**Transformer on Load:**

We know that when the transformer is loaded, the current I_{2} starts to flow through the secondary winding of the transformer to the load producing the potential difference of V_{2}. The main thing is that the output power of the transformer depends on the power factor and it is totally determined by the loaded nature.

We know that

**W = V _{2}I_{2}cosФ**

Where

**cosФ = Power factor**

The magnitude and phase of I_{2} are also determined by the loaded nature. When the load is the resistive one, I_{2} is in phase with the V_{2} which means that the resistive loads have a power factor of 1. When the load is inductive, I_{2} lags the V_{2}. When the load is capacitive, I_{2} leads the V_{2}.

There is one thing named magnetomotive force (mmf) which depends on the output current and turns of the transformer.

**We can say that mmf = N _{2}I_{2}**

Therefore, there exist a secondary magnetomotive force N_{2}I_{2} due to which secondary winding sets up its own flux Ф_{2}. This flux opposes the main flux Ф of the transformer produced in the core due to the **magnetizing** component of the no-load current I_{o} of the transformer due to which magnetomotive force N_{2}I_{2} is called demagnetizing ampere-turns.

The flux produced due to the magnetomotive force Ф_{2} momentarily reduces the main flux Ф produced in the transformer due to which primary winding induced emf also E_{1} reduces. Hence, the vector difference **V _{1} – E_{1} **increases due to which the primary winding of the transformer draws more current from the AC power supply. This additional current drawn by the primary winding of the transformer is due to the load connected to the transformer hence called load component of the primary current denoted as I

_{2}

^{’}. This current I

_{2}

^{’}is in antiphase with the I

_{2} and this current I

_{2}

^{’}sets up its own flux Ф

_{2}

^{’}which opposes the flux Ф

_{2}produced due to the magnetomotive force in the secondary winding of the transformer due to the load setting up on the output. And this flux Ф

_{2}

^{’}helps the main flux Ф in the transformer. This flux Ф

_{2}

^{’}neutralizes the flux Ф

_{2}produced by the current I

_{2 }and the ampere-turns or magnetomotive force due to flux Ф

_{2}

^{’}, N

_{1}I

_{2}

^{’ }balance the ampere-turns or magnetomotive force due to flux Ф

_{2}which is N

_{2}I

_{2}. Due to which the net flux in the core of the transformer is gain maintained on the constant level.

The main thing to remember is that when the transformer is on any load condition which is from load to no-load, the flux in the core is practically constant. The load component current I_{2}^{’} neutralizes the changes in the load. As practically the transformer core flux is constant due to which core loss is constant for all the loads, the transformer is called constant flux machine.

As ampere-turns are balanced, we can write,

**N _{2}I_{2} = N_{1}I_{2}^{’}**

**I _{2}^{’} = (N_{2}/N_{1})*I_{2}**

Therefore,

**I _{2}^{’} = KI_{2}**

Thus, when the transformer is loaded, the primary winding current I_{1} has two components:

- The no-load current I
_{o}which lags V_{1}by angle Фo. It has two components I_{m}and I_{c}which are the magnetizing and active components, respectively. - The load component of primary winding current I
_{2}^{’}which is in antiphase with the I_{2 }current and phase is determined with the nature of the load.

Hence, the primary current is the vector sum of the I_{o} and I_{2}^{’}.

Therefore,

**I _{1} = I_{o} + I_{2}^{’}**

The verification of the ratios of the current can be determined as follow:

As no-load current I_{o} is very small, so we can neglect I_{o }and we can write

**I _{1 }= I_{2}^{’}**

Balancing the ampere-turns,

**N _{1}I_{2}^{’} = N_{1}I_{1} = N_{2}I_{2}**

Therefore,

**N _{2}/N_{1} = I_{1}/I_{2} = K**

Under full load conditions when I_{o} is very small compared to the full load currents, the ratio of primary winding current I_{1} to the secondary winding current I_{2} is constant.

**Transformer on no Load:**

Actually, in the transformer, the iron core causes hysteresis **losses** and eddy currents losses as the transformer are subjected to the alternating flux. While the transformer is being designed the efforts are made to put to keep these losses minimum by,

- Using high-grade material like silicon steel to reduce hysteresis losses.
- Manufacturing the transformer core in the form of laminations or stacks of thin laminations to reduce or minimize the eddy currents losses.

Apart from this, there are iron losses in the transformer. And primary winding of the transformer has got certain resistance which subjects to small primary copper loss. Thus, the primary winding current under the no-load condition of the transformer has to supply the core losses i.e. hysteresis loss and eddy current loss and a small amount of primary copper loss. This current is denoted by I_{o} known to be the no-load current.

Now the no-load input current I_{o} has two components:

- A purely reactive component I
_{m}which known as the magnetizing component of no-load current which is required to produce the flux in the transformer. It is also known as the wattles component. - An active component I
_{c}which supplies total losses under the no-load condition of the transformer is known to be the power component of the no-load current. This is also called watchful or core loss component of I_{o}.

The total no-load current I_{o} is the vector addition of I_{m} and I_{c}.

Therefore,

**I _{o} = I_{m} + I_{c}**

In the transformer, due to winding resistance, no-load current I_{o} is no longer at 90^{o} with respect to V_{1 }but it lags V_{1} by angle Фo which is less than 90^{o} due to which cosФo is called no-load power factor of the transformer.

The magnetizing component I_{m} lagging V_{1} is

**I _{m} = I_{o}sinФo**

And the core loss component which is in phase with V_{1} is

**I _{c} = I_{o}cosФo**

The magnitude of the no-load current is

**I _{o} = (I_{m}^{2} + I_{c}^{2})^{1/2}**

While

Фo = no-load primary power factor angle

The total input power on no-load is denoted as W_{o} and is given by,

**W _{o} = V_{1}I_{o}cosФo = V_{1}I_{c}**

The main thing to notice here is that the current I_{o} is very small which is about 3-5% of the full rated current of the transformer on load. Hence primary copper loss is negligibly small due to which it is called the core loss component of I_{o}. Hence, input power W_{o} on no-load always represents the core losses in the transformer and are constant for all the loads.

**W _{o} = P_{i }= V_{1}I_{o}cosФo = V_{1}I_{c} = Core losses **

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