Question

A simple generator has a square armature 5.0 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 light bulb rated at 12.0 V and 25.0 W.
At what rate should the generator rotate to provide 12.0 V to the bulb? Consider the resistance of the wire on the armature.

Answer 1

Answer:

$25.39Hz$

Explanation:

#First, we need to determine the actual emf required. The generator's internal resistance will cause a voltage drop inside the generator/

Internal resistance is defined using the formula:

$R=\rho L/A\\\rho=1.68\times10^-^8 \Omega m\\L=0.050m\times4\times 60=12.0m\\A=\pi r^2=\pi d^2/4=\pi(0.00059m)^2/4=2.734\times10^-^7m^2\\\\R=(1.68\times10^-^8 \Omega \ m\times 12.0m)/2.734\times10^-^7m^2\\=0.7374\ \Omega$

#The bulb is rated 12.0V,25.0W

Current, $I=25.0W/12.0V=2.083A$

Therefore, the voltage drop in the generator is calculated as:

$2.083A\times0.7374\Omega=1.5360V$  

Actual EMF required is thus   1.536V+12.0V=13.536V

#peak voltage is $13.536V\sqrt 2=19.143V$

#For a generator, by Faraday's Law

$E_m_a_x=NBA\ \omega\\19.143=60\times 0.800T\times (0.05m)^2\ \omega\\\\\omega=159.525rad/s$

$f=\omega/2\pi\\=159.525/2\pi=25.39Hz$

#The rate of the generator is 25.39Hz