Electrical Power and Machines - Ideal Transformer
-
Upload
dhanis-paramaguru -
Category
Documents
-
view
226 -
download
0
Transcript of Electrical Power and Machines - Ideal Transformer
-
8/9/2019 Electrical Power and Machines - Ideal Transformer
1/14
Ideal Transformer[Chapter 9]
-
8/9/2019 Electrical Power and Machines - Ideal Transformer
2/14
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
-
8/9/2019 Electrical Power and Machines - Ideal Transformer
3/14
-
8/9/2019 Electrical Power and Machines - Ideal Transformer
4/14
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
-
8/9/2019 Electrical Power and Machines - Ideal Transformer
5/14
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
-
8/9/2019 Electrical Power and Machines - Ideal Transformer
6/14
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
-
8/9/2019 Electrical Power and Machines - Ideal Transformer
7/14
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
-
8/9/2019 Electrical Power and Machines - Ideal Transformer
8/14
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
-
8/9/2019 Electrical Power and Machines - Ideal Transformer
9/14
Ideal Transformer
• 0deal transformer model
a = N / / N *
E * = E / / a
I / = I * / a
-
8/9/2019 Electrical Power and Machines - Ideal Transformer
10/14
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
-
8/9/2019 Electrical Power and Machines - Ideal Transformer
11/14
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
-
8/9/2019 Electrical Power and Machines - Ideal Transformer
12/14
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
-
8/9/2019 Electrical Power and Machines - Ideal Transformer
13/14
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/''
-
8/9/2019 Electrical Power and Machines - Ideal Transformer
14/14
Ideal Transformer
• Tutorials
7roblems6 &-/, &-), &-, &-8