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A stepping mo...

Question

A stepping motor has N = 150, P = 5.08 mm/rev; If n = 2250 pulses then the distance travelled in x-direction and the pulse frequency for a travel speed of 406.4 mm/min _______.

A

200

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B

300

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C

400

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D

250

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Solution

The correct option is A 200
$distance travelled, X = p (\frac{n}{N})$

X = (5.08)(2250)/150 = 76.2 mm

From which, RPM = 80

f = (150)(80)/(60) = 200 Hz

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