Losses in transformers include core losses due to hysteresis and eddy currents, and copper losses due to winding resistances. Core losses depend on material, flux density, frequency, and core dimensions. Copper losses depend on current and load. Total losses equal core losses plus copper losses. Efficiency is calculated as the output power divided by the input power. Regulation is the percentage change in output voltage with load. An auto-transformer uses a single winding with a variable point to provide multiple voltages. It has lower costs but also lower safety due to lack of isolation between primary and secondary.

1. Losses
101 / 102 Basic Electrical Technology Dept of E & E, MIT Manipal
• Core Loss (Pi)
 Hysteresis Loss [material, max flux density, frequency, core volume]
 Eddy Current Loss [material, max flux density, frequency, core thickness]
 Constant Flux Unit, hence constant losses
 Minimized using high grade core material, using laminations
• Copper Loss
 Winding Resistances [primary, secondary]
 Current / load dependent, hence variable losses
Total Losses = Core Loss + Copper
Loss

2. Efficiency
101 / 102 Basic Electrical Technology Dept of E & E, MIT Manipal
%100x
Input
Output
% η =
2222222 cos)(cos)(cos θθ VAratingxxIVθIVerOutput Pow FL ===
cueFL WxRIx
e
RIRIRIsCopper Los 2
2
2
2
2
2
2
22
2
21
2
1
==



=



 +=
%x
]W[xP]θVArating[x
θVAratingx
% η
cui
100
cos)(
cos)(
2
2
2
++
=
%x
LossesTotalOutput
Output
100
+
= %x
Input
LossesTotalInput
100
−
=
Where x = fraction of load
I2 = xI2FL
Where Wcu = I2FL
2
R2e = full load copper loss

3. Voltage Regulation
101 / 102 Basic Electrical Technology Dept of E & E, MIT Manipal
• Voltage drop across primary and secondary windings
• Load nature is the deciding factor
• Term used to identify the characteristic of voltage change with loading
%100x
NLV
VNLV
VR%
L−
=
VNL = Terminal voltage on no load
VL = Terminal voltage on load

4. Example
101 / 102 Basic Electrical Technology Dept of E & E, MIT Manipal
A 2 KVA, 230/115V, 50 Hz single phase transformer has an impedance of
(0.1+j0.2)Ω referred to LV side. Determine the efficiency and regulation
when it is supplying full load at 0.8 lagging pf. Assume the constant losses
to be 50W.
Ans: η = 95.22%, VL = 111.535V, Reg = 3.01%

5. Auto Transformer
101 / 102 Basic Electrical Technology Dept of E & E, MIT Manipal
Primary winding
N1 turns
Supply side Load side
Secondary winding
N2 turns
Variable point
• One winding wound over the entire core
• Secondary voltage can be varied using the variable point
•One winding Transformer
 Part of primary winding forms the secondary winding

6. Types
101 / 102 Basic Electrical Technology Dept of E & E, MIT Manipal
LOAD
LOAD
Primary
N1
Secondary
N2
N1 > N2 : Step Down type
N1 < N2 : Step Up type

7. Principle of operation
101 / 102 Basic Electrical Technology Dept of E & E, MIT Manipal
V2
V1
LOAD
I1
I2
I2 - I1
Power Input = V1I1
Power Output = V2I2
Neglecting Losses, leakage reactance & magnetizing
current
1N
2N
2I
1I
1V
2V
==

8. Salient Points
• One winding Transformer
101 / 102 Basic Electrical Technology Dept of E & E, MIT Manipal
• Lesser copper requirement, Copper saving, lower copper losses
• Lower resistance & leakage reactance, higher efficiency
• Smaller in size, lower in cost compared to two winding type
• Lower impedance, chances of short circuit on secondary
• No electrical isolation between pry & sec, risky during high voltages
Advantages:
Disadvantages:
Applications:
Voltage Boosting
Power System interconnection
Voltage Variation

9. Example
A single phase, 200/115V, single winding transformer is supplying a load of
1.15KVA at 115V. Find the current distribution at all parts of the winding,
assuming losses to be zero.
101 / 102 Basic Electrical Technology Dept of E & E, MIT Manipal
LOAD
V1=200V
V2=115V
I2=10A
I2 - I1= 4.25A
I1=5.75A
I1=5.75A

10. Example
A single phase, 200/115V, single winding transformer is supplying a load of
1.15KVA at 115V. Find the current distribution at all parts of the winding,
assuming losses to be zero.
101 / 102 Basic Electrical Technology Dept of E & E, MIT Manipal
LOAD
V1=200V
V2=115V
I2=10A
I2 - I1= 4.25A
I1=5.75A
I1=5.75A