Transfers electric power from one circuit

to another without a change of frequency.

It accomplishes this by electromagnetic

induction where the two electric circuits

are in mutual inductive influence of each

other.

The physical basis oftransformer between twocircuits linked by acommon magnetic flux.

It consists of twoinductive coils which areelectrically separatedbut magnetically linkedthrough a path of lowreluctance.

Simple Elements of

a Transformer

Two coils having

mutual inductance

Laminated steel

core

Two types of Transformer

Core-type

Shell-type

Spiral-core or wound-core type

Core-type (L-type)

The windings surround a considerable

part of the core

Does not heat up easily

Bulky

Shell-type (E-I core)

The core surrounds a considerable

portion of the winding

Heat up easily

Compact

Where: Eave = average induced emf in coil

N = number of turns in coil

t = time for flux to reach its maximum value

Φ = flux density in maxwells; weber- drop 10 ‾ ⁸

Volts 10NE 8

AVE

t

Volts 10

41

NE 8

AVE

f

Volts 10N4E 8

AVE

f

Since for a sine wave, the effective voltage E

is equal to 1.11 times the average voltage

Eave,

Volts 10N44.4

E 11.1E

8

ave

f

Primary Induced Voltages

Secondary Induced Voltages

Where: Np = number of primary turns

Ns = number of secondary turns

Volts 10N44.4E 8pp

f

Volts 10N44.4E 8ss

f

The 2300 volt primary winding of a 60 cycle

transformer has 4800 turns. Calculate (a) the

mutual flux, (b) the number of turns in the

230 volt secondary winding.

The maximum flux in the core of a 60 cycle

transformer that has 1230 primary turns and46 secondary turns is 3.76x10⁶ maxwells.

Calculate the primary and secondary induced

voltages.

A single phase transformer has 400 primary and

1000 secondary turns. The net cross-sectional

area of the core is 60 cm². If the primary

winding be connected to a 50 Hz supply at

520 V, calculate (a) the maximum value of

the flux density in the core in Wb/m² (b) the

voltage induced in the secondary winding.

The primary and secondary induced voltages

are related to each other by the ratio of the

number of primary and secondary turns.

S

p

S

p

N

N

E

E

Volts 10N44.4E 8pp

f

Volts 10N44.4E 8ss

f

The volts per turn in a 25 cycle 2400/230 volt

transformer is 8. Calculate (a) the primary

and secondary turns, (b) the maximum flux

in the core.

Since transformer is stationary, it is extremely

efficient because the only losses are those

that occur in the copper windings and those

iron losses.

sssppp PFIEPFIE

sspp IEIE

P

S

S

p

I

I

E

E

P

S

S

p

I

I

N

N

The primary and secondary currents of a

transformer were measured and found to be

3.8 and 152 A, respectively. If the secondary

load voltage is 116 V, what is the primary

voltage?

The turn ratio, induced-voltage ratio and the

current ratio is called the ratio of

transformation, and is represented by the

symbol .

P

S

S

p

S

p

I

I

N

N

E

Ea

a

Step –down transformer

When the primary voltage is reduced to a lower

secondary voltage

> 1

2300/230 volt transformer

Step-up transformer

When the primary voltage is raised.

