A single uniform underground cable linking A to B, 50 km long, has a fault in it at distance d km from end A. This is caused by a break in the insulation at X so that there is a flow of current through a fixed resistance R into the ground. The ground can be taken to be a very low resistance conductor. Potential differences are all measured with respect to the ground, which is taken to be at 0 V.
[Figure: Cable diagram showing points A, X, and B, with distances d and (50-d), and resistance R to ground at point X]
In order to locate the fault, the following procedure is used. A potential difference of 200V is applied to end A of the cable. End B is insulated from the ground, and it is measured to be at a potential of 40V.
a) What is the potential at X? Explain your reasoning. [2]
b) What is
(i) the potential difference between A and X? [1]
(ii) the potential gradient along the cable from A to X (i.e. the volts/km)? [1]
c) The potential applied to end A is now removed and A is insulated from the ground instead. The potential at end B is raised to 300 V, at which point the potential at A is measured to be 40 V.
(i) What is the potential at X now? [1]
(ii) Having measured 40 V at end B initially, why is it that 40 V has also been required at end A for the second measurement? [2]
d) What is the potential gradient along the cable from B to X? [1]
e) The potential gradient from A to X is equal to the potential gradient from B to X.
(i) Explain why this is true [2]
(ii) From the two potential gradients that you obtained earlier, deduce the value of d. [1]