# Transformer ON Load Condition

When the transformer is on loaded condition, the secondary of the transformer is connected to load. The load can be resistive, inductive or capacitive. The current I_{2} flows through the secondary winding of the transformer. The magnitude of the secondary current depends on the terminal voltage V_{2} and the load impedance. The phase angle between the secondary current and voltage depends on the nature of the load.

**Contents:**

- Operation of the Transformer on Load Condition
- Phasor Diagram of Transformer on Inductive Load
- Steps to draw the phasor diagram
- Phasor Diagram of Transformer on Capacitive Load
- Steps to draw the phasor diagram at capacitive load

## Operation of the Transformer on Load Condition

The Operation of the Transformer on Load Condition is explained below

- When secondary of the transformer is kept open, it draws the no-load current from the main supply. The no-load current induces the magnetomotive force N
_{0}I_{0}and this force set up the flux Φ in the core of the transformer. The circuit of the transformer at no load condition is shown in the figure below.- When the load is connected to the secondary of the transformer, the I
_{2}current flows through their secondary winding. The secondary current induces the magnetomotive force N_{2}I_{2}on the secondary winding of the transformer. This force set up the flux φ_{2}in the transformer core. The flux φ_{2}oppose the flux φ, according to Lenz’s law

- As the flux φ
_{2}opposes the flux φ, the resultant flux of the transformer decreases and this flux reduces the induces EMF E_{1}. Thus, the strength of the V_{1}is more than E_{1}and an additional primary current I’_{1}drawn from the main supply. The additional current is used for restoring the original value of the flux in the core of the transformer so that the V_{1}= E_{1}. The primary current I’_{1}is in phase opposition with the secondary current I_{2}. Thus, it is called the primary counter balancing current. - The additional current I’
_{1}induces the magnetomotive force N_{1}I’_{1}. And this force set up the flux φ’_{1}. The direction of the flux is same as that of the φ and it cancels the flux φ_{2}which induces because of the MMF N_{2}I_{2}

- As the flux φ

- When the load is connected to the secondary of the transformer, the I

- The phasor difference between V
_{1}and I_{1}gives the power factor angle ϕ_{1}of the primary side of the transformer. - The power factor of the secondary side depends upon the type of load connected to the transformer.
- If the load is inductive as shown in the above phasor diagram, the power factor will be lagging, and if the load is capacitive, the power factor will be leading.The total primary current I
_{1}is the vector sum of the current I_{0 }and I_{1}’. i.e

## Phasor Diagram of Transformer on Inductive Load

The phasor diagram of the actual transformer when it is loaded inductively is shown below

### Steps to draw the phasor diagram

- Take flux ϕ a reference
- Induces emf E
_{1}and E_{2 }lags the flux by 90 degrees. - The component of the applied voltage to the primary equal and opposite to induced emf in the primary winding. E
_{1}is represented by V_{1}’. - Current I
_{0}lags the voltage V_{1}’ by 90 degrees. - The power factor of the load is lagging. Therefore current I
_{2}is drawn lagging E_{2}by an angle ϕ_{2}. - The resistance and the leakage reactance of the windings result in a voltage drop, and hence secondary terminal voltage V
_{2}is the phasor difference of E_{2}and voltage drop.

V_{2} = E_{2} – voltage drops

I_{2 }R_{2} is in phase with I_{2} and I_{2}X_{2} is in quadrature with I_{2}.

- The total current flowing in the primary winding is the phasor sum of I
_{1}’ and I_{0}. - Primary applied voltage V
_{1}is the phasor sum of V_{1}’ and the voltage drop in the primary winding. - Current I
_{1}’ is drawn equal and opposite to the current I_{2}

V_{1} = V_{1}’ + voltage drop

I_{1}R_{1} is in phase with I_{1} and I_{1}X_{I} is in quadrature with I_{1}.

- The phasor difference between V
_{1}and I_{1}gives the power factor angle ϕ_{1}of the primary side of the transformer. - The power factor of the secondary side depends upon the type of load connected to the transformer.
- If the load is inductive as shown in the above phasor diagram, the power factor will be lagging, and if the load is capacitive, the power factor will be leading. Where I
_{1}R_{1}is the resistive drop in the primary windings

I_{2}X_{2}is the reactive drop in the secondary winding

Similarly

## Phasor Diagram of Transformer on Capacitive Load

The Transformer on Capacitive load (leading power factor load) is shown below in the phasor diagram.

### Steps to draw the phasor diagram at capacitive load

- Take flux ϕ a reference
- Induces emf E
_{1}and E_{2 }lags the flux by 90 degrees. - The component of the applied voltage to the primary equal and opposite to induced emf in the primary winding. E
_{1}is represented by V_{1}’. - Current I
_{0}lags the voltage V_{1}’ by 90 degrees. - The power factor of the load is leading. Therefore current I
_{2}is drawn leading E_{2} - The resistance and the leakage reactance of the windings result in a voltage drop, and hence secondary terminal voltage V
_{2}is the phasor difference of E_{2}and voltage drop.

V_{2} = E_{2} – voltage drops

I_{2 }R_{2} is in phase with I_{2} and I_{2}X_{2} is in quadrature with I_{2}.

- Current I
_{1}’ is drawn equal and opposite to the current I_{2} - The total current I
_{1 }flowing in the primary winding