be shown here:
-The easiest thing is here to use the formulas of $R_s$ and $X_{Ls}$ from before:+
  
 +=== with the Euler representation ===
 +Given the formula $\underline{Z} = {{U}\over{I}}\cdot e^{j\cdot \varphi}$ the following can be derived:
 \begin{align*}  \begin{align*} 
-{{1}\over{R_p}} + {{1}\over{j\cdot X_{Lp}}} &=& {{1}\over{R_s + j\cdot X_{Ls}}} \\ +{{1}\over{\underline{Z}^{\phantom{A}}}} &= {{I}\over{U}}\cdot e^{-j\cdot \varphi} \\ 
-{{1}\over{R_p}} - j {{1}\over{X_{Lp}}}      &=& {{R_s - j\cdot X_{Ls}}\over{R_s^2 + X_{Ls}^2}} \\ +                                        &= {{1}\over{Z}}\cdot e^{-j\cdot \varphi} &&= {{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}}  \\ 
-                                            &=& {{Z \cdot \cos \varphi - j\cdot Z \cdot \sin \varphi }\over{Z^2}}\\ +                                        &= {{1}\over{Z}}\cdot \left( \cos(\varphi) - {\rm j}\cdot \sin(\varphi) \right) &&= {{1}\over{R_p}} - {{\rm j}\over{X_{Lp}}}  \\
-                                            &=& {{\cos \varphi - j \cdot \sin \varphi }\over{Z}}\\+
 \end{align*} \end{align*}
 +
 +Therefore, the following can be concluded:
 +\begin{align*} 
 +{{1}\over{Z}}\cdot \cos(\varphi)        &= {{1}\over{R_p}}           &&\rightarrow && R_p    &= {{Z}\over{\cos(\varphi)}}  \\
 +- {\rm j}\cdot \sin(\varphi)            &= - {{\rm j}\over{ X_{Lp}}} &&\rightarrow && X_{Lp} &= {{Z}\over{\sin(\varphi)}}  \\
 +\end{align*}
 +
 +=== with the calculated values of the series circuit ===
 +Another way is to use the formulas of $R_s$ and $X_{Ls}$ from before.
 +
 +\begin{align*} 
 +{{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}} &=& {{1}\over{R_s + {\rm j}\cdot X_{Ls}}} \\
 +{{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 - {\rm j}\cdot Z \cdot \sin \varphi }\over{Z^2}} \\
 +                                                  &=& {        {\cos \varphi - {\rm j} \cdot \sin \varphi }       \over{Z}} \\
 +\end{align*}
 +
 +Therefore 
  
 Now, the real and imaginary part is analyzed individually. First the real part: Now, the real and imaginary part is analyzed individually. First the real part:
Zeile 888: Zeile 909:
 \begin{align*}  \begin{align*} 
 {{1}\over{R_p}}   &=& {{\cos \varphi}\over{Z}}  \\ {{1}\over{R_p}}   &=& {{\cos \varphi}\over{Z}}  \\
-\rightarrow R_p   &=& {{Z}\over{\cos \varphi}} &=& {{46 ~\Omega}\over{\cos 60°}} = \boldsymbol{92 ~\Omega}+\rightarrow R_p   &=& {{Z}\over{\cos \varphi}} &=& {{46 ~\Omega}\over{\cos 60°}} 
 \end{align*} \end{align*}
  
 \begin{align*}  \begin{align*} 
 {{1}\over{X_{Lp}}}  &= {{\sin \varphi}\over{Z}} & \\ {{1}\over{X_{Lp}}}  &= {{\sin \varphi}\over{Z}} & \\
-\rightarrow X_{Lp}  &= {{Z}\over{\sin \varphi}} = {{46 ~\Omega}\over{\sin 60°}} = 53.1 ~\Omega \\ +\rightarrow X_{Lp}  &= {{Z}\over{\sin \varphi}} = {{46 ~\Omega}\over{\sin 60°}} \\ 
-\rightarrow L_p     &= {{46 ~\Omega}\over{2\pi \cdot 50~\rm Hz \cdot \sin 60°}} &= \boldsymbol{169 ~\rm mH}+\rightarrow L_p     &= {{46 ~\Omega}\over{2\pi \cdot 50~\rm Hz \cdot \sin 60°}} 
 \end{align*} \end{align*}
  
