Transformers are necessary in power systems to step up voltage for efficient transmission and step down voltage for safe distribution. They operate on the principle of electromagnetic induction and have a core made of laminated steel, with coils of wire wound around the core. The ratio of the number of turns in the primary and secondary windings determines the voltage transformation ratio. Power losses in transformers include copper losses from winding resistance and iron losses from hysteresis and eddy currents in the core. Efficiency is highest when load matches the ratio of iron to copper losses.

1. Transformers
Workshop on Basic Electrical Engineering held at
VVCE, Mysuru, on 30-April-2016
R S Ananda Murthy
Associate Professor
Department of Electrical & Electronics Engineering,
Sri Jayachamarajendra College of Engineering,
Mysore 570 006
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2. Learning Outcomes
After completing this lecture the student should be able to –
State why transformers are necessary in power systems.
Describe the principle of operation of transformers.
Describe the construction of core and shell types of
single-phase transformers.
Calculate the induced E.M.F. in a transformer.
Define voltage regulation of a transformer.
Describe various power losses in a transformer.
Find the efficiency of a transformer.
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3. Photograph of a Power Transformer
Power transformers are used at transmission level.
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4. Photograph of a Distribution Transformer
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5. Necessity of Transformers in Power Systems
Transformers are necessary to step up from generation
voltage which is typically 11 kV to transmission voltage
levels which are typically 220 kV or 400 kV.
They are also required to step down voltage from
transmission levels which are typically 220 kV or 400 kV to
distribution levels which are typically 11 kV, 415 V
(three-phase) or 230 V (single-phase).
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6. Core and Shell Types
Core Type
Cross-section of core type transformer
L.V. Winding H.V. Winding
L.V. Winding H.V. Winding
Shell Type Cross-section of shell type transformer.
Core
Yoke
Yoke
Core
Laminations
Yoke
Yoke
Core
Core
Core
Laminations
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7. Core Type Transformer
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8. Shell Type Transformer
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9. Principle of Operation of Transformer
Load
When supply is given to the primary winding having T1 turns,
an alternating flux is established in the transformer core. Due to
this, there will be change of flux linkage in the primary and also
in the secondary winding. Then, according to Faraday’s law,
voltage is induced in both the primary and secondary windings.
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10. Induced E.M.F. in Transformer
When the primary applied voltage is sinusoidal, the mutual flux
φ also varies sinusoidally. So, let
φ = Φmax sinωt (1)
where ω = 2πf and f = supply frequency. Let the flux linkage of
the primary and secondary be
λ1 = φT1 and λ2 = φT2 (2)
According to Faraday’s law, when φ changes, voltage is
induced in both primary and secondary. The voltage induced in
the primary is given by
e1 = −
dλ1
dt
= −T1
d
dt
(Φmax sinωt) = ωT1Φmax sin(ωt −90◦
) (3)
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11. Induced E.M.F. in Transformer
Similarly the voltage induced in the secondary is given by
e2 = −
dλ2
dt
= −T2
d
dt
(Φmax sinωt) = ωT2Φmax sin(ωt −90◦
) (4)
This shows that both e1 and e2 lag behind the mutual flux φ by
90◦. The R.M.S. values of e1 and e2 are given by
|E1| =
2πfΦmax T1
√
2
= 4.44fΦmax T1 (5)
|E2| =
2πfΦmax T2
√
2
= 4.44fΦmax T2 (6)
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12. Voltage Transformation Ratio
In a practical transformer, the voltage drop in primary and
secondary winding impedances is very small — typically less
than 5% of the rated voltage. So, by applying K.V.L. to the
primary and secondary we get
V1 ≈ E1 and V2 ≈ E2 (7)
Hence we can write
|V1|
|V2|
≈
|E1|
|E2|
=
T1
T2
= a12 (8)
This is known as voltage transformation ratio.
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13. Current Transformation Ratio
Due to good magnetic material used for the core, it can be
shown that the reluctance of the core is almost zero. This
means that
F1 = F2 =⇒ |F1| = T1|I1| = |F2| = T2|I2|
where F1 is the primary M.M.F. and F2 is the secondary M.M.F.
Therefore,
|I1|T1 ≈ |I2|T2 =⇒
|I1|
|I2|
=
T2
T1
= a21 =
1
a12
(9)
This shows that current transformation ratio is the reciprocal of
voltage transformation ratio.
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14. Voltage Regulation
Voltage regulation of a transformer is defined as
Regulation =
|V2NL|−|V2FL|
|V2FL|
×100%
where V2NL = secondary terminal voltage when there is no
load, V2FL = secondary terminal voltage when there is full
load.
Ideally the regulation should be zero which means that the
secondary terminal voltage should not vary from no-load to
full-load.
Typically the regulation of practical transformers would be
less than 5%.
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15. Power Losses in Transformer
Copper Loss (Pcu) in primary and secondary windings.
Iron Loss (Pi) which has two components
Hysteresis loss Ph = khtfB1.6
max
Eddy current loss Pe = kef2t2B2
max
where Pe = eddy current loss, Ph = hysteresis loss, t =
thickness of laminations used in the core. kh = hysteresis loss
constant, and ke =eddy current loss constant which are
constants for a particular material used for the core.
Bmax = Φmax /Ai is the maximum flux density of in the core.
This shows that by using very thin laminations, iron loss can be
reduced and by using good conducting material like copper or
aluminium for the windings, copper loss can be reduced.
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16. Copper Loss (Pcu)
This is nothing but I2R loss occurring in the primary and
secondary winding resistances. Since I2R power loss is
proportional to square of current in the windings, it is obvious
that this power loss varies with load.
If Pcu(r) is the copper loss in the transformer at full-load current,
then, at any multiple x of full load, the copper loss will be
Pcu = x2Pcu(r).
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17. Iron Loss (Pi)
Maximum value of flux in the transformer core is given by
Φmax ≈
|V1|
4.44fT1
(10)
Since the primary supply voltage |V1| and frequency f are
constant, Φmax will also be constant.
The iron loss in a transformer is given by
Pi = Pe +Ph = kef2
t2
B2
max +khtfB1.6
max (11)
This shows that, at constant frequency and supply voltage
the iron loss Pi will remain practically constant even if the
load varies.
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18. Efficiency of Transformer
The efficiency of transformer at any multiple x of full-load is
given by
η =
Pout
Pin
=
x|V2r |·|I2r |cosθ
x|V2r |·|I2r |cosθ +Pi +x2Pcu(r)
(12)
which is same as
η =
x|Sr |cosθ
x|Sr |cosθ +Pi +x2Pcu(r)
where x = any multiple of full load, |Sr | = rated volt-amperes of
the transformer, cosθ = power factor of the load, Pi = iron loss
of the transformer which is constant, and Pcu(r) = copper loss in
the transformer at rated load.
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19. Condition for Maximum Efficiency
At maximum efficiency, dη/dx = 0. Applying this condition we
get the condition for maximum efficiency as
Pi = x2
Pcu(r) =⇒ x =
Pi
Pcu(r)
This gives the multiple x of full load at which efficiency will be
maximum. In this equation, for full load we have to take x = 1,
for half-load x = 0.5, for quarter load x = 0.25, and so on.
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20. License
This work is licensed under a
Creative Commons Attribution 4.0 International License.
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