A square coil with sides of 0.36 meters wrapped 40 times is spun at some frequen

Question

A square coil with sides of 0.36 meters wrapped 40 times is spun at some frequency in a uniform magnetic field of 0.17 Tesla to form a generator. It is then connected to a transformer. The primary side has 188 turns and the secondary side has 356 turns. The secondary output of the first transformer is then connected to another transformer. The second transformer has 10 turns on the primary side and the secondary side has 17 turns. The maximum secondary EMF after for the second transformer is 749 volts. At what frequency was the coil spun in Hz?

Explanation / Answer

when a coil of area A and with turns N rotate in a magnetic field B with an angular speed of (omega)w

we get emf as E = BANwsin(wt)

max emf E= BANw = 0.17*(0.36)^2 * 40 *w = 0.88128w (let say E1)

after first transformer action

E1/E2 = 188/356 ; E2 = 1.894 E1

after second transformer action

E2/E3 = 10/17 , E3 = 1.7 E2 = 1.7(1.894 E1) = 3.22 E1

given E3 as 749 V

749 = 3.22E1 ; E1 = 232.67

but E1 = 0.88128 w

by this we get w = 264 rad/s

frequency = w/(2*pi) = 42 Hz ( approx)

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