Time Overcurrent Relay 51

[jflan13]

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?

[seaborgium]

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.

[jflan13]

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.

[EricGoetz92]

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.

[kammersm21]

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.

Thank you for laying this all out. It was very helpful.

[young_dad]

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?

The tap value is interchangeable with pickup ie. the relay's timing (51 is timed overcurrent) will begin when the CT secondary current is at least equal to the tap setting (pickup, 5A).

With a CT ratio of 500:5, you can see right away that 500 primary amps will cause 5 secondary amps, which is the pickup value of this 51 device.

If it were not so obvious, you could simplify the CT ratio 500:5 = 100:1 and multiply that by the tap setting (5x100).