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1 Complex numbers 1

This applies to the following complex numbers:
\(Z_1 = 3+j4\)
\(Z_2 = 2-j\)
\(Z_3 = j7\)

The following parts of the task are to be calculated:

  • \(Z_1 + Z_2\)
  • \(Z_1 - Z_3\)
  • Polar form of \(Z_1\), \(Z_2\) and \(Z_3\)
  • \(Z_1 \cdot Z_2\)
  • \(\frac {Z_1}{Z_3}\)
  • Pointer diagram of \(Z_1 + Z_2\) and \(Z_1 \cdot Z_2\)

1.1 Lösung:
  • \(Z_1 + Z_2\)
    \begin {align} Z_1 + Z_2 &= (3 + j4) + (2 - j) \nonumber \\ &= 5 + j3 \nonumber \end {align}

  • \(Z_1 - Z_3\)

    \begin {align} Z_1 - Z_3 &= (3 + j4) - (0 + j7) \nonumber \\ &= 3 - j3 \nonumber \end {align}

  • Polar form of \( Z_1, Z_2\) and \(Z_3 \)

    \begin {align} Z_1 &= 5 \cdot \mathrm {e}^{\mathrm {j}53.13^\circ } \nonumber \\ Z_2 &= 2,236 \cdot \mathrm {e}^{\mathrm {j}-26,57^\circ } \nonumber \\ Z_3 &= 7 \cdot \mathrm {e}^{\mathrm {j}90^\circ } \nonumber \end {align}

  • \(Z_1 \cdot Z_2\)

    \begin {align} Z_1 \cdot Z_2 &= (3 + j4) \cdot (2 - j) \nonumber \\ &= 10 + j5 \nonumber \end {align}

  • \(\frac {Z_1}{Z_3} \)

    \begin {align} \frac {Z_1}{Z_3} = \frac {3 + j4}{j7} \nonumber \end {align}

    (Expand with the conjugate value of the denominator, i.e. -j7)

    \begin {align} \frac {3 + j4}{j7} \cdot \frac {-j7}{-j7} = \frac {(3 + j4) \cdot (-j7)}{(j7) \cdot (-j7)} \nonumber \end {align}

    Note: \(j^2 = -1\)

    \begin {align} \frac {Z_1}{Z_3} &= \frac {28 - j21}{49} = \frac {28}{49} - j \frac {21}{49} = \frac {4}{7} - j \frac {3}{7} \nonumber \\ &= \frac {4}{7} - j \frac {3}{7} \nonumber \end {align}

  • Pointer diagram of \(Z_1 + Z_2 \) and \(Z_1 \cdot Z_2 \)

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2 Complex numbers 2

This applies to the following complex numbers:
\(K_1 = 10 + j40\)
\(K_2 = 50 \cdot e^{-j40^o}\)
\(K_3 = 250 \cdot e^{j\frac {\pi }{4}}\)

The following parts of the task are to be calculated:

  • \(K_1 - K_2\), Result in polar coordinates
  • \(K_3 + K_2\), result in Cartesian coordinates
  • \(K_1 \cdot K_3\), result in polar coordinates
  • \(K_1 + K_2\), graphically in the pointer diagram
  • \((K_1 - K_2)^2\)
  • \(\sqrt {K_1 + K_3}\)

2.1 Lösung:

  • \(K_1 - K_2\), Result in polar coordinates

    \begin {align} 10 + \mathrm {j} \cdot 40 - 50 \cdot \mathrm {e}^{-\mathrm {j}40^\circ } = ? \nonumber \end {align}

    \(\underline {K}_2\) convert to Cartesian coordinates:

    \begin {align} 50 \cdot \mathrm {e}^{-\mathrm {j}40^\circ } &= 50 \cdot \cos (-40^\circ ) + \mathrm {j} 50 \cdot \sin (-40^\circ ) \nonumber \\ &= 50 \cdot 0,766 + \mathrm {j} 50 \cdot (-0,6427) \nonumber \\ &= 38,302 - \mathrm {j} 32,1393 \nonumber \end {align}

    \(\underline {K}_2 - \underline {K}_2\) Subtract in Cartesian coordinates:

