Balanced Three Phase Load System

[JosefMTD]

Homework Statement

f = 60 Hz
pf = 0.8 (lagging)

Homework Equations

Z(L) = 2 (pi) * f * L
Given z = x + iy
tan(angle) = y/x
pf = cos(angle)

The Attempt at a Solution

I changed the Wye-connected resistors to an equivalent Delta-connected resistors:

The new resistors become 3R in the equivalent Delta-connected resistors

Next step is calculating Z(L)
Z(L) = j*2 (pi) * 60 Hz * 2 H = 754j ohm

1/Zphase = 1/Z(L) + 1/Z(R)
1/Zphase = 1/754j ohm + 1/3R ohm = -j/754 ohm + 1/3R ohm
1/Zphase = (754 ohm - 3Rj)/754 ohm*3R ohm

Since V is not given value in this problem, I choose to use an arbitrary value of x with angle of 0 degree, thus making it a constant.

Iphase = Vphase/Zphase = x*(754 ohm -3Rj)/754 ohm*3R ohm

It's known that power factor is 0.8

cos(angle) = 0.8
angle = arc cos(0.8) = +/- 37

Since known that the current is lagging, the angle is lower than 0 degree, thus making the appropriate angle is -37 degree.

Since I know the angle of phase I, I put it in the formula to find the angle of a complex number:
tan(angle) = Im(complex variable)/Re(complex variable)
tan(-37 degree) = [(x/754 ohm*3R)(-3Rj)]/[(x/754 ohm*3R)(754)] = -3R / 754
- 0.75 = -3R / 754
R = 754/4 Ohm = 188.5 ohm

Am I doing it right?
Thank you in advance for helping me to solve this!

 

[JosefMTD]

Sorry for not using the latex:

Calculating total impedance in a phase
{\frac{1}{Z_p}} = {\frac{1}{Z_L}} + {\frac{1}{Z_R}}
{\frac{1}{Z_p}} = {\frac{1}{754 j\Omega}} + {\frac{1}{3R}}
{\frac{1}{Z_p}} = {\frac{-j}{754 \Omega}} + {\frac{1}{3R}}
{\frac{1}{Z_p}} = {\frac{754\Omega - 3Rj}{754 j\Omega * 3R}}

Making the i = V/Z
I = V*{\frac{754\Omega - 3R}{754 j\Omega * 3R}}
I = {\frac{V}{754 j\Omega * 3R}}*(754\Omega - 3Rj)

Finding the angle of I:
\tan{\theta} = ({\frac{V}{754 j\Omega * 3R}}*-3R)/({\frac{V}{754 j\Omega * 3R}}*754\Omega)
\tan{\theta} = {\frac{-3R}{754 \Omega}}