Question

A simple generator has a square armature 6.0 cm on a side. The armature has 85 turns of 0.59-mm-diameter copper wire and rotates in a 0.65-T magnetic field. The generator is used to power a lightbulb rated at 12.0 V and 25.0 W. At what rate should the generator rotate to provide 12.0 V (rms) to the bulb? Consider the resistance of the wire on the armature.

Answer 1

Answer:

f=15.5 Hz

Explanation:

Let's determine the internal resistance:

$R=\frac{(p*L)}{A}$

ρ = 1.68*10^-8 Ω m

$L=0.060m*4*60 = 14.4m$

$A=\pi*r^2\\A= \pi*(5.9x10^-4m/2)^2=2.734*10^-7m^2$

$R=(1.68*10^-8)*(14.4m)/(2.734*10^-7m^2)= 0.884$Ω

Since the bulb is rated at 12.0 V and 25.0 W,

Current

$I=\frac{25W}{12.0v}=2.08 A$

Therefore, voltage drop inside generator =

$V=(2.08 A)*(0.88)=2.35v$

Actual EMF required is

$E_{mf}=12.0v+2.35v=14.35v$

Note that this is an RMS value.  

The peak voltage is

$v_{peak}=14.15v*\sqrt{2} =20.29v$

For a generator, by Faraday's Law,

$E_{(max)}=N*B*A*w$

$20.29v=(60)*(0.650T)*(0.06m)^2$*ω

ω$=144.5\frac{rad}{s}$

f=ω/(2π)=

f=144.5 rad/s/(2π)

f=23.001 Hz