---
title: "IDMT settings for sectionalised radial feeder A, B, C"
subject: "Switchgear and Protection"
chapter: 7
key: "semester_6/switchgear_and_protection:sectionalised-feeder-idmt-abc"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/sectionalised-feeder-idmt-abc
site: BCT Engineering
---

# IDMT settings for sectionalised radial feeder A, B, C

Switchgear and Protection · Chapter 7: Protective Relays

- Asked in: 2074 Bhadra, 8 marks
- Video: [Time Graded Protection System_Radial Feeder](https://www.youtube.com/watch?v=CLsp8VaA3vE) — ELECTRICAL PERSPECTIVE (ELECTRICAL PERSPECTIVE)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/sectionalised-feeder-idmt-abc
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-7

## Question

IDMT relays for protection of sectionalised radial feeder shown in figure below and the method of selection of their settings are given below. Time setting multiplier is 1.
[Figure: Source – bus A – CB_A, CT_A, relay R_A – bus B – CB_B, CT_B, relay R_B – bus C – CB_C, CT_C, relay R_C – feeder end.]

| Relay Point | CT ratio | Fault current | Current setting |
|---|---|---|---|
| A | 400/5 | 4000 A | 100% |
| B | 300/5 | 3000 A | 75% |
| C | 200/5 | 2000 A | 50% |

IDMT Characteristics:

| PSM | 2 | 4.2 | 6.5 | 8.8 | 10 | 12.4 | 13.3 | 16.3 | 18 | 20 |
|---|---|---|---|---|---|---|---|---|---|---|
| Time in sec | 8 | 4.2 | 3.8 | 3.1 | 2.8 | 2.6 | 2.5 | 2.4 | 2.3 | 2.2 |

## Short answer

Relay C TMS 0.1 gives 0.22 s; B TMS 0.234 gives 0.585 s; A TMS 0.310 gives 0.869 s, based on grading margin 0.5 s and IDMT curve. Pick-up: C 100 A, B 225 A, A 400 A; PSM values: C 20, B 8.89 (or fault at C), A 7.5 (or fault at B).

## Answer

The relays are graded so that C (farthest from source) trips first, B backs up C, and A backs up B, with a fixed time margin.

**Assumptions:** curve given for TMS = 1; TMS of relay C = 0.1 (fastest relay); grading margin = 0.5 s; CT secondary 5 A; linear interpolation of the table.

```text
Pick-up = Current setting × CT primary
PSM = I_fault / Pick-up ;  t = TMS × t(curve)
```

#### Relay C

```text
Pick-up = 0.5 × 200 = 100 A
PSM = 2000/100 = 20 → 2.2 s
t_C = 0.1 × 2.2 = 0.22 s
```

#### Relay B

```text
Pick-up = 0.75 × 300 = 225 A
Fault at C: PSM = 2000/225 = 8.89
  t = 3.1 − (8.89−8.8)/(10−8.8) × 0.3 = 3.078 s
Required t_B = 0.22 + 0.5 = 0.72 s
TMS_B = 0.72/3.078 = 0.234
Fault at B: PSM = 3000/225 = 13.33 → 2.499 s
t_B = 0.234 × 2.499 = 0.585 s
```

#### Relay A

```text
Pick-up = 1.0 × 400 = 400 A
Fault at B: PSM = 3000/400 = 7.5
  t = 3.8 − (7.5−6.5)/(8.8−6.5) × 0.7 = 3.496 s
Required t_A = 0.585 + 0.5 = 1.085 s
TMS_A = 1.085/3.496 = 0.310
Fault at A: PSM = 4000/400 = 10 → 2.8 s
t_A = 0.310 × 2.8 = 0.869 s
```

| Relay | Pick-up | TMS | Back-up time | Own-fault time |
|---|---|---|---|---|
| C | 100 A | 0.10 | – | 0.22 s |
| B | 225 A | 0.234 | 0.72 s (fault at C) | 0.585 s |
| A | 400 A | 0.310 | 1.085 s (fault at B) | 0.869 s |

**Answer:** TMS_C = 0.1, TMS_B ≈ 0.234, TMS_A ≈ 0.31; times for faults at their own sections are 0.22 s, 0.585 s and 0.869 s. If a different TMS for C or margin is used, the same steps apply.

## Check yourself

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

1. C: pick-up 100 A, fault 2000 A (PSM 20, 2.2 s), TMS 0.1. Its time is:
   A. 2.2 s
   B. 0.22 s
   C. 0.5 s
   D. 0.72 s
   Answer: B. 0.1 × 2.2 = 0.22 s. (Key point: Relay C time)

2. B: pick-up 0.75 × 300 = 225 A. For a fault at C (2000 A), PSM is about:
   A. 8.89
   B. 13.3
   C. 20
   D. 4.4
   Answer: A. 2000/225 = 8.89. (Key point: Relay B PSM for fault at C)

3. B must back up C with a 0.5 s margin. Required t_B for a fault at C is:
   A. 0.22 s
   B. 0.5 s
   C. 0.72 s
   D. 1.085 s
   Answer: C. 0.22 + 0.5 = 0.72 s. (Key point: Required time for B)

4. With curve time 3.078 s at PSM 8.89, TMS_B is:
   A. 0.72/3.078 = 0.234
   B. 0.22/3.078
   C. 0.5/3.078
   D. 0.72/2.2
   Answer: A. TMS = required time / curve time = 0.234. (Key point: TMS of B)

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