A fault current of 2000A is passing on the primary side of 400/5 CT on...
The pickup value of the relay is 5A, since the relay setting is 50%
Therefore, The operating current of the relay is
5 x 50/100 = 2.5 A
View all questions of this testA fault current of 2000A is passing on the primary side of 400/5 CT on...
Given Information:
Fault Current on Primary Side of CT = 2000A
CT Ratio = 400/5
Plug Setting of Relay = 50%
To find:
Pickup Setting Multiplier (PSM) of the Relay.
Solution:
Step 1: Calculation of Secondary Current of CT
Secondary Current (Isec) = Primary Current (Iprim) / CT Ratio
Isec = 2000 / (400/5) = 125A
Step 2: Calculation of Pickup Current of Relay
Pickup Current (Ipickup) = Plug Setting of Relay x Isec
Ipickup = 50/100 x 125 = 62.5A
Step 3: Calculation of PSM
PSM = Fault Current / Ipickup
PSM = 2000 / 62.5
PSM = 32
Step 4: Calculation of Actual Plug Setting of Relay
Actual Plug Setting = PSM / CT Ratio
Actual Plug Setting = 32 / (400/5)
Actual Plug Setting = 0.4 x 32
Actual Plug Setting = 12.8%
Step 5: Verification
The actual plug setting is less than the pickup setting (50%), which is expected for an inverse time overcurrent relay. Hence, the answer is option (c) 10.
Therefore, the Pickup Setting Multiplier (PSM) of the Relay is 32 and the actual plug setting is 12.8%.