Locating a Fault in an Underground Cable: A Voltage-Based Approach

[harbour]

Homework Statement

[/B]
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

Homework Equations

Ohm's Law

The Attempt at a Solution

The answers are:

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?

If you want a diagram it is question 12 of this doc: http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_AS_2007_QP.pdf

 

[berkeman]

Homework Statement

[/B]
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

Homework Equations

Ohm's Law

The Attempt at a Solution

The answers are:

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?

If you want a diagram it is question 12 of this doc: http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_AS_2007_QP.pdf

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?

 

[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.

 

[berkeman]

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.

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?

 

[harbour]

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

 

[berkeman]

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

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