Underground cables faulty point detected

[tellmesomething]

Homework Statement

This doubt is extended from a certain question, I had a uniform underground cable connecting two telegraph offices. There was a fault at a point on the cable which made it have contact with the earth causing a current to flow between the earth and the faulty point because of the obvious potential difference. I also know that some heat was lost in this path from the faulty point to earth implying some resistance. I have some questions regarding this question

Relevant Equations

None

So assuming this wire is probably not ideal since its used in a real life situation, theres possible resistances in the wire before the faulty point as well as after the faulty point. Theres also resistance connecting the faulty point and the ground.

Now does current stop flowing in the wire after this incident? Is ot because the circuit is "broken now" like theres no way of retrieving the charge that goes into Earth I.e it isnt a closed circuit?

 

[berkeman]

Is this a twisted pair cable that has one or both conductors ground faulted somewhere in the middle of the cable run? Or are they implying just a single conductor with an Earth return path? Is there a diagram given with the problem?

 

[tellmesomething]

Is this a twisted pair cable that has one or both conductors ground faulted somewhere in the middle of the cable run? Or are they implying just a single conductor with the Earth return path implied? Is there a diagram given with the problem?

Heres The whole problem with the diagram

A uniform underground cable AB connects two telegraph offices which are 8km apart. Due to a fault at a place C the conductor of the cable comes in contact with the earth. Current flowing to the earth at fault location faces $10 \Omega$ resistance. To locate the fault when an engineer applies 20V between the end A and the earth, his colleague measures 5 volt between the end B and the earth. Moreover when his colleague applies 30V between the end B and rhe earth, the engineer measures 5V between the end A and the earth. What is the distance of the fault location along the cable from the end A?

 

[berkeman]

Hmm. It still seems underspecified to me, but maybe some things can be neglected. I still need to know the gauge of the wire (or its resistance per km), and the resistance of the Earth per km.

Can you upload a diagram showing the resistances and voltages mentioned in the problem statement to see if that is enough to solve this? Just label the distances along the wire "x" and "8km-x".

 

[tellmesomething]

Hmm. It still seems underspecified to me, but maybe some things can be neglected. I still need to know the gauge of the wire (or its resistance per km), and the resistance of the Earth per km.

Can you upload a diagram showing the resistances and voltages mentioned in the problem statement to see if that is enough to solve this? Just label the distances along the wire "x" and "8km-x".

I can do that but before that i d like to understand the question a bit more....does the current stop flowing after theres a fault discovered? And before connecting the ends a and b to the ground

 

[berkeman]

I can do that but before that i d like to understand the question a bit more....does the current stop flowing after theres a fault discovered?

"Stop" is such a harsh word!

No, it's more like a voltage divider. For a single conductor scenario, you would have a driving voltage source at one end, driving a current into the resistance of the 8km long wire with its resistance to the detector, with an 8km long Earth return with its resistance. So the voltage divider there is "output resistance of driver" + "wire resistance" dividing into the "resistance of the Earth return".

With the ground fault in the middle somewhere, that divides off some current that depends on the ground fault resistance (given as $10\Omega$) and the relative resistances of the distances of the 2 sections of transmission wire and of Earth return.

 

[tellmesomething]

"Stop" is such a harsh word!

No, it's more like a voltage divider. For a single conductor scenario, you would have a driving voltage source at one end, driving a current into the resistance of the 8km long wire with its resistance to the detector, with an 8km long Earth return with its resistance. So the voltage divider there is "output resistance of driver" + "wire resistance" dividing into the "resistance of the Earth return".

With the ground fault in the middle somewhere, that divides off some current that depends on the ground fault resistance (given as $10\Omega$) and the relative resistances of the distances of the 2 sections of transmission wire and of Earth return.

I dont understand how the current lost to the earth is taken back if that makes any sense? Like how is the circuit completed? The current path to earth feels redundant

 

[berkeman]

Um, give me a couple minutes to make a crude drawing...

(unless somebody else beats me to it!)

 

[berkeman]

Something like this (can you see how the problem is underspecified?)...

 

[haruspex]

Something like this (can you see how the problem is underspecified?)...

View attachment 349526

Is it not reasonable to assume that both the resistance through the wire and the resistance through the Earth are proportional to distance?

Edit: no.. see post #24.

 

[berkeman]

Is it not reasonable to assume that both the resistance through the wire and the resistance through the Earth are proportional to distance?

Yes. Is that one of the simplifying things that makes the question not underspecified? I haven't worked through the problem in detail myself. I was just trying to give them a starting framework to begin to write the equations. Hopefully things can start to cancel out...

 

[haruspex]

Yes. Is that one of the simplifying things that makes the question not underspecified? I haven't worked through the problem in detail myself. I was just trying to give them a starting framework to begin to write the equations. Hopefully things can start to cancel out...

I get an answer on that basis.

 

[tellmesomething]

Yes. Is that one of the simplifying things that makes the question not underspecified? I haven't worked through the problem in detail myself. I was just trying to give them a starting framework to begin to write the equations. Hopefully things can start to cancel out...

we've been told that unless the distance between the connecting wires in the ground is 30 to 40km we can ignore the resistance. Considering its very much less than that I think we are supposed to ignore the resistance you specified inside the ground..and in place of that just place a conducting wire..

 

[haruspex]

we've been told that unless the distance between the connecting wires in the ground is 30 to 40km we can ignore the resistance. Considering its very much less than that I think we are supposed to ignore the resistance you specified inside the ground..and in place of that just place a conducting wire..

That is just a special case of assuming the resistance ratios are the same, so it is not necessary to ignore ground resistance.

 

[tellmesomething]

That is just a special case of assuming the resistance ratios are the same, so it is not necessary to ignore ground resistance.

