Unit-3
Transformers
In a magnetic circuit, this field is represented by magnetomotive force. It is analogous to the electromotive force in electrical circuit. This field is responsible to “set up” certain flux, which in turn gives rise to certain flux density B. Note that, here H is cause and B is its effect. The amount of flux which can be setup in a material is determined by an inherent property of the material, called as permeability, denoted by µ.
B = µH
When external magnetic field H is applied to a material, all the domains align in a particular direction, setting up net flux in the material. Due to domain alignment B (i.e. the magnitude of B) increases. However, after a certain value of B, the slope of B − H curve starts reducing as shown in Figure below
Fig: Magnetization curve
B-H Curve:
We cannot measure B and H directly. Further if we have transformer, we only have terminal measurements are with us. Hence, it is required to process the signals to get values of B and H. From Faraday’s law, V = N dφ/dt . Also, B is directly proportional to flux φ.
Fig: B-H curve
The lag or delay of a magnetic material known commonly as Magnetic Hysteresis. The Magnetic Hysteresis loop above, shows the behaviour of a ferromagnetic core graphically as the relationship between B and H is non-linear. Initially both B and H will be at zero, point 0 on the magnetisation curve.
If the magnetisation current, i is increased in a positive direction to some value than the magnetic field strength H increases linearly with i and the flux density B also increases (curve from point 0 to point a) as it heads towards saturation.
Now if the magnetising current in the coil is reduced to zero, the magnetic field circulating around the core also reduces to zero. However, the coils magnetic flux will not reach zero due to the residual magnetism present within the core and this is shown on the curve from point a to point b.
To reduce the flux density at point b to zero we need to reverse the current flowing through the coil. The magnetising force which must be applied to null the residual flux density is called a “Coercive Force”. This coercive force reverses the magnetic field re-arranging the molecular magnets until the core becomes unmagnetized at point c.
An increase in this reverse current causes the core to be magnetised in the opposite direction and increasing this magnetisation current further will cause the core to reach its saturation point but in the opposite direction, point d on the curve.
This point is symmetrical to point b. If the magnetising current is reduced again to zero the residual magnetism present in the core will be equal to the previous value but in reverse at point e.
Again, reversing the magnetising current flowing through the coil this time into a positive direction will cause the magnetic flux to reach zero, point f on the curve and as before increasing the magnetisation current further in a positive direction will cause the core to reach saturation at point a.
Then the B-H curve follows the path of a-b-c-d-e-f-a as the magnetising current flowing through the coil alternates between a positive and negative value such as the cycle of an AC voltage. This path is called a Magnetic Hysteresis Loop
I Ø transformers and electrostatics
- Types of transformers
Acc to input supply : I phase and 
phase
Acc to construction : core and shell type
Acc to 0/P : step up and step down
Construction of transformers (study only for MCQs)
- Laminated steel core
Material used for core is (silicon steel) it is used for its (high permeability) and (low magnetic reluctance) 
magnetic field produced is very strong
The core is formed of (stacks of laminated thin steel sheets) which are electrically isolated from each other. They are typically (0.35 to 0.5 mm thick)
We can used 2 ‘L’ shaped sheets or 2 
shaped sheets for laminations
Construction and types
- There a 2 types of winding
- Concentric or cylindrical
- Sandwiched type
- Cylindrical
L.V. = low voltage H.V. = high voltage are mounted on same limb to minimum leakage .
L.V. Winding placed inside and H.V. Winding placed outside with (proper insulation between the winding as it is easy to insulated L.V. Winding) than H.V. Winding.
2. Sandwiched
The H.V. And L.V. Winding are divided into no. Of small coils and there small windings are interleaved.
(the top and bottom winding are L.V. Coils because they are close to core)
- Transformer tank : wholes assembly of winding and core placed inside the Transformer tank (sheet metal tank) which is filled with Transformer oil or insulating oil which acts as an (insulator or coolant ) MCQ
- Transformer oil : (The function of oil is to remove efficiently the heat generated in core and in winding)
- Moisture should not be allowed which creeps the insulation which achieved by closed Transformer tank.
