Unit-IV: SINGLE PHASE TRANSFORMER

Transformer is a static device which transfers the A.C electrical power from one circuit to
another circuit without changing the frequency, but voltage levels are changed according to
requirement.
The transformer consists of two windings called as Primary Winding and Secondary Winding. The
winding which is connected to ac supply is called primary winding and the winding which is connected
to load is called secondary winding. The symbol of transformer is as shown below.

If the secondary voltage (V2) is greater than primary voltage (V1), then the transformer is called
step up transformer and if the secondary voltage (V 2) is less than primary voltage (V 1), then the
transformer is called step down transformer. In another way, if the secondary winding turns (N 2) is
greater than primary winding turns (N 1), then the transformer is called step up transformer and if the
secondary winding turns (N2) is less than primary winding turns (N 1), then the transformer is called step
down transformer.

Operating or Working principle of Transformer:

The transformer works on mutual induction principle. The transformer consists of two windings
and are placed on laminated core as shown in figure.(1).

Figure (1)
When an AC supply of V1 volts is connected to primary winding, an alternating flux is set up in
the core. This alternating flux is linked with the secondary winding, an emf will induced across
secondary winding called secondary emf (E2) and the same flux is linking with the primary winding
also, it produces an emf called primary emf (E1). Both the primary and secondary emf directions are
opposite to supply voltage (V1) according to Lenz’s law.

Ideal Transformer:
 A transformer is said to be an ideal transformer when it obeys the following conditions (or)
properties
(i) The resistance of the winding should be zero
(ii) There should not any leakage flux in the transformer core i.e the total flux produced in the core
links with primary and as well as secondary.
(iii) The total losses in the transformer should be zero i.e the input power is equal to output power.
i.e since the winding resistance and magnetic leakage flux is zero, Copper and Iron (Core)
losses are zero respectively. So the efficiency of the idle transformer is 100%.
Since in ideal transformer the output (VA) equal to input (VA) i.e

E 2 I2 = E 1 I1 

but the EMF is proportional to no .of turns (N) i.e E 1  E2 Since it is a ideal transformer,
the primary and secondary winding resistance is zero so E1 V E2 V

 Transformation Ratio (K) =

Transformation ratio is the ratio of secondary turns to primary turns and is represented by K.
EMF equation of 1-Ph Transformer:
When an AC voltage is applied to primary winding of transformer, a sinusoidal flux (Φ m) as shown in
fig.( ) is setup in the transformer core which links with both primary and secondary windings.
Let Φm = Maximum flux in wb = Bm*A
f = Frequency of the supply in Hzs.
N1 = No. of turns in Primary winding.
N2 = No. of turns in Secondary winding.
E1 = EMF across Primary winding.
E2 = EMF across Secondary winding.
According to faraday’s law of electromagnetic
induction principle, the average induced emf per turn is dΦ/dt.
Where dΦ = Change in flux from +Φm to – Φm
= +Φm – (– Φm)
= 2 Φm
and dt = Time required to change in flux from +Φm to – Φm
= 1/2f
Average induced emf per turn =

= 4 Φm f volts

But we know that for sine wave, Form factor =

RMS value of EMF per turn = 1.11 (4 Φm f ) volts

= 4.44 Φm f volts
Since the primary and secondary windings has N1 and N2 no. of turns respectively
RMS value of EMF in primary winding (E1) = 4.44 Φm f N1 volts
Similarly RMS value of EMF in secondary winding (E2) = 4.44 Φm f N2 volts
Note: if Bm is maximum flux density, A is area of the transformer core, then flux Φm = Bm A and EMF (E) = 4.44 Bm A f N.
Prob: A 1-ph transformer has 400 primary and 1000 secondary turns. The net x-sectional area of core
is 60cm2. If the primary winding be connected to a 50Hz supply at 520V, calculated peak value
of flux density in the core, voltage induced in secondary winding.
Sol: Given that

No. of turns at primary winding N1 = 400 ; No. of turns at primary winding N1 = 1000
Area of core A =60cm2 = 60x10-4 m2 ; Primary voltage V1 = 520 V.
We know that transformation ratio (K) =

Voltage induced in secondary voltage E2 = V1 = 520 x

= 208V
Emf induce at primary winding E1 = 4.44Φm f N1 = 4.44(Bm x A) f N1 ( because Φm = Bm x A)
Maximum flux density Bm = E1 / 4.44 A f N1
= 520 / 4.44. 60x10-4.50.400
= 0.976 wb/m2