Okay. Would that mean R1/R2= ratio of resistance in the ground since area of cross section is same so length is the only parameter which is same as well..

 

[haruspex]

Okay. Would that mean R1/R2= ratio of resistance in the ground since area of cross section is same so length is the only parameter which is same as well..

Yes. You can leave the Earth symbol out of the diagram. The Earth is fully represented by the extra "wires".

 

[tellmesomething]

Yes. You can leave the Earth symbol out of the diagram. The Earth is fully represented by the extra "wires".

I see okay. Also one doubt, theres ia already a current in the circuit A to B when I connect the A to earth will I not already have a current flowing through it? Why did I have to connect a voltage source ?

 

[berkeman]

Why did I have to connect a voltage source ?

The problem specifies connecting two different voltage sources as part of the measurements by the technicians to locate the fault. Draw the 2 diagrams corresponding to those two measurements...

To locate the fault when an engineer applies 20V between the end A and the earth, his colleague measures 5 volt between the end B and the earth. Moreover when his colleague applies 30V between the end B and rhe earth, the engineer measures 5V between the end A and the earth. What is the distance of the fault location along the cable from the end A?

 

[haruspex]

I see okay. Also one doubt, theres ia already a current in the circuit A to B when I connect the A to earth will I not already have a current flowing through it? Why did I have to connect a voltage source ?

There is no current until a voltage is applied. In normal operation, a voltage is applied when someone tries to send a signal. During the test for the fault, no-one is trying to send a signal. The only voltage being applied is the 20V or 30V used in the test.

 

[tellmesomething]

There is no current until a voltage is applied. In normal operation, a voltage is applied when someone tries to send a signal. During the test for the fault, no-one is trying to send a signal. The only voltage being applied is the 20V or 30V used in the test.

What i meant is even without applying the voltage theres already a potential difference between the terminal A and the earth no? So current would have flown without the additational battery being there..?

 

[tellmesomething]

Some questions:
i)There was a current in the circuit AB before there was a fault. As soon as there was a fault the current passing through AB flowed down to the earth at C to equalize the potential difference.. my question is did all the current coming in flow down the earth? If so why? Why is it not going to B like it did before?

ii) How did you decide on the resistors in the cable (red and green)? What part of the question hinted at that? Why cant there be no resistors ( red and green)? Why can it not be an ideal underground cable?

iii) What is the significance of applying a voltage between earth and terminal A? If they connected a cable directly to earth current coming in through A would still flow to earth since theres an already existing potential difference..

iv) Why does the current flow back up to C? Ater connecting the cable didnt the potnetial at C and earth became equal?

v) After coming out of C why would the current not flow through B ? Is it because its an open circuit?

 

[berkeman]

What i meant is even without applying the voltage theres already a potential difference between the terminal A and the earth no?

No.

When there is no signal being sent and no test technicians putting test voltages in, there is no potential difference anywhere in the circuit.

 

[Tazerfish]

When I first looked at this problem, I thought it was about wave dynamics and reflections/distortions caused to a dynamic signal

Ground has always been a very confusing concept for me. To now have current flowing through two different grounds is pretty antithetical to the entire idea ^^.

So, as Berkeman and Haruspex have done, I wouldn't draw in "ground" anywhere.
The earth is just another cable lying parallel to the telegraph line in this exercise.

Consider the two scenarios:
(1) The technician at A is applying a voltage, B is measuring.
Think about it, if B is measuring voltage, then there is no current flowing from the line to ground at B. No current flows through the voltage measurement device.
Plus, there isn't any parallel path or load connected at B; the ends are just open.
Consequently, there can't be any current in the line segment leading from the fault to B or in the "earth segment" between the fault and B.
No current, no voltage drop.
So the voltage B's measuring is actually the voltage drop across our 10 $\Omega$ fault at C.
(2) When B is applying a voltage and A is measuring, likewise, they just end up measuring the voltage drop across the fault at C.

If you feel like that already helped, go think some more on the problem. In case it didn't: when measuring the voltage drop across the fault, it's like a voltage divider. It's a simple circuit now with the same current flowing through everything.
The fraction of the total voltage drop across the fault is the same as the fraction of the total resistance.

 

[haruspex]

Ground has always been a very confusing concept for me. To now have current flowing through two different grounds is pretty antithetical to the entire idea ^^

Yes, I now think my suggestion that the resistance through ground would be proportional to distance is quite wrong. Most of the resistance would be in the immediate vicinity of the contacts. In terms of the diagram in post #9, that makes RE1 and RE2 equal.
We do have to take them as negligible to solve the problem.

 

[Tazerfish]

Yes, I now think my suggestion that the resistance through ground would be proportional to distance is quite wrong. Most of the resistance would be in the immediate vicinity of the contacts. In terms of the diagram in post #9, that makes RE1 and RE2 equal.
We do have to take them as negligible to solve the problem.

Yeah, I recall once calculating the voltage drop across someone's feet when lightning strikes. We made a bunch of unrealistic assumptions, treating the conductivity of the ground as constant and isotropic.

Then, when you force current into a point, it spreads out radially, with the current density dropping off with an inverse square law. That way nearly all the resistance exists in the direct vicinity of the fault, just like you described.

But I think that's way too complicated. For the homework problem, a linear model for the "through the dirt–resistance" seems reasonable.

Also, I can't imagine the dirt being a better conductor of electricity than the line. If we assume the dirt-resitance is negligible, then we have to conclude that the 25V remaining drop when applying 30V happens in the transmission line, corresponding to a transmission line resistance of 50 Ohm.
I just looked up typical transmission line resistance values. 50 Ohm are roughly equivalent to 500km medium transmission line or 10,000km HV transmission line. That doesn't seem right.
I'd assume most of the resistance is from the dirt path.

But who knows what's going through the heads of the exercise-makers