A 2-wire DC distributor 200 metres long is uniformly loaded with 2A/metre. The resistance of a single wire is 0·3 Ω/km. If the distributor is fed at one end, then the voltage drops up to a distance of 150 m from the feeding point will be:

Concept of Uniformly Loaded Distributor Fed at One End:

Fig shows the single line diagram of a 2-wire DC distributor A B fed at one end A and loaded uniformly with i amperes per metre length.

• It means that at every 1 m length of the distributor, the load tapped is i amperes. Let l metres be the length of the distributor and r ohm be the resistance per metre run.
• Consider a point C on the distributor at a distance x metres from the feeding point A as shown in Fig.
• The current at point C is = (il − ix) A = i (l − x) A
   

Now, consider a small length dx near point C.

Its resistance is r dx and the voltage drop (dv) over length dx is given by,

dv = i (l − x) r dx = i r(l − x) dx

The total voltage drop (V) in distributor up to point C is given by,

V = \(\int_0^x ir(l-x)dx=ir(lx-\frac{x^2}{2})\)

Application:

We have,

Current loading (i) = 2 A/m

Resistance of distributor per metre run (r) = 2 × 0·3/1000 = 0·0006 Ω

Length of distributor (l) = 200 m

Now, Voltage drop upto a distance x metres from feeding point (V) is given by,

V = \(ir(lx-\frac{x^2}{2})\)

For, x = 150 m

Voltage Drop = \(2\times 0.0006(200\times 150 - \frac{150^2}{2})=22.5 \ V\)