---
title: "Varley loop test for cable faults"
subject: "Power System Analysis I"
chapter: 2
key: "semester_4/power_system_analysis_i:varley-loop-test"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/varley-loop-test
site: BCT Engineering
---

# Varley loop test for cable faults

Power System Analysis I · Chapter 2: Overhead and Underground Transmission

- Asked in: 2070 Bhadra, 6 marks
- Video: [Varley Loop Test || Power System Analysis](https://www.youtube.com/watch?v=A5VHZaW2vvI) — Rojib EEE Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/varley-loop-test
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-2

## Question

How can underground cable faults be located by the Varley loop test method? Explain with necessary figures and mathematical expressions.

## Short answer

The Varley loop test links faulty and sound cores at the far end. With bridge ratio arms P and Q, first measure loop resistance RL = (Q/P)S1. Then with battery to earth, rebalance to get S2; fault distance from test end X = (P·RL – Q·S2)/(P+Q). When P=Q, X = (RL – S2)/2.

## Answer

The Varley loop test is a Wheatstone bridge method for locating earth faults and short-circuit faults in underground cables. It differs from the Murray loop in that the ratio arms $P$ and $Q$ are fixed, a variable resistor $S$ is adjusted for balance, and the loop resistance is measured in a separate step.

#### Connections
```text
                battery + switch K
                       |
           P           B           Q
   A ----/\/\----------+---------/\/\---- C
   |                                      |
   +-----------------(G)------------------+
   |                                      |
   S (variable)                           |
   |                                      |
   D  <- K position 1                     |
   |                                      |
 faulty core (X to fault)          sound core (Y)
   |                                      |
   +======== link at far end =============+
   K position 2: earth (current enters through the fault)
```

1. Link the faulty core and the sound core at the far end to form a loop.
2. Connect the fixed ratio arms $P$ and $Q$, with the variable resistor $S$ between A and the test end D of the faulty core. Connect the galvanometer across A and C.
3. The battery is connected to the P–Q junction B. A two-way switch K connects its other terminal either to D, the test end of the loop (position 1), or to earth (position 2).

#### Step 1: loop resistance (K at position 1)
The bridge is an ordinary Wheatstone bridge, with $S$ in one arm and the whole loop $R_L = X + Y$ in the other. Adjust $S$ to $S_1$ for balance:

$$
\frac{P}{Q} = \frac{S_1}{R_L} \;\Rightarrow\; R_L = \frac{Q}{P}S_1
$$

#### Step 2: fault position (K at position 2, battery to earth)
The current now enters the loop through the fault. One arm is $S + X$ (variable resistor plus faulty core up to the fault), and the other arm is $Y = R_L - X$. Rebalance with $S = S_2$:

$$
\frac{P}{Q} = \frac{S_2 + X}{R_L - X}
$$

$$
P R_L - P X = Q S_2 + Q X
\;\Rightarrow\;
X = \frac{P R_L - Q S_2}{P + Q}
$$

With $P = Q$ (the usual case):

$$
X = \frac{R_L - S_2}{2}
$$

#### Distance to the fault
If $r$ is the conductor resistance per metre:

$$
x = \frac{X}{r} \quad\text{or}\quad x = \frac{X}{R_L}\times 2L
$$

where $L$ is the cable length (both cores of equal size).

#### Example
With $P = Q$, $R_L = 2\ \Omega$ for a 2 km cable (4 km of loop), and $S_2 = 0.6\ \Omega$:

$$
X = (2 - 0.6)/2 = 0.7\ \Omega,\qquad x = 0.7/2\times 4 = 1.4\ \text{km}
$$

The variable resistor allows finer balance than the Murray loop. The fault resistance is in the battery circuit, so it does not affect the result.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The Varley loop test is based on
   A. A Wheatstone bridge with fixed ratio arms P and Q and a variable resistor S
   B. A potentiometer with a sliding contact
   C. A Schering bridge
   D. Time-domain reflection of pulses
   Answer: A. It is a Wheatstone bridge method; P and Q are fixed and S is varied to balance. (Key point: Principle)

2. With switch K at position 1 (loop), what is found at balance?
   A. Fault resistance
   B. Loop resistance R_L = (Q/P) S1
   C. Distance to the fault directly
   D. Insulation resistance
   Answer: B. P/Q = S1/R_L gives the loop resistance R_L = (Q/P) S1. (Key point: Step 1)

3. Why does the fault resistance not affect the Varley loop result?
   A. It is in the battery circuit, not in a bridge arm
   B. It is made zero by the link
   C. It is cancelled by P = Q
   D. It is included in S
   Answer: A. The fault lies in the battery path, so it does not alter the bridge balance. (Key point: Fault resistance effect)

4. With P = Q and K at position 2, the resistance X of the faulty core up to the fault is
   A. (R_L + S2)/2
   B. R_L - S2
   C. (R_L - S2)/2
   D. S2 - R_L
   Answer: C. From P/Q = (S2+X)/(R_L-X) with P=Q, X = (R_L - S2)/2. (Key point: Fault position)

## Related in Power System Analysis I

- [Murray loop test for cable faults](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Amurray-loop-test) — Power System Analysis I, chapter 2, asked 5×
- [Murray loop test for open and short circuit faults](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Amurray-loop-open-short-circuit) — Power System Analysis I, chapter 2, asked 5×
- [Cable fault location techniques and Murray loop](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Acable-fault-techniques-murray) — Power System Analysis I, chapter 2
- [Locating faults in underground cables](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Acable-fault-location-method) — Power System Analysis I, chapter 2, asked 3×

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- [Advantages of closed loop with mathematical support](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Acl-advantages-math) — Control Systems, chapter 1
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