---
title: "Time-current grading for relays X, Y, Z"
subject: "Switchgear and Protection"
chapter: 7
key: "semester_6/switchgear_and_protection:idmt-grading-xyz"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/idmt-grading-xyz
site: BCT Engineering
---

# Time-current grading for relays X, Y, Z

Switchgear and Protection · Chapter 7: Protective Relays

- Asked in: 2072 Magh, 7 marks
- Video: [C2 Overcurrent Protection: Overcurrent Time Grading](https://www.youtube.com/watch?v=wl16y_DhTTg) — iMooX at
- 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-grading-xyz
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-7

## Question

IDMT characteristics of the relay are shown in figure below. Design the time-current grading of the system as given below. TSM at the relay Z is 0.1. Time setting multiplier is 1.
[Figure: Source – relay X – relay Y – relay Z along a radial feeder.]

| Relay Point | C.T. ratio | Fault current | Current Setting |
|---|---|---|---|
| X | 400/5 | 5000 A | 125% |
| Y | 200/5 | 4000 A | 125% |
| Z | 200/5 | 2000 A | 100% |

IDMT Characteristic:

| PSM | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 |
|---|---|---|---|---|---|---|---|---|---|---|
| Time in sec | 8 | 4 | 3.6 | 3 | 2.8 | 2.6 | 2.4 | 2.2 | 2 | 2 |

## Short answer

Relay Z: pick-up 200 A, PSM=10, TMS=0.1 yields 0.28 s. Relay Y: pick-up 250 A, for fault at Z PSM=8, required 0.78 s → TMS=0.26; own fault 0.572 s. Relay X: pick-up 500 A, for fault at Y required 1.072 s → TMS=0.357; own fault 1.0 s. Grading margin 0.5 s.

## Answer

Time-current grading means setting each IDMT relay so that the relay nearest the fault trips first, and every relay towards the source waits one grading margin longer.

**Assumptions:** the given curve is for TMS = 1; the TMS of relay Z is 0.1; grading margin between adjacent relays = 0.5 s (CB time + relay overshoot + errors); CT secondary = 5 A; linear reading of the curve between points.

**Formulas**

```text
Pick-up current (primary) = Current setting × CT primary
PSM = Fault current / Pick-up current
t = TMS × t(TMS=1, read from curve)
```

#### Relay Z (farthest from source)

```text
Pick-up = 1.00 × 200 = 200 A
PSM     = 2000 / 200 = 10   → t(TMS=1) = 2.8 s
t_Z     = 0.1 × 2.8 = 0.28 s
```

#### Relay Y (back-up for Z)

For a fault at Z (2000 A), Y must wait 0.5 s longer than Z.

```text
Pick-up = 1.25 × 200 = 250 A
PSM (fault at Z) = 2000 / 250 = 8 → 3.0 s
Required t_Y = 0.28 + 0.5 = 0.78 s
TMS_Y = 0.78 / 3.0 = 0.26
Fault at Y (4000 A): PSM = 4000/250 = 16 → 2.2 s
t_Y = 0.26 × 2.2 = 0.572 s
```

#### Relay X (back-up for Y)

```text
Pick-up = 1.25 × 400 = 500 A
PSM (fault at Y) = 4000 / 500 = 8 → 3.0 s
Required t_X = 0.572 + 0.5 = 1.072 s
TMS_X = 1.072 / 3.0 = 0.357
Fault at X (5000 A): PSM = 5000/500 = 10 → 2.8 s
t_X = 0.357 × 2.8 = 1.0 s
```

#### Grading summary

| Relay | Pick-up (A) | TMS | Own-fault PSM | Own-fault time |
|---|---|---|---|---|
| Z | 200 | 0.10 | 10 | 0.28 s |
| Y | 250 | 0.26 | 16 | 0.572 s |
| X | 500 | 0.357 | 10 | 1.00 s |

```text
Source ──X──────Y──────Z────── load
time:   1.0 s   0.57 s  0.28 s   (own-fault times)
For the same fault, each relay is 0.5 s slower
than the next relay downstream.
```

**Answer:** TMS_Z = 0.1, TMS_Y = 0.26, TMS_X ≈ 0.357 (set 0.36); the operating times for faults at their own points are 0.28 s, 0.572 s and ≈ 1.0 s. A fault at Z is cleared by Z in 0.28 s; Y would back it up at 0.78 s and X later still, so discrimination is achieved.

## Check yourself

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

1. Plug setting multiplier (PSM) is:
   A. Pick-up current / fault current
   B. Fault current / pick-up current
   C. CT ratio / fault current
   D. TMS × fault current
   Answer: B. PSM = fault current divided by primary pick-up current. (Key point: PSM definition)

2. Z has pick-up 200 A, fault 2000 A, TMS 0.1. Its operating time is:
   A. 2.8 s
   B. 0.28 s
   C. 0.5 s
   D. 0.1 s
   Answer: B. PSM = 10 gives 2.8 s at TMS 1, so t = 0.1 × 2.8 = 0.28 s. (Key point: Relay Z time)

3. Y must back up Z by 0.5 s. For a fault at Z, PSM = 8 giving 3.0 s. TMS_Y is:
   A. 0.78 × 3.0
   B. 0.78/3.0 = 0.26
   C. 0.5/3.0
   D. 0.28/3.0
   Answer: B. Required time is 0.28 + 0.5 = 0.78 s, so TMS = 0.78/3.0 = 0.26. (Key point: Relay Y TMS)

4. In time-current grading the relay nearest the fault:
   A. Operates last
   B. Operates first, with upstream relays slower by a grading margin
   C. Has the highest TMS
   D. Is set identically to others
   Answer: B. Upstream relays act as back-up and wait one margin longer. (Key point: Grading principle)

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