How to Determine Capacitance for Active Power in a Three-Phase System?

[gruba]

Homework Statement

In the given AC circuit, electromotive forces form a symmetric and direct three-phase system.
Angular frequency [itex]\omega[/itex], magnitudes [itex]U_{AB},U_{BC},U_{CA}=U[/itex],
resistance [itex]R[/itex], inductance [itex]L[/itex] and coefficient of inductive coupling [itex]k\neq 1[/itex]
are given.
Find capacitance [itex]C[/itex] such that the three-phase generator develops only the active power and find that active power.

2. The attempt at a solution
[itex][/itex]
Total active power is given by [tex]P=3\cdot\frac{\left(\frac{U}{\sqrt 3}\right)^2}{R}[/tex]
where [itex]\frac{\left(\frac{U}{\sqrt 3}\right)^2}{R}[/itex] is the active power of one phase, and
[itex]\frac{U}{\sqrt 3}[/itex] is the voltage of one phase.

Generator needs to develop only the active power, so the reactive power must be zero,
[tex]Q=\mathfrak{I}(\underline{U_{AB}}{\underline{I_1}}^{*}+\underline{U_{BC}}{\underline{I_2}}^{*}+\underline{U_{CA}}{\underline{I_3}}^{*})=0[/tex]

[tex]L_{12}=+kL\Rightarrow \underline{U_{AB}}=j\omega L\underline{I_1}+j\omega kL\underline{I_2},\underline{U_{BC}}=j\omega L\underline{I_2}+j\omega kL\underline{I_1},\underline{U_{CA}}=\frac{\underline{I_3}}{j\omega C}[/tex]
[tex]\Rightarrow \underline{I_1}=\frac{\underline{U_{AB}}-k\underline{U_{BC}}}{j\omega L(1-k^2)},\underline{I_2}=\frac{\underline{U_{BC}}-k\underline{U_{AB}}}{j\omega L(1-k^2)}, \underline{I_3}=j\omega C\underline{U_{CA}}[/tex]

First, we find [itex]\underline{U_{AB}}{\underline{I_1}}^{*}[/itex].
[tex]\underline{I_1}=\frac{\underline{U_{AB}}-k\underline{U_{BC}}}{j\omega L(1-k^2)}\cdot \frac{-j\omega L(1-k^2)}{-j\omega L(1-k^2)}=\frac{-j\omega L(1-k^2)\underline{U_{AB}}+j\omega L(1-k^2)k\underline{U_{BC}}}{{\omega}^2L^2(1-k^2)^2}=j\frac{k\underline{U_{BC}}-\underline{U_{AB}}}{\omega L(1-k^2)}[/tex]
[tex]\Rightarrow \underline{{I_1}^{*}}=-j\frac{k\underline{U_{BC}}-\underline{U_{AB}}}{\omega L(1-k^2)}[/tex]

[tex]\underline{U_{AB}}{\underline{I_1}}^{*}=-j\frac{kU^2-U^2}{\omega L(1-k^2)}=j\frac{U^2}{\omega L(1+k)}[/tex]

We get [itex]\underline{U_{BC}}{\underline{I_2}}^{*}=\underline{U_{AB}}{\underline{I_1}}^{*}[/itex]

[tex]\underline{U_{CA}}{\underline{I_3}}^{*}=-j\omega CU^2[/tex]

Now, [tex]\underline{U_{AB}}{\underline{I_1}}^{*}+\underline{U_{BC}}{\underline{I_2}}^{*}+\underline{U_{CA}}{\underline{I_3}}^{*}=j\frac{U^2(2-{\omega}^2LC(1+k))}{\omega L(1+k)}[/tex]

[tex]Q=\frac{U^2(2-{\omega}^2LC(1+k))}{\omega L(1+k)}[/tex]
If [itex]Q=0[/itex], then [tex]C=\frac{2}{{\omega}^2L(1+k)}[/tex]

In my book, it says that the solution is [tex]Q=U^2\left(\frac{2+k}{\omega L(1-k^2)}-\omega C\right)=0\Rightarrow C=\frac{2+k}{{\omega}^2L(1-k^2)}[/tex]

Is it a mistake in the book, or in my solution?

 

[KataKoniK]

It is possible that there is a mistake in your solution. It is always a good idea to double check your calculations and make sure that your equations are correct. If you are confident in your calculations, then it is possible that there is a mistake in the book. It is always a good idea to consult with a colleague or an expert in the field to verify your solution and find any potential errors.