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Transformer
• Electromagnatic induction
• Voltage induced in a coil
Transformer
max44.4 fNE
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]
Transformer
• Applied voltage and induced voltage
max44.4 fNgE
Transformer
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
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
= 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
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
Transformer
Transformer
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.
Solution(a)
fNgE
44.4max
wb 005.0906044.4120
.5mwb
(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.
(d)
The inductance is
mH
Hf
XL m
6.79)602(
30
0796.02
304
120m
m IEX
(c)The inductive reactance is
Transformer
Transformer
A Tarnsformer
Transformer