electrical power and machines - ideal transformer

8/9/2019 Electrical Power and Machines - Ideal Transformer

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Ideal Transformer[Chapter 9]


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Introduction

• Transformers are one of the most useful electrical devices

provides a change in voltage and current levels

provides galvanic isolation between different electrical circuits

changes the apparent magnitude value of an impedance


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Applied Voltage

• Consider a coil connected across an C

voltage source

the coil and source resistances are negligible

the induced voltage E must equal the source

voltage! "#$ a sinusoidal C flux Φ must exist to generate the

induced voltage on the N turns of the coilo Φmax varies in proportion to E g

o placing an iron core in the coil will not change the flux Φ

magneti%ation current I m drives the C fluxo the current is &'( out-of-phase and lagging with respect

to the voltage

o with an iron core, less current is needed to drive the C

flux


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Induced Voltages

Example

• a coil, having )''' turns, links an C flux with a peak valueof * m+b at a frequency of ' %

• calculate the rms value of the induced voltage

• what is the frequency of the induced voltage.

Example

• a coil, having &' turns, is connected to a /*' #, ' % sourcethe rms magneti%ation current is )

• calculate the peak value of the flux and the mmf

• find the inductive reactance and the inductance of the coil


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Elementary Transformer•

Consider an air-core coil excited by an C source E g

draws a magneti%ation current I m

produces a total flux Φ

• second coil is brought close to

the first a portion Φm/ of the flux couples the second

coil, the mutual flux

an C voltage E * is induced

the flux linking only the first coil is called

the leakage flux, Φf/

• 0mproved flux coupling concentric windings, iron core

weak coupling causes small E *

• Consider an air-core coil

the magneti%ation current I m produces both fluxes Φm/

and Φf/ the fluxes are in-phase

the voltages E g and E * are in phase

terminal orientation such that the

coil voltages are in-phase are said

to possess the same polarity


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Ideal Transformer

• n ideal transformer

transformer has no losses core is infinitely permeable

all fluxes link all coils

there are no leakage fluxes

• #oltage relationship consider a transformer with two coils of

N / and N * turns a magneti%ing current I m creates a flux

Φm

the flux varies sinusoidally and has a

peak value of Φmax

the induced voltages are

from these equations, it can be deduced

that

the ratio of the primary and secondary

voltages is equal to the ratio of the

number of turns

o E / and E * are in-phase

o polarity marks show the terminalon each coil that have a peak

positive voltage simultaneously


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Ideal Transformer

• Current relationship let a load be connected across the

secondary of an ideal transformer

current I * will immediately flow

I * = E * 1 Z

coil voltages E / and E * cannot

change when connected to a fixed

voltage source and hence fluxΦm

cannot change

current I * produces an mmf

mmf * = N * I *

if mmf 2 acts along, it would

profoundly changeΦm

Φm

can only remain fixed if the

primary circuit develops a mmf which

exactly counterbalances mmf *

current I / must flow such that

o I / and I * must be in-phase

o when I / flows into the positive

polarity marking of the primary, I *

flows out of the positive polarity

marking of the secondary


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Ideal Transformer

• 0deal transformer model

a = N / / N *

E * = E / / a

I / = I * / a


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Ideal Transformer

Example

a not so ideal transformer has *'' turns in the primary coil and /' turns in

the secondary coil2 the mutual coupling is perfect, but the magneti%ation

current is / 2 the primary coil is connected to a )3' #, ' % source2

calculate the secondary rms voltage, peak voltage

Example

for the transformer above, a load is connected to the secondary coil that

draws 3' of current at a '23 lagging pf2 calculate the primary rms current

and draw the phasor diagram


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Impedance Ratio

• Transformers can also be used to

transform an impedance

the source sees the effective impedance

Z x = E /1 I /

on the other side, the secondary winding ofthe transformer sees the actual impedance

Z = E *1 I *

the effective impedance is related to the

actual impedance by


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ShiftingImpedances

• 0mpedances located on the

secondary side of a

transformer can be relocated

to the primary side

the circuit configuration

remains the same 4series or

shunt connected5 but the

shifted impedance values are

multiplied by the turns ratio

squared

• 0mpedance on the primary

side can be moved to the

secondary side in reversemanner

the impedance values are

divided by the turns ratio

squared


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Shifting Impedances

• 0n general, as an impedance is

shifted across the transformer

the real voltage across the impedance

increases by the turns ratio

the actual current through the

impedance decreases by the turns

ratio

the required equivalent impedance

increases by the square of the turns

ratio

Exampleusing the shifting of impedances calculate

the voltage E and current I in the circuit,

knowing that the turns ratio is /6/''


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Ideal Transformer

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