# Three-Phase system problem

Tags:
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?

#### Attached Files:

• ###### circuit.PNG
File size:
1.7 KB
Views:
42
2. Apr 17, 2016