    \begin {align} \underline {K}_1 - \underline {K}_2 &= (10 + \mathrm {j} \cdot 40) - (38,302 - \mathrm {j} 32,1393) \nonumber \\ &= 10 + \mathrm {j} \cdot 40 - 38,302 + \mathrm {j} 32,1393 \nonumber \\ &= -28,3 + \mathrm {j} 72,139 \nonumber \end {align}

    Result in polar coordinates:

    \begin {align} \underline {K}_1 - \underline {K}_2 &= \sqrt {\Re ^2+\Im ^2} \cdot \mathrm {e}^{\mathrm {j}\arctan (\frac {\Im }{\Re })} \nonumber \\ &= 77,49 \cdot \mathrm {e}^{\mathrm {j}111,42^\circ } \nonumber \end {align}

  • \(K_3 + K_2\), Result in Cartesian coordinates

    \begin {align} K_3 &= 250 \cdot e^{j\frac {\pi }{4}} = 176,78 + j176,78 \nonumber \\ K_2 &= 50 \cdot e^{-j40^\circ } = 38,3 - j32,15 \nonumber \\ K_3 + K_2 &= (176,78 + j176,78) + (38,3 - j32,15) \nonumber \\ K_3 + K_2 &= 215,08 + j144,63 \nonumber \end {align}

  • \(K_1 \cdot K_3\), Result in polar coordinates

    \begin {align} \underline {K}_1 &= 10 + \mathrm {j}40 \nonumber \\ &= 41,23 \cdot \mathrm {e}^{\mathrm {j}75,96^\circ } \nonumber \\ \underline {K}_1 \cdot \underline {K}_3 &= 10 + \mathrm {j}40 \cdot 250 \cdot \mathrm {e}^{\mathrm {j}45^\circ } \nonumber \\ &= 10307,75 \cdot \mathrm {e}^{\mathrm {j}120,96^\circ } \nonumber \end {align}

  • \(K_1 + K_2\), graphically in the pointer diagram

    PIC

  • \((K_1 - K_2)^2\)

    Result from part a): \begin {align} K_1 - Z_2 = 77,49 \cdot \mathrm {e}^{\mathrm {j}111,42^\circ } \nonumber \end {align}

    Rule for the square of a complex number: \begin {align} (K_1 - K_2)^2 &= 77,49^2 \cdot \mathrm {e}^{\mathrm {j}2 \cdot 111,42^\circ } \nonumber \\ &= 6004 \cdot \mathrm {e}^{-\mathrm {j}137^\circ } \nonumber \end {align}

  • \(\sqrt {K_1 + K_3}\)

    \begin {align} K_1 + K_3 = 186,78 + j216,78 \nonumber \end {align}

    Conversion to polar form:

    \begin {align} K_1 + K_3 = 286,09 \cdot \mathrm {e}^{\mathrm {j}49,32^\circ } \nonumber \end {align}

    The square root of a complex number in polar form is:

    \begin {align} \sqrt {r \cdot \mathrm {e}^{\mathrm {j}\varphi }} &= \sqrt {r} \frac {\varphi }{2} \nonumber \\ \sqrt {286,09} \cdot \mathrm {e}^{\mathrm {j}\frac {49,32^\circ }{2}} &= 16,91 \cdot \mathrm {e}^{\mathrm {j}24,66^\circ } \nonumber \end {align}

3 Pointer diagrams 1

Let the voltage be \(\hat {U} = 325\ V \cdot \mathrm {e}^{\mathrm {j}30^\circ }\).

  • Draw the phasor diagram for the voltage \(\hat {U}\) in the complex number space.
  • Calculate the real and imaginary parts of the complex voltage \(\hat {U}\).
  • Explain the significance of the phase angle in relation to alternating voltage.

