Electrician Red Seal · Question
A protective relay for a 13.8 kV feeder has a current transformer (CT) ratio of 600:5. If the relay is set to trip at 2.5 A primary equivalent, what is the actual primary current (in Amperes) that will cause the relay to trip?
The CT ratio indicates that for every 600 A on the primary side, there are 5 A on the secondary side (to the relay). To find the primary current equivalent for
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Question: A protective relay for a 13.8 kV feeder has a current transformer (CT) ratio of 600:5. If the relay is set to trip at 2.5 A primary equivalent, what is the actual primary current (in Amperes) that will cause the relay to trip?
Answer options: ✅ 300 A
- 2.5 A
- 1500 A
- 600 A
Correct answer: 300 A
Explanation: The CT ratio indicates that for every 600 A on the primary side, there are 5 A on the secondary side (to the relay). To find the primary current equivalent for a given secondary current, use the ratio: Primary Current / Secondary Current = CT Ratio. So, Primary Current = Secondary Current * (Primary CT Value / Secondary CT Value) = 2.5 A * (600 / 5) = 2.5 A * 120 = 300 A. Therefore, the relay will trip when the primary current reaches 300 A.
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