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.