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Locating phase-to-phase cable fault

Short answer

Use the Murray loop test with one sound core. Link the sound core and one faulty core at the far end, connect a Wheatstone bridge with ratio arms P and Q, and apply the battery between P–Q junction and the other faulty core. At balance, fault distance from test end is x = (Q/(P+Q)) × 2L.

Explain one of the techniques to locate phase to phase fault on underground power cable.

A phase-to-phase (short-circuit) fault between two cores of a multi-core cable can be located by the Murray loop test, a Wheatstone bridge method that needs one sound core.

Connections

   +--------------(G)--------------+
   |                               |
   c ---/\/\------+------/\/\----- a
   |     P        |        Q       |
   |           battery             |
   |              |                |
   |           core 2              |
 sound core 3                faulty core 1
   |                               x  <- fault (core 1 to core 2)
   |                               |
   +====== link at far end ========+
  1. Core 1 and core 2 are shorted together at the fault. Core 3 is sound.
  2. At the far end, link core 1 to sound core 3 to form a loop.
  3. Connect ratio arm PP to the test end c of core 3 and ratio arm QQ to the test end a of core 1, and connect the galvanometer across a and c.
  4. Connect the battery between the junction of PP and QQ and core 2 (the other shorted core). Core 2 takes the place of earth in the earth-fault test. Current then flows through the fault into the loop.
  5. Adjust PP and QQ for zero galvanometer deflection.

Calculation

Let RxR_x be the resistance of core 1 from the test end to the fault, and RLR_L the whole loop resistance (core 1 + core 3). The arm next to PP is RL−RxR_L - R_x and the arm next to QQ is RxR_x. At balance:

PQ=RL−RxRx  ⇒  Rx=QP+QRL\frac{P}{Q} = \frac{R_L - R_x}{R_x} \;\Rightarrow\; R_x = \frac{Q}{P+Q}R_L

For cores of equal size and cable length LL:

x=QP+Q×2Lx = \frac{Q}{P+Q}\times 2L

Example

A 3 km cable balances at P=2.5 ΩP = 2.5\ \Omega and Q=1 ΩQ = 1\ \Omega:

x=13.5×6=1.714 kmx = \frac{1}{3.5}\times 6 = 1.714\ \text{km}

The fault resistance lies in the battery circuit, so it does not affect the balance. The Varley loop test can be used in the same way when fine adjustment by a series resistor is preferred.

✓ Answer checked by Claude Opus 5.5 (high effort)

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