Prob: A 25KVA, 1-Ph, 50Hz, 6600/600V transformer has a maximum value of flux in the core is
0.08wb, find the no. of turns in each winding.
Sol: Given that 6600/600V i.e Primary voltage V1 = E1 = 6600V
 Secondary voltage V2 = E2 = 600V and Maximum flux Φm = 0.08wb
We know that Primary EMF E1 = 4.44 Φm f N1

No. of turns at primary winding N1 = = 372 turns

Also Secondary EMF E2 = 4.44 Φm f N2

No. of turns at secondary winding N2 =

= 34 turns
Construction of T/F :
The transformer has two different types of constructions. Those are
(i) Core type transformer (ii) Shell type transformer
Core type construction: In core type construction, the magnetic core is built up with laminations in the
form of L-shape limbs/strips or C & I shape limb as shown in fig. (2) and then the L-shaped limbs or C
& I shape limb are joined as rectangular frame as shown in fig.(3).

In core type transformer, the primary and secondary windings placed on each limb to reduce the
leakage flux as shown in fig.(3). Here a part of core is surrounded by the winding and it has only one
magnetic path.
Shell type shell: In shell type shell, the magnetic core is built up with laminations in the form of E & I-
shape limbs/strips or F & L-shape limbs/strips as shown in fig. (4) and then the E & I-shaped limbs or F
& L-shape limbs/strips are joined as rectangular frame as shown in fig.(5).
In shell type transformer, the primary and secondary windings are wound on central limb only as shown
in fig.(5). Here the winding is surrounded by core and the total magnetic flux is splitted into two equal
halves.

Differences between core type and shell type transformer:

Core type Transformer Shell Type Transformer
1. 1.

2. It consists of three limbs and flux in two

2. It consists of two limbs and flux in both
outer limbs is the half of the flux in central
limbs is same.
limb.
3. A part of core is surrounded by winding. 3. The winding is surrounded by the core.
4. It consists of single magnetic circuit. 4. It consists of two magnetic circuits.
5. Its construction is difficult. 5. Its construction is easy.
Transformer on DC :
A transformer cannot works on DC supply. When a DC supply is given to the primary, a flux of
constant magnitude will be set up in the core. Since the flux is not moves in transformer core, the
primary induced EMF E1 which opposes the primary voltage is zero. If the impedance of primary
winding is Z1, then the primary current I 1= (V1-E1)/Z1whcih is very high. Thus it will produce more heat
losses i.e I12R1 and the insulation on primary winding by melt. This result in short circuit of primary
winding turns and primary winding burns. That is why DC supply is never applied to a transformer.

Transformer on No Load:
A transformer is said to be on No Load when the secondary of the transformer open circuited
and secondary current I2 is zero. Under no load condition, the transformer primary draws a minimum
current called no load current I0. This no load current is usually 5% to 7% of rated current because of
this reason, the copper losses (I02R1) are low and are neglected. Since the secondary winding is opened
the secondary copper losses (I22R2) are zero. Therefore the no load current I 0 lags the primary voltage V 1
by an angle ϕ0 as shown in fig.(6).

The no load current I0 has two components:

(i) Active or working component Iw and is in-phase with the supply
voltage V1. This current Iw supplies iron or core losses and small
copper losses which are negligible.
(ii) Reactive or Magnetizing component I µ which produces the flux Φm
and is in phase with the flux Φm.
The fig.(6) shown the vector diagram for transformer on no load. From the vector diagram, Working
component of current Iw = I0 Cosϕ0 and the Magnetizing component of current Iµ = I0 Sinϕ0.

No load current I0 =

No load input power P0 = V1 I0 Cosϕ0 watts

Transformer on Load:
When the transformer secondary is connected to a load, a current I 2 flows through the load as
shown in fig.(7). The secondary current I 2 depends on terminal voltage V2 and load impedance. The
phase angle (ϕ2) between the I2 and V2 depends on nature of the load i.e whether the load is resistive or
inductive or capacitive. The operation of transformer on load is explained as follows:
When the transformer is on no load, the transformer draws no load current I o from the supply.
This no load current setup flux Φm in the core.

