## 28/02/2021

### Murray Loop Test

The Murray loop test is useful to find out location of underground fault for earth fault and short circuit.

Earth fault

Figure shows the circuit diagram for location of underground fault in the cable.

Let

AB = Sound cable ( without fault )

CD = Faulty cable

P = Variable resistance

Q = Variable resistance

B = Battery ( DC Voltage )

G = Galvanometer

l = length of each cable

2r = Total resistance of sound cable and faulty cable

K1 = Switch

d = Distance of faulty cable from point E =  ?

• Let us consider that the earth fault occurs at a distance x from point E.
•  The far end C of faulty cable is joined in series with the sound cable by low resistance link.
• There are two variable resistance P and Q connected to cable end A and E respectively.
• This forms PQXR Wheatstone bridge. The four arm of the Wheatstone bridge is

P = Variable resistance

Q = Variable resistance

X = Resistance of cable from point E to point D

R = Resistance of the cable from test end to point D

• The variable resistance P and Q are varied till the galvanometer shows zero deflection after switch K and K1 is closed. For balance conditions

PX = RQ

P / Q = R / X

( P + Q ) / Q = ( R + X ) / X

Here R + x = 2r

( P + Q ) / Q = 2r / X

X = 2r ( P + Q ) / Q …….. ( 1 )

If l is the length of each cable, the resistance per meter length of cable = r / l

Distance of faulty cable from point E is

d = X / ( r / l ) ……… ( 2 )

From equation ( 1 ) and ( 2 )

d = { 2r ( P + Q ) / Q }  / ( r / l )

d = { 2r ( P + Q ) l / r Q }

d = { 2( P + Q ) l /  Q }

d = 2l { ( P + Q )  /  Q }

d = { ( P + Q )  /  Q } × 2l

d = { ( P + Q )  /  Q } × Loop length

d = { ( P + Q )  /  Q } × Length of faulty cable and length of sound cable

• The fault resistance does not affect the balancing of the bridge circuit.
• If the fault resistance is high, the sensitivity of the bridge is reduced.

Short circuit fault

• The Murray loop test is used to find out location of fault point in the underground cable for short circuit fault.
• Let the P, Q , R and X are the arms of the Wheatstone bridge.
• Note that the fault resistance is in the battery B circuit not in the bridge circuit.
• The balancing of bridge is done by vary the ratio arms P and Q till the galvanometer shows zero reading after closing the switch K and K1.

Let

P = Variable resistance

Q = Variable resistance

X = Resistance of cable from point E to point D

R = Resistance of the cable from test end to point D

PX = RQ

P / Q = R / X

( P + Q ) / Q = ( R + X ) / X

Here R + x = 2r

( P + Q ) / Q = 2r / X

X = 2r ( P + Q ) / Q

X =  { ( P + Q ) / Q } × Loop length

X = { ( P + Q )  /  Q } × Length of faulty cable and length of sound cable

• The location of fault is find out by adjusting ratio arms P and Q.

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