Performance under no-load condition with phasor diagram.

 When voltage V1 is applied at primary keeping secondary open, no load current I0 will flow,

part of which will set-up flux in core and remaining will be spent in supplying core loss

(neglecting very small amount of copper loss in the primary winding). Since secondary

current is zero at no load, the magnitude of primary current at no load I0 is very small which

is about 3-5% of rated full load current. Because magnitude of Magnetising component of

no load current Im is significantly small due to high permeability of core material (it is lagging

behind by an angle 900

from V1 & in phase withφm) and loss component of no load current is

also very small due to low iron loss in the laminated core of the transformer (it is in phase

with the V1).

V1≈ −E1, E1 will be approximately equal, slightly less, but opposite to the applied voltage V1

(neglecting primary circuit voltage drop due to Io).

Induced emfs due to the linking of same changing flux, E1 and E2 are in same phase and

lagging behind by an angle 900

from φm i.e. opposite to V1.

No load current I0 is divided into two components Im and Iw.

Performance under load condition with phasor diagram.

A transformer gets loaded when we try to draw power from the secondary. It will be

explained how the primary reacts when the secondary is loaded. It will be shown that any

attempt to draw current/power from the secondary, is immediately responded by the

primary winding by drawing extra current/power from the source. We shall also see that

mmf balance will be maintained whenever both the windings carry currents. Together with

the mmf balance equation and voltage ratio equation, invariance of Volt-Ampere (VA or

KVA) irrespective of the sides will be established.

We have seen that the secondary winding becomes a seat of emf E2 and ready to deliver

power to a load if connected across it when primary is energized. Under no load condition

power drawn is zero as current drawn is zero for ideal transformer. However when loaded,

the secondary will deliver power to the load and same amount of power must be sucked in

by the primary from the source in order to maintain power balance. We expect the primary

current to flow now. Here we shall examine in somewhat detail the mechanism of drawing

extra current by the primary when the secondary is loaded.

Therefore flux produced by the secondary clearly opposes the primary flux fulfilling the

condition set by Lenz’s law. If the transformer is loaded by closing the switch S, current will

be delivered to the load. Since the secondary winding carries current it produces flux in the

opposite direction in the core and tries to reduce the original flux. However, KVL in the

primary demands that core flux should remain constant no matter whether the transformer

is loaded or not V1 + E1=0 . Such a requirement can only be met if the primary draws a

definite amount of extra current in order to nullify the effect of the mmf produced by the

secondary.

If I

2

is the magnitude of the secondary current and I2

’ is the additional current drawn by the

primary then following relation must hold good:

Magnetising amper-turn = Demagnetising amper-turn

To draw the phasor diagram under load condition, let us assume the power factor angle of

the load to be θ2 lagging. Therefore the load current phasor I2, can be drawn lagging the

secondary terminal voltage E2 by θ2 as shown in the figure.

The reflected current magnitude can be calculated from the relation I2

’== I2/a and is shown

directed 180° out of phase with respect to I2. Now,