< 1

13200/66000 volt transformer

a

a

s

s

p

p

Z

E

Z

E 22

ss

s

pp

p

jxr

E

jxr

E 22

2222s ps pp sp s EjxErEjxEr

Real to real Imaginary to imaginary

22s pp s ErEr

s

p

s

p

r

r

E

E2

2

s

p2

s

p

r

r

E

E

s

p2

r

ra

22s pp s EjxEjx

s

p

s

p

x

x

E

E2

2

s

p2

s

p

x

x

E

E

s

p2

x

xa

Referred to the primary side

0V ZIV seqppp az

0V ZIV seqppp az

loaddropapplied VVV

0Vxxrr θIV ss2

ps2

ppp aaja

0VjXR IV seqpeqppp a

Referred to the primary side

Referred to the secondary side

0V ZIV seqssp aa z

0V ZIV seqssp aa z

loaddropapplied VVV

0V

xx

rr θI

Vs

2

ps

2

pss

p

aj

aa

0VjXR IV

seqseqssp

a

Referred to the secondary side

0V ZIV

szeqssp

a

Unity power factor

0V Z0IV szeqppp a

0V90X 0I0R 0IV seqppeqppp a

2eqpp2

eqppsp XIRIVV a

Lagging power factor

0V ZIV szeqppp a

0V90X I0R IV seqppeqppp a

2eqpps2

eqppsp XIsinVRIcosVV aa

Leading power factor

0V ZIV szeqppp a

0V90X I0R IV seqppeqppp a

2eqpps2

eqppsp XIsinVRIcosVV aa

Where: VNL= VP

VF = Vrated

100V

V-VRegulation Voltage

FL

FLNL

A 25 kVA 2300/230 volt distribution

transformer has the following resistance and

leakage-reactance values: rp = 0.8 ohm, xp =

3.2 ohms, rs = 0.009 ohm, xs = 0.03 ohm.

Calculate the equivalent values of

resistance, reactance, and impedance in (a)

in primary side, (b) in secondary side.

Calculate the equivalent resistance and

reactance drops for a secondary load current

of 109 A: (a) in primary side, (b) in secondary

side.

Calculate the percent regulation: (a) for

unity power factor, (b) for a lagging power

factor of 0.8, (c) for a leading factor of

0.866.

A 10 kVA, 2400/240 V single-phase transformer

has the following resistances and reactances.

Find the primary voltage required to produce

240 V at the secondary terminals at full load

when the power factor is 0.8 lagging.

rp = 3 Ω

xp = 15 Ω

rs = 0.03 Ω

xs = 0.15 Ω

Given: 1-10 kVA transformer

2400/240 V

rp=3.66 Ω xp=3.66 Ω

rs=0.0282 Ω xs=0.0413 Ω

Find: (a) Reqp (g) V.R. at 0.8 pf lagging

(b) Xeqp (h) V.R. at unity pf

(c) Zeqp (i) V.R. at 0.8 pf leading

(d) Reqs

(e) Xeqs

(f) Zeqs

Core Loss It is the constant loss.

Where:

kh = proportionality constant that depends upon the volume of the core

ke = proportionality constant that depends upon the thickness of lamination, resistivity of steel

f = frequency

β = flux density available

226.1c

ehc

losscurrent eddy loss hysteresisloss core

P

PPP

PPP

fkfk eh

Hysteresis Loss

Purely magnetic

Results due to the tiny magnetic particles that

produces a kind of molecular friction as they

tend to change alignment with the rapid

reversals of alternating current

Eddy-current loss

Electromagnetic in character

Caused by the flow of currents in the iron in

exactly the same way as in the transformer

windings.

6.1hP fkh

22eP fke

From

Therefore,

10N44.4E 8 f

f

Ek

f

E

NA

f

44.4

10

N44.4

10EA

8

8

Substituting the value of the flux density:

22

6.0

6.1

1

2

2

6.1

226.1cP

Ekf

Ek

f

Ekfk

f

Efkk

fkfk

eh

eh

A 110 V, 30 cps distribution transformer was

tested at its core loss was found to be 480 W.

of the 480 W, 360 W was due to hysteresis

loss. If the applied voltage and frequency are

doubled, what will be its core loss assuming

that there has no considerable changes in

flux.

Voltage ratings must be known.

1. From the circuit shown, the variac should

be set to its minimum output value

2. Adjust the output of the variac until the

reading of the voltmeter is equal to the

rated voltage of the low side.

3. Take the readings of the meters.

_________ I

_________ V

_________ P

oc

oc

oc

Core loss of the transformer

Copper Loss

Variable Loss

eqs2

s

eqp2

p

s2

sp2

pcu

RI

RI

rIrIP

kVA ratings and voltage ratings must be

known.

1. From the circuit shown, the variac should

be set to its minimum output value

2. Adjust the output of the variac until the

reading of the ammeter is equal to the

rated current of the high side.

3. Take the readings of the meters.

_________ I

_________ V

_________ P

sc

sc

sc

Copper loss of the transformer

From the short-circuit test,

22eqpeqpeqp

sc

sceqp

2sc

sceqp

RZX

I

VZ

I

PR

100PPP

P

100PP

P

100P

PE

cucout

output

losses totaloutput

output

input

output

Given: 1-10 kVA transformer

2400/240 V

Reqp = 6.48 ohms

Xeqp = 9.37 ohms

Pc = 120 W

Find:

