Real Transformer on -

Load
 A transformer which experiences leakage effects, i.e. not all the flux
produced by the primary mmf threats the secondary windings
 The flux produced by the demagnetising mmf and the compensating mmf
equally have a leakage effect
 The magnetic core is not perfect, i.e. not100% permeable
 Both the primary and the secondary windings experience losses,
precisely copper losses
 Effect of load current considered on both secondary and primary
 Real Transformer on Load
A real transformer is characterised by the
following:
 The effect of leakage flux is considered.
𝟏 𝟐  The effect of primary and secondary windings resistances is
considered
 Therefore 𝑉 ≠ 𝐸
 All core losses considered, i.e. (hysteresis and eddy current
losses) in the core and winding Cu losses
𝑹𝟏 𝑿𝟏 𝑹𝟐 𝑿𝟐
𝑰𝟏 𝑰𝟐 A simplified model depicting the characteristics of a transformer

𝑰𝟎 𝑰𝟐 = 𝑰𝑳 𝒁𝑳 The characteristics of the load current 𝐼

𝑰𝒄 𝑰𝝁
(phase angle, magnitude) flowing in the secondary
𝒗𝟏 𝑹𝒄 𝒗𝟐
𝑿𝟎 𝑬𝟏 𝑬𝟐 when the transformer is loaded depend upon the
~ characteristics of the load impedance connected.

𝑍 = 𝑍 = 𝑅 =𝑅 Pure resistive load

𝑳 =0
𝝋𝟐 = 𝟗𝟎𝟎
Pure capacitive load

𝑍 − 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑖𝑣𝑒 𝑙𝑜𝑎𝑑
𝑍 − 𝑖𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑒 𝑙𝑜𝑎𝑑
Pure Inductive load 𝝋𝟐 = 𝟗𝟎𝟎

The magnitude will depend upon the ratio of load impedance to terminal voltage or current drawn
 Transformer on-
Load
With Winding
Resistance
𝑅 and 𝑅 represent the resistances of
a transformer on the primary and
𝟏
secondary windings respectively
𝟏 𝑳𝒐𝒂𝒅
𝑊ℎ𝑒𝑛 𝑍 = 𝑅
𝑊ℎ𝑒𝑛 𝑍 = 𝑍 = 𝑅 − 𝑗𝑋

𝑊ℎ𝑒𝑛 𝑍 = 𝑍 = 𝑅 + 𝑗𝑋

Primary windings induced emf and winding resistive

drop

𝜑 < 90
𝑳 𝑳
𝑳 𝜑 < −90
 Example #: 5-1
A single phase transformer supplies a current of 80A at 80% power factor lagging. The
transformer has 1000 turns in the primary windings and 250 turns in the secondary windings.
Assuming that the transformer draws a primary current of 30A at 0.7 power factor lagging,
determine:
a. the hysteresis angle of displacement
b. The no-load current
c. Draw a phasor diagram

Solution

⎖
𝟐

⎖
𝟐

⎖
𝟎
 Example #: 5-2

A single phase transformer supplies a current of 80A at 70% power factor leading, at 240V, 50Hz. The
transformer has 1000 turns in the primary windings and 200 turns in the secondary windings. The respective
primary and secondary windings resistances are 0.2Ω and 0.5Ω. Assuming that the transformer draws a no-
load current of 2A at a hysteresis angle of displacement of 80 degrees. determine:
a. the current drawn from the primary 𝐼
b. the voltage on the primary 𝑉
c. Draw a phasor diagram

Solution
 Transformer on-
With Primary and Secondary Winding Load
Reactance

For a lagging power factor load

⎖
 Example #: 5-3

A single phase transformer supplies a current of 80A at 70% power factor lagging, at 240V, 50Hz.
The transformer has 800 turns in the primary windings and 200 turns in the secondary windings.
The respective primary and secondary windings inductances are 0.021H and 0.007H. Assuming that
the transformer draws a no-load current of 1.4A at a hysteresis angle of displacement of 85
degrees and the primary and secondary resistances negligible, determine:
a. the current drawn from the primary 𝐼
b. the voltage on the primary 𝑉
c. Draw a phasor diagram
 With Primary and Secondary Winding
Transformer on-
Impedance Load

