Finding the Resistance of a Fault in a Pair of Transmission Lines

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SUMMARY

The discussion focuses on determining the resistance of a fault (R5) in a pair of transmission lines with differing resistivities. Using a potential difference of 1.5 V, the forum participants derived equations based on the measured currents across various terminals (AC, BD, AB, and CD). The fault is identified as R5, which represents the insulation failure between the cables. The fault occurs at a distance of 15 meters from points A and B, indicating that the insulation failure affects the current flow in a predictable manner.

PREREQUISITES
  • Understanding of Ohm's Law and basic circuit theory
  • Familiarity with resistive wire characteristics
  • Knowledge of voltage dividers and potentiometers
  • Ability to formulate and solve equations involving resistors in parallel
NEXT STEPS
  • Study the principles of resistive wire and its characteristics
  • Learn how to analyze circuits with faults using Kirchhoff's laws
  • Explore the concept of voltage dividers in more depth
  • Investigate methods for measuring resistance in practical applications
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in troubleshooting transmission line faults will benefit from this discussion.

Seneka
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Homework Statement


https://www.bpho.org.uk/user/pages/05.past-papers/06.as-challenge/_general/BPhO_AS_2010_QP.pdf

12. A combination of resistors shown below represents a pair of transmission lines with a fault in
the insulation between them. The wires have a uniform resistance, but do not have the same
resistance as each other. The following procedure is used to find the value of the resistance R5.
upload_2018-10-27_9-59-11.png
A potential difference of 1.5 V is connected in turn across various points in the arrangement.

With 1.5 V applied across terminals AC a current of 37.5 mA flows
With 1.5 V applied across terminals BD a current of 25 mA flows
With 1.5 V applied across terminals AB a current of 30 mA flows
With 1.5 V applied across terminals CD a current of 15 mA flows

a) Write down four equations relating the potential difference, the resistor values and the
currents.

b) Determine the value of resistor R5.

c) If the ends C and D are connected together, what would be the resistance measured between
A and B?

d) If the length AC (and also BD) is 60 metres of resistive wire, how far from A (or C) does the
fault occur?

2. Homework Equations

The Attempt at a Solution


I was able to do all the questions up to part d after which I just blanked. I don't know where or how to approach this question and I'm not sure if it is a gap in my knowledge. Please help!
 

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What are the characteristics of resistive wire? Do you recall voltage dividers? How about potentiometers?
 
gneill said:
What are the characteristics of resistive wire? Do you recall voltage dividers? How about potentiometers?

Do you want me to recall R=(resistivity X length) / Area
V out = V in X (R1/R1+R2)
Can you explain the question to me please. I don't understand what a fault is and how the circuit above models the situation. Is R5 supposed to represent the fault?
 
Yes R5 is the fault. There are two parallel cables with differing resistivities. Somewhere along their length a fault has developed in the insulation that separates the cables. The important thing to know is that the fault occurs at the same location (distance along the cable) for both cables.
 
gneill said:
Yes R5 is the fault. There are two parallel cables with differing resistivities. Somewhere along their length a fault has developed in the insulation that separates the cables. The important thing to know is that the fault occurs at the same location (distance along the cable) for both cables.

Okay I'm starting to see where this is going. Since AC and BD are equal. R5 has split that length in the same ratio so R1/R2=R3/R4 and you can solve it that or way or because R1+R3=25 and R3+R4= 75 you can see that the fault is a quarter of the way/15m from A/B. I get that but what is the purpose of R5. What is a fault?
 
Seneka said:
Okay I'm starting to see where this is going. Since AC and BD are equal. R5 has split that length in the same ratio so R1/R2=R3/R4 and you can solve it that or way or because R1+R3=25 and R3+R4= 75 you can see that the fault is a quarter of the way/15m from A/B. I get that but what is the purpose of R5. What is a fault?
A fault is any unexpected situations that arise in a circuit that were not part of the system's design or part of normal operation. In this case the insulation between the cables has deteriorated or failed in some fashion causing an unwanted bridging of the cables at the location of the fault.
 
gneill said:
A fault is any unexpected situations that arise in a circuit that were not part of the system's design or part of normal operation. In this case the insulation between the cables has deteriorated or failed in some fashion causing an unwanted bridging of the cables at the location of the fault.

Okay and the purpose of R5 was to be able to link the resistors to formulate equations.
 
Seneka said:
Okay and the purpose of R5 was to be able to link the resistors to formulate equations.
More or less. R5 represents the fault. When insulation fails it usually doesn't produce a zero-resistance path. Typically it's some charred (carbonized) material with perhaps water infiltration, too.
 
gneill said:
More or less. R5 represents the fault. When insulation fails it usually doesn't produce a zero-resistance path. Typically it's some charred (carbonized) material with perhaps water infiltration, too.
Ah okay thanks!
 

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