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--- electrical_engineering_2:polyphase_networks [2023/06/12 10:37]
ott
+++ electrical_engineering_2:polyphase_networks [2024/06/18 02:38]
mexleadmin [Excercises]
@@ 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
@@ Zeile 34: / Zeile 34: @@
 Out of the last formula we derived the following instantaneous voltage $u(t)$
 \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*}
@@ Zeile 201: / Zeile 200: @@
 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>
@@ Zeile 411: / Zeile 410: @@
 The phase currents are given as the currents through a single line: $I_1$, $I_2$, $I_3$. \\
 The potential of the star point is called **neutral** $\rm N$ </WRAP>
-  * **Star-voltages** $U_\rm Y$ (alternatively: phase-to-neutral voltages, line-to-neutral voltages, in German: //Sternspannungen//): the voltages of the lines can be also measured or used referring to the neutral potential.
+  * **Star Voltages** $U_\rm Y$ (alternatively: phase-to-neutral voltages, line-to-neutral voltages, in German: //Sternspannungen//): the voltages of the lines can be also measured or used referring to the neutral potential.
 <WRAP> <imgcaption imageNo13 | Example of an Three-Phase System></imgcaption> {{drawio>ExampleThreePhaseSystem.svg}} </WRAP>
@@ Zeile 434: / Zeile 433: @@
 The phase voltage is therefore ${{1}\over{\sqrt{3}}} \cdot 400~\rm V \approx 230~\rm V$. The following two simulations show these voltages.
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 ==== 7.2.3 Load and Power in Three-Phase Systems ====
@@ Zeile 467: / Zeile 466: @@
 </panel>
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+<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgzCAMB0l5AWAnC1b0DYrTAdiWBhrgBxIIYlgCMuuWATNSAnCAKyQcCmAtNdQBQADxC8M1JCAFTeyBNIYMW0kiAAKACwCGAZ24AdXQDUA9gBsALtoDm3XSLEMkWakrHUyi9is8gAytYATkZmVrb2gjZiGAhqYAgK4nHgGFzpggDuTikJSQxgynlQWU6x4Ik58ZWQggAOToXgbo1FlRDMtQ28DCkFyj19LuDSJdm87EjK-WKEXDNdZQqe+eW98aOLcsMMJFzbjJAbnaUTUyC7+8iMe2Nik8orYteqClsPFyln07cdJd0fBhsb4XJp-WpA3AxPpNZJqIFqBQAGQYgkh0PhtzhF3YWGRYDRkCh2KeJLSKiRQnR2OKNKBKgAcqcXgjnsMnrVxiysWB2D8MuNYtUkry2m9BAAlDGvGLskhvUaccAMDDQSgK9LQdiS6UzcQ7JpcZhKwqq9VQJywchgJDsaiQAoESTsCAwbVS7GXWU3dKKrimtXyi1yaCeEj4B2IUgYG2QHxu05CmUgjmJmEDQGGtOY-aA26cpafNQ9crFAuDeHDCsVcXjFXLIO8W6puvpkDwrN1ppekg5u49MGVJv+mqCIJiW69fbNoNG2rjjvKTFqLjagDG7Yuvc3hV9MHgkHQR-QHl6gcIGF2NqYuHcQPn7b7293FqBgnME-mrE-W5X2DgzA0GqKp4C4pC4BQSCIP2M7LLsMoFouihIYhfZuH25a-JUwp3Dhz6jtk+FwThtR4PscyggMhRIcwtGnNRlFVIxpG4OR5KkgxTy0dIzJyiK7GzvR8GkvWCGCGRsx8lu5HwV63FCOMDFegO-IlBJw7KtOI7LKMQjqQxtKijWuk8Yp8G0qJZanL89IabsK7iaxO70jh8mOVwz4ua5umOHIFCKDJtEFL4ahaHohi6AAwgArkEQTcAAdpYDiiE2PhuB0VDSJQIUgEipjaAAJkYACWCVGAA6gAnhFkWmAlCXcGulglfVRgGAYCVlQAKpo8V8GF+hGAAYqYsW8BVJXxUY-hVboljcAAth1CWZCVliaDNlhBGVNhGDFcWJclvmsLRGA+LwNquDlDYgAAkr5NrpbgECXdutB4tIyyPQQFzuG9jAMD4ywKKiqU2lITAiuQFy0LlD3g0gUIqhd14XDdqj3T9zAMC9sxBrjn3UAoBKI1ChTEmjYBgFwt0I7MSCAby+NM4iX0gEyqXoi+-BQcqEAg+AnNOL0zSvUwRTUETagAKq+Uo8JST0Us4sDmNy1zAg4rIkig5evig8LyvTPaHi2rDAvs9QRtKJDkDi3DQLSyActdpLAwzF65aAsCHwtvc5zrAH0xZql7AkD4QJJOwePordGv3DlTDMBMLqwwogtuL5uKK8S7BpyqtPswU2cRziuax2n8cnRBf2yAgkyMYLNAnYzf0DAg+B-Wrfhy6Y0iW-MiDhiAsSap0IwtFwUhcINEVhNYdiCP3atD3EWBj9g3EQFgHkWoE2ghCYFiL5E-f+QqrBI6P352tA0ybv5AuXyTLAz2-IxSauRQaDoQ1RbFeKSUHBAA noborder}} </WRAP>
 <callout title="Voltages - Currents - True Power - Apparent and Reactive Power">
@@ Zeile 541: / Zeile 540: @@
 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 related by: $U_{\rm L} = \sqrt{3}\cdot U_{\rm S}$ (equal to the asymmetric load).
