The contacts of an oil circuit breaker move 15 inches from arcing contact parting to the point where the manufacturer assures interruption. At what average velocity must the contacts open to achieve interruption in 5 cycles on a 60-hertz basis?
My thought process on this question was to divide the frequency by the amount of cycles allotted for interruption.
60 hertz/5 cycles = 1/12 of a second.
Divide the distance needed to travel by the maximum time of interruption.
15 inches / (1/12) of a second = 180 inches in 1/12 of a second to achieve manufacturer's guaranteed interruption.
Calculate into feet.
180 inches / 12 inches = 15 feet/second.
Did I approach this problem correctly?
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