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Question:

Determine the time of operation of a 5 amp. , 3 sec. overcurrent relay having a current setting of 150 percent and a time setting multiplier of 0.8 connected to supply circuit through a 500 /5 current transformer when the circuit carries a fault current of 5000 Amp.(If plug setting multiplier comes to be less then 5 , then time operation is 2.5 second and if plug setting multiplier comes to be less then 5, then time of operation is 3.5 secondes.)

Answers:

prince_s

Expert

Karma Points: 981

Date Posted: 7/10/2008 3:18:12 AM Status: Live

Asker's Rating: None Provided Moderator's Rating: N/A

Response:

Relay rated secondary current of C.T =5A

Pickup current =5*1.50=7.5

Fault Relay current =

Plug Settig Multiplier(P.S.M)=

In the problem you are both P.S.M are less than 5. but one should be greater than 5.that must be for 2.5sec because P.S.M curve inversly propotional to time

So if P.S.M< 5 then t=3.5sec

P.S.M>5 then t=2.5sec

From the above condition if P.S.M is 6.67 then time of operation is 2.5

so the Relay time of operation is

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