1. The document discusses voltage drop calculations for DC distribution lines under different loading and feeding conditions. It provides formulas to calculate the voltage drop and power loss for lines fed at one end or both ends. 2. Examples are given for calculating the voltage drop and maximum voltage drop for a 200m line fed at one end and loaded uniformly with 2A/m. Also for a 1000m line fed at both ends with 220V and loaded with 0.5A/m, it shows how to find the maximum and minimum voltages. 3. Key formulas introduced are for voltage drop up to a point x on a line fed at one end as V=ir(Lx-x^2/2)

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