1. Consider a generator that rotates its 200 turn, 0.13 m diameter coil at 3550 rpm…
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Question “1. Consider a generator that rotates its 200 turn, 0.13 m diameter coil at 3550 rpm…”
1. Consider a generator that rotates its 200 turn, 0.13 m
diameter coil at 3550 rpm in a 0.65 T field.
a. Calculate the peak voltage of the generator.
2. A precision laboratory resistor is made of a coil of wire.
The coil is 1.5 cm in diameter, 3.75 cm long, and has 500
turns.
a.
What is its self-inductance in mH? | ||
b. What average emf in V is induced if the 12.0 A current
through it is turned on in 5.00 ms (one-fourth of a cycle for 50 Hz
AC)?
Answer
1)Peak voltage=BAN
B=magnetic field=.65T
A=area of cross section of coil= =3.14X=.01327m2(r=radius=.13/2=.065m
N=numbers of turns =200
f
f=frequency of rotation=3550rpm=3550/60=59.17Hz
Peak voltage=BAN=.65X.01327X200X371.59=641.03volt
2
A/l
500 = N=numbers of turns
m2 (r=1.5/2=.75cm)
l=length=3.75cm=.0375m
L=
A/l=4
X
x
X1.77X
/.0375=1.48mH
b)induced EMF e=Ldi/dt
di=change in current=12-0=12A
dt=time=5ms=.005
L=1.48mH=.00148H
e=Ldi/dt=.00148×12/.005=3.552volt
Conclusion
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