---
title: "Minimum TMS of relay A (16000 A fault)"
subject: "Switchgear and Protection"
chapter: 7
key: "semester_6/switchgear_and_protection:minimum-tms-relay-a"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/minimum-tms-relay-a
site: BCT Engineering
---

# Minimum TMS of relay A (16000 A fault)

Switchgear and Protection · Chapter 7: Protective Relays

- Asked in: 2077 Chaitra, 6 marks
- Video: [PSM and TMS Numericals](https://www.youtube.com/watch?v=1Za43V_-9tM) — Basic Electrical
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/minimum-tms-relay-a
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-7

## Question

As shown in figure, the relay at B has a plug setting of 125% and TMS of 0.2. The circuit breakers take 0.20 sec to clear the fault and the relay error in each case is 0.15 sec. For a plug setting of 200% on the relay A, determine the minimum TMS on that relay for it not to operate before the circuit breaker at B has cleared the fault. A relay operating time curve is shown in table.
[Figure: Bus A – CB – CT 800/5 with relay R2 – bus B – CB – CT 800/5 with relay R1 – fault F, If = 16000 A.]

IDMT characteristic:

| PSM | 2 | 3.5 | 5 | 6.5 | 8 | 10 | 12 | 14 | 16 |
|---|---|---|---|---|---|---|---|---|---|
| Operating Time (sec) | 10 | 6 | 5 | 4 | 3.15 | 3 | 2.9 | 2.5 | 2 |

## Short answer

Relay B: PSM=16000/(1.25×800)=16, t=0.2×2=0.4 s. Required t_A includes breaker time 0.2 s and errors ±0.15 s each: t_A=0.4+0.2+0.15+0.15=0.9 s. Relay A pickup 1600 A, PSM=10, t(TMS=1)=3 s, so minimum TMS_A=0.9/3=0.3.

## Answer

Relay A is the back-up for relay B. It must not trip until relay B has operated **and** breaker B has cleared the fault, allowing for errors in both relays.

**Data:** CT 800/5 for both; relay B: PS 125%, TMS 0.2; relay A: PS 200%; If = 16000 A; CB time 0.2 s; relay error 0.15 s each.

#### Relay B

```text
Pick-up = 1.25 × 800 = 1000 A
PSM = 16000/1000 = 16 → t(TMS=1) = 2 s
t_B = 0.2 × 2 = 0.4 s
```

#### Required time of relay A

Relay B may be 0.15 s slow and relay A may be 0.15 s fast, so:

```text
t_A = t_B + CB time + error(B) + error(A)
    = 0.4 + 0.2 + 0.15 + 0.15 = 0.9 s
```

#### TMS of relay A

```text
Pick-up = 2.0 × 800 = 1600 A
PSM = 16000/1600 = 10 → t(TMS=1) = 3 s
TMS_A = 0.9 / 3 = 0.3
```

**Answer:** minimum TMS of relay A = 0.3 (operating time 0.9 s). Any smaller TMS could let relay A trip before breaker B has cleared the fault.

## Check yourself

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

1. Relay B (PS 125%, CT 800/5, TMS 0.2) sees 16000 A. Its operating time is:
   A. 0.4 s
   B. 2 s
   C. 0.2 s
   D. 0.9 s
   Answer: A. Pick-up = 1000 A, PSM = 16 gives 2 s at TMS 1, so t_B = 0.2 x 2 = 0.4 s. (Key point: Relay B time)

2. Which sum gives the minimum operating time of back-up relay A?
   A. t_B + CB time + error(B) + error(A) = 0.9 s
   B. t_B + CB time = 0.6 s
   C. t_B + error(A) = 0.55 s
   D. CB time + errors only = 0.5 s
   Answer: A. A must wait for B to operate (possibly 0.15 s slow), the breaker to clear, and A may be 0.15 s fast: 0.4+0.2+0.15+0.15 = 0.9 s. (Key point: Required time of relay A)

3. Relay A has a 200% plug setting on CT 800/5. Its PSM at 16000 A is:
   A. 16
   B. 20
   C. 10
   D. 8
   Answer: C. Pick-up = 1600 A, so PSM = 16000/1600 = 10, giving 3 s at TMS 1. (Key point: PSM of relay A)

4. The minimum TMS of relay A is:
   A. 0.45
   B. 0.3
   C. 0.9
   D. 0.2
   Answer: B. TMS_A = 0.9/3 = 0.3; any smaller value could let A trip before breaker B has cleared. (Key point: Minimum TMS)

## Related in Switchgear and Protection

- [Relay 1 time and relay 2 TMS (3000 A)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aidmt-300-1-3000a) — Switchgear and Protection, chapter 7
- [Percentage differential relay: 125 MVA transformer](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Apercentage-differential-125mva) — Switchgear and Protection, chapter 7
- [Voltage-monitored overcurrent relay timing](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Avoltage-monitored-overcurrent) — Switchgear and Protection, chapter 7
- [IDMT settings for 20 MVA transformer and feeder](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Atransformer-feeder-idmt-20mva) — Switchgear and Protection, chapter 7, asked 2×
- [IDMT relays across 132/11 kV transformer (3500 A)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aidmt-132-11kv-3500a) — Switchgear and Protection, chapter 7
- [Earth fault protection: 20 MVA alternator](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aearth-fault-20mva-5-ohm) — Switchgear and Protection, chapter 7

## Related in other subjects

- [SP, DP and TP miniature circuit breakers](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Asp-dp-tp-mcb) — Industrial Power Distribution and Illumination, chapter 3
- [Weir waterway and crest levels, 16000 m3/s](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Aweir-waterway-crest-q16000) — Irrigation and Drainage Engineering, chapter 5
- [Sine bar principle, setting and error](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Asine-bar-setting-and-error) — Metrology, chapter 5
- [Fault level at end of two 200 km lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-two-stations-132kv-200km) — Power System Analysis II, chapter 3, asked 2×