a. E at H.L. w/ 0.8 pf

b. E at F.L. w/ 0.8 pf

c. E w/ 25% O.L. w/ 0.8 pf

100RIPkI

kIE

cosE k let

100RIPcosIE

cosIEE

100P

PE

eqp2

pcp

p

pp

eqp2

pcppp

ppp

input

output

cuc

eqp2

pc

eqp2

pc

eqp2

pp2

eqp2

pcp2

2eqp

2pcp

eqpppeqp2

pcp

p

PP

RIP

RkIkP

0RkI2IkRkIkPIk

0)RIP(kI

)RI20)(k(kI-)(k)RIP(kI

dI

dE

Maximum efficiency occurs when Pc = Pcu

Since, to have a maximum efficiency, Pc =

Pcu,

100P2P

P

100P

PE

cmaxout

maxout

input

maxoutput

eqs

cs

ceqs2

s

ccu

R

PI

PRI

PP

Multiplying Vs/1000 at both sides,

cu

cRmax

eqsFL

cFLsss

eqs

c

FL

FLsss

P

PkVAkVA

RI

P

1000

IV

1000

IV

R

P

I

I

1000

V

1000

IV

2

Given: 50 kVA, 4600/230 V

O.C.T.: Voc = 230 V, Poc = 285 W

S.C.T.: Vsc = 150 V, Psc = 615 W, Isc = 10.87 A

Find:

a. kVAmax at 0.84 pf

b. Maximum efficiency

Percent Impedance

Percent Reactance

Percent Resistance

100V

ZI100

V

ZI%Z

s

eqss

p

eqpp

100V

XI100

V

XI%X

s

eqss

p

eqpp

100V

RI100

V

RI%R

s

eqss

p

eqpp

Where:

Ip = rated primary current

Vp = rated primary voltage

Is = rated secondary current

Vs = rated secondary voltage

%Z = percent impedance

Zeqp = equivalent impedance referred to the primary side

Zeqs = equivalent impedance referred to the secondary side

%X = percent reactance

Xeqp = equivalent reactance referred to the primary side

Xeqs = equivalent impedance referred to the secondary side

%R = percent resistance

Reqp = equivalent resistance referred to the primary side

Reqs = equivalent resistance referred to the secondary side

A 500 kVA, 13200/2400 V, 60 Hz single-phasetransformer has 4% reactance and 1%resistance. The leakage reactance andresistance of the low voltage winding are0.25 ohm and 0.055 ohm respectively. Thecore loss is 2000 W.

a. Find the leakage reactance and resistanceof the high voltage winding.

b. Calculate the efficiency of this transformerat full load and with 80% power factor.

c. What will be the maximum efficiency withthe same power factor?

It is the ratio of the energy delivered

by the transformer in a 24-hr period

to the energy input in the same

period of time.

)tP(P outhrs 24in output

100

100PPP

PA.D.E.

hrs 24in cu hrs 24in chrs 24in out

hrs 24in output

hrW

hrW

rating)(t)VA )(load of size(P hrs 24in output

)(24)P(P chrs 24in c

)(t)P()load of size(P FLcu2

hrs 24in cu

A 5 kVA, 2300/230, 60 cycle standard

distribution transformer was tested with the

following results: short-circuit test input =

113 W, open-cicuit test input = 40 W. If this

transformer operates with the following

loads: 1 ½ times rated kVA, pf = 0.8, 1 hr; 1

¼ times rated kVA, pf = 0.8, 2 hrs; rated kVA,

pf = 0.9, 3 hrs; ½ times rated kVA, pf = 1, 6

hrs; ¼ times rated kVA, pf = 1, 8 hrs; no

load, 4 hrs. Calculate its all-day efficiency.

It is generally used in low power applications

where a variable voltage is required. The

autotransformer is a special type of power

transformer. It consists of only one winding.

By tapping or connecting at certain points

along the winding, different voltages can be

obtained.

Step-down Autotransformer Step-up Autotransformer

2

1

2

1

N

N

V

Va

a

1

N

N

I

I

1

2

1

2

11input IVP

a

11PP inputdtransforme

dtransformeinputconducted P-PP

Power

transformed

Power

conducted

An autotransformer of given physical

dimensions can handle much more load

power than an equivalent two-winding

transformer.

An autotransformer is to be used to supply a

load of 80 A at 160 V from a 220 V line. If the

transformer is ideal, determine (a) the turns

ratio between the series and the common

windings, (b) current of the series winding,

(c) power conducted to the load, (d) kVA

rating as an ordinary two winding

transformer, (e) kVA rating as an

autotransformer.