When the transformer is on load, a current I2 flows through the secondary winding of
transformer. This secondary current setups its own flux Φ 2 in the core, but the direction of this flux Φ 2 is
opposite for the flux set up by the no load current I 0 according to Lenz’s law. Since the flux Φ 2 is
opposite for the flux Φm, the resultant flux (Φm – Φ2) decreases and causes the reduction in self induced
EMF (E1). This causes the additional current I2’ from the supply (called as load component of primary
current). This additional current crates a flux Φ 2’ which is equal in magnitude but opposite to flux Φ 2.
Now the flux in magnetic core is only Φm because the fluxes Φ2 and Φ2’ cancels each other.

From the above discussion it is clear that the flux in the transformer core is constant when the
load on the transformer is varies, hence the core or iron losses are constant irrespective of load
variations. Now the total primary current is the vector sum of I 0 and I2’. The phasor diagrams for
various loads on transformer are as shown below.

To the draw the vector diagram, the flux is taken as reference. The EMFs E 1 and E2 are opposite to
supply voltage V1 according to Lenz’s law. The no-load current I 0 lags the voltage V1 by angle ϕ0. The
additional current I2’ is opposite to current I2. Therefore the total current I1 is the sum of I0 and I2’. When
the transformer is loaded with resistive load, the secondary current I 2 is in phase with secondary voltage
V2. For inductive load, the secondary current I 2 is lags the secondary voltage V 2 by an angle ϕ2 and for
capacitive load, the secondary current I2 is leads the secondary voltage V2 by an angle ϕ2 The phase
angle between the voltage V1 and current I1 is ϕ1.
Losses and Effiency of Transformer:
Since the transformer is a static device, the mechanical losses in transformer are zero. Therefore
the transformer has only Iron losses and Copper losses.
Iron losses (Wi): These losses are also known as constant losses or core losses because these losses
are depend on flux but flux is constant irrespective of load variations in transformer. The Iron losses are
divided into Hysteresis and Eddy current.
Hysteresis losses: The hysteresis losses are given by an empherical formula ie.
Wh Bm1.6 f V watts or
Wh KhBm1.6 f V watts
Where Kh = hysteresis loss constant and is depends on type of material used for transformer core.
Bm = Maximum flux density ; V = volume of core in m3 and f = Frequency in Hzs.
The hysteresis losses are minimized by design the T/F core using the silicon/CRGO steel material.
Eddy current losses: The Eddy current losses are given by an empherical formula ie.
We Bm2 f 2 V2 t2 watts or We Ke Bm2 f 2 V2 t2 watts
Where Ke = Eddy current loss constant and t =Thickness of the transformer core laminations.
Bm = Maximum flux density ; V = volume of core in m3 and f = Frequency in Hzs.
Generally, the thickness of laminations of T/F core may varies from 0.35mm to 0.5 mm for 50Hz. The
eddy current losses are reduced by laminating the transformer core. The total iron losses are reduced by
designing the T/F core with silicon/CRGO steel laminations. Practically, the Iron losses of transformer
are determined by conducting Open Circuit (O.C) test.
Effect of Variations of Frequency & Supply Voltage on Iron Losses:
Let Bm = Maximum flux density;
V = volume of core in m3 and t = thickness of laminations
f = Frequency in Hz.
Kh = hysteresis loss constant and Ke = eddy current loss constant
The iron losses of a transformer are Wi = Wh + We ------------- (1)
The hysteresis losses are given as Wh Bm1.6 f V
Wh KhBm1.6 f V
Wh KhBm1.6 f
The Eddy current losses are given as We Bm2 f 2 V2 t2
 We Ke Bm2 f 2 V2 t2
We Ke Bm2 f 2
Iron losses Wi = KhBm1.6 f + Ke Bm2 f 2 ----------------- (2)

We know that E1 = V1 = 4.44 Bm A f N1  Bm =

(2)  Wi = Kh( )1.6 f + Ke ( )2 f 2

= Kh( )1.6 V11.6( )1.6 f + Ke ( )2 V12( )2 f 2

= KhV11.6f 1-1.6 + Ke V12 f2

Wi = KhV1.6f -0.6
+ Ke V2 ------------------ (3)
From the above equation it is clear that,
(i) when the supply voltage varies, both hysteresis and eddy current losses varies.
(ii) when the frequency varies, hysteresis losses varies but eddy current losses independent of
frequency variations.

