# 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.

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

### Right Answer is: 1.92 sec

#### SOLUTION

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

Pulse setting Multiplier PSM is given as

PSM = Fault current/ Relay current setting × CT Ratio

= 4000/6.25 x 80 = 8 Amp

After calculating the value of PSM we have to find out the total time of operation of the relay with the help of the Time / PSM curve and for the standard 2.2-sec curve the operating time for PSM = 8  is 3.2 sec

Now in the above question, the value of TSM is given which is 0. 6 sec

Then the actual operating time of the relay is PSM × TSM = 3.2 x 0.6= 1.92 sec

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