# Determine the time of operation of a relay of rating 5 A, 2.2s IDMT and having a relay setting of 125%, TMS = 0.6. It is connected to the supply circu

85 views
in General
closed

Determine the time of operation of a relay of rating 5 A, 2.2s IDMT and having a relay setting of 125%, TMS = 0.6. It is connected to the supply circuit through a CT 400 : 5 ratio. The fault current is 4000 A.

 PSM 2 5 8 10 Time 10 8 3.2 2.5

1. 1.89 s
2. 1.87 s
3. 1.93 s
4. 1.92 s

by (48.1k points)
selected

Correct Answer - Option 4 : 1.92 s

Concept:

The plug setting multiplier of a relay is defined as the ratio of secondary fault current to the relay current setting.

PSM = Secondary fault current/Relay current setting

Calculation:

Relay CT ratio = 400/5

Therefore, pickup value of relay = 5 A

Since the relay setting is 125%

The operating current of relay = 5 × 1.25 = 6.25 A

Given that, primary fault current, Ipf = 4000 A

Therefore, the secondary fault current is

${I_{sf}} = {I_{pf}} \times \frac{1}{{CT.\;Ratio}} = \frac{{4000}}{{80}} = 50$

$PSM = \frac{{4000}}{{6.25\; \times \;80}}$

⇒ PSM = 8

For the standard IDMT curve of 2.2 sec, the operating time for PSM 8 is 3.2 sec

Since, TMS = 0.6

Therefore, actual operating time = 0.6 × 3.2 = 1.92 sec