Could someone tell me why they divide by 4? Thanks.
Thanks. When we talk about losses and efficiency, specifically the copper loss, we are talking about both the primary copper losses and secondary copper losses, right? NCEES 528 only considered the primary copper losses, what about the secondary?At 1/2 full-load, the load losses (I^2*R) is 1910WNow, find the load losses at full-load. If R is constant and I goes up by two, then (I^2) * R will go up by (2^2) or 4.
At full-load, the load losses are 4*1910W = 7640W
Total Loss = Core Loss + Load Loss = 460W + 7640W = 8100W
Both the primary and secondary resistances are represented in the 10.9 ohm value in the problem. When they say it's referred to the primary side, all that means is that the total of both as read from the primary is 10.9 ohms.Thanks. When we talk about losses and efficiency, specifically the copper loss, we are talking about both the primary copper losses and secondary copper losses, right? NCEES 528 only considered the primary copper losses, what about the secondary?
This solution that NCEES has been bugging, so I approached it from an another angle to see if I would get the same answer. Let me know what you think.At 1/2 full-load, the load losses (I^2*R) is 1910WNow, find the load losses at full-load. If R is constant and I goes up by two, then (I^2) * R will go up by (2^2) or 4.
At full-load, the load losses are 4*1910W = 7640W
Total Loss = Core Loss + Load Loss = 460W + 7640W = 8100W
Could someone tell me why they divide by 4? Thanks.
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