Question

A 2-wire D.C distributor AB is fed from both ends. At the feeding point A the voltage is maintained at 240 V and at B is 245 V. The total length of the distributor is 200 meters and loads are tapped off as under: 25A at 50 meters from A; 50A at 75 meters from A; 30A at 100 meters from A;40A at 150 meters from A. If the resistance per Km of one conductor is 0.3$\Omega$, calculate: i) The currents in the various sections of the distributor ii) The minimum voltage and the point at which it occurs. iii) The power dissipated in the distributor

Answer 1

Answer:

238.66 V

Explanation:

Lets call Tap Point C , D E & F  respectively for 50 m , 75 m , 100 m , & 150 m

Resistance per km (1000m) = 0.3 Ω

Resistance of 50 m for Both conductors ( to & fro) AC = 2 *  (50/1000) * 0.3

= 0.03  Ω

Resistance of CD =  2 * ((75-50)/1000) * 0.3 = 0.015 Ω

Resistance of DE =  2 * ((100-70)/1000) * 0.3 = 0.015 Ω

Resistance of EF =  2 * ((150-100)/1000) * 0.3 = 0.03 Ω

Resistance of EB =  2 * ((200-150)/1000) * 0.3 = 0.03 Ω

Let Say Current  from End A  = IC

Then Current in AC = Iₐ A

Current in CD = Iₐ - 25 A

Current in DE = Iₐ - 75 A

Current in EF = Iₐ - 105 A

Current in FB = Iₐ - 145 A

Vₐ  - Iₐ  *  0.03 - (Iₐ - 25)*0.015 - (Iₐ - 75)*0.015 - (Iₐ - 105)*0.03 - (Iₐ - 145)*0.03 = Vb

=> 240 - Iₐ *0.015 (2 + 1 +  1 + 1 + 2) + 0.015(25 + 75 + 210 + 290) = 245

=> - Iₐ *0.015 (7) + 0.015(600) =  5

dividing by 5

=> - Iₐ *0.003 (7) + 0.003(600) =  1

multiplying by 1000

=> - Iₐ *3 (7) + 3(600) =  1000

=> -21 * Iₐ = -800

=> Iₐ = 38.1 A

Then Current in AC = 38.1 A

Current in CD = 13.1 A

Current in DE = -36.9 A

Current in EF = -66.9 A

Current in FB = -106.9 A

-ve sign indicates that current is being feedded from B

Minimum Potential = 240 - (38.1)*0.03 - (13.1)(0.015)

= 240 - 1.143 - 0.1965

= 240 - 1.3395

= 238.66 V