Giao trinh bai tap english for students of physics – vol 1

Transformers – Introduction
 Transferring electrical energy from one circuit to another through
time-varying magnetic field.

ECE430

 Applications: both power and communications fields.
 In transmission, distribution, and utilization of electrical energy: step-

Power Circuits and Electromechanics

up or step-down voltages at a fixed frequency (50/60 Hz), at power of
hundreds of watts to hundreds of megawatts.

Dr. Nam Nguyen-Quang
Fall 2009

 In communications, transformers can be used for impedance
matching, DC isolation, and changing voltage levels at power of a few
watts over a very wide frequency range.

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 This course concerns only power transformers.
Lecture 4

1

Ideal transformer


i1

wound as shown. Ignore losses, stray

+
v1
–

capacitance, and leakage flux.

N1

i2
+
v2
–

N2

 Magnetic permeability is infinite or zero
reluctance.

d
dt

2

Ideal transformer (cont.)

 Consider a magnetic core with two coils

v1 t   N 1

Lecture 4

v 2 t   N 2

d
dt



v1 t  N 1

a
v2 t  N 2

v1 N1

a
v2 N2

i1

i1
N
1
 2 
i2
N1
a



i2
+

v1

v2

v1 t i1 t   v2 t i2 t   0

–

v1 N1

a
v2 N 2

+

+

v1

v2

–
N1:N 2
i1

i1 N 2 1


i2 N 1 a

v1 t i1 t   v 2 t i2 t 

Ideal

i2

–

–
N1:N 2

a is called turns ratio.
 Total mmf is given by

Ideal

+

 It can be shown, for an ideal transformer

mmf  N1i1  N 2 i2  R  0

i1 t 
N

1
 2 
i2 t 
N1
a

k 1
Lecture 4

L2
i1
v
1

 2 
i2
v1
a
L1

3

Impedance-changing property of ideal transformer



L1 N 22  L2 N 12

Lecture 4

4

Impedance matching

 Consider an ideal transformer with resistive load accross winding 2

 The impedance-changing property can be used for maximizing power

v2
 RL
 By Ohm’s law,
i2
 Subsituting v2  v1 a and i2  ai1

transferring between to windings, or matching impedances.

i1

Ideal

+

2

i2
+

v1

v2

–

–

N 
v1
 a 2 R L   2  R L
i1
 N1 

RL

 An ideal transformer is placed between power source (impedance Zo )
and load (impedance ZL). Turns ratio is so selected that

N1:N 2

Z o  N 1 N 2  Z L
2

 Ex. 3.7: Two ideal transformer (each of ratio 2:1) and one resistor R
are used to maximize power transfer. Find R.

 The discussion can easily be extended to systems with complex load.
It can be verified that

V1  N 2

I 1  N 1

2

Load resistance 4  together with R referred to the input side is (R +

2

 V2  N 2 

 Z L  a 2 Z L
 
 I 2  N1 

Lecture 4

4(2)2)(2)2. For maximum power transfer,

10  4 R  64
5



R  13.5 

Lecture 4

6

1

Power transformer

More pictures

 Two windings mounted on a magnetic
core, minimizing leakage flux.
 “Primary” winding (N1 turns) connected
to power supply, “secondary” winding (N2

Small power

turns) connected to load circuit.

Small 3-phase

Control

Cast resin

 Assuming an ideal transformer: no leakage flux, winding resistances
are neglected, magnetic core has infinite permeability, and is lossless.
 Let v1(t) = Vm1cost is the voltage applied to the primary winding, it can
be shown that

Vm1  2fN 1 max

or

V1  4.44 fN 1 max

Lecture 4

Example

110 kV, oil

10 kV, oil

7

500 kV, oil

Lecture 4

8

Equivalent circuit of transformer with linear core

 Ex. 3.8: Given N1, N2, core cross-sectional area, mean core length, B-

 Consider now a transformer with leakage flux and winding resistances.

H curve, and the applied voltage. Find maximum flux density, and

Equivalent directly derived from physical model is simple but somewhat useless.

required magnetizing current.

The equations on the secondary side is mulplied by a (= N1/N2) and i2 is replaced

V1  4.44 fN 1 max

 max

Hence,

where

by i2/a, to derive a more useful equivalent circuit.

V1  230 V, f  60 Hz, N1  200

i1

230

 4.32  10 3 webers
4.44  60  200

+
v1

L1 – aM

R1

i2
+

2
a2R2 a L2 – aM

+
RL

v2

+
i1

v1

aM

i2/a

av2

a2RL

4.32 10 3
Therefore,
Bm 
 0.864 webers/m 2
0.005
The required H m  0.864  300  259 At/m , the peak value of

 L1 – aM is termed the leakage inductance of winding 1, a2L 2 – aM is termed

magnetizing current is (259)(0.5)/200 = 0.6475 A. Hence, Irms = 0.46 A is

“referred” (to side 1) leakage inductance of winding 2. aM is the magnetizing

–

–

–

N1:N 2

–

inductance, and its associated current is called magnetizing current.

the magnetizing on the primary side.
Lecture 4

9

Equivalent circuit of transformer with linear core (cont.)