(To increase cooling surface are tubes or fins are provided)
- Conservator tank : above tank T/F tank there is one small tank in which same empty space is always provided above the oil level. (this space is required for oil to expand or contract due to temperature change) MCQ
However during contraction outside air can have moisture which will deteriorate the insulating properly of oil.
- Breather : the air goes in or out through the breather (To reduce the moisture content of their air . Same drying agents such as (silica gel or) calcium chloride) is used in the breather (The dust particles present in air are also removed by breather)
- Buccholz Relay : (for incipient (slowly increasing) faults
There is pipe connecting rain tank and conservator. On the pipe a protective device called Buccholz Relay is mounted.
When the Transformer is about to be faulty and draw range current the oil becomes very hot and decompose.
During this process different types of gases are liberated.(The Bucchoz Relay get operated by these gases) and gives an alarm to the operator. ɡȴ the fault continues to persist then there lay will trip off main circuit breaker to protect the Transformer.
- Explosion Vent :
An explosion Vent or relief value is the bent up pipe filled on the main tank.
(The explosion vent consist of aluminium of oil ) when the T/F becomes faulty the cooling oil get decomposed and various types of gases are liberated
(ɡȴ the gas pressure exodus certain level then the aluminium of oil (diagram) in explosion vent will burst) to release pressure. The will save main tank from getting damaged.
- Single phase transformer - for MCQs
(Symbol and principal of operation)
It is a static device which can transfer electrical energy from one ac circuit to another ac circuit without change in frequency
It can increase or decrease the voltage but with corresponding decrease or increase in current
It works on Principle 
“Mutual Induction”
A major application of transformer is to increase voltage before transmitting electrical energy over a long distance through conductors and to again reduce voltage at place where it is to be used.
Symbol
- Primary winding : (ac supply side) For MCQs
The winding which is connected to supply is called primary side
2. Secondary winding : (load side) for MCQs
The winding which is connected to load is called as secondary winding
Principle of operation
- Working : For MCQs
It works on principle of mutual Induction ie “when 2 coil are inductively coupled and if current in one coil is changed uniformly then an emf get induced in another coil”
- When alternating voltage V1 is applied to primary winding am alternating current I1 flows in it producing alternating flux in the core.
- As per Faradays Law of Electromagnetic Induction 1 an emf E1 is induced in the primary winding E = N1
- The emf induced in the primary winding is nearly equal and opporite to applied voltage V1
- Assuming leakage flux to be negligible almost whole flux produced in primary winding links with secondary winding thence emf induced in the secondary winding E2 = N2
EMF equation
As primary winding excited by the a sinusoidal alternating voltage an alternating current flows in the winding producing an alternating varying flux Ø
Ø = Øm sin wt
As per Faradays Law of Electromagnetic Induction emf E1 is induced
E1 = N1 
E1 = N1 
(Ø Sin wt)
= - N1 Øm w Cos wt
Sin (90-wt)
= - Sin (wt-90)
E1 = N1 Øm w Sin (wt-90)
w = 
E1 = 
N1 Ø m Sin (wt-90)
Max value of E1 = E max
Is when Sin (wt-90) = 
1
E1 max = 
N1 Ø m
Hence rms value of induced EMF in primary winding
E1 rms = 
= 
E1 = 4.44 F Ø m N1
Similarly RMS value of induced EMF in secondary wdg Is
E2 = 4.44 F Ø m N2
Transformer on No Load:
When an actual transformer is put on load we get losses (iron loss) in core and (copper loss) in primary and secondary windings. In NO LOAD (primary current is not purely reactive) the input current has to supply iron loss to core and small amount of copper loss in primary.
The no-load power is W0 = V1 I0 cos Φ0
I0 – no load primary input current
Cos Φ0 – power factor at no-load condition
I0 lags V1 by Φ0 < 900
The phasor is shown below.
Phasor for No-Load Transformer
From above figure we can conclude that I0 has two components Iw(in phase with V1) Iµ in quadrature with V1
Iw = I0ws φ0 ( iron loss component )
Iµ = I0 cos φ0 ( magnetising component )
I0 = I2µ + I2w
The no-load primary current I0 is very small as compared to full-load primary current. And hence, no-load cu loss is very small. So, primary input is practically equal to iron loss of transformer.