3.1 Lösung:

  • Pointer diagram for voltage \(\hat {U}\)

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  • The real and imaginary parts of the voltage \(\hat {U}\) can be calculated using the following formulas:

    \begin {align} U_{\text {Re}} &= \hat {U} \cdot \cos (\varphi ) = 325 \cdot \cos (30^\circ ) \approx 281,46\ V \nonumber \\ U_{\text {Im}} &= \hat {U} \cdot \sin (\varphi ) = 325 \cdot \sin (30^\circ ) = 162,5\ V \nonumber \end {align}

  • The phase angle \(\varphi \) describes the time lag between the voltage and the current in an alternating current circuit. A positive phase angle indicates that the voltage lags behind the current (inductive load), while a negative phase angle indicates that the current lags behind the voltage (capacitive load).

4 Complex AC calculation 1

The specific values for the resistance \(R\), the coil \(L\) and the capacitor \(C\) are given.
\(R=10\ \Omega \), \(L=10\ mH\), \(C=10\ pF\)
The impedances of the three components are to be calculated at the following frequencies:
\(1\ \mu Hz\),\(1\ mHz\),\(1\ Hz\),\(1\ kHz\),\(1\ MHz\)

Enter the results in a table.

4.1 Lösung:

  • \(\bf Frequency\) \(\bf X_R\) \(\bf X_C\) \(\bf X_L\)
    \(1\ \mu Hz\) \(10\ \Omega \) \(-j 15,92\ M\Omega \) \(j 6,28\)
    \(1\ mHz\) \(10\ \Omega \) \(-j 15,92\ k\Omega \) k
    \(1\ Hz\) \(10\ \Omega \) \(-j 15,92\ \Omega \) k
    \(1\ kHz\) \(10\ \Omega \) \(-j 15,92\ m\Omega \)
    \(1\ MHz\) \(10\ \Omega \) \(-j 15,92\ \mu \Omega \)

5 Complex AC calculation 2

An impedance has a value of \(\underline {Z} = (300+\mathrm {j}400)\ \Omega \) at a frequency of \(f = 1\ kHz\).

  • Determine the apparent resistance (magnitude) of the impedance.
  • Determine the admittance \(\underline {Y} = \frac {1}{\underline {Z}}\) and the apparent conductance.
  • To realise the impedance \(\underline {Z}\), specify a circuit consisting of two components connected in series and dimension this circuit.
  • To realise the impedance \(\underline {Z}\), specify a circuit consisting of two components connected in parallel and dimension this circuit.

In addition, a voltage \(u(t)=\hat {U} \cdot \cos (2\pi f t + \varphi _U)\) with \(\hat {U}=5\ V\), \(\varphi _U=\pi /4\) and \(f=1\ kHz\) is applied to the impedance \(\underline {Z} = (300+\mathrm {j}400)\ \Omega \).

  • Determine the complex amplitude \(\underline {\hat {U}}\) of the applied voltage.
  • Calculate the complex amplitude \(\underline {\hat {I}}\) of the current through the impedance.
  • Specify the time dependence of the current \(i(t)\).
  • Represent the impedance in polar form and specify the angle \(\varphi = arg\{\underline {Z}\}\). Compare the angle of the impedance with the phase shift between voltage and current. What do you notice?

5.1 Lösung:
  • \begin {align} |\underline {Z}| &= \sqrt {Re^2+Im^2} \nonumber \\ &= \sqrt {300^2+400^2}\ \Omega \nonumber \\ &= 500\ \Omega \nonumber \end {align}
  • \begin {align} \underline {Y} &= \frac {1}{\underline {Z}} = \frac {1}{(300+\mathrm {j}400)\ \Omega } \nonumber \\ &= \frac {1}{(300+\mathrm {j}400)\ \Omega } \cdot \frac {(300-\mathrm {j}400)\ \cancel {\Omega }}{(300-\mathrm {j}400)\ \cancel {\Omega }} \nonumber \\ &= \frac {1}{300^2+400^2} \cdot (300-\mathrm {j}400)\ S \nonumber \\ &= (0.0012-\mathrm {j}0.0016)\ S \nonumber \\ |\underline {Y}| &= \frac {1}{|\underline {Z}|} \nonumber \\ &= \frac {1}{500}\ S \nonumber \end {align}

  • Series connection:
    Real part \(\rightarrow \) Ohmic resistance
    Imaginary part \(\rightarrow \) coil or capacitor?
    Reactance coil: \(X_L=2\pi f \cdot L > 0\)
    Reactance capacitor: \(X_C=-\frac {1}{2\pi f\cdot C} < 0\)

