3. Consider an AC senerator with a square coil 25.0 cm on a side, with 800 tums

Question

3. Consider an AC senerator with a square coil 25.0 cm on a side, with 800 tums of wire rotating with 3600 s.(b) Find the average emf in 4.5 periods. (When you convert rpm to rad, keep the n) rpm in a uniform magnetic field or o 300 T. At time r-o-o.(a) Find the emf at time 1/240 s and Loop (seern end-on) Plux decreasing most rapidly largest positive emf. Flux increasing most rapidly largest negative emf Flux at its most positive value, emf is zero, Flux at its most negative value, emf is Slip rings Brush

Explanation / Answer

3. for the given problem

a = 25 cm

n = 800 turns

w = 36000 rpm

B = 0.3 T

t = 0, phi = 0

a. flux through the coil at time t = phi

phi = n*B*a^2*cos(phi)

hence

d(phi)/dt = -nBa^2*sin(phi)*w

hence

EMF = nBa^2*w*sin(phi)

phi = wt

V = nBa^2*w*sin(wt)

V(1/120) = 0 V

V(1/240) = 0 V ( because sin(wt) is 0 for these points)

b. average emf in 4.5 periods = integrate(nBa^2*w*sin(wt)dt) for half time period / half time period

Vav = nBa^2*(1 - cos(w*pi/w)) / (pi/w) = nBa^2*2*w/pi = 800*0.3*2*0.25^2*600*2 = 36000 V

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