   - 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 **(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>
@@ Zeile 566: / Zeile 565: @@
 </panel>
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 <callout title="Voltages - Currents - True Power - Apparent and Reactive Power">
@@ Zeile 711: / Zeile 710: @@
 </panel>
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+<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgzCAMB0l5AWAnC1b0DYrTAdiWBhrgBxIIYlgCMuuWATNSAnCAKyQcCmAtNdQBQADxC8M1JCAFTeyBNIYMW0kiAAKACwCGAZ24AdXQDUA9gBsALtoDm3XSLEMkWakrHUyi9is8gAytYATkZmVrb2gjZiGAhqYAgK4nHgGFzpggDuTikJSQxgynlQWU6x4Ik58ZWQggAOToXgbo1FlRDMtQ28DCkFyj19LuDSJdm87EjK-WKEXDNdZQqe+eW98aOLcsMMJFzbjJAbnaUTUyC7+8iMe2Nik8orYteqClsPFyln07cdJd0fBhsb4XJp-WpA3AxPpNZJqIFqBQAGQYgkh0PhtzhF3YWGRYDRkCh2KeJLSKiRQnR2OKNKBKgAcqcXgjnsMnrVxiysWB2D8MuNYtUkry2m9BAAlDGvGLskhvUaccAMDDQSgK9LQdiS6UzcQ7JpcZhKwqq9VQJywchgJDsaiQAoESTsCAwbVS7GXWU3dKKrimtXyi1yaCeEj4B2IUgYG2QHxu05CmUgjmJmEDQGGtOY-aA26cpafNQ9crFAuDeHDCsVcXjFXLIO8W6puvpjwtBanGZewYXfOCIJiW69fbNoNG2qD+FNTFqLjagDGIFny+Vvpg8Eg6G36CcTBwCFxcbQcdwEB6cAHy5z17XFqBgnMQ-mrGffbn2DgxoQ0CmDFI1AxgkvKkHcTZGkG1CVC2q4zIB07KAWJA3vBfYCjkqRcPE5IFlBDbLJQMpIdhRrtDhgh4PscyggMhQIaMzBCOMdE0VUrG1JRszkqSLFPIx0jMnKIrceOpy8Y29ZERRuBUXyaGzLs8n8UxCk3j0TRehxMlvmAhA6dxDHSVRim0qKNYMQJzEmZUJYKGWpy-Hp4HviUnHxHpJEWUIbnuVgyGYV5jhyBQijGYxBS+GoWh6IYugAMIAK5BEE3AAHaWA4ohNj4bgdO4UFQg2IBIqY2gACZGAAlqlRgACLcOERhxaYqWpdwC6WJVLVGAYBipZklWWJoRiBEE1U2E1SUpelmXPJAjEYD4vA2q4hFFQAkkFNo5Wesz+bQeLSMsW0EBc7jLSQjAMD4ywKKiWU2lITAiuQFy0JFICbQ9SBQiqS02tMa2qJ9J3MP+55gEG-6HVB4AnVChTEgD4BgFwG2gyj-0kMwumIkdIBMll6KFPskj+hFt3gITe52dQ55MEUgEfQAqkFSjwnJPRM9dN3A6zRMCDisiSHdGDKJTDDU1z0z2h4tpvRAlPUFLShPZA9PvUCMNqKzdZgi06n8mBgLAh8MEgus9znJ2WXsCQPhAkk7C7eiRX8-chFMMwEwum9ChK-dHsc8S7C+yqaP4wUQV2w7SoTC7vtu0FCC4HdzjPJMrGUzQydIGD50p0910sw5-rkvwLTUeWAhFOXNfsV2dnl708nVx2M4N628n+ZppR4TK-dV0Fi0o8WJoIMwAdBbgOOozEWC6YrwNfdIPiLxc-p6ejohMBUOML5UEsEtk8T0r5dz+bpfmnxkogoCjyyICjhXL449prwgjz0gkvOwznpjSCXvMRA4YQCxE1J0EYLQuBSC4NFfQoQLDWDsIIABvNgFxCwOA7A-EIBYCwhaQI2gQgmCQREBwADDwqDRluKE2DHbQHhKuKhitXQsAgIrD8uQOCQLtCMeBsVErJTShlQQQA noborder}} </WRAP>
 <callout title="Voltages - Currents - True Power - Apparent and Reactive Power">
@@ Zeile 850: / 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%>
-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.
