As shown in Fig.(a), let SB be the distributor with A as the mid point. Total impedance of the distributor is = (0.4 + j 0.6) Ω.
Let the voltage VB at point B be taken as the reference voltage.
∴ VB = (240 + j 0)V ; IB = 100 (0.8 − j 0.6) = 80 − j 60 A
Drop in section AB = (80 − j 60) (0.2 + j 0.3) = (34 + j12) V
VA = VB + drop over AB = (240 + j 0) + (34 + j12) = (274 + j12) V
The phase difference between VA and VB is = tan–1 (12/274) = 2°30′
The load current IA has lagging power factor of 0.6 with respect to VA. It lags VA by an angle φ = cos –1(0.6) = 53°8′
Hence, it lags behind VB by an angle of (53°8′ − 2°30′) = 50°38′ as shown in the vector diagram of Fig.(b).
IA = 100 (cos 50°38′ − jsin 50°38′) = (63.4 − j77.3) A
I = IA + IB = (80 − j 60) + (63.4 − j77.3) = (143.4 − j137.3) A
Drop in section SA = (143.4 − j137.3) (0.2 + j 0.3) = (69.87 + j 15.56)
VS = VA + drop in section SA
= (274 + j12) + (69.87 + j15.56) = 343.9 + j 27.6 = 345 ∠5°28′ V
Hence, supply voltage is 345 V and lead VB by 5°28′