Copper losses (Wcu): These losses occur due to resistance of the transformer windings. Let R 1 and R2 are
internal resistance of primary and secondary windings, I 1 and I2 are full currents of primary and
secondary windings.
The primary winding copper losses are given by I12R1
The secondary winding copper losses are given by I22R2.
Total copper losses (Wcu) = I12R1 + I22R2  Full load copper losses
Total copper losses at x load (Wcu) = x2 (I12R1 + I22R2)
The copper losses are also known as variable losses because they are proportional to square of the load
current. Practically, these losses are determined by conducting Short Circuit (S.C) test on transformer.
Total losses: The total losses in T/F are the sum of iron losses and full load copper losses i.e
Total losses = Iron losses (Wi) + full load copper losses (Wcu)  For full load
Total losses for x load = Wi + x2 full load copper losses
Effiency (η): The Effiency of the transformer are defined as the ratio of output power (P 0) to input
power (Pi) i.e

Effiency (η) =

Output Power = x V2 I2 Cosϕ = x (VA rating of transformer) power factor

Input Power = Output power + Total losses
= x (VA rating of transformer) p.f + Iron losses + x2 full load Copper losses
Effiency(η)=
Condition for max. Effiency(ηmax):
The Effiency of the transformer are defined as the ratio of output power to input power i.e

Effiency (η) =
Input = V1I1CosΦ1 , Iron losses = Wh + We and Total copper losses Wcu = I12R01

Effiency (η) =

1
To get max. Effiency, differentiating above equation on both sides w.r.t to I and equate it to zero i.e

2
i 1 01
W =I R

i.e for maximum Effiency, the condition is full load copper losses(variable losses) are equal to iron

losses(Constant losses)

Maximuneffiency (ηmax) =

Referred Values:
(i) All parameters referred to secondary side:
If R1 is the primary winding resistance and is transferred to
secondary, let this transferred resistance as R1’ as shown in fig.(11).
I1 and I2 are the full load primary and secondary currents. Now
copper losses of primary winding is equal to copper losses of
secondary winding i.e
I12 R1 = I22 R1’
  R1’ = K2 R1
Total equivalent resistance referred to secondary is R02 = R2 + R1’
= R2 + K 2 R 1
Where Transformation Ratio (K) =
Similarly
Reactance transformed to secondary is X1’ = K2 X1
Total equivalent resistance referred to secondary is X02 = X2 + X1’ = X2 + K2 X1
Impedance transformed to secondary is Z1’ = K2 Z1
Total equivalent impedance referred to secondary is Z02 = Z2 + Z1’ = Z2 + K2 Z1
or
Z02 =
(ii) All parameters referred to primary side:
If R2 is the secondary winding resistance and is transferred to primary, let this transferred
resistance as R2’ as shown in fig.(12). I1 and I2 are the full load primary and secondary currents. Now
copper losses of secondary winding is equal to copper losses of primary winding i.e
I22 R2 = I12 R2’
 R2’ = R2/ K2
Total equivalent resistance referred to primary is R01 = R1 + R2’

= R1 + R2/ K2
Where Transformation Ratio (K) =

Similarly
Reactance transformed to primary is X2’ = X2/ K2
Total equivalent resistance referred to primary is X01 = X1 + X2’ = X1 + X2/ K2

Impedance transformed to primary is Z2’ = Z2/ K2

Total equivalent impedance referred to primary is Z01 = Z1 + Z2’ = Z1 + Z2/ K2
or
Z01 =

EQUIVALENT CIRCUIT OF TRANSFORMER:

The equivalent circuit of a transformer is used to predetermine the behavior or performance of
the transformer under various operating conditions. The transformer primary and secondary winding has
resistance (R1 & R2), leakage reactance (X1 & X2). Under no load condition, the transformer draws no
load current I0 and this current is resolves into working component of current I w and magnetizing
component of current Iµ. The working component of current Iw produces resistance of R0 and
magnetizing component of current Iµ produces resistance of X0. These R0 and X0 components are
connected in parallel across the primary circuit as shown in fig.(4).

Fig. 4: Equivalent ckt of T/F

Since the effect of no load current I0 is negligible on R1+jX1 impedance, so the no load parallel branch is
moved towards supply voltage V1 as shown below

The above figure represents the exact equivalent circuit of 1-ph transformer.
(i) Equivalent circuit referred to primary:
If all the secondary quantities are referred to the primary, we get the equivalent circuit of the
transformer referred to the primary as shown in Fig. (5).
i. Let R2’, X2’ and ZL’ are the resistance, reactance and load impedance of secondary winding
referred to primary and are determined by dividing R2, X2 and ZL with K2
i.e R2’ = R2 / K2 , X2’ = X2 / K2 and ZL’ = ZL / K2
ii. Similarly voltage and EMF are divided by K and current is multiplied by K. i.e V 2’ = V2 / K ,
E2’ = E2 / K and I2’ = K I2.
 Fig.5: Equivalent circuit of the transformer referred to the primary
we know that R01 = R1 + R2’ & X01 = X1 + X2’, the above citcuit diagram can redrawn as shown below