    PIC

    \begin {align} R &= 300\ \Omega \nonumber \\ L &= \frac {X_L}{2\pi \cdot f} \nonumber \\ &= \frac {400\ \Omega }{2\pi \cdot 1\ kHz} \nonumber \\ &= 63.66\ mH \nonumber \\ \underline {Z} &= R+\mathrm {j}X_L = R+\mathrm {j}\ 2\pi \cdot f \cdot L \nonumber \end {align}

  • Parallel connection:
    \(\rightarrow \) Addition of guiding values:

    \begin {equation} \underline {Y} = G + \mathrm {j}B = (0.0012-\mathrm {j}0.0016)\ S \qquad (\text {aus b)}) \nonumber \end {equation}

    \(B<0 \rightarrow \) Coil, inductive reactance
    \(B>0 \rightarrow \) Capacitor, capacitive reactance

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    \begin {align} R &= \frac {1}{G} \nonumber \\ &= \frac {1}{0.0012\ S} \nonumber \\ &= 833\ \Omega \nonumber \\ B &= -\frac {1}{2\pi \cdot f \cdot L} \nonumber \\ L &= \frac {1}{2\pi \cdot f \cdot B} \nonumber \\ &= 99.47\ mH \nonumber \end {align}

  • \begin {align} \underline {U} &= \hat {U} \cdot \mathrm {e}^{\mathrm {j} \varphi _\mathrm {U}} = 5\ V \cdot \mathrm {e}^{\mathrm {j} \pi /4} = 5\ V \cdot \mathrm {e}^{\mathrm {j} 45^\circ } \nonumber \\ &= 5\ V \cdot (\cos (45^\circ )+\mathrm {j}\sin (45^\circ )) \nonumber \\ &= 5\ V \cdot \frac {\sqrt {2}}{2} (1+\mathrm {j}) \nonumber \\ &\approx (3,53 + \mathrm {j} 3,53) \ V \nonumber \end {align}

6 RMS value 1

Given a sinusoidal voltage

\begin {equation} u(t) = 150\ V \cdot \sin (100\pi t) \nonumber \end {equation}

where \(t\) is given in seconds and the voltage \(u(t)\) is given in volts.

Calculate the following for this voltage:

  • The arithmetic mean \(\bar {u}\) of the voltage over one period.
  • The RMS value \(U_{RMS}\) of the voltage.

6.1 Lösung:

  • The arithmetic mean of a function u(t) over a period T is given by: \begin {equation} \bar {u} = \frac {1}{T} \int _0^T u(t) dt \nonumber \end {equation}

    Since \(u(t)\) represents a sine function and the sine wave is symmetrical over one period, the positive and negative parts of the wave cancel each other out. Therefore, the arithmetic mean for each sine wave is always zero:

    \begin {equation} \bar {u} = 0 \nonumber \end {equation}

  • The RMS value of a sinusoidal voltage \(u(t)\) is calculated using the following formula: \begin {equation} U_{Eff} = \frac {\hat {U}}{\sqrt {2}} \nonumber \end {equation}

    In this case, \(\hat {U} = 150\ V\) is the amplitude of the voltage.

    \begin {equation} U_{Eff} = \frac {150\ V}{\sqrt {2}} \approx \frac {150\ V}{1.414} \approx 106.1\ V \nonumber \end {equation}