Q1>. A 2500/200 V transformer draws a no-load primary current of 0.5 A and absorbs 400 W. Find magnetising and loss currents.
Sol: Iron-loss current = No load input(W) / Primary voltage
= 400/2500 = 0.16 A
I20 = I2w + I2µ
Iµ = √I20 – I2w
= √ (0.5)2 – (0.16)2
Iµ = 0.473 A
Q2>. A 1-φ transformer has 1000 turns on primary and 200 on secondary. The no load current is 4 amp at p.f of 0.2 lagging. Find primary current and pf when secondary current is 280 A at pf of 0.6 lagging.
Sol : cos-1 0.6 = 53.130 (sin φ = 0.8)
I2 = 280/-53.130A
Φ = cos-1 0.2 = 78.50
Sin φ = 0.98
I1 = I0 + I’2
I’2 = (I2/K) ( -53.130
K = N1/N2 = 1000/200 = 5
I’2 = 280/5 (-53.130
I’2 = 56(-53.130
I1 = I0 + I’2
= 4(0.20 – j0.98) + 56(0.6 – j0.8)
= 0.80 – j3.92 + 33.6 – j44.8
I1 = 34.4 – j48.72
I1 = 59.64 ( -54.770
I lags supply voltage by 54.770
Transformer ON – LOAD :
When secondary winding is loaded then the current in it is setup(IQ). The current I2 is in phase with V2 if load is non-inductive it lags if load is inductive & leads if load is capacitive.
An mmf is set up at secondary and hence φ2 flux is induced which is in opposition to main flux φ due to I0. The flux φ2 weaken φ and hence primary back emf E1 is reduced. An additional primary current I’2 is introduced exactly at instance when I2 is introduced. So, that magnetic effect of I2 is neutralized by I’2.
Φ2 = φ’2
N2 I2 = N1 I’2
I’2 = N2/N1 x I2 = K2 I2
The phasor is shown below,
(a) (b)
Fig. Phasor for Load Transformer
From fig (a) (non – inductive load)
I2 is in phase with E2(=V2)
I’2 is antiphase with but equal in magnitude
I1 is vector sum of I’2 and I0, but lags V1
From fig (b), Inductive load
I2 lags V2
I’2 is not in phase with I2 but equal in magnitude
I1 is vector sum of I’2 & I0 but lags V1
I’2/I2 = I1/I2 = N2/N1 = K
Q>. A 1- φ transformer with ratio of 440/110-V takes a no-load current of 6 A at 0.3 pf lagging. If secondary supplies 120 A at pf of 0.8 lagging. Find current taken by primary.
Sol>.
Cos φ2 = 0.8
Φ2 = 36.540
Cos φ0 = 0.3
Φ0 = 72.540
K = V2/V1 = 110/440 = ¼
I’2 = KI2 = 120 x ¼ = 30 A
I0 = 6A
Angle between I0 & I’2
= 72.54 – 36.54
= 35.670
From vector diagram,
I1 = √(62 + 302 + 2 x 6 x 30 cos 35.67)
I1 = 35.05 A
An ideal transformer has no losses i.e. its winding have no magnetic leakage and no ohmic resistance. Hence, an ideal transformer has only two purely inductive coils wound on a loss-free core.
Fig . Ideal Transformer
For above transformer when secondary is open and primary is having input sinusoidal voltage V1. An alternating current flows due to difference in potential. As primary coil is purely inductive so, Iµ current is drawn through it. This current is very small and logs V1 by 900.
The current Iµ produces magnetic flux φ and hence are in same phase. The flux is linked with both the windings and hence, self induced emf is produced E1 which is equal and opposite of V1. Similarly E2 is induced in secondary which is mutually induced emf E2 is proportional to rate of change of flux and number of secondary windings.
The phasor is shown below.
Phasor for Ideal Transformer
Rest all phasors are discussed in the sections below.
