DC distributed load(Uniformly loaded), fed at one end or both ends

1. Attention Students: We missed this in a hurry: Please go
through this
DC DISTRIBUTED LOAD
1. Fed at one end (Uniformly loaded) :
Let ‘L’m be the length of distributor and r ohm be the resistance/m
Run. Consider a point C on the distributor at a distance x metres from the feeding
pointA. The current at point C is = iL – ix = i(L-x) Amp.
This is true and it can be understood if we consider the follwing example:
Current flowing in DE will 10x5 – 10x2 = 30A

2. Now consider a small length dx near point C. Its resistance is rdx and the voltage drop
over length dx is
dv = i(L-x)rdx = ir(L – x )
Total voltage drop in the feeder uptopoint C
v= = ir(Lx – x2/2)
The voltage drop upto point B can be obtained by putting x = l
VAB = ir(L2 – L2/2) = irL2/2 =( iL)(rL)/2 = IR/2
Loss = = = i2rL3/3 Watts
2. DC distributed load fed at both the ends(VA = VB)(Uniformly loaded)
Current supplied from each feeding point = iL/2
Consider a point C at a distance x meters from the feding point A. Then current at point C is
= iL/2 – ix = i(L/2 – x )

3. Consider a small length dx near point C. Its resistance is rdx and the voltage drop over length dx
is dv = i(L/2 - x)rdx
Voltage drop upto point C = (ir/2)(Lx –x2)
Obviously point of minimum potential will be the mid point.
Maximum voltage drop (when x =L/2) = ir/2(L2/2 – L2/4) = irL2/8 = iLrL/8 = I R/8
Minimum potential (at mid- point) = V – IR/8
Power loss in loss in the distributor fed from both the ends be calculated firstly for half of the
line and then can be doubled.
Loss =2 ∫ ( L / 2 − x)( L / 2 − x )iirdx = i2rL3/12 Watts
Problems :
1. A 2 wire dc distributor 200m long is uniformly loaded with 2A/m. Resistance of
single wire is 0.3 ohm/km. If the distributor is fed at one end, calculate:
(i) The voltage drop upto a distance of 150m from the feeding point.
(ii) The maximum voltage drop
Solution:
I = 2A/m, r = 2x0.3/1000 = 0.0006 ohm/m
(i) Vx = ir (Lx – L2/2) = 2x0.0006(200x150 – 1502/2) = 22.5V
(ii) I = iL = 400A; R = rL = 0.12
Total voltage drop = IR/2 = 24V
2. A 2 wire dc distributor cable 1000m long is loaded with 0.5A/m. Resistance of each
conductor is 0.05 ohm/km. Calculate the maximum voltage drop if the distributor is
fed from both ends with equal voltages of 220V. what is the minimum voltage and
when it occurs.

4. Solution:
Current loading: i = 0.5A/m
r = 2x0.05x1000 = 0.1x10-3 ohm
L = 1000m
I = iL = 500A
R = rL = 0.1 ohm]
Maximum voltage drop = IR/8 = 500x0.1/8 = 6.25V
Minimum voltage occurs at the mid point and its value is = V –IR/8 = 220 – 6.25 =
213.75V