Plug setting multiplier
- It is the ratio of fault current in relay coil which comes from current transformer to the pick up current of relay.
- Plug setting Multiplier =Fault Current in relay coil/Pick up current
- = Fault current in relay coil / Rated Secondary current of CT * Current setting
- Pick up current is multiplication of rated secondary current of current transformer and current setting of relay.
- For example, suppose a relay coil is connected to 400/5 current transformer and set at 150 % with a primary fault current of 4000A.
Pick up value = Rated secondary current of CT * Current setting
= 5 * 1.5 (here = 130/100 = 1.3 )
= 7.5 A
Fault current in Relay coil = Primary fault current * ( primary current of CT/ Secondary current of CT)
Fault current in Relay coil = 3000 * (5/400)
= 37.5A
plug setting multiplier = Fault current in relay coil / Pick up current
= 37.5 / 7.5
= 5
Time setting Multiplier
- A electrical relay is provided with control of adjusting the time of operation of relay. this types of adjustment is known as time setting multiplier.
- The time setting dial is calibrated from 0 to 1 steps in each having of 0.05 sec. these figures are multipliers to be used to covert the time derived from time/P.S.M curve into the actual operating time.
- Thus if the time setting is 0.2 and the time obtained from the time/P.S.M curve is 4 seconds, then actual relay operating time is:
Relay operating time = time setting * Time obtained from time/P.S.M curve
= 0.2 * 4
= 0.8 Seconds
MCQS on Plug setting Multiplier & Time setting Multiplier
1. If Current Transformer primary current is 5A and current setting is 120 % what is Pick up value of relay ?
(A) 5.5 A
(B) 6 A
(C) 7A
(D) 6.5 A
Answer: (B) 6 A
Explanation :
Here Primary current of current transformer = 5 A
Current setting = 120 % = 120/100 = 1.2
Pick up value of relay = Rated Secondary current of CT * Current Setting
= 5 * 1.2
= 6 A
2. If Current transformer ratio is 600/5 and primary fault current is 3000 A , what is Fault current in relay coil ?
(A) 25 A
(B) 30 A
(C) 40 A
(D) 35 A
Answer: (A) 25 A
Explanation :
Here Current transformer ratio = 600 / 5
Primary fault current = 3000 A
Fault current in relay coil = ?
Fault current in relay coil = primary fault current / CT ratio
= 3000 * 5 /600
= 25 A