Transformer Equivalent Circuit :
The basic transformer and its equivalent circuit both are shown below,
Fig. Equivalent Transformer Circuit
Iµ - magnetising component of current
Iw = working component
R0 – Non- inductive resistance
I0 – No load current
X0 = E1/I0. R0 = E1/Iw
E2/E1 = N2/N1 = K
E’2 = E2/K = E1
V’2 = V2/K
I’2 = K I2
The total equivalent circuit is again given as,
But the above circuit is exact equivalent but harder to solve so, it can be further simplified as,
Z = Z1 + Zm || ( Z’2 + Z’L )
= Z1 + Zm(Z’2 + Z’L)/Zm + (Z’2 + Z’L)
Z’2 = R’2 + jX’2
Zm = impedance of exciting circuit
V1 = I1[ Z1 + Zm(Z’2 + Z’L) / Zm + (Z’2 + Z’L) ]
Core or Iron Loss :
This includes both hysteresis loss and eddy current loss. Iron or core loss is found from open circuit test.
Hysteresis Loss(Wh) = nB1.6max f V Watt
Eddy current Loss(We) = P B2max f2 t2 Watt
(2). Copper Loss :
This is due to ohmic resistance of the transformer windings.
Total cu loss = I21 R1 + I22 R2
= I21 R01 + I22 R02
Cu loss α I2
The value of cu loss is found from short-circuit test.
Voltage Regulation:
It can be explained in terms of various parameter of transformer.
(1). When transformer has constant primary voltage.
In this case secondary voltage decreases
0V2 = No-load secondary terminal voltage
0V2 = E2 = EK1 = KV1
V2 = secondary voltage on full-load.
Voltage Regulations is change in secondary terminal voltage from no-load to full load per unit full load voltage.
% regn down = 0V2 – V2/0V2 x 100
% regn up = 0V2 – V2/V2 x 100
(2). Voltage regulation in terms of primary values.
The secondary no-load terminal voltage as referred to primary is E’2 = E2/K = E1 = V1
For secondary full load voltage referred to primary is V’2 = V2/K
% regn = V1 – V’2/V1 x 100
% regn = I1 R01 cosφ + I1 X01 sinφ / V1 x 100
(3). As the transformer is loaded, to maintain a constant output voltage, the primary voltage should be increased. Here the regulation is given as
% regn = V’1 – V1 / V1 x 100
It can also be defined as the change in primary voltage from no-load to full-load at given power factor to maintain a constant output per unit primary voltage.
Q. A – 100 KVA transformer has 500 turns on primary and 80 turns on secondary. The primary and secondary resistances are 0.3 and 0.01 Ω respectively and the corresponding leakage reactances are 1.1 and 0.035 Ω. The supply voltage is 2400 V. Find
(i). Equivalent impedance referred to primary
(ii). Voltage regulation and the secondary terminal voltage for full load having pf 0.8 lagging?
Sol. Equivalent impedance referred to primary
Z01 = √R201 + X201 = R01 + jX01
R01 = R1 + R2/K2 = 0.3 + 0.01/K2 = 0.69 Ω
K = 80/500 = 4/25
X01 = X1 + X2/K2 = 1.1 + 0.035/(0.16)2 = 2.467 Ω
Z01 = 0.69 + j2.46
(ii). Secondary terminal voltage Z02 = K2 Z01
Z02 = 0.018 + j 0.063
= 0.065 ( 74.050
No-load secondary voltage = KV1
= 0.16 x 2400 = 384 V
I2 = 100 x 103/384 = 260.42 A
Full load voltage drop referred to secondary
= I2 (R02 cosφ – X02 Sinφ)
Cosφ = 0.8
Φ = 36.860
Sinφ = 0.6
= 260.42(0.018 x 0.8 – 0.063 x 0.6)
= - 6.094 V
% regn = -6.094/384 x 100
= -1.587
Secondary terminal voltage on-load
= 384 – (-6.094)
= 390.09 V
Efficiency:
Basically, efficiency is defined as
n = output/input
But for transformer there are small amount of losses so the improved way to find efficiency is
n = output/output + losses
n = output/output + cu loss + iron loss
Or n = Input – losses/Input
= 1 – Losses/Input