7 Power calculation 1

Text

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  • Reading the phase angle values of voltage and current. Describing the behaviour.
  • Specifying the period duration, frequency and angular frequency.
  • Calculate the instantaneous power at time t = 12.5 ms.
  • Calculate the active power and reactive power.
  • Calculate the apparent power using the conjugate complex current.
7.1 Lösung:
  • The voltage has a phase angle of \(\varphi _\mathrm {u}=0\) and the current has a phase angle of \(\varphi _\mathrm {i}=\pi /3\).
    The lagging of the current indicates an inductive behaviour.
  • The period T is \(50\ ms\).
    The frequency is: \begin {equation} f=\frac {1}{T}=\frac {1}{50\ ms}= 20\ Hz \nonumber \end {equation} The angular frequency is: \begin {equation} \omega = 2\pi \cdot f = 2\pi \cdot 20\ Hz = 125,66\ Hz \nonumber \end {equation}
  • The instantaneous power at time t = 12.5 ms is: \begin {align} p(t) &= u(t) \cdot i(t) = 12\ V \cdot \sin (2\pi \cdot 20\ Hz\cdot 0,0125\ s) \cdot 2\ A \cdot \sin (2\pi \cdot 20\ Hz\cdot 0,0125\ s+\frac {\pi }{3}) \nonumber \\ p(t) &= 12\ W \nonumber \end {align}
  • The active power is: \begin {align} P &= U \cdot I \cdot \cos (\varphi ) = \frac {\hat {U}}{\sqrt {2}} \cdot \frac {\hat {I}}{\sqrt {2}} \cdot \cos (\varphi ) = \frac {12\ V}{\sqrt {2}} \cdot \frac {2\ A}{\sqrt {2}} \cdot \cos (60^o) \nonumber \\ P &= 6\ W \nonumber \end {align} The reactive power is: \begin {align} Q &= U \cdot I \cdot \sin (\varphi ) = \frac {\hat {U}}{\sqrt {2}} \cdot \frac {\hat {I}}{\sqrt {2}} \cdot \sin (\varphi ) = \frac {12\ V}{\sqrt {2}} \cdot \frac {2\ A}{\sqrt {2}} \cdot \sin (60^o) \nonumber \\ Q &= 10,392\ var \nonumber \end {align}
  • The apparent power is: \begin {align} \underline {S} &= \underline {U} \cdot \underline {I}^* = \frac {\underline {\hat {U}}}{\sqrt {2}} \cdot \frac {\underline {\hat {I}}^*}{\sqrt {2}} = \frac {12\ V}{\sqrt {2}} \cdot e^{j(2\pi \cdot 20\ Hz)} \cdot \frac {2\ A}{\sqrt {2}} \cdot e^{j(2\pi \cdot 20\ Hz+\frac {\pi }{3})} \nonumber \\ \underline {S} &= 12\ VA \cdot e^{j\frac {\pi }{3}} = 6\ W + j10,392\ var \nonumber \end {align}

8 Three-phase current 1

A three-phase motor has the following nominal values:

  • Nennspannung \(U_N = 400\ V\)
  • Nennstrom \(I_N = 10\ A\)
  • Calculate the voltage across each winding of the motor in star connection \(U_\star \).
  • Calculate the voltage across each winding of the motor in delta connection \(U_\mathrm{d}elta \).
  • Calculate the current through each winding \(I_\star \) in star connection.
  • Calculate the current through each winding \(I_\Delta \) in delta operation.

8.1 Lösung:
  • Voltage across each winding of the motor \(U_Y \) in star operation: \begin {align} U_Y = \frac {U_N}{\sqrt {3}} = \frac {400\ V}{\sqrt {3}} \approx 230.94\ V \nonumber \end {align}
  • Voltage across each winding of the motor \(U_\Delta \) in delta operation: \begin {align} U_\Delta = U_N = 400\ V \nonumber \end {align}
  • Current through each winding \(I_Y \) in star mode: \begin {align} I_Y = I_N = 10\ A \nonumber \end {align}
  • Current through each winding \(I_\Delta \) in delta connection: \begin {align} I_\Delta = \sqrt {3} \cdot I_Y = \sqrt {3} \cdot 10\ A \approx 17.32\ A \nonumber \end {align}

9 Three-phase current 2

It is a symmetrical three-phase system with the following voltages:

\(\underline {U}_1=230~V\), \(\underline {U}_2=230~V\cdot \underline {a}^2\), \(\underline {U}_3=230~V\cdot \underline {a}\)

The impedances all have a value of \(Z=10\Omega \)

The following applies to all parts of the exercise: Solve the exercises with the least possible effort!

  • Draw the three-phase and single-phase ESBs in star and delta connections and label the ECDs correctly!
  • Calculate the magnitude of the consumer voltages for both the star and delta connections!
  • Calculate the complex consumer voltages in the delta connection!
  • Draw the phasor diagrams of the consumer voltages for the star and delta connections!
  • Calculate the complex load currents for the star connection and the magnitude of the load currents for the delta connection!
  • Calculate the apparent power for the star and delta connections!

