em ps lec 3 t
DESCRIPTION
electro magnetic lecter noteTRANSCRIPT
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Transformer
• Electromagnatic induction
• Voltage induced in a coil
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Transformer
max44.4 fNE
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Transformer
• Where
• E=effective voltage induced[V]
• f= frequency of the flux [Hz]
• N=number of turns on the coil
• Φmax = Peak value of the flux [Wb]
• 4.44 = a constant [exact value = 2π/√2]
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Transformer
• Applied voltage and induced voltage
max44.4 fNgE
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Transformer
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TRANSFORMER EQUATION.Average rate of change of flux =2m / (1/2f)
= 4f m webers/second
As flux change from positive (+) maximum to m
(maximum) in ½ cycle
i.e
f
fT
2
1
2
1
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average e.m.f induced per turn = 4fm Volts
As for sinusoidal wave the r.m.s or effective
value is 1.11 times the average value,
r.m.s e.m.f/turn = 1.11 x 4fm.
E = 4.44 N1fm Volts
Also
If = instantaneous value of flux in Webers
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= m sin 2ft
induced e.m.f per turn
e.m.f = d/dt Volts
e.m.f = 2fm (Cos 2ft) Volts
•The maximum value of e.m.f /turn
Also
If = instantaneous value of flux in Webers
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e.m.f/turn(maximum) = 2fm Volts
and r.m.s value of induced
e.m.f/turn = 0.707 x 2fm
= 4.44fm Volts
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Transformer
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Transformer
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ExampleA coil having 90 turns is connected to a
120V, 60Hz source. If the effective value
of the magnetizing current is 4A.
Calculate the following: -
The peak value of flux.The peak value of m.m.f.The inductive reactance of the coil.The inductance of the coil.
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Solution(a)
fNgE
44.4max
wb 005.0906044.4120
.5mwb
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(b)The peak current is
AIpeakIm
64.5441.12)(
The peak mmf
64.590.. NIfmm
A6.507
The flux is equal to 5 mwb at the instant
when the coil m.m.f. is 507.6A.
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(d)
The inductance is
mH
Hf
XL m
6.79)602(
30
0796.02
304
120m
m IEX
(c)The inductive reactance is
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Transformer
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Transformer
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A Tarnsformer
Transformer