Maximum power factor of three phase receiver

[gruba]

Homework Statement

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.

Homework 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.

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?

 

[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?

 

[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))?