---
title: "Operating time of two relays (250/1, 3000 A)"
subject: "Switchgear and Protection"
chapter: 7
key: "semester_6/switchgear_and_protection:idmt-250-1-3000a"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/idmt-250-1-3000a
site: BCT Engineering
---

# Operating time of two relays (250/1, 3000 A)

Switchgear and Protection · Chapter 7: Protective Relays

- Asked in: 2076 Bhadra, 6 marks
- Video: [Over current relay solved numerical problem](https://www.youtube.com/watch?v=LUQuUmMAERU) — Techspecial Information
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/idmt-250-1-3000a
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-7

## Question

Determine the time of operation of two relays assuming that both the relays have the characteristics as shown in the following table. The time multiplier setting of relay 1 is 0.18 and the time grading margin between the relays is 0.4 second.
[Figure: CB – CT 250/1 with Relay 2 (plug setting = 150%) – CB – CT 250/1 with Relay 1 (plug setting = 125%) – fault, If = 3000 A.]

| PSM | 2 | 3.6 | 5 | 9.23 | 10 | 15 | 20 |
|---|---|---|---|---|---|---|---|
| Time in sec | 10 | 6 | 3.9 | 2.9 | 2.8 | 2.2 | 2.1 |

## Short answer

Relay 1: pickup 312.5 A, PSM=3000/312.5=9.6, interpolated t(TMS=1)=2.852 s, t1=0.18×2.852=0.513 s. Relay 2: pickup 375 A, PSM=8, t(TMS=1)=3.191 s, required t2=0.513+0.4=0.913 s, TMS2=0.913/3.191=0.286.

## Answer

Relay 1 is nearer the fault, so it trips first; relay 2 must wait 0.4 s more. Times between table points are found by linear interpolation.

#### Relay 1

```text
Pick-up = 1.25 × 250 = 312.5 A
PSM = 3000/312.5 = 9.6
Between 9.23 (2.9 s) and 10 (2.8 s):
t = 2.9 − (9.6 − 9.23)/(10 − 9.23) × 0.1 = 2.852 s
t1 = 0.18 × 2.852 = 0.513 s
```

#### Relay 2

```text
Pick-up = 1.5 × 250 = 375 A
PSM = 3000/375 = 8
Between 5 (3.9 s) and 9.23 (2.9 s):
t = 3.9 − (8 − 5)/(9.23 − 5) × 1.0 = 3.191 s
t2 = t1 + 0.4 = 0.913 s
TMS2 = 0.913 / 3.191 = 0.286
```

| Relay | Pick-up | PSM | t at TMS=1 | TMS | Operating time |
|---|---|---|---|---|---|
| 1 | 312.5 A | 9.6 | 2.852 s | 0.18 | 0.513 s |
| 2 | 375 A | 8 | 3.191 s | 0.286 | 0.913 s |

**Answer:** t1 ≈ 0.513 s, t2 ≈ 0.913 s (TMS of relay 2 ≈ 0.286).

## Check yourself

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

1. For relay 1 (125%, CT 250/1, 3000 A fault), the PSM is:
   A. 8
   B. 9.6
   C. 12
   D. 10
   Answer: B. Pick-up = 1.25 x 250 = 312.5 A; PSM = 3000/312.5 = 9.6. (Key point: PSM of relay 1)

2. PSM 9.6 lies between 9.23 (2.9 s) and 10 (2.8 s). What method gives the time at TMS = 1?
   A. Linear interpolation, about 2.85 s
   B. Take the nearest value, 2.8 s
   C. Average of 2 and 20, 6 s
   D. Extrapolate from 5, 3.9 s
   Answer: A. The table has no 9.6 entry, so interpolate between neighbours to get about 2.852 s. (Key point: Interpolation)

3. What is the operating time of relay 2 when relay 1 operates in 0.513 s and the margin is 0.4 s?
   A. 0.4 s
   B. 0.913 s
   C. 0.113 s
   D. 3.191 s
   Answer: B. t2 = t1 + margin = 0.513 + 0.4 = 0.913 s. (Key point: Relay 2 operating time)

4. Relay 2 has PSM 8 (3.191 s at TMS 1). Its required TMS is about:
   A. 0.286
   B. 0.18
   C. 0.913
   D. 2.9
   Answer: A. TMS2 = required time / time at TMS 1 = 0.913/3.191 = 0.286. (Key point: TMS of relay 2)

## Related in Switchgear and Protection

- [Operating time of two relays (300/1, 6000 A)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aidmt-300-1-6000a) — Switchgear and Protection, chapter 7
- [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
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- [Time-current grading for relays X, Y, Z](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aidmt-grading-xyz) — Switchgear and Protection, chapter 7
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