Question 2.45: A three-phase star-connected load consumes a total of 12 kW ...

Total power consumed = 12 kW
Per phase power consumed = 4 kW
So,                                              V_{Ph}  I_{Ph} \cosΦ = 4000  W

I_{Ph}=\frac{4000}{V_{Ph} \cosΦ}

or,                          I_{Ph}=\frac{4000}{(400 / \sqrt{3})  ×  0.8} =21.6  A

I_{Ph}=\frac{  V_{Ph}}{  Z_{Ph}}

Z_{Ph}=\frac{ V_{Ph}}{ I_{Ph}}= \frac{231}{21.6}=10.7  Ω.      Power factor,   \cos Φ = 0.8,     \sin Φ= 0.6

R = Z_{Ph} \cosΦ = 10.7  ×  0.8 = 8.56  Ω
X = Z_{Ph} \sin Φ = 10.7  ×  0.6 = 6.42  Ω

X = 6.42
2π fL = 6.42

L =\frac{6.42}{2  ×  3.14  ×  50}=20.4  ×  10^{-3} H

= 20.4  mH