EE212 Passive AC Circuits. Lecture Notes 3 Transformers. Magnetic Circuit. What is the relationship between magnetic flux and magnetomotive force, F m ? What is the relationship between electric current, i , and magnetomotive force? Magnetic field strength, H = F m / l.
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Magnetic field strength, H = Fm / l
EE 212 20102011
EE 212 20102011
i
a
di
L
vab = L
dt
b
Magnetic Circuit
Consider a coil around a magnetic core. If a current i flows through the coil, a magnetic flux is generated in the core.
= N iin webers (Wb)
N = number of turns in the coil
Rm = constant known as reluctance
(depends on the magnetic path of the flux)
Direction of flux by
RightHand Rule
l = length of magnetic path
A = crosssection area
m = permeability
Fingers curled around coil
– direction of current
Thumb – direction of flux
EE 212 20102011
i1
a
vab
L1
b
f2
c
vcd
L2
i2
d
Coupled Circuits
Circuits that affect each other
by mutual magnetic fields
± depending on whether the fluxes add or oppose each other
L1, L2: self inductance
The flux f2 generated by
current i2 in Coil 2
induces a voltage in Coil 1,
and viceversa.
EE 212 20102011
I2
M
a
c
L1
Vab
Vcd
L2
b
d
Coupled Circuits in Phasors
If input signals are sinusoidal waveforms,
coupled circuits can be in phasor representation
EE 212 20102011
EE 212 20102011
A current i entering a dotted terminal in one coil induces a voltage M with a positive polarity
at the dotted terminal of the other coil.
f1
i1
a
L1
b
f2
c
L2
I1
i2
a
c
d
L1
L2
d
b
I2
Dot Convention
Dots are placed at one end of each coil, so that currents entering the dots produce fluxes that add each other.
The dots provide information on how the coils are wound with respect to each other.
(currents entering the dots
produce upward fluxes)
(+) if both currents enter the dotted terminals (or
the undotted terminals).
EE 212 20102011
EE 212 20102011
if
+
+
e1
N1
N2
e2
v1


secondary
winding
primary
winding
e2 = N2
If
+
+
E2
E1
V1


N1
N2
V1 = E1
If
f
E1
E2
Transformer
When a voltage V1 is applied to the primary winding, an emf e2 is induced in the secondary winding. The induced emf lags the inducing flux by 900.
Faraday’s Law
EE 212 20102011
Instrument Transformers:
CT (current transformer)
PT or VT (voltage transformer)
EE 212 20102011
= a (turns ratio)
=
= a
=
Ideal Transformer
Coupling Coefficient, k = 1, i.e. thesame flux f goes through both windings
e1 = N1
and e2 = N2
=>
Turns ratio (a or n) is also known as the transformation ratio.
No Losses No voltage drops in the windings: V1 =  e1
Instantaneous powers in primary and secondary are equal
(i.e. all the energy from the primary is transferred to the secondary winding).
e1 i1 = e2 i2 Therefore,
EE 212 20102011
I2
+
V2
ZL
e2
e1
V1

N1
N2
Transformer Loading(Lenz’s Law: effect opposes the cause)
Primary current I1 = I’1 + If
EE 212 20102011
I2
M
a
c
L1
Vab
Vcd
L2
b
d
Ideal TransformerV2 = V1/a
I2 = a I1
Z1= a2Z2 where Z2 is the load impedance
+
+
●
●


EE 212 20102011
i1
R1
R2
+
+
e1
v1
ZL
e2
v2


N2
N1
Actual Transformer
EE 212 20102011
Eddy Current Loss:
emf induced in core generates eddy currents which circulate in the core material, generating heat.
laminations (silica sheets between core layers) –
to reduce eddy current, and minimize loss
EE 212 20102011
Hysteresis Loss:
The direction of the magnetic flux in the core changes every cycle. Power is consumed to move around the magnetic dipoles in the core material, and energy is dissipated as heat.
Hyst. loss (vol. of core) x (area of hyst. loop)
EE 212 20102011
EE 212 20102011
EE 212 20102011
EE 212 20102011
EE 212 20102011
Functions of Transformer Oil
EE 212 20102011
EE 212 20102011
Polemounted
Singlephase Transformer
Threephase Transformer
EE 212 20102011
For single phase transformer:
Rated primary current = Rated VA / Rated Pr Voltage
When rated current flows through a transformer, it said to be fully loaded.
The actual current through a transformer varies depending on the load connected at different times of the day.
EE 212 20102011
Represent inductively coupled circuits by a conductively connected circuit
Equivalence in terms of loop equations
Assume a load impedance is connected to the secondary.
turns ratio,
= a
R2
R1
M
+
+
V2
V1
L11
L22
I2
I1


N1
N2
KVL at Loop 1:
V1 + R1I1 + jwL11I1  jwM I2 = 0
KVL at Loop 2:
R2I2 + V2+ jwL22I2  jwM I1 = 0
inductively coupled circuits
The loop equations are:
EE 212 20102011
L1
R1
V1
 V2
I1
I2
a2L2
a2R2
+
+
aM
V1
I1
aV2