Condition for maximum efficiency :
For n to be maximum dn/dI1 = 0
(Ww) cu loss = I21 R01 or I22 R02
Iron loss = Hysteresis loss + Eddy current loss
= Wn + We = Wi
n = Input – losses/Input
Primary Input = V1I1 Cosφ1
n = V1 I1 Cosφ1 – losses/V1 I1 cos φ1
n = V1 I1 cos φ1 – I21 R01 – Wi / V1I1 cosφ1
= 1 – I1R01/V1cosφ1 – Wi/V1I1cosφ1
Differentiating w.r.t I1 both sides of above equation we have
Dn/dI1 = 0 – R01/V1cosφ1 + Wi/V1I21 Cosφ1
For max value dn/dI1 = 0
R01/V1cosφ1 = Wi/V1I21 cosφ1
Wi = I21 = I21 R01
Hence,
Wi = Wcu
Iron loss = copper loss
The value of output current for maximum efficiency will be
I2 = √Wi/R02
The maximum efficiency can also be given as,
nmax = full load x √ Iron loss / F.L .cu loss
Or
nmax = R’ x full load KVA x pf / R’ x full load KVA x pf + Wi + Wcu x 100
R’ = ratio of actual to full load KVA
Wi = iron loss (KW)
Wcu = copper loss (KW)
Q1). In a 50 KVA, 2200/200 V, 1-φ transformer, the iron and full-load copper losses are 400 W and 450 W respectively. Calculate n at unity power factor on (i). Full load (ii). Half-full load?
Sol. (i). Total loss = 400 + 450 = 850 W
F.L output at unity power factor = 50 x 1
= 50 KVA
n = 50 / 50 + .850 = 50/50.850 = 0.98 = 98%
(ii). Half full load, unity pf
= 50 KVA/2 = 25 KVA
Cu loss = 400 x (1/2)2 = 100 W
Iron loss is same = 450 W
Total loss = 100 + 450 = 550 W
n = 25/25 + 0.55 = 25/25.55 = 0.978 = 97.8 %
Q>. A 40 KVA 440/220 V, 1- φ, 50 Hz transformer has iron loss of 300 W. The cu loss is found to be 100 W when delivering half full-load current. Determine (i) n when delivering full load current at 0.8 lagging pf (ii) the percentage of full-load when the efficiency will be max.
Sol. Full load efficiency at 0.8 pf
= 40 x 0.8/(40 x 0.8) + losses
Full load cu loss = (440/220)2 x 100
= 400 W
Iron loss = 400 + 300
= 700 W
n = 40 x 0.8/(40 x 0.8) + 0.7 = 97.8 %
(ii). KVA for maximum / F.L KVA = √ iron loss / F.L cu loss
= √300/400 = 0.866
Transformers.
It is a transformer in which only one winding is common it both primary and secondary .because it has one windy it uses loss copper and is cheaper.
It can be step -up and step dawn auto transformer . The circuit for both is shown below.
Diagram:-
AB-Primary winding
BC- Secondary windings
Negating iron loss and no-load current-
=
=
=K
The current in CB(
) is vector difference (
-
) 
>
An auto transformer when compared to as ordinary 2- winding transformer having same output, given higher efficiency and is smaller in size with superior voltage regulation.
FOR AUTO TRANSFORMER:
From above fig (a)& (b)
Weight of copper in AC 
(N1-N2)I1
Weight of copper in BC
(N2(I2-I1)
Total weight of copper 
(N1-N2) I1+N2(I2-I1) 
auto transformer
For 2- winding transformer:
Weight of 
on primary 
N1I1
Weight of 
on primary 
N2I2
Total weight of 
N1I1+N2I2
=
= 
= 1-
=1-k
= K , 
=
Weight 
in auto transformer =(1-k)*(weight of 
in ordinary transformer)
Saving = Wo-Wa
= W0-(1-k)W0
Saving = Kwo
Power transformer inductively =(1-k) input hence, saving will increase at K approaches unity.
Q. An auto transformer suppliers load of 4KW at 100v at unity pf. IF the applied primary voltage is 220v. Calculating power transferred to load (a) Inductively (b) conductively.
Soln, Power transferred inductively = Input(1-k)
Power transferred conductively = K* Input
K= 
=
Input= Output =4KW
Inductively transferred power =4(
)
=3.82KW
Conductively transferred power = 
*4
= 0.182 kw