Lecture 4

10

Transformer under sinusoidal steady-state conditions

 There are losses in the magnetic core due to hysteresis and eddy current.

 For steady-state operation, impedances and phasors can be used in

These losses are very difficult to calculate analytically. The sum of these losses

the equivalent circuit.

represents the total loss in the magnetic circuit of the transformer, and depends
+

be placed in parallel with the magnetizing inductance aM to account for them.
i1

R1

L1 – aM

2
a2R2 a L2 – aM

+
v1

V1
Ideal

+
Rc1

(aM)1

–

v2

–

–

I1

ja2xl2

I2 a
Rc1

jXm1

+

I2
+

aV 2

V2

–

–

Ideal

ZL

N1:N 2

where

RL

N1:N 2

 Real load RL and its associated voltage and current can be retained by

  L1  aM   xl1  Leakage reactance of winding 1
 aM   X m1  Magnetizing reactance referred to winding 1
 L2  M a   xl 2  Leakage reactance of winding 2
 a 2 L2  aM   a 2 xl 2  Leakage reactance of winding 2 referred to side 1

referring them back to the secondary side, using an ideal transformer.
Lecture 4

a2R2

–

i2
+

av2

jxl1

R1

only upon the value of B m. They are called core or iron losses. A resistance can

11

Lecture 4

12

2


Transformer under steady-state (cont.)

Approximate equivalent circuit

 All quantities can be referred to winding 1
jxl1

R1
+

 Magnetizing branch makes computation somewhat difficult, hence this

a2R2

I1

branch is moved to the terminals of winding 1, yielding an approximate

ja2xl2

V1

Rc1

equivalent circuit, with no serious numerical error introduced.

+

I2 a
a2Z

jXm1

aV2

L

–

R1
+

–

ZL

jXm1/a 2

+

V2

–

I1

V1 Rc1

–

I2 a
jXm1

R1eq  R1  a 2 R2

aV2

x1eq  xl1  a 2 xl 2

–

13

Open- and short-circuit tests of transformers

+

a2ZL

–

Lecture 4

aV2

jx1eq

R1eq
+

I2
Rc1/a2

a2ZL

–

jxl2

R2

aI1

V1 a

+

I2 a
jXm1

–

jxl1/a2

R1/a2
+

ja2xl2

a2R2

I1

V1 Rc1

 Or they can be referred to winding 2

jxl1

Lecture 4

14

Open-circuit test

 Parameters in equivalent circuit can be determined by two simple

 The test is performed with all instrumentation on the LV side with the

tests: open-circuit test and short-circuit test.

HV side being open-circuited. Rated voltage is applied to LV side. Voc,

 In power transformers, the windings are called high-voltage (HV) and

Ioc, and Poc are measured with the meters.

low-voltage (LV) windings.

A

Rc 

I oc

W

Voc

V oc

Hence,

IR

V

Rc
LV

IX

Voc2
Poc

IR 

Voc
Rc

I oc  I R  I X

I oc

I X  I oc2  I R2

Voc

V
X m  oc
IX

Xm

HV

IR

IX

Rc

 Rc and Xm are the values referred to the LV side.

Open-circuit test

Equivalent circuit
Lecture 4

15

Short-circuit test

Lecture 4

16

Example

 All the instrumentation is on the HV side. Rated current is supplied to

 Ex. 3.9: Given OC and SC tests’ readings. Find equivalent circuit

HV side. Vsc, Isc, and Psc are measured with the meters.

parameters referred to the HV side.

A

Xm

I sc

W

Req

From OC test

Xeq

Vsc

Vsc

Rc 

220 2
50
2

I X  12  0.227  0.974 A

V

IR 

 968 

Xm 

220
 0.227 A
968

220
 225.9 
0.974

From SC test
HV

Req 

Psc
I sc2

Z eq 

LV

V sc
I sc

Req 
X eq  Z eq2  Req2

Z eq 

 Req and Xeq are referred to the HV side.
Lecture 4

60

17 2

 0.2076 

15
 0.882 
17

X eq  0.882 2  0.2076 2  0.8576 
17

Lecture 4

18

3


Efficiency and voltage regulation

In-class quiz

 Efficiency is defined as the ratio of output power to input power.



 Problem 3.22 and 3.23.

Pout
Pout
Pout

100% 
 100%
Pin Pout  losses
Pout  Pc  Pi

Losses are copper loss Pc and iron losses Pi.
 Alternatively, if input power is known,



Pin  Pc  Pi
 100%
Pin

 Voltage regulation is defined as

% voltage regulation 

Vno load  Vload
 100%
Vload

Lecture 4

19

Lecture 4

20

4