Question

A 3-wire d.c. distributor, 250 m long, is supplied at end P at 500/250 V and is loaded as under :

Positive side : 20 A, 150 m from P ; 30 A, 250 m from P

Negative side : 24 A, 100 m from P ; 36 A, 220 m from P

The resistance of each outer wire is 0·02 Ω per 100 m and the cross-section of the middle wire is one half that of the outer. Find the voltage across each load point

Answer 1

Solution : Pi GA A loom B 30A P side it hom 820A F TOA E IVA Id CA kvom som er tom 6AC Zana 836Azum P2. -lle 2 GOA 36 H side Cross section of the Middle is Half Ceasily, given of the outer that of oulen Reststance - 2x of middle Resistance SO =2x0.02 loom w/10om w/10om =0.04 - hlow, Voltage of Current (A) | Doply) Resistance (2) section 50 105 0.03 150 X 0.02 100 P, A 0.6 30 AB 100 r 0.02 -0.02 Too 0.168 ) هي CD 70x 6.04 = 0.028 100 50 x 0.04 14 16.208 0.02 IT 100 CS Scanned with CamScanner

Oul 10 EF 100 K0:04 Ovoy a 100 30 0.36 GO CPR CM 30x0.04 00012 700 GP2 roox v.valliloo -00cm Now Positive side voltage - al point AD= 250 - Voltage doop P,A - Voltage drop in DE tvollage drop in EF = 250 -1.5 -Q-28 +004 = 248.62 volts point BME voltage vollage drop inns - at Voltage al pont A doop In CD tiroltage Voltage drop in CM = 248.62 -0.6 +0.168 -0.36 = 247.528 volts sede - Now, Negative Nollage at potht EG = 250 - voltage drop in FF dropen GP2 -tortage = 250 - Coou) - 1.2 = 245.4 Vorts CS Scanned with CamScanner

a OSS Now vollage (H = Voltage arooss FP2 - Voltage drop In EF - Voltage doop en ED Voltage dropln.co - Vollege doop en HG lloltage drop en Gilz - 250- (0.4) -(0:28) - 0.168 0012 -(12 720 X0:02 TOO ¥36) = 247.65 llolts CS Scanned with CamScanner