E = NBAW sin(wt) (1) Thus Emax = NBAW Peak output of generator (2) %3D You are given or can determine the following: Emax in Volts f: in rpm (rev per minute -convert to rev per sec) where: w = 2f A = TTr2 (in m²) B: magnetic field (in T) Thus can determine the number of turns N of the coil from Eq. (2).
E = NBAW sin(wt) (1) Thus Emax = NBAW Peak output of generator (2) %3D You are given or can determine the following: Emax in Volts f: in rpm (rev per minute -convert to rev per sec) where: w = 2f A = TTr2 (in m²) B: magnetic field (in T) Thus can determine the number of turns N of the coil from Eq. (2).
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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An alternator is an AC generator, so from there, we are given a car alternator which produces about 14 V peak output to charge the car's 12-V battery. If an alternator coil is 17 cm in diameter is spinning at 1200 revolutions per minute in a 0.19T magnetic field, how many turns must it have to produced a 14V peak output?
I am unfamiliar with the derivations, so need someoone who can help me with understandig symbology to the values better.
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