We shall consider the follow two cases:

1.When transformer is on no-load.

2.When transformer is on load.

**Transformer on no-load**. A transformer is said to be on *no-load *if its secondary side is open and primary is connected to a sinusoidal alternating voltage *V _{1} *The alternating applied voltage will cause

*flow of alternating current in the primary winding which will create alternating flux.*This primary input current

*(I*under no-load conditions supply:

_{0})- i.
*Iron losses in the core (i.e.*hysteresis loss and eddy current loss). - ii. A very
*small amount of copper losses in primary*(there being no copper loss in the secondary as it is open). - iii. Thus I
_{0}^{0}behind V_{1}*,*but lags it by an angle ɸ_{0}< 90^{0}

No-load power input, P_{0} = V_{1}I_{0 }cos ɸ_{0}

where cos ɸ_{0} = primary power factor under no-load conditions.

As is evident from Fig. 19, primary current I_{0}* *has the following two components:

*i. **Active *or *working *or *iron loss component I _{w}, *This component is in

*phase with*V

_{1}

*and mainly supplies the iron loss plus small quantity of primary copper loss.*

I_{w} = I_{0} cos ɸ_{0 }… (i)

*ii. **Magnetising component **I*_{m}*. *This component is *in quadrature with *V_{1}* *and its function is to *sustain the alternating flux in the core. *It is *wattless. *

The following points are *worthnoting : *

- The no-load primary current
*I*is_{w}*very small as compared to the full-load primary current.* - As
*10*is very small, the no-load primary copper loss is negligibly small which means that*no-load primary input is practically equal to the iron loss in the**trans**former.* - Since, it is primarily the core loss which is responsible for shift in the current vector, angle ɸ
_{0}is known as**hysteresis angle of advance.**

**Example 7. ***A 3300/300 V single phase transformer gives 0.6 A and 60 W as ammeter wattmeter readings when supply is given to the low voltage winding and high voltage winding is kept open, find:*

*(i) **Power factor for no-load current*

*(ii) **Magnetising component.*

*(iii) **Iron loss component.*

**Example 8**. *Find **(i) **active and reactive components of no-load current; and **(ii) **no-load current **of a 440/220**V single-phase transformer if the power input on no- load to the high voltage winding **is 80 W and **power factor of no- load current is 0.3 lagging. *

**Solution**. Primary, . E_{1} = 440 V

Secondary, E_{2} = 220 V

Power factor, cos ɸ_{0} = 0.3 (lagging)

No-load losses, P_{0} = 80 W

**Example 9**. *A 3300/220 V, 30 kVA, single-phase transformer takes a no-load current of *1.5 *A *when *the low voltage winding is kept open. The iron loss component is equal to 0.4 A find: *

*(i) **No-load input power. *

*(ii) **Magnetising component and power factor of no-load current: *

**Solution**. Rating of transformer = 30 kVA

Primary, E_{1} = 3300 V

Secondary, E_{2} = 220 V

No-load current, I_{0} = 1.5 A

Iron loss component, I_{m} = 0.4 A

(i) **No-load input power, **

P_{0} = V_{1}I_{0} cos ɸ_{0} = V_{1}I_{w}

= 3300 × 0.4 = **1320 W. (Ans.) **

**Transformer on load**. *The transformer is said to be loaded when the secondary circuit of a transformer is completed through an impedance or load. *The magnitude and phase of secondary current I_{2} with respect to secondary terminal voltage will depend upon the characteristic of load, *i.e. *current I_{2}* *will be in phase, lag behind and lead the terminal voltage V_{2}* *respectively when the load is purely resistive, inductive and capacitive.

The secondary current I_{2 }sets up its own ampere-turns (= N_{2}I_{2}*) *and creates its own flux ɸ_{2 }opposing the main flux ɸ_{0} created by no-load current I_{0}*, *The opposing secondary flux ɸ_{2} weakens the primary flux ɸ_{0} momentarily hence primary counter or back e.m.f. E_{1}* *tends to be reduced. V_{1} gains the upper hand over E_{1} momentarily and hence causes more current to flow in primary. Let this additional primary current be I_{2}‘. It is known as load component of primary current. The additional primary m.m.f. N_{1}I_{2}‘ sets up its own flux ɸ_{2}‘ which is in opposition to ɸ_{2} (but is in the same direction as ɸ_{0}) and is equal to it in magnitude. Hence they cancel each other. Thus we find that the magnetic effects of secondary current I_{2} are immediately neutralised by the additional primary current I_{2}’ which is brought into existence exactly at the same instant as I_{2},

Fig. 20. An ideal transformer on load.

From above discussion it can be concluded that:

- Whatever be the load conditions, the
*net flux passing through the core is approximately the same as at no-load. .* - Since the core flux remains constant at all loads, the core
*loss almost remains constant under different loading conditions.*

The total primary current is the vector sum of I_{0} and I_{2}‘ ; the current I_{2}‘ is in antiphase with I_{2} and k times in magnitude.

The total primary current can be shown as below:

The vector diagrams for transformer on non-inductive, inductive and capacitive loads are

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