# Transformer on load condition

A transformer is said to be on load when the primary is connected to the alternating voltage source and its secondary is connected to the load. When a transformer is operated at no load condition, the current from the secondary winding to the load will not flow.

But when the transformer is connected to the load, the secondary current flows from the winding to the load. Let us now look at what happens, when the load is connected to the transformer.

## Operation of Transformer on load condition

When the secondary circuit of a transformer is connected through an impedance or load, the transformer is said to be loaded. The current I_{2} is set up in the secondary winding that flows through the load.

The magnitude and phase of this secondary current I_{2} with respect to secondary terminal voltage V₂ will depend upon the characteristics of the load, i.e., resistive, inductive, and capacitive.

If the load is resistive, the current I_{2} will be in-phase with V_{2}. Similarly the current will lag behind and lead the terminal voltage if the load is inductive and capacitive respectively.

When the transformer is on no load, as shown in the above figure, it draws no-load current I_{0}, from the supply mains. The no-load current I_{0} sets up an MMF (N_{1}I_{0}), which produces flux Φ_{0} in the core.

When an impedance is connected across the secondary terminals, as shown in the Figure below, current I_{2} flows through the secondary winding. The secondary current I_{2} sets up its own MMF(N_{2}I_{2}) and hence creates a secondary flux Φ_{2}.

The secondary flux Φ_{₂} opposes the main flux set up by the exciting current I_{0}, according to Lenz’s law. The opposing secondary flux Φ_{2} weakens the main flux momentarily, so primary back emf E_{1} tends to be reduced.

At this period, the applied voltage V_{1} will be more than the back emf E_{1}. Thus the primary winding draws more current from the supply. It again causes an increase in back emf E_{1}, and it adjusts itself such that there is a balance between applied voltage V_{₁} and back emf E_{₁}.

Let the additional primary current be I’_{2}, called as **load component of the primary current**. The current I’_{2} is in phase opposition with secondary current I_{2} and is also called the **counter-balancing current**.

The additional current I’_{2} in the primary winding sets up an MMF (N_{1}I’_{2}) producing flux Φ’_{2} in the same direction as that of the main flux Φ_{0}. It cancels the flux Φ_{₂} produced by secondary MMF (N_{₂}I_{2}) being equal in magnitude.

Hence during the load condition, only the main flux exists in the transformer core. Therefore, whatever might be the load condition, the flux passing through the core is almost constant as that of no load condition.

As *Φ _{₂} = Φ’_{2}*

When the transformer is on load, the primary winding has two currents.

- No load current I
_{0} - Counterbalancing or load component of primary current I’
_{2}

Thus the total primary current I_{1} is the phasor sum of both currents I_{0} and I’_{2}.

Since the no-load current is very small compared to I’_{2}, the total primary current is approximately equal to I’_{2}

where K is called the transformation ratio.

From equation(2), it can be understood that the primary and secondary currents are inversely proportional to their respective turns.

## Phasor diagram of a transformer on load

The phasor diagram of a transformer on load is drawn below for resistive, inductive, and capacitive loads. To draw the phasor diagram,

- let us consider the flux Φ
_{0}as the reference phasor. - Draw the V
_{1}phasor leading the flux Φ_{0}by 900. - Draw the induced EMFs E
_{1}and E_{2}phasor opposite to the voltage phasor V_{1}. - Draw the no-load primary current I0 by lagging V
_{1}by an angle Φ_{0}. - Draw the secondary current I
_{2}. **For different load conditions**,- Resistive Load – I
_{2}is drawn in phase with V_{2}. - Inductive Load – I
_{2}is drawn lagging V_{2}by an angle Φ_{2}. - Capacitive Load – I
_{2}is drawn leading V_{2}by an angle Φ_{2}.

- Resistive Load – I
- Draw the load component of primary current I’
_{2}, opposite and equal to the current phasor I_{2}. - Draw the primary current I
_{1}lagging the voltage phasor V_{1}by an angle Φ_{1}. I1 is also the vector sum of I_{0}and I’_{2}.

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