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

  • Thread starter Thread starter harbour
  • Start date Start date
AI Thread Summary
The discussion revolves around locating a fault in a 50 km underground cable using a voltage-based approach. When a 200 V potential is applied at end A, the potential at point X is determined to be 40 V, as no current flows from X to B due to the lack of a load at B. Insulating end A and applying 300 V at B also results in 40 V at A, indicating the same current flows through the fault, maintaining consistent voltage gradients. The participants clarify that "insulate" means no current can pass through that point, and the voltage at A is monitored while adjusting B's voltage. The calculations for the fault's location yield approximately 19 km from end A.
harbour
Messages
4
Reaction score
0

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
 
Physics news on Phys.org
harbour said:

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

And I get (160V/260V)*50km = 19.23km. Good job! :-)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top