---
title: "IDMT relays across 132/33 kV transformer (2700 A)"
subject: "Switchgear and Protection"
chapter: 7
key: "semester_6/switchgear_and_protection:idmt-132-33kv-2700a-graph"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/idmt-132-33kv-2700a-graph
site: BCT Engineering
---

# IDMT relays across 132/33 kV transformer (2700 A)

Switchgear and Protection · Chapter 7: Protective Relays

- Asked in: 2075 Bhadra, 4 marks
- 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-132-33kv-2700a-graph
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-7

## Question

It is given that fault current level at 33 kV side is 2700 A, CT ratio at 33 kV side is 200:1 and 132 kV side is 100:1 (figure below). If both the relays R1 and R2 are set for 100% plug setting, determine the operating time for both the relays when time grading margin of 0.6 second is given and TMS for relay R1 is 0.15. (PSM/Operation Time graph is given below.)
[Figure: 132 kV bus – CB – CT 100/1 with relay R2 – 132/33 kV transformer – 33 kV bus – CB – CT 200/1 with relay R1 – fault, IF = 2700 A.]
[Graph: operating time (seconds, 0–12) against PSM (0–20) for TMS = 1, a falling inverse curve; read approximately as PSM 2 → about 11.5 s, 4 → about 5.5 s, 6 → about 4 s, 10 → 3 s, 14 → about 2.4 s, 20 → about 2.1 s.]

## Short answer

R1 operating time ≈ 0.37 s, R2 ≈ 0.97 s, TMS of R2 ≈ 0.255. From graph, R1 PSM=13.5 gives t=2.475 s at TMS=1; R2 PSM=6.75 gives t=3.81 s. Fault current on 132 kV side is 675 A.

## Answer

R1 (33 kV) is the primary relay; R2 (132 kV) backs it up with a grading margin of 0.6 s.

```text
R1: pick-up = 1.0 × 200 = 200 A
    PSM = 2700/200 = 13.5
    From graph (between 10→3 s, 14→2.4 s): t ≈ 2.475 s
    t_R1 = 0.15 × 2.475 = 0.371 s

R2: I(132 kV) = 2700 × 33/132 = 675 A
    pick-up = 1.0 × 100 = 100 A
    PSM = 675/100 = 6.75
    From graph (between 6→4 s, 10→3 s): t ≈ 3.81 s
    t_R2 = 0.371 + 0.6 = 0.971 s
    TMS_R2 = 0.971 / 3.81 = 0.255
```

**Answer:** t_R1 ≈ 0.37 s and t_R2 ≈ 0.97 s, with TMS of R2 ≈ 0.255. These values depend on how the graph is read, so they are approximate (±0.05 s).

## Check yourself

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

1. R1: CT 200/1, 100% plug, fault 2700 A. PSM is:
   A. 13.5
   B. 27
   C. 6.75
   D. 2.7
   Answer: A. 2700/200 = 13.5. (Key point: R1 PSM)

2. With curve time about 2.475 s and TMS 0.15, t_R1 is:
   A. 2.475 s
   B. 0.371 s
   C. 0.15 s
   D. 0.971 s
   Answer: B. 0.15 × 2.475 = 0.371 s. (Key point: R1 time)

3. R2 sees 2700 × 33/132 = 675 A with pick-up 100 A. PSM is:
   A. 27
   B. 6.75
   C. 13.5
   D. 2.7
   Answer: B. 675/100 = 6.75. (Key point: R2 PSM)

4. R2 needs 0.371 + 0.6 = 0.971 s, and the graph gives about 3.81 s. TMS_R2 is:
   A. 0.255
   B. 0.6
   C. 0.971
   D. 0.15
   Answer: A. 0.971/3.81 = 0.255. (Key point: R2 TMS)

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