Cable Calculation

[Blueday]

Hi,

I was looking through old exam papers and came across the following question;

A 2-core distributor cable of uniform csa is loaded as follows:
Point B = 30A
Point C = 90A
Point D = 62A

The cable is supplied at End A at 229V and at End E at 231V. The resistance of the cable is 0.01 ohms/100metres (lead & return). The following are the distances involved:
Point A to B = 200m
Point B to C = 200m
Point C to D = 200m
Point D to E = 1 000m

Calculate (i) the current in each section
(ii) voltage at load C


Can this be solved by using Kirchhoffs Laws/Superposition or is there another way to solve. I don't remember ever doing anything like this at college.

Many thanks in advance.

 

[xraytek]

I would be interested to see this one get solved, as I have forgotten how to do it .

If I remember correctly the sum of all voltage drops in a closed circuit equate to zero.
You are given the current at each point as well as cable resistance

so you create an equation something like this:

(I*R)+(I*R)+(I*R)=0

Thats as much as I can remember, so can anyone pick it up from here?