Determine the location of a fault on a cable in which two


Question 1. (a) Using the Murray Loop test, determine the distance to an earth fault on one of the cores of a uniform three-core underground cable. Ra = 2 ohms, Rb = 1 ? and the cable length is 300 m.

(b) If the cable measurement is accurate to ±1% what length of excavation would be required to locate the fault?

Question 2. A cable run consists of 100 m of 120 mm2 three-core cable jointed to 100 m of 240 mm2 cable. The ratio of the potentiometer resistances in a Murray Loop test, Rb/(Rb + Ra), is 1/3.

For an earth on one core of the cables, determine the location of the fault.

Question 3. A Varley Bridge is connected to a faulty three-core copper cable by two identical copper leads of resistance Rl.

(a) Show that for the initial reading (connection to earth);

2Rx = 2Rc - Ri .............................. (1)

where Rc is the resistance of the cable core

Ri is the initial reading of the bridge

Rx is the cable resistance to the fault from the bridge and for the final reading:

2Rc = Rf - 2Rl .............................. (2)

where Rl is a lead resistance

Rf is the final reading resistance.

Then by substituting (2) in (1) and rearranging the equation, show:

Rx + Rl = ( Rf - Ri)/Rf (Rc + Rl)

(b) By multiplying the rhs brackets and collecting terms, show the effect of the leads is given by:

Rx = ((Rf-Ri)/Rf)Rc - RlRi/Rf

i.e. = effect with no leads - ratio of initial and final readings × lead resistance

(c) Determine the distance to the fault by modifying the expression in (b) and using

Rx/Rc = x/L

where x is the cable distance to the fault and L is the length of a cable core.

Derive an expression for x, the distance to the fault.

(d) Using R = ρL/A

where ρ is the resistivity,

L is the length

and A is the cross-sectional area of a cable core

show that Rl = RcAcLl/AlLc (where Lc = L)

and by substituting for Rl show that the distance to the fault is given by:

x =[(Rf - Ri)/Rf - Ri/Rf AcLl/AlL]L

(e) Determine the distance to a fault on a 200 m, 120 mm2 copper cable if copper 10 mm2 test leads of length 10 m are used and Ri/Rf = 0.2.

Question 4. FIGURE 1 shows a Wheatstone Bridge connection used to determine the location of a fault on a cable in which two cores are of different cross-sectional area (the Murray Fisher method). Ignoring test lead resistance, show that:

(a) For the first reading position, the resistance to the fault, Rx, is given by

Rx =  (Rb1/(Ra1 + Rb1))*(Rc1 + Rc2)

(b) For the second reading position and substituting for Rc1, the distance to the fault, x, is given by:

x = Rb1/(Ra1 + Rb1) × ((Ra2 + Rb2)/Rb2) × L

x = Rb1/Rb2 * [(Ra2 + Rb2)/(Ra1 + Rb1)] x L

(c) If the cable tested is 100 m in length, Ra1 is 1.5 ohms, Rb1 is 1.0 ohm and Ra2/Rb2 = 0.8, calculate the distance to the fault.

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FIG. 1

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Electrical Engineering: Determine the location of a fault on a cable in which two
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