Unit 5
Single Phase Transformer
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
- 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
- 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
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)
= - N1Øm w Cos wt
Sin (90-wt)
= - Sin (wt-90)
E1 = N1Øm 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
Voltage and current ratios
Voltage ratio is the ratio of primary voltage to secondary voltage in voltage transformer.
Current ratio is the ratio of primary current to secondary current in current transformer.
VP/VS = IS/IP
As voltage ratio is equal to turn ratio so
NP/NS = IS/IP
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
Q3)When operated at 120 V in the primary of an iron core transformer, the current in the primary is 5 amps. Find the current in the secondary if the voltage is stepped up to 400 V.
Sol: VP/VS = IS/IP
120/400 = IS/5
IS = 1.5A
Q4) A transformer with 400 turns on the primary and 60 turns on the secondary draws Solution: 0.5 amps from a 120 V line. Find Is.
Sol: NP/NS = IS/IP
400/60 = IS/0.5
IS = 3.34A
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
Reference’s
1.Theory and Problems of Basic electrical Engineering : NagrathKothri
2.Electrical technology by B.L.Theraja, A.K.Theraja: S chand publication
3.Basic Electrical Engineering : V K Mehta
4.Basic Electrical Engineering : S K Sahdev: Pearson publication
5.Electrical Safety, Fire Safety engineering S.Rao, Khanna Publication