(ii) Equivalent circuit referred to secondary:

If all the primary quantities are referred to the secondary, we get the equivalent circuit of the
transformer referred to the secondary as shown in Fig. (6). Note that when primary quantities are
referred to secondary, resistances/reactance/impedances are multiplied by K 2, voltages are multiplied by
K and currents are divided by K. i.e

Fig.6: Equivalent circuit of the transformer referred to the secondary

We know that R1’ = R1* K2 and X1’ = X1* K2 , the above citcuit diagram can redrawn as shown below
Voltage regulation of Transformer (or) Expression for voltage regulation from equivalent
ckt of T/F:
When a transformer is loaded with constant supply voltage (Primary voltage), the terminal
voltage (Secondary voltage) changes depending upon the load and its power factor.
The voltage regulation of transformer is defined as ‘ the change in secondary voltage from no load to
full load with respect to full load voltage’.
Let the secondary voltage at no load is 0V2 = E2 and V2 = Secondary voltage on full load.
then Voltage Regulation =

Consider the approximate equivalent circuit of T/F referred to secondary and its vector diagram as
shown below

From the vector diagram oc = oa + ab + bc

E2 = V2 +
= V2 +
E2 - V2 =
Secondary voltage V2 = E2 -
 % age Voltage Regulation

+ for lagging or inductive loads and – for leading or capacitive loads

For lagging or inductive loads, the voltage regulation is +ve and for leading or capacitive loads, the

voltage regulation may be –ve.

Condition for Maximum and zero voltage regulation:

(i) Condition for maximum voltage regulation:

The expression for voltage regulation of T/F is Voltage Regulation =

--(1)
to get the condition for maximum voltage regulation, differentiate the equation (1) w.r.t to Φ 2 and equate
it to zero. i.e
(Voltage regulation) =0

=0

=0

- I2 =0

TanΦ2 =  Φ2 =

i.e the voltage regulation of T/F is maximum, when the load angle Φ2 =

(ii) Condition for zero voltage regulation:

The zero voltage regulation of T/F may obtained for leading or capacitive loads,
Voltage Regulation = =0
I2 =0

TanΦ2 =  Φ2 =

i.e the voltage regulation of T/F is maximum, when the load angle Φ2 =

Why Transformer rating in KVA?

The copper losses of a transformer depend on current and iron losses on voltage. Hence, the
total losses of a transformer depend on volt-ampere but not on the phase angle between the voltage and
current i.e the total losses of transformer are independent of load power factor. That is why the rating of
the transformer is in KVA but not in KW.
Magnetizing Current:
When the ac power source is connected to a transformer, a current flows in the primary
winding, even when the secondary winding is open-circuited. This is the current required to produce the
flux in ferromagnetic core. This current is known as magnetization current.
It consists of these components:
1)The magnetization current Im :The current required to produce the flux in the core of the transformer.
2)The core-loss current Ie :The current required to make up for the hysteresis and eddy current losses.
For any coil, E = N*d(φ)/dt. Always.
If E is the applied voltage φ will be the flux produced.
If φ is the applied flux to the coil, E will be the emf produced.
Lets look at this step by step.
1. When there is no load on the secondary side (i.e. No load condition) E is the emf applied to the
primary coil, then the coil will produce(φ) flux in the core.
You can see that, if E is sinusoidal, φ will be co sinusoidal.
The coil will consume that much current as required to produce the φ flux.
since,φ = N*I / reluctance of core we can calculate what will be the current required to produce the flux
φ, This current is the magnetizing current.
But since there is co sinusoidal flux in the core, the same rule applies to the secondary coil and there
will be Emf induced in the secondary coil given by
E2 = N2*d(φ)/dt.
2) When you connect load to the secondary, then there will be secondary current.
The secondary current flowing in the secondary coil will produce flux φ2 which will be in opposition
with the flux previously being produced by the primary coil (by virtue of the primary magnetizing
current). So, in effect the net flux in the core will reduce to φ- φ2.
But Since Emf E is still being applied to primary coil, it demands that the flux linkage of primary coil
still be φ. So what happens is the current in primary coil increases so that it now produces the flux: φ +
φ 2 so that the net flux linkage of the coil (that is flux in the core) becomes: φ + φ2 – φ2 = φ again.
The additional current required in primary coil to restore the flux will not be equal to the current in the
secondary unless the no. of turns are same in both of the coil. Hence, the secondary current that flows
when load is connected to the secondary coil will be reflected in the primary coil on top of the already
present magnetizing current (not as a replacement for it).

