# Homework Help: A Short-Circuit Question

1. Nov 1, 2014

### harbour

1. The problem statement, all variables and given/known data

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
In order to locate the fault, the following procedure is used. A potential difference of 200 V 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 40 V.

a) What is the potential at X? Explain your reasoning.

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.

b) What is the potential at X now?

c)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?

d) The potential gradient from A to X is equal to the potential gradient from B to X. Explain why this is true

2. Relevant equations

Ohm's Law

3. The attempt at a solution

a) 40V
b) 40V
c) So that X is at the same potential and then the same current flows into the ground through R
d) Because the same currents flowed along AX and BX

What I don't understand is the physics behind it. Why is there no current from X to B in question a? Why must there be the same current when applying 200V at A and when applying 300V at B?

2. Nov 1, 2014

### Staff: Mentor

Welcome to the PF.

There are a couple key concepts here. First, when you insulate B from ground and apply the voltage at A with respect to ground, current flows to the fault and into the ground. No current flows from X to B, because there is no load at B for the current to go through. The voltage at B is the same as it is at X in this situation, since no current flows from X to B (hence, no V=IR drop).

When you insulate A and drive a voltage into B with respect to ground, you get the same situation -- makes sense?

The second key concept is that they are setting the A and B voltages in this test to give the same voltage drop across the resistive fault at X. What does that mean about the 2 test currents that are involved? And then what does that imply about the "voltage gradients" in V/km? Why?

3. Nov 1, 2014

### harbour

Does 'insulate' mean that no current can passed through this point; i.e. the wire is disconnected? Then I completely understand that.

The second part might just be a wording of the question that caught me. When they say they apply a potential difference at B, do they mean they gradually increase it until they get 40V at A; because I thought they question meant they applied an arbitrary voltage (200V then 300V) and still got the same potential at both ends.

4. Nov 1, 2014

### Staff: Mentor

They must monitor Va while increasing Vb. That's how they got the Vx=40V for both test cases. Glad it makes more sense to you now. What do you get for d now?

5. Nov 1, 2014

### harbour

I get 160/d = 260/(50 - d), which gives me 19km.

6. Nov 1, 2014

### Staff: Mentor

And I get (160V/260V)*50km = 19.23km. Good job! :-)