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• Electromagnatic induction

• Voltage induced in a coil

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max44.4 fNE

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• 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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• Applied voltage and induced voltage

max44.4 fNgE

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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

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

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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.

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

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A Tarnsformer

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