Effect of nonlinear B-H curve of magnetic core material:

B-H curve :
The curve plotted between flux density (B) and magnetizing force (H) of a material is called B-H Curve.
The shape of curve is non-linear. This indicates that relative permeability of a material is not constant
but it varies.

.e., μr =
B-H curves are very useful to analyze the magnetic circuit. If value of flux density and dimension of
magnetic circuit is known than from B-H curve total ampere turn can be easily known.

Magnetic hysteresis
The phenomenon of lagging behind of induction flux density (B) behind the magnetizing force (H) in
magnetic material is called magnetic hysteresis.
 Hysteresis loop is a four quadrant B – H graph from where the hysteresis loss, coercive force and
retentively of magnetic material are obtained.
 To understand hysteresis loop, we suppose to take a magnetic material to use as a core around which
insulated wire is wound.
 The coils is connected to the supply (DC) through variable resistor to vary the current I. We know that
current I is directly proportional to the value of magnetizing force (H).
 When supply current I = 0, so no existence of flux density (B) and magnetizing force (H). The
corresponding point is o in the graph above.
When current is increased from zero value to a certain value, magnetizing force and flux density both
are set up and increased following the path o to a.
 For a certain value of current, flux density becomes maximum (Bm). The point indicates the magnetic
saturation or maximum flux density of this core material. All element of core material get aligned
perfectly.
When the value of current is decreased from its value of magnetic flux saturation, H is decreased along
with decrement of B not following the previous path rather following the curve a to b.
 The point b indicates H = 0 for I = 0 with a certain value of B. This lagging of B behind H is called
hysteresis.
 The point b explains that after removing of magnetizing force (H), magnetism property with little
value remains in this magnetic material and it is known as residual magnetism (Br) or residual flux
density.
 If the direction of the current I is reversed, the direction of H also gets reversed. The increment of H in
reverses direction following path b – c decreases the value of residual magnetism that gets zero at point
c with certain negative value of H. This negative value of H is called coercive force (Hc)
 Now B gets reverses following path c to d. At point‘d’, again magnetic saturation takes place but in
opposite direction with respect to previous case. At point‘d’, B and H get maximum values in reverse
direction.
 If decrease the value of H in this direction, again B decreases following the path d. At point e, H gets
zero valued but B is with finite value.
 The point e stands for residual magnetism (-Br) of the magnetic core material in opposite direction
with respect to previous case.
 If the direction of H again reversed by reversing the current I, then residual magnetism or residual
flux density (-Br) again decreases and gets zero at point ‘f’ following the path e to f.
 Again further increment of H, the value of B increases from zero to its maximum value or saturation
level at point a following path f to a.
 Hard and soft material hysteresis loop are given below.

Harmonics in magnetization current:

 Non-linear devices generate harmonic currents.
 The transformer cores have a non-linear behavior in the saturated part of the magnetizing curve,
This non-linear behavior generates harmonic currents.
 Wave-shape of the magnetizing current depends on the applied voltage waveform and the B-H
curve of the core material.
 Since induced voltage is expressed by e = N(dφ/dt), phi being the flux and N being the turns,
Flux wave-shape is decided by the voltage wave-shape.
Thus, if the voltage is sinusoidal, the flux will also be sinusoidal. If the voltage is square, the flux will
be trapezoidal

 Now, the Flux Density (B) is proportional to Flux for a given core area.
 B is related to H by the B-H curve of the core material.
 H is related to the magnetizing current by the number of turns with fixed core length.
 Thus, if the B-H curve is straight line (linear), then the magnetizing current wave-shape will be
sinusoidal for a sinusoidal applied voltage.
 However, in practical cases, we design voltage transformers for economy and to reduce size and
price, we push up the working value of B well above the linear region of the B-H curve.
 Hence we get a non-sinusoidal magnetizing current waveform even if the applied voltage is
sinusoidal.
 A non-sinusoidal magnetizing current implies harmonic content in it.
 If you design the transformer with low values of B such that it is in the linear region of the B-H
curve, then you will get sinusoidal magnetizing current with sinusoidal applied voltage.