---
title: "IDMT grading: R1, R2 with 1000/5 CTs"
subject: "Switchgear and Protection"
chapter: 7
key: "semester_6/switchgear_and_protection:idmt-r1-r2-1000-5"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/idmt-r1-r2-1000-5
site: BCT Engineering
---

# IDMT grading: R1, R2 with 1000/5 CTs

Switchgear and Protection · Chapter 7: Protective Relays

- Asked in: 2071 Bhadra, 8 marks; 2080 Chaitra, 6 marks (asked 2 times)
- Video: [IDMT Relay setting calculation|TIME GRADATION|RELAY CO-ORDINATION](https://www.youtube.com/watch?v=k45WmROaFEY) — ELECTRICAL TECHNOLOGY and INDUSTRIAL PRACTICE
- 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-r1-r2-1000-5
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-7

## Question

Two relays R1 and R2 are connected in two sections of a feeder as shown in figure. CTs are of ratio 1000/5 A. The plug setting of relay R1 is 100% and R2 is 125%. The operating time characteristic of the relays is given below. The time multiplier setting of the relay R1 is 0.3. The time grading scheme has a discriminative time margin of 0.5 s between relays. A three phase short circuit at F results in a fault current of 5000 A. Find the actual operating times of R1 and R2. What is the time multiplier setting (TMS) of R2?

| Plug setting multiplier | 2 | 4 | 5 | 8 | 10 | 15 | 20 |
|---|---|---|---|---|---|---|---|
| Time in seconds for a time multiplier of 1 | 10 | 5 | 4 | 3 | 2.8 | 2.6 | 2.4 |

[Figure: Bus A – CB – CT 1000/5 with relay R2 – bus B – CB – CT 1000/5 with relay R1 – fault F (5000 A) – bus C.]

## Short answer

Relay current = 25 A. R1 PSM=5 gives 4 s at TMS=1, so t1=0.3×4=1.2 s. R2 pick-up=6.25 A, PSM=4 gives 5 s at TMS=1. Required t2=1.2+0.5=1.7 s, so TMS=1.7/5=0.34. R1 operates in 1.2 s, R2 in 1.7 s, TMS of R2 is 0.34.

## Answer

Each relay's operating time is found from its plug setting multiplier (PSM) using the given curve, then multiplied by the time multiplier setting (TMS). R2 (upstream) must operate 0.5 s after R1.

**Data:** CT ratio 1000/5 (ratio 200); relay rated current 5 A; fault current = 5000 A; R1: PS = 100%, TMS = 0.3; R2: PS = 125%; discrimination margin = 0.5 s.

#### Relay current
Both relays see the same fault current (the fault is beyond R1 and fed through R2):

```text
Relay current = 5000 / 200 = 25 A (secondary)
```

#### Relay R1
```text
Pick-up current = 100% × 5 = 5 A
PSM = 25 / 5 = 5
Time at TMS = 1 (from table, PSM 5) = 4 s
Actual time t₁ = 0.3 × 4 = 1.2 s
```

#### Relay R2
```text
Pick-up current = 125% × 5 = 6.25 A
PSM = 25 / 6.25 = 4
Time at TMS = 1 (from table, PSM 4) = 5 s
Required time t₂ = t₁ + 0.5 = 1.2 + 0.5 = 1.7 s
TMS of R2 = 1.7 / 5 = 0.34
```

#### Results

| Relay | PS | PSM | t (TMS=1) | TMS | Operating time |
|---|---|---|---|---|---|
| R1 | 100% | 5 | 4 s | 0.3 | 1.2 s |
| R2 | 125% | 4 | 5 s | 0.34 | 1.7 s |

**Answer:** R1 operates in **1.2 s**, R2 operates in **1.7 s**, and the TMS of R2 is **0.34**.

## Check yourself

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

1. What is the secondary relay current for 5000 A with CT 1000/5?
   A. 25 A
   B. 5 A
   C. 50 A
   D. 12.5 A
   Answer: A. CT ratio is 200, so 5000/200 = 25 A. (Key point: Relay current)

2. What is the plug setting multiplier of R1 (PS 100%, 5 A relay)?
   A. 2.5
   B. 4
   C. 5
   D. 10
   Answer: C. Pick-up 5 A, PSM = 25/5 = 5, giving 4 s at TMS 1, so t1 = 0.3 x 4 = 1.2 s. (Key point: PSM of R1)

3. What is the PSM of R2 with plug setting 125%?
   A. 5
   B. 3.2
   C. 6.25
   D. 4
   Answer: D. Pick-up = 1.25 x 5 = 6.25 A, PSM = 25/6.25 = 4, which gives 5 s at TMS 1. (Key point: PSM of R2)

4. What TMS must R2 have for 0.5 s discrimination from R1 (t1 = 1.2 s)?
   A. 0.30
   B. 0.34
   C. 0.24
   D. 0.50
   Answer: B. R2 must operate in 1.2 + 0.5 = 1.7 s, so TMS = 1.7/5 = 0.34. (Key point: TMS of R2)

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