- #1

- 208

- 1

## Homework Statement

Given symmetric three phase system (see attachment) of phase voltages with angular frequency [itex]\omega=100 rad/s[/itex], [itex]R=5\omega L=100\Omega[/itex]. Find capacitance of capacitor [itex]C[/itex] such that power factor of three phase receiver has maximum value.

## Homework Equations

Power factor is given by [itex]\cos\phi=\frac{P}{S}[/itex], where [itex]P[/itex] is active, and [itex]S[/itex] is apparent power.

Alternatively, [itex]\cos\phi=\frac{\mathfrak{R}{(\underline{S}})}{|\underline{S}|}[/itex], where [itex]\underline{S}=P+jQ[/itex] is complex apparent power and [itex]Q[/itex] is reactive power.

## The Attempt at a Solution

After transformation of

**Y**capacitors to [itex]\Delta[/itex] (see attachment), [itex]C_1=\frac{C}{3}[/itex].

Now we have a [itex]\Delta[/itex] connection of impedance [itex]\underline{Z}[/itex] which is a parallel of [itex]R,j5\omega L[/itex] and [itex]C_1[/itex]. Let [itex]\underline{Z_1}=R+j5\omega L[/itex]. From given data we can find that [itex]L=0.2 H[/itex]. This gives [itex]\underline{Z_1}=100(1+j)\Omega[/itex]. Now [tex]\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.[/tex] Now we have a three phase system with receiver in [itex]\Delta[/itex] connection (see attachment). After [itex]\Delta[/itex] to

**Y**transformation (see attachment), we get new impedance [tex]\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.[/tex]

Let [tex]\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.[/tex]

Now we have a clean

**Y**receiver connection with impedance [itex]\underline{Z_3}[/itex] (see attachment).

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