Three-Phase system problem

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1. Apr 12, 2016

gruba

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

2. The attempt at a solution

Total active power is given by $$P=3\cdot\frac{\left(\frac{U}{\sqrt 3}\right)^2}{R}$$
where $\frac{\left(\frac{U}{\sqrt 3}\right)^2}{R}$ is the active power of one phase, and
$\frac{U}{\sqrt 3}$ is the voltage of one phase.

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

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

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

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

We get $\underline{U_{BC}}{\underline{I_2}}^{*}=\underline{U_{AB}}{\underline{I_1}}^{*}$

$$\underline{U_{CA}}{\underline{I_3}}^{*}=-j\omega CU^2$$

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

$$Q=\frac{U^2(2-{\omega}^2LC(1+k))}{\omega L(1+k)}$$
If $Q=0$, then $$C=\frac{2}{{\omega}^2L(1+k)}$$

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

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

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2. Apr 17, 2016

Staff: Admin

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

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