Transcript Magnetism and Transformers I

Lesson 25:
Magnetism and Transformers
1
Learning Objectives
• Analyze the relationship between the transformation
ratio, voltage ratio, current ratio, and impedance
ratio.
• Construct a circuit equivalent of a transformer and
calculate primary and secondary voltage, current and
polarity.
• Explain the relationship between the power
developed in the primary and secondary of a
transformer.
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Transmission of Power
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Transformer Overview
• A transformer is a magnetically coupled circuit
whose operation is governed by Faraday’s Law.
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Faraday’s Experiment #3:
Mutually Induced Voltage

• Voltage is induced across Coil 2 when i1 is changing.
• When i1 reaches steady state, voltage across Coil 2
returns to zero.
− NOTE: The coil to which the source is attached is the
primary and the coil attached to the load is the secondary.
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Transformer Overview
• A time-varying current in the primary windings
induces a magnetic flux (field) in the iron core.
• The flux flows through the core and induces a current
the secondary windings.
• Thus power flows via the magnetic field without the
windings being electrically connected.
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Winding Direction
• The polarity of ac voltages can be changed by
changing the direction of the windings.
0º phase
shift
180º phase
shift
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Iron-Core Transformers
• Two basic types of iron-core transformers are the
core type and the shell type.
• In both, the core is constructed of laminated sheets
of steel to reduce eddy current losses.
Core Type
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Shell Type
Iron-Core Transformers
• We will consider the ideal transformer which
−
−
−
−
Neglects coil resistance.
Neglects core losses.
Assumes all flux is confined to the core.
Assumes negligible current required to establish core flux.
• Transformer operation is governed by Faraday’s
Law.
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Transformation Ratio
• According to Faraday’s Law, voltage (e) is directly
related to the number of turns (N) in the primary or
secondary windings:
dm
eN
dt
• For each winding we can write:
dm
dm
e pri  N pri
and esec  N sec
dt
dt
• Because the flux (m) is the same through both
windings, we can write:
E pri N pri

Esec N sec
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Transformation Ratio
• “The ratio of the primary voltage to secondary voltage is equal
to the ratio of primary turns to secondary turns.”
E pri
Esec

N pri
N sec
• This ratio is called the transformation ratio (or turns ratio)
and given by the symbol a.
a
N pri
N sec

E pri
Esec
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Step-up and Step-down
• Transformers are used to change or “transform”
voltage.
• Step-Up transformer:
− The secondary voltage is higher than the primary voltage.
− There are fewer primary windings than secondary
windings (a < 1).
• Step-Down transformer:
− The secondary voltage is lower than the primary voltage.
− There are more primary windings than secondary
windings (a > 1).
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Step-up and Step-down
Step-Up
Step-Down
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Example Problem 1
Suppose the transformer depicted below has 4000 turns on its primary winding
and 1000 turns on its secondary.
a. Determine it’s turns ratio (a). Is it step-up or step-down?
b. If the primary voltage epri = 480 sin t, what is it’s secondary voltage?
a
N pri
N sec

4000
 4 => Step-Down since a > 4
1000
Remember, this is a Step-Down
for voltage, but a step-up for
current.
a
E pri
Esec
Esec 
E pri
a

480V
120V
4
=> esec = 120V sin (Ѡt)
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Current Ratio and Power
• Because we are considering an ideal transformer,
power in equals power out (Pin = Pout).
Sin  Sout
E pri I pri  Esec I sec
I pri
I sec
Esec 1


E pri a
• If the voltage is stepped up, then the current is stepped
down, and vice versa.
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Impedance
• The impedance of the primary of an ideal transformer
is the transformation ratio (a) squared times the
impedance of the load (secondary winding) and is
derived as below:
− NOTE: If the load is capacitive or inductive, the reflected
impedance is also capacitive or inductive.
Vg
VL

Np
Ns
a
and
Ip
Is

Ns 1

Np a
Recall from Ohm's Law:
V
V
Zp  g
and
ZL  L
Ip
Is
Rearranging the equations:
Z p  a2Z L
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Example Problem 2
For the figure below Eg = 120 V 0º, the turns ratio is 6:1, and ZLD = 100100j. The transformer is ideal. Find:
N
a. load voltage
b. load current
c. generator current
d. Active power to the load
a)
E pri
Esec

Esec 
b) I sec 
N pri
N sec

120V 0 6

Esec
1
c)
120V 0
 20V 0
6
Esec
20V 0

 141mA45
Z sec 100  j100
I pri / g
I sec/ LD

I pri 
1
a
I sec 141mA45

 23.5mA45
a
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d) Psec/ LD  ( I sec ) 2 RLD  (141mA) 2 (100 )  1.99W
Note that Ohm’s law and regular power equations are used here.
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The Dot Convention
• The direction of the windings is not obvious looking at
a transformer, therefore we use the dot convention.
• Dotted terminals have the same polarity at all instants
of time. Used for phase shifting (180).
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Example Problem 3
For the figure below Ig = 25 30ºmA, the turns ratio is 4:1, and VLD =
600ºV. The transformer is ideal. Find:
a.
b.
c.
d.
Load current
Load impedance
Generator voltage
Real and Reactive load power
a)
b)
c)
d)
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Example Problem 4
For the figure below i1 = 100 sin (ωt) mA and the transformer is ideal.
Determine the secondary currents i2 and i3.
Ip
Is

Ns 1

Np a
I1 1
  I 2  a * I1  0.25*100mA  25mA
I2 a
i2  25mA sin(t  180)
Why -180⁰
I2 1
  I 3  a * I 2  2 * 25mA  50mA
I3 a
i3  50mA sin(t )
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Note the dot convention here, which
switches the polarity 180⁰
Power Transformer Ratings
• Just like ac motors and generators, power
transformers are rated in terms of voltage and
apparent power.
− For example, a transformer is rated at 2400/120 volt, 48
kVA has:
S

recall:  I  
E

• On the primary winding, the current rating is
48,000 VA / 2400 V = 20 A.
• On the secondary winding, the current rating is
48,000 VA / 120 V = 400 A.
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Example Problem 5
A 7.2 kV, a=0.2 transformer has a secondary winding rated current of 3 A. What
is its kVA rating?
I pri
I sec

1
I
3A
 I pri  sec 
 15 A
a
a
0.2
Sin  E pri * I pri
Sin / g  I pri / g * E pri / g  15 A * 7.2kV  108kVA
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Transmission of Power
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Transformers
Mitsubishi 500 MVA Single-Phase Auto-Transformers
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QUESTIONS?
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