# Maximum power factor of three phase receiver

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1. May 16, 2016

### gruba

1. The problem statement, all variables and given/known data
Given symmetric three phase system (see attachment) of phase voltages with angular frequency $\omega=100 rad/s$, $R=5\omega L=100\Omega$. Find capacitance of capacitor $C$ such that power factor of three phase receiver has maximum value.

2. Relevant equations
Power factor is given by $\cos\phi=\frac{P}{S}$, where $P$ is active, and $S$ is apparent power.
Alternatively, $\cos\phi=\frac{\mathfrak{R}{(\underline{S}})}{|\underline{S}|}$, where $\underline{S}=P+jQ$ is complex apparent power and $Q$ is reactive power.

3. The attempt at a solution
After transformation of Y capacitors to $\Delta$ (see attachment), $C_1=\frac{C}{3}$.
Now we have a $\Delta$ connection of impedance $\underline{Z}$ which is a parallel of $R,j5\omega L$ and $C_1$. Let $\underline{Z_1}=R+j5\omega L$. From given data we can find that $L=0.2 H$. This gives $\underline{Z_1}=100(1+j)\Omega$. Now $$\underline{Z}=\frac{\underline{Z_1}\cdot (-jX_{C_1})}{\underline{Z_1}+ (-jX_{C_1})}=\frac{300(3+j(3-2\cdot 10^4 C))}{2\cdot 10^8C^2-6\cdot 10^4C+9}\Omega.$$ Now we have a three phase system with receiver in $\Delta$ connection (see attachment). After $\Delta$ to Y transformation (see attachment), we get new impedance $$\underline{Z_2}=\frac{\underline{Z}}{3}=\frac{100(3+j(3-2\cdot 10^4 C))}{2\cdot 10^8C^2-6\cdot 10^4C+9}\Omega.$$
Let $$\underline{Z_3}=\underline{Z_2}+jX_L=\underline{Z_2}+j20=\frac{300+160(3-2\cdot 10^4C+25\cdot 10^6C^2)}{2\cdot 10^8C^2-6\cdot 10^4C+9}\Omega.$$

Now we have a clean Y receiver connection with impedance $\underline{Z_3}$ (see attachment).

Question: We are not given any values for voltage, current or power, so how to express power factor $\cos\phi$ without knowing any of those values?

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• ###### Y_connection.PNG
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2. May 17, 2016

### gruba

EDITED:

Last line:

$$\underline{Z_3}=\underline{Z_2}+jX_L=\underline{Z_2}+j20=\frac{300+j160(3−2⋅10^4C+25⋅10^6C^2)}{2⋅10^8C^2−6⋅10^4C+9}\Omega=\frac{20(15+j8(3−2⋅10^4C+25⋅10^6C^2))}{2⋅10^8C^2−6⋅10^4C+9}\Omega$$

Power factor can be expressed by $\cos\phi=\frac{\mathfrak{R}(\underline{S})}{\sqrt{P^2+Q^2}}$ where $\underline{S}$ is complex apparent power, $P$ is active and $Q$ is reactive power. Power factor has a maximum value when reactive power tends to zero. Since we know only the impedance, we can look at the imaginary part of impedance $Z_3$. If we introduce a function $$f(C)=\frac{160(25\cdot 10^6C^2-2\cdot 10^4C+3)}{2⋅10^8C^2−6⋅10^4C+9}$$, minimum value of $f(C)$ is $\frac{-40}{3}$ at $C=3\cdot 10^{-4}F.$ So, maximum power factor is for $C=0.0003F$.

Question: Is this correct?

Last edited: May 17, 2016
3. May 18, 2016

### gruba

EDIT:
If we find a minimum of function $f(C)$ we get that $C=0.0003F$, but if we solve $f(C)=0$ (phase resonance), we get two possible solutions (quadratic equation):
$C=0.0002F$ or $C=0.0006F$.
Question: Which equation is correct to solve in order to determine $C$, $f(C)=0$ (phase resonance) or $f'(C)=0$ (finding minimum value of a function $f(C)$)?