Answer :
Answer:
f=15.5 Hz
Explanation:
Let's determine the internal resistance:
[tex]R=\frac{(p*L)}{A}[/tex]
ρ = 1.68*10^-8 Ω m
[tex]L=0.060m*4*60 = 14.4m[/tex]
[tex]A=\pi*r^2\\A= \pi*(5.9x10^-4m/2)^2=2.734*10^-7m^2[/tex]
[tex]R=(1.68*10^-8)*(14.4m)/(2.734*10^-7m^2)= 0.884[/tex]Ω
Since the bulb is rated at 12.0 V and 25.0 W,
Current
[tex]I=\frac{25W}{12.0v}=2.08 A[/tex]
Therefore, voltage drop inside generator =
[tex]V=(2.08 A)*(0.88)=2.35v[/tex]
Actual EMF required is
[tex]E_{mf}=12.0v+2.35v=14.35v[/tex]
Note that this is an RMS value.
The peak voltage is
[tex]v_{peak}=14.15v*\sqrt{2} =20.29v[/tex]
For a generator, by Faraday's Law,
[tex]E_{(max)}=N*B*A*w[/tex]
[tex]20.29v=(60)*(0.650T)*(0.06m)^2[/tex]*ω
ω[tex]=144.5\frac{rad}{s}[/tex]
f=ω/(2π)=
f=144.5 rad/s/(2π)
f=23.001 Hz
The angular speed of the generator is 1.125 x 10⁶ rad/s.
Resistance of the copper wire
The resistance of the copper wire is calculated as follows;
[tex]R = \frac{\rho L}{A} \\\\R = \frac{(1.68 \times 10^{-8} ) \times (4 \times 0.06)}{\pi (0.295 \times 10^{-3})^2} = 0.0147 \ ohms[/tex]
Current in the armature
I = P/V
I = 25/12
I = 2.1
Voltage drop in the generator
V = IR
V = 2.1 x 0.0147
V = 0.031 V
Total emf of the generator
E = 0.031 + 12
E = 12.031 V
Peak voltage
[tex]V_o =12.031 \times \sqrt{2} \\\\v_0 = 17 \ V[/tex]
Angular speed of the generator
[tex]V_o= NBA \omega\\\\\omega = \frac{V_0}{NBA} \\\\\omega = \frac{17}{85 \times \times 0.65 \times \pi \times (0.295 \times 10^{-3})^2} \\\\\omega = 1.125 \times 10^6 \ rad/s[/tex]
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