9.1 Lösung:

  • ECDs of the star connection:
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    ECDs of the triangular circuit:

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  • The amount of consumer voltage in a star connection is equal to the source voltage: 230 V. The amount of consumer voltage in a delta connection is \(\sqrt {3}\) times greater than in a star connection: \(230 V\cdot \sqrt {3}=398.37 V\approx 400 V\).

  • The complex consumer voltage in the delta connection is calculated from the differences in the star voltages.

    \begin {align} \underline {U}_\mathrm {L1L2}&=\underline {U}_\mathrm {L1}-\underline {U}_\mathrm {L2}=230 V-230 V\cdot \underline {a}^2 \notag \\ &=230 V(1-(-\frac {1}{2}-j\frac {\sqrt {3}}{2}))=\sqrt {3}\cdot 230 V(\frac {3}{2}+j\frac {1}{2}) \notag \\ &=\sqrt {3}\cdot 230 V\cdot e^{j\cdot 30^\circ } \notag \end {align} \begin {align} \underline {U}_\mathrm {L2L3}&=\underline {U}_\mathrm {L2}-\underline {U}_\mathrm {L3}=230 V\cdot \underline {a}^2-230 V\cdot \underline {a} \notag \\ &=230 V(-\frac {1}{2}-j\frac {\sqrt {3}}{2}-(-\frac {1}{2}+j\frac {\sqrt {3}}{2}))=\sqrt {3}\cdot 230 V(-j) \notag \\ &=\sqrt {3}\cdot 230 V\cdot e^{-j\cdot 90^\circ } \notag \end {align} \begin {align} \underline {U}_\mathrm {L3L1}&=\underline {U}_\mathrm {L3}-\underline {U}_\mathrm {L1}=230 V\cdot \underline {a}-230 V \notag \\ &=230 V(-\frac {1}{2}+j\frac {\sqrt {3}}{2}-1)=\sqrt {3}\cdot 230 V(-\frac {\sqrt {3}}{2}+j\frac {1}{2}) \notag \\ &=\sqrt {3}\cdot 230 V\cdot e^{j\cdot 150^\circ } \nonumber \end {align}

  • Pointer diagrams of the consumer voltages for star and delta connections:

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  • The current is calculated using Ohm’s law. The following applies to star connections:

    \begin {align} \underline {I}_1&=\frac {\underline {U}_\mathrm {L1}}{\underline {Z}}=\frac {230 V}{10\ohm }=23 A \notag \\ \underline {I}_2&=\frac {\underline {U}_\mathrm {L2}}{\underline {Z}}=\frac {230 V\cdot \underline {a}^2}{10\ohm }=23 A\cdot \underline {a}^2 \notag \\ \underline {I}_3&=\frac {\underline {U}_\mathrm {L3}}{\underline {Z}}=\frac {230 V\cdot \underline {a}}{10\ohm }=23 A\cdot \underline {a} \notag \end {align}

    With delta current, a distinction is made between the phase current and the delta current, i.e. the current through the consumers. The current in the consumer phases can be calculated using the voltage across the consumers in the delta circuit:

    \begin {align} I_\Delta &=\frac {U\Delta }{Z}=\frac {400V}{10\ohm }=40 A \notag \end {align}

    The current in the strand is again \(\sqrt {3}\) greater than in the consumer strands: \begin {align} I_\mathrm {str}=\sqrt {3}\cdot I_\Delta =\sqrt {3}\cdot 40 A=69,28 A \notag \end {align}

    This value corresponds to three times the value of the current in a star connection.

  • The general power equation is as follows:

    \begin {align} S=\sqrt {3}\cdot U_\Delta \cdot I_\mathrm {str} \notag \end {align}

    The current required depends on the circuit in question. As calculated in task 5, the phase current in the delta circuit is three times greater than in the star circuit.

