×
Follow Us
Results 1 to 4 of 4

Time Overcurrent Relay 51

 Jump to latest post
    #1
  1. Join Date
    Jun 2016
    Posts
    28
    Reputation

    Time Overcurrent Relay 51

    A time overcurrent relay (51) served from a 500:5 current transformer is set on tap 5, time dial 10. The primary current necessary for the relay to pickup and start the timing cycle is:

    A 5000 amperes (5 x 10 x 100)
    B 2500 amperes (5 x 500)
    C 500 amperes (5 x 100)
    D 1000 amperes (10 x 100)

    Not sure how to do this, can someone help?

  2. #2
  3. Join Date
    Aug 2017
    Posts
    5
    Reputation
    Quote Originally Posted by jflan13 View Post
    A time overcurrent relay (51) served from a 500:5 current transformer is set on tap 5, time dial 10. The primary current necessary for the relay to pickup and start the timing cycle is:

    A 5000 amperes (5 x 10 x 100)
    B 2500 amperes (5 x 500)
    C 500 amperes (5 x 100)
    D 1000 amperes (10 x 100)

    Not sure how to do this, can someone help?
    500:5 = 100:1, you would need 5A on the secondary for the relay to pickup which means you need 500A. Assuming this is not a parallel feeder, 500A would be the max the CT could protect, which makes it a dead give away.

  4. #3
  5. Join Date
    Jun 2016
    Posts
    28
    Reputation
    Quote Originally Posted by seaborgium View Post
    500:5 = 100:1, you would need 5A on the secondary for the relay to pickup which means you need 500A. Assuming this is not a parallel feeder, 500A would be the max the CT could protect, which makes it a dead give away.
    When you put it that way it seems so simple and I feel a little sheepish, thank you for the help.

  6. #4
  7. Join Date
    Mar 2017
    Posts
    19
    Reputation
    Quote Originally Posted by jflan13 View Post
    A time overcurrent relay (51) served from a 500:5 current transformer is set on tap 5, time dial 10. The primary current necessary for the relay to pickup and start the timing cycle is:

    A 5000 amperes (5 x 10 x 100)
    B 2500 amperes (5 x 500)
    C 500 amperes (5 x 100)
    D 1000 amperes (10 x 100)

    Not sure how to do this, can someone help?
    So for any sort of over current relay, the pick-up in secondary amps is the tap, so to find the primary amps you just multiply it by the CT ratio (in this case 100). so it would be 500 Amps primary.

    To take this one step further, if you had an under or over voltage relay, the same principle would apply. Let's say the tap for an under voltage relay was set at 82V and the PT ratio was 14400:120. So the relay would pick-up at 82V secondary, which translates to 9840V primary.

Subscribe

Login or register to leave a reply!


Share this thread

Related Topics

  1. Cable hi-pot withstanding time, why 15 minutes?
    By moon1958moon in forum Electrical Testing Talk
    Replies: 31
    Last Post: April 16, 2019, 06:37 PM
  2. Time Domain Reflectometer
    By adamwilson in forum NETA Level 3 Exam
    Replies: 4
    Last Post: July 26, 2018, 12:17 PM
  3. Proper connection for overcurrent relay with parallel CT's
    By marvinray in forum NETA Level 4 Exam
    Replies: 10
    Last Post: October 21, 2017, 08:07 AM
  4. How to read the Time Current Characterstics
    By mask1290 in forum Electrical Testing Talk
    Replies: 1
    Last Post: January 24, 2017, 06:41 PM

Tags for this Thread



Related Content


Follow us


Explore TestGuy


NETA Certification Training


NICET Electrical Power Testing


Help and Support