Question

A single-phase a.c. distributor 500 m long has a total impedance of (0.02 + j 0.04) Ω and is fed from one end at 250V. It is loaded as under :

(i) 50 A at unity power factor 200 m from feeding point.

(ii) 100 A at 0.8 p.f. lagging 300 m from feeding point.

(iii) 50 A at 0.6 p.f. lagging at the far end.

Calculate the total voltage drop and voltage at the far end.

Answer 1

First Method

Current in section AD (Fig.) is the vec

tor sum of the three load currents.

∴ current in AD

= 50 + 100 (0.8 − j 0.6) + 50 (0.6 − j 0.8)

= 160 − j100

Impedance of section AD

= (200/500) (0.02 + j 0.04)

= (0.008 + j 0.016) W

Second Method

We will split the currents into their active and reactive components as under :

50 × 1 = 50 A ; 100 × 0.8 = 80 A; 50 × 0.6 = 30 A

These are shown in(a). The reactive or wattless components are

50 × 0 = 0 ; 100 × 0.6 = 60 A ; 50 × 0.8 = 40 A

These are shown in Fig.(b). The resistances and reactances are shown in their respective figures.

Drops due to active components of currents are given by taking moments

= 50 × 0.008 + 80 × 0.012 + 30 × 0.02 = 1.96 V

Drops due to reactive components = 60 × 0.024 + 40 × 0.04 = 3.04

Total drop = 1.96 + 3.04 = 5 V

This is approximately the same as before.

Third Method

The centre of gravity (C.G.) of the load is at the following distance from the feeding end

cos φav = 0.8 ; sin φav = 0.6

Drop = 200(0.013 × 0.8 + 0.026 × 0.6) = 5.2 V

This is approximately the same as before.