    The following applies to star connections: \begin {align} S=\sqrt {3}\cdot U\cdot I=\sqrt {3}\cdot 400 V \cdot 23 A=15.934 VA \notag \end {align}

    And for the delta connection, the following applies: \begin {align} S=\sqrt {3}\cdot U\cdot I=\sqrt {3}\cdot 400 V \cdot 3\cdot 23 A=47.804 VA \notag \end {align}

10 Multiphase systems 1

Given is an unbalanced system with the following voltages:

\(\underline {U}_\mathrm {L1}=230 V\), \(\underline {U}_\mathrm {L2}=460 V\cdot e^{-j45^\circ }\), \(\underline {U}_\mathrm {L1}=345 V\cdot e^{j120^\circ }\)

  • Draw the phasor diagram of the original system!
  • Set up the decomposition equations for the voltages!
  • Calculate the voltages of the transformed systems!
  • Draw the phasor diagrams of the transformed systems!

10.1 Lösung:

  • Pointer diagram of the original system:

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  • \begin {align} \underline {U}_{\mathrm {L1}}&=230 V=\underline {U}_{0}+\underline {U}_{1}+\underline {U}_{2} \notag \\ \underline {U}_{\mathrm {L2}}&=460 V\cdot e^{-j45^\circ }=\underline {U}_{0}+\underline {a}^2\cdot \underline {U}_{1}+\underline {a}\cdot \underline {U}_{2} \notag \\ \underline {U}_{\mathrm {L3}}&=345 V\cdot e^{j120^\circ }=\underline {U}_{0}+\underline {a}\cdot \underline {U}_{1}+\underline {a}^2\cdot \underline {U}_{2} \notag \end {align}

  • For the zero system:

    \begin {align} \underline {U}_0&=\frac {1}{3}\cdot (\underline {U}_{\mathrm {L1}}+\underline {U}_{\mathrm {L2}}+\underline {U}_{\mathrm {L3}}) \notag \\ &=\frac {1}{3}\cdot (230 V+460 V\cdot e^{-j45^\circ }+345 V\cdot e^{j120^\circ }) \notag \\ &=\frac {1}{3}\cdot (230V+325,27V-j325,27V-172,5V+298,78V) \notag \\ &=127,59V-j8,83=127,9\cdot e^{-j3,56^\circ } \notag \end {align}

    For the positive system:

    \begin {align} \underline {U}_{1}&=\frac {1}{3}\cdot (\underline {U}_{\mathrm {L1}}+\underline {a}\cdot \underline {U}_{\mathrm {L2}}+\underline {a}^2\cdot \underline {U}_{\mathrm {L3}}) \notag \\ &=\frac {1}{3}\cdot (230V+(-\frac {1}{2}+j\frac {\sqrt {3}}{2})\cdot (325,27V-j325,7V)+(-\frac {1}{2}-j\frac {\sqrt {3}}{2})\cdot (-172,5V+298,78V)) \notag \\ &=\frac {1}{3}\cdot (230V+119,42V+j444,54V+345V) \notag \\ &=231,48V+j148,18V=274,84\cdot e^{j32,63^\circ } \notag \end {align}

    For the negative system:

    \begin {align} \underline {U}_{2}&=\frac {1}{3}\cdot (\underline {U}_{\mathrm {L1}}+\underline {a}^2\cdot \underline {U}_{\mathrm {L2}}+\underline {a}\cdot \underline {U}_{\mathrm {L3}}) \notag \\ &=\frac {1}{3}\cdot (230V+(-\frac {1}{2}-j\frac {\sqrt {3}}{2})\cdot (325,27V-j325,7V)+(-\frac {1}{2}+j\frac {\sqrt {3}}{2})\cdot (-172,5V+298,78V)) \notag \\ &=\frac {1}{3}\cdot (230V+(-444,33V-j119,16V)+(-172,5V-298,78V)) \notag \\ &=-128,95V-j139,28V=189,8\cdot e^{j132,79^\circ } \notag \end {align}

  • Pointer diagrams of the transformed systems:

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    Note: Creating the pointer diagrams is a good way to check the results calculated in step 3. The paths of the pointers taken from the original system must end where the pointer of the transformed system leads. If the end points do not match, an error has been made in one of the calculations.

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