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# Murray Loop Test and Locating Faults

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Hi there,

Q.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 ohm 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?

Q.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.

https://brainmass.com/engineering/electrical-engineering/murray-loop-test-locating-faults-516389

#### Solution Preview

Q.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 ohm 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?

Solution: (a) Murray loop test is based on the Wheatstone bridge principle.
Wheatstone bridge:

Ra Rc
Ammeter

Rb Rd

V

Wheatstone principle: When current through the ammeter is zero (balanced Wheatstone bridge), various resistors are related as follows: Ra/Rb = Rc/Rd
Murray loop test:
Length L = 300m Three core cable
A B
C Rc D
Ra E E Rd G F
x m (300-x) m
Earth fault
...

#### Solution Summary

The expert examines Murray loop tests and locating faults. A cable measurement accuracy is determined.

\$2.19

## Varley bridge/ Wheatstone bridge

The Varley loop test is a method of fault location in telecom and pilot wire circuits where fairly high circuit resistances are encountered. This solution determines the distance to the fault.

3. A Varley Bridge is connected to a faulty three-core copper cable by two identical copper leads of resistance R1.
(a) Show that for the initial reading (connection to earth);
2Rx = 2R, — Ri (1)
where R, 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
2R, = R f — 2R1 (2)
where R1 is a lead resistance
Rf is the final reading resistance.

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

(see the PDF for correct calculations)

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

(see the PDF for correct calculations)

i.e. = effect with no leads — ratio of initial and final readings

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

(see the PDF for correct calculations)

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 = pi, A
where p is the resistivity,
L is the length
and A is the cross-sectional area of a cable core

(see the PDF for correct calculations)

and by substituting for R1 show that the distance to the fault is given by:

(see the PDF for correct calculations)

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

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