Question

A d.c. distributor is fed at both the ends. At feeding point A, the voltage is maintained at 235 V and at B 236 V. The total length of the distributor is 200 m and loads are tapped off as under: 20 A at 50 m from A, 40 A at 75 m from A, 25 A at 100 m from A, 30 A at 150 m from A The resistance per kilometre of one conductor is 0.4 Ω. Calculate the minimum voltage and the point at which it occurs.

Answer 1

Resistance/metre single = 0.4/1000 = 4 × 10−4 Ω 

Resistance/metre double = 8 × 10−4 Ω 

Voltage drop on both conductors of 200-meter long distributor is 

= 8 × 10−4 [50i + 26 (i − 20) + 25(i − 60) + 50(i − 85) + 50(i − 115)] 

= 8 × 10−4 (200i − 12,000) volt 

This drop must be equal to the potential difference between A and B. 

∴ 8 × 10−4 (200i − 12,000) = 236 − 237 = −1 

∴ i = 53.75 A 

Current in section AC = 53.75 A 

Current in section CD = 53.75 − 20 = 33.75 A 

Current in section DE = 53.75 − 60 = − 6.25 A 

Current in section EF = 53.75 − 85 = − 31.25 A
Current in section FB = 53.75 − 115 = − 61.25 A 

Obviously, minimum voltage occurs at point D 

i.e. 75 metre from point A (or 125 m from B) 

Voltage drop across both conductors of the distributor over the length AD is 

= 8 × 10−4 (50 × 53.75 + 25 × 33.75) = 2.82 V 

∴ potential of point D = 236 − 2.82 = 233.18 V