𝑰 𝟏 𝑿𝟏

𝑬 𝟏 𝑰 𝟏 𝒁𝟏
𝑰𝟎 𝑰𝟏 𝑹 𝟏
𝑰𝒄 𝑰𝝁 𝑬𝟏
𝑹𝒄
𝑰𝟏

𝑰𝒐

𝑰𝒄 𝛷
𝑰𝝁

𝑰𝟐
𝑬𝟐
𝑉 𝒁𝑳
𝑽𝟐
𝑰𝟐 𝑹 𝟐
𝑰 𝟐 𝒁𝟐
𝑰 𝟐 𝑿𝟐
𝑍 = 𝑅 + 𝑗𝑋 – primary winding impedance The load can be presented as:
𝑍 = 𝑅 + 𝑗𝑋 – secondary winding impedance 𝑆 , 𝑆 , 𝑃 , 𝑃 𝑍 𝑍 𝑤𝑖𝑡ℎ 𝑝. 𝑓.
𝑉 = 𝐼 𝑅 + 𝑗𝑋 +𝐸 𝐸 = 𝐼 𝑅 + 𝑗𝑋 +𝑉 𝑉 = 𝐼 (𝑅 ±𝑗𝑋 )
 Example #: 5-3

A single phase transformer supplies a load of 1000KW at 80% power factor lagging, at 400V, 50Hz. The
transformer has 800 turns in the primary windings and 200 turns in the secondary windings. The respective
primary and secondary windings impedances are 8 < 60 Ω and 4Ω . Assuming that the transformer draws a
no-load current of 3A at a hysteresis angle of displacement of 88 degrees. determine:
a. the current drawn from the primary 𝐼
b. the voltage on the primary 𝑉
c. Draw a phasor diagram
 Example #: 5-4
Group discussion – 4
exercise Group #: 1 minutes

Draw a phasor diagram for a leading power

factor load Group #: 2
Draw a phasor diagram for a unit power factor
load
Group #: 3
Draw a phasor diagram for a pure capacitive
load Group #: 4
Draw a phasor diagram for a pure inductive
load
Group #: 5
Draw a phasor diagram for an inductive load
neglecting the no-load current
 Transformer Equivalent
Resistance
Equivalent Circuit with
The two resistances can be transferred to one of the two windings, resistances referred to the
which simplifies calculations since lumped on one side (winding) primary windings

Total resistance as referred to the

primary windings: 𝑅 = 𝑅 + 𝑹⎖ 𝟐
The procedure implies that the resistance from one of the windings
is being moved to the other winding 𝑹𝟐
𝑅 =𝑅 + 𝟐
When resistance from secondary is moved it becomes a referred 𝑲
resistance Equivalent Circuit with resistances
referred to the secondary windings
In addition the total resistance also becomes a referred resistance
being the equivalence of the entire resistance referred to the
windings
The resistance 𝑅 in the primary windings is equivalent to 𝑲𝟐 𝑹𝟏
It is thus called the equivalent primary resistance referred to the
secondary windings - 𝑹⎖ 𝟏
𝑹𝟐
The resistance 𝑅 in the secondary windings is equivalent to Total resistance as referred to the
𝑲𝟐
It is thus called the equivalent secondary resistance referred to secondary windings:
𝑅 = 𝑅 + 𝑲𝟐 𝑹 𝟏
the primary windings - 𝑹⎖ 𝟐 𝑅 = 𝑅 + 𝑹⎖ 𝟏
 Transformer Equivalent
Impedance

 
𝑍 = 𝑅 +𝑋

𝑍 = 𝑅 + 𝑗𝑋 𝑍 = 𝑅 + 𝑗𝑋

   
𝑍 = 𝑅 +𝑋 𝑍 = 𝑅 +𝑋

𝑋
𝑋⎖ =
𝐾 𝑋⎖ =𝐾 𝑋

𝑋 = 𝑋 + 𝑋⎖ =𝑋 +

𝑋 = 𝑋 + 𝑋⎖ = 𝑋 +𝐾 𝑋

 
𝑍 = 𝑅 +𝑋
 Example #: 5-5

A 60kVA, 4000V/200V, 50Hz has the primary and secondary winding impedances of 𝑍 = 8 + 𝑗6Ω
anf 𝑍 = 4 + 𝑗3Ω.
a. The equivalent resistance as referred to the primary and secondary
b. The equivalent reactance as referred to the primary and secondary
c. The equivalent impedance as referred to the primary and secondary
d. The total winding copper losses