+. 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}}
+#@HiddenEnd_HTML~71111,Result~@#
-</collapse>
+. 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:
@@ Zeile 878: / Zeile 874: @@
 \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 $\underline{Z} = {{U}\over{I}}\cdot e^{j\cdot \varphi}$. \\
+There are multiple ways to solve this problem. Two ways shall be shown here:
+=== with the Euler representation ===
+Given the formula $\underline{Z} = {{U}\over{I}}\cdot e^{j\cdot \varphi}$ the following can be derived:
+\begin{align*}
+{{1}\over{\underline{Z}^{\phantom{A}}}} &= {{I}\over{U}}\cdot e^{-j\cdot \varphi} \\
+                                        &= {{1}\over{Z}}\cdot e^{-j\cdot \varphi} &&= {{1}\over{R_p}} + {{1}\over{{\rm j}\cdot X_{Lp}}}  \\
+                                        &= {{1}\over{Z}}\cdot \left( \cos(\varphi) - {\rm j}\cdot \sin(\varphi) \right) &&= {{1}\over{R_p}} - {{\rm j}\over{X_{Lp}}}  \\
+\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*}
-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 the calculated values of the series circuit ===
-The easiest thing is here to use the formulas of $R_s$ and $X_{Ls}$ from before:
+Another way is to use the formulas of $R_s$ and $X_{Ls}$ from before.
 \begin{align*}
@@ Zeile 888: / Zeile 902: @@
                                                   &=& {        {\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:
@@ Zeile 893: / Zeile 909: @@
 \begin{align*}
 {{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*}
 \begin{align*}
 {{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*}
-</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~@#
+. 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 ^
 | 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 ~@#
+. 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:
@@ Zeile 933: / Zeile 976: @@
 apparent power:
 \begin{align*}
-Q &= U \cdot I \\
+S &= U \cdot I \\
   &= 230{~\rm V} \cdot 5{~\rm A}  \\
   &= 1150 {~\rm VA}
@@ Zeile 1029: / 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">
-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>
@@ Zeile 1212: / Zeile 1255: @@
 </WRAP></WRAP></panel>
+#@TaskTitle_HTML@#7.2.2 Motor on 3-Phase System I#@TaskText_HTML@#
+A three-phase motor is connected to an artificial three-phase system and can be configured in wye or delta configuration.
+  * The voltage measured on a single coil shall always be $230 ~\rm V$.
+  * The current measured on a single coil shall always be $10 ~\rm A$.
+  * The phase shift for every string is $25°$
+  - The motor shall be in wye configuration. \\ Write down the string voltage, phase voltage, string current, phase current, and active power
+  - The motor shall be in delta configuration. \\ Write down the string voltage, phase voltage, string current, phase current, and active power
+  - Compare the results
+#@TaskEnd_HTML@#
+#@TaskTitle_HTML@#7.2.3 Motor on 3-Phase System II#@TaskText_HTML@#
+A three-phase motor is connected to a three-phase system with a phase voltage of $400 ~\rm V$. The phase current is $16 ~\rm A$ and the power factor $0.9$. \\
+Calculate the active power, reactive power, and apparent power.
+#@TaskEnd_HTML@#
+#@TaskTitle_HTML@#7.2.4 Motor on 3-Phase System III#@TaskText_HTML@#
+A symmetrical and balanced three-phase motor of a production line shall be configured in a star configuration and provide a power of $17~\rm kW$ with a power factor of $0.75$. The voltage on a single string is measured to be $135 ~\rm V$. \\
+Calculate the string current.
+#@TaskEnd_HTML@#
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