8.MET August 2018 Q.8(B)

A ring-main, 900m long, is supplied at a point A at a p.d. of 220V. At a point B, 240m from A, a load of 45A is drawn from the main, and at a point C, 580m from A, measured in the some direction, a load of 78A is taken from the main. If the resistance of the main (lead and return) is 0.25 ohm per kilometre, calculate the current which will flow in each direction round the main from the supply point A and the potential difference across the main, at the load where it is lowest

              If you are noticing some error in problems kindly comment below.Thanks

Given

Total length = 900 m
R = 0.25 ohm / Km

To find

1. Current
2. Potential difference

Solution

Total load current = 45 + 78
                             = 123 A

Length BC = 580 - 240
                  = 340 m

Consider current in AC section  = I

Current in AB section = 123 - I
Current in BC section = 123 - I - 45
                                    = 78 - I

Resistance in Section = Resistance x Distance

Distance AC = 900 -580 = 320 m

Resistance AC = 0.25 x (320 / 1000)
                         = 0.08 ohm

Resistance AB = 0.25 x (240 / 1000)
                         = 0.06 ohm

Resistance BC = 0.25 x (340 / 1000)
                        = 0.085 ohm

As per Kirchoff's Law Total voltage = Sum of voltage at each sections

V at AC = V at AB + V at BC

I x 0.08 = ((123 - I) x 0.06) + ((78 - I) x 0.085)
            I = 62.27 A

1. Current in section AC, I = 62.27 A
    Current in section AB = 123 - I
                                       = 123 - 62.27
                                       = 60.73 A
    Current in section BC = 78 - I
                                       = 78 - 62.27
                                       = 15.73 A
   
2. Potential difference = Total PD - PD at the section

    PD at C = 220 - (62.27 x 0.08)
                  = 215 V
    PD at B = 220 - (60.73 x 0.06)
                 = 216.36 V

    PD at the load C is lowest = 215 V