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A fault current of 2000A is passing on the primary side of 400/5 CT on the secondary side of the C.T. an inverse time over current relay is connected whose plug setting is set at 50%. The PSM will be
  • a)
    25
  • b)
    50
  • c)
    10
  • d)
    30
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
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
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Most Upvoted Answer
A 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%.
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A fault current of 2000A is passing on the primary side of 400/5 CT on the secondary side of the C.T. an inverse time over current relay is connected whose plug setting is set at 50%. The PSM will bea)25b)50c)10d)30Correct answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2026 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A fault current of 2000A is passing on the primary side of 400/5 CT on the secondary side of the C.T. an inverse time over current relay is connected whose plug setting is set at 50%. The PSM will bea)25b)50c)10d)30Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A fault current of 2000A is passing on the primary side of 400/5 CT on the secondary side of the C.T. an inverse time over current relay is connected whose plug setting is set at 50%. The PSM will bea)25b)50c)10d)30Correct answer is option 'C'. Can you explain this answer?.
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