Question

A simple generator has a square armature 9.0 cm on a side. The armature has 95 turns of 0.59-mm-diameter copper wire and rotates in a 0.800-T magnetic field. The generator is used to power a lightbulb rated at 12.0 V and 25.0 W. At what frequency (Hz) should the generator rotate to provide 12.0 V to the bulb?

Answer 1

Answer:

f = 3.102 Hz

Explanation:

In this case you have that the required voltage is the maximum induced emf produced by the rotating generator.

In order to calculate the frequency of rotation of the generator that allows one to obtain 12.0V you use the following formula:

$emf_{max}=NBA\omega$    (1)

N: turns of the armature = 95

B: magnitude of the magnetic field = 0.800T

A: area of the square armature = (9.0cm)^2 = (0.09m)^2 = 8.1*10^-3 m^2

emf_max = 12.0V

w: angular frequency

you solve the equation (1) for w:

$\omega=\frac{emf_{max}}{NBA}=\frac{12.0V}{(95)(0.800T)(8.1*10^{-3}m^2)}\\\\\omega=19.49\frac{rad}{s}$

Then, the frequency is:

$f=\frac{\omega}{2\pi}=\frac{19.49rad/s}{2\pi}=3.102Hz$