Attention Students: We missed this in a hurry: Please go
through this
DC DISTRIBUTED LOAD
  1. Fed at one end (Uniformly l...
Now consider a small length dx near point C. Its resistance is rdx and the voltage drop
        over length dx is

       ...
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

...
Solution:

Current loading: i = 0.5A/m

r = 2x0.05x1000 = 0.1x10-3 ohm

L = 1000m

I = iL = 500A

R = rL = 0.1 ohm]

Maxim...
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Dc Distributed Load

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

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  • mam, please can you tell me the second question is from which book ??
    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.
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Dc Distributed Load

  1. 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. 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. 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. 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

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