R1 + jwL11 jwM
 jwMR2 + jwL22
=
Consider following substitutions:
M a·ML2a2·L2R2a2·R2
V2a·V2I2
conductively connected circuit
KVL at Loop 1:
V1 + R1I1 + jwL1I1 + jwaM (I1I2/a) = 0
KVL at Loop 2:
jwaM (I2/a  I1) + jw a2L2.I2/a + jw a2R2.I2/a + aV1 = 0
Let L11= L1 + aM
and L22= L2 + M/a
The loop equations remain the same:
EE 212 20102011
a2X2
X1
a2R2
R1
+
I1
Ie
If
a2ZL
Ic
V1
Xf
Rc

ZL = load impedance
Ie = excitation current
If = magnetizing current
Xf = magnetizing
reactance
equivalent circuit referred to the primary side
R1, R2= primary, secondary winding resistance
X1, X2= primary, secondary leakage reactance
I1, I2= primary, secondary current
Rc = core loss resistance (equivalent resistance contributing to core loss)
Ic = core loss equivalent current
EE 212 20102011
a)
b)
j0.92/a2 W
j0.0090 W
0.72/a2 W
0.0070W
j44.6429 W
308.642 W
j0.92W
j0.0090a2W
0.72W
0.0070a2 W
j44.6429 a2W
308.642a2 W
A 50kVA, 2400/240V, 60Hzdistribution transformer has a leakage impedance of (0.72 + j0.92) Win the high voltage (HV) side and (0.0070 + j0.0090) W in the low voltage (LV) winding. At rated voltage and frequency, the admittance of the shunt branch of the equivalent circuit is (0.324 – j2.24)x102S [siemens]when viewed from the LV side. Draw the circuit:
a) viewed from the LV side
b) viewed from the HV side
Turns ratio, a = 2400/240 = 10
EE 212 20102011
Xe
Re
Xf
Rc
Xe
Re
Xe
Re = R1 + a2R2
Xe = X1 + a2X2
where, R2 and X2 are referred
to the primary side
Re and Xe obtained from Short Circuit Test
Rc and Xf obtained from Open Circuit Test
For transformers operating close to full load
For transformers in a large power network analysis
EE 212 20102011
P = copper loss = I2 Re Re =
Ze = Xe =
+
R1 ≈ a2 R2 ≈
Vsupply
X1 ≈ a2 X2 ≈

HV
LV
W
V
A
Xe
Re
I
+
V
Xf
Rc

voltage source and increase voltage
gradually to get rated current reading
on the Ammeter
Equivalent circuit referred to HV side
EE 212 20102011
P = core loss = Rc =
+
Ie = I , Ic =
Vsupply
If = Xf =

HV
LV
W
V
A
I
+
Ie
If
V
Ic
Rc
Xf

Equivalent circuit referred to LV side
EE 212 20102011
Efficiency, h = Power Output / Power Input
Power Output = VL IL cos = IL2 RL
Power Input = Power Output + Cu losses + Core losses
Cu Loss – varies with load current
Core Loss – depends on voltage (usually a constant for practical purposes)
Transformer efficiency is maximum when Cu loss = core loss
EE 212 20102011
designed to have max. h at less than full load
EE 212 20102011
Find the equivalent circuit of a 50kVA, 2400/240V transformer. The following readings were obtained from short circuit and open circuit tests:
EE 212 20102011
EE 212 20102011
 no current, and therefore no voltage drop in transformer
 secondary voltage, Vsec = rated voltage
 voltage drop in transformer (Vdrop) increases
EE 212 20102011
Voltage Regulation (continued)
To maintain rated voltage at the secondary,
At no load, primary voltage required, Vpr = rated voltage
At a certain load, required Vpr = rated voltage + Vdrop in transformer
Voltage regulation is the change in primary voltage required to keep the secondary voltage constant from no load to full load, expressed as a percentage of rated primary voltage.
Load p.f. has a big effect on voltage regulation.
EE 212 20102011
EE 212 20102011
The allowable voltage variation at the customer load point is usually ± 5% of the rated (nominal) value. Control measures are taken to maintain the voltage within the limits.
EE 212 20102011
EE 212 20102011
From J.D. Irwin, “Basic Engineering Circuit Analysis”, 3rd ed. Macmillan
EE 212 20102011
From Jackson et al. “Introduction to Electric Circuits” 8th ed., Oxford
EE 212 20102011
Assume N1 = 200
and N2 = 100
Let Vsource = 120 V
Then what is V2?
Assume I2 is 15 A. Then Sload = ?, and Ssource = ?
EE 212 20102011
What is the node equation at the tap?
Node equation: I1 + IZY = I2 therefore, IZY =
From transformer action:
N1I1 = N2IZY
therefore,
That is, only part of the load current is due to the magnetomotive force that the current in the primary coil exerts on the secondary coil. The remaining portion is direct conduction of the source current to the load.
EE 212 20102011