17. 0/1 points I Previous Answers SerPSET9 31.P.045. My Notes In a 260-turn auto

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17. 0/1 points I Previous Answers SerPSET9 31.P.045. My Notes In a 260-turn automobile alternator, the magnetic flux in each turn is DB 2.50 x 10-4 cos at, where DB is in webers, is the angular speed of the alternator, and t is in seconds. The alternator is geared to rotate five times for each engine revolution. The engine is running at an angular speed of 1.00 x 103 rev/min (a) Determine the induced emf in the alternator as a function of time. (Assume s in V.) (b) Determine the maximum emf in the alternator. What are the maximum values of the sine and cosine functions? V Need Help? Read tt Watch It Submit Answer Save Progress Practice Another Version

Explanation / Answer

N = 260 , phi = 2.5*10^-4*cos(wt)

n = 5 , wE = 1*10^3 rev/min

wA =5*fE = 5*10^3 rev/min

wA= 5*1000*2*3.14/60 = 523.33 rad/s

from faradays law induced emf V = -d(phi)/dt

V = N*wA*(2.5*10^-4)sin(Wat)

V = (260*523.33*2.5*10^-4)sin(523.3t)

V = (34V)sin(523.3*t)

(b) for maximum value sin(523.3*t) = 1

Vmax = 34 V

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