Calculate the peak voltage of a generator that rotates its 195 turn, 0.100 m dia

Question

Calculate the peak voltage of a generator that rotates its 195 turn, 0.100 m diameter coil at 3600 rpm in a 0.840 T field.

=____ V

At what velocity will the peak voltage of a generator be 480 V, if its 525 turn, 8.00 cm diameter coil rotates in a 0.250 T field?

=____ rpm

A 70 turn, 10.0 cm diameter coil rotates at an angular velocity of 8.00 rad/s in a 1.35 T field, starting with the plane of the coil parallel to the field. Assume that the positive max emf is reached first.

(a) What is the peak emf?

=____ V

(b) At what time is the peak emf first reached?

=____ s

(c) At what time is the emf first at its most negative?

=____ s

(d) What is the period of the AC voltage output?

=____ s

Explanation / Answer

E = BANsin = BANsin(t)

peak when sin(t) = 1

peak emf =  BAN = 0.84 * pi *( 0.1/2)^2 * 195*3600*2pi /60 rad /s

= 485 V answer

now we want

480 = BAN = 0.25 * pi *( 0.08/2)^2 * 525*w = 480

w= 727.565 rad/s =  727.565 =6947.73 RPM

last Q

a) peak emf = BAN = 1.35*pi( 0.1/2)^2*70*8 = 5.93 V

b) T = 2pi /w = 2pi / 8 = 0.785398163 sec

first time peak = T /4 = 0.785398163 /4 =0.196349541 sec

c) T/2 =0.392699081

d ) period =T = 0.785398163

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