-</collapse>+#@HiddenEnd_HTML~71112,Solution ~@#
  
 +#@HiddenBegin_HTML~71113,Result~@#
 +For the series circuit:
 +\begin{align*} 
 +R_s      &= {23 ~\Omega} \\
 +L_s      &= {127 ~\rm mH} \\
 +\end{align*}
  
-<button size="xs" type="link" collapse="Loesung_7_1_1_3_Endergebnis">{{icon>eye}} Result for 3. </button><collapse id="Loesung_7_1_1_3_Endergebnis" collapsed="true"> +For the parallel circuit: 
 +\begin{align*}  
 +R_p      &= {92 ~\Omega} \\ 
 +L_p      &= {169 ~\rm mH} \\ 
 +\end{align*} 
 +#@HiddenEnd_HTML~71113,Result~@#
  
 +3. Calculate the real, reactive, and apparent power based on the equivalent components for both circuits from 2. . \\
 +
 +#@HiddenBegin_HTML~71114,Solution~@#
 +The general formula for the apparent power is $\underline{S} = U \cdot I \cdot e^{\rm j\varphi}$. \\ By this, the following can be derived:
 +\begin{align*} 
 +\underline{S} &= U \cdot I  \cdot e^{\rm j\varphi} \\
 +              &= Z \cdot I^2 \cdot e^{\rm j\varphi}     &&= \underline{Z} \cdot I^2 \\
 +              &= {{U^2}\over{Z}} \cdot e^{\rm j\varphi} &&= {{U^2}\over{\underline{Z}^{*\phantom{I}}}} \\
 +\end{align*}
 +
 +These formulas are handy for both types of circuits to separate the apparent power into real part (real power) and complex part (apparent power):
 +  - for **series circuit**: $\underline{S} =\underline{Z} \cdot I^2 $ with $\underline{Z} = R + {\rm j} X_L$ 
 +  - for **parallel circuit**: $\underline{S} ={{U^2}\over{\underline{Z}^{*\phantom{I}}}} $ with ${{1} \over {\underline{Z}^{\phantom{I}}} }   = {{1}\over{R}} + {{1}\over{{\rm j} X_L}} \rightarrow {{1} \over {\underline{Z}^{*\phantom{I}}} }   = {{1}\over{R}} + {{\rm j}\over{ X_L}} $ 
 +\\ 
 +Therefore: 
 ^                  ^ series circuit ^ parallel circuit ^ ^                  ^ series circuit ^ parallel circuit ^
 | active   power   | \begin{align*} P_s &= R_s \cdot I^2 \\ &= 23.0 ~\Omega \cdot (5{~\rm A})^2 \\ &= 575 {~\rm W} \end{align*} | \begin{align*} P_p &= {{U_p^2}\over{R_p}} \\ &= {{(230{~\rm V})^2}\over{92~\Omega}} = 575 {~\rm W}   \end{align*} | | active   power   | \begin{align*} P_s &= R_s \cdot I^2 \\ &= 23.0 ~\Omega \cdot (5{~\rm A})^2 \\ &= 575 {~\rm W} \end{align*} | \begin{align*} P_p &= {{U_p^2}\over{R_p}} \\ &= {{(230{~\rm V})^2}\over{92~\Omega}} = 575 {~\rm W}   \end{align*} |
-| reactive power   | \begin{align*} Q_s &= Z_{Ls} \cdot I^2 \\ &= 39.8 ~\Omega \cdot (5{~\rm A})^2 \\ &= 996 {~\rm Var} \end{align*} | \begin{align*} Q_p &= {{U_p^2}\over{Z_{Lp}}} \\ &= {{(230{~\rm V})^2}\over{53.1 ~\Omega}} = 996 {~\rm Var}   \end{align*} | +| reactive power   | \begin{align*} Q_s &= X_{Ls} \cdot I^2 \\ &= 39.8 ~\Omega \cdot (5{~\rm A})^2 \\ &= 996 {~\rm Var} \end{align*} | \begin{align*} Q_p &= {{U_p^2}\over{X_{Lp}}} \\ &= {{(230{~\rm V})^2}\over{53.1 ~\Omega}} = 996 {~\rm Var}   \end{align*} | 
-| apparent power   | \begin{align*} S_s &= \sqrt{P_s^2 - Q_s^2} \\ &= I^2 \cdot \sqrt{R_s^2 + Z_{Ls}^2} \\ &= 1150 {~\rm VA} \end{align*} | \begin{align*} S_p &= \sqrt{P_s^2 - Q_s^2} \\ &= U^2 \cdot \sqrt{{{1}\over{R_p^2}} + {{1}\over{Z_{Lp}^2}}} \\ &= 1150 {~\rm VA}   \end{align*} |+| apparent power   | \begin{align*} S_s &= \sqrt{P_s^2 - Q_s^2} \\ &= I^2 \cdot \sqrt{R_s^2 + X_{Ls}^2} \\ &= 1150 {~\rm VA} \end{align*} | \begin{align*} S_p &= \sqrt{P_s^2 - Q_s^2} \\ &= U^2 \cdot \sqrt{{{1}\over{R_p^2}} + {{1}\over{X_{Lp}^2}}} \\ &= 1150 {~\rm VA}   \end{align*} |
  
-</collapse>+#@HiddenEnd_HTML~71114,Solution ~@#
  
 +4. Check the solutions from 3. via direct calculation based on the input in the task above. \\
  
-<button size="xs" type="link" collapse="Loesung_7_1_1_4_Endergebnis">{{icon>eye}} Result for 4. </button><collapse id="Loesung_7_1_1_4_Endergebnis" collapsed="true"> +<button size="xs" type="link" collapse="Loesung_7_1_1_4_Endergebnis">{{icon>eye}} Solution </button><collapse id="Loesung_7_1_1_4_Endergebnis" collapsed="true"> 
  
 active power: active power:
Zeile 928: Zeile 976:
 apparent power: apparent power:
 \begin{align*}  \begin{align*} 
-Q &= U \cdot I \\+S &= U \cdot I \\
   &= 230{~\rm V} \cdot 5{~\rm A}  \\   &= 230{~\rm V} \cdot 5{~\rm A}  \\
   &= 1150 {~\rm VA}   &= 1150 {~\rm VA}
Zeile 977: Zeile 1025:
  
 \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 1024: Zeile 1072:
 <button size="xs" type="link" collapse="Loesung_7_1_3_1_Rechnung">{{icon>eye}} Result for 1.</button><collapse id="Loesung_7_1_3_1_Rechnung" collapsed="true">  <button size="xs" type="link" collapse="Loesung_7_1_3_1_Rechnung">{{icon>eye}} Result for 1.</button><collapse id="Loesung_7_1_3_1_Rechnung" collapsed="true"> 
  
-The active power is $P = 1.80 kW$. \\ +The active power is $P = 1.80 ~\rm kW$. \\ \\ 
-The apparent power is $S = U \cdot I = 220V \cdot 20A = 4.40 kVA$. \\ +The apparent power is $S = U \cdot I = 220 ~\rm V \cdot 20 ~\rm A = 4.40 ~\rm kVA$. \\ \\ 
-The reactive power is $Q = \sqrt{S^2 - P^2} = \sqrt{(4.40 kVA)^2 - (1.80 kW)^2} = 4.01 kVar$ \\ +The reactive power is $Q = \sqrt{S^2 - P^2} = \sqrt{(4.40 ~\rm kVA)^2 - (1.80 ~\rm kW)^2} = 4.01 ~\rm kVar$ \\ \\ 
-The power factor is $\cos \varphi = {{P}\over{S}} = {{1.80 kW}\over{4.40 kVA}} = 0.41$.+The power factor is $\cos \varphi = {{P}\over{S}} = {{1.80 ~\rm kW}\over{4.40 ~\rm kVA}} = 0.41$.
  
 </collapse> </collapse>
Zeile 1090: Zeile 1138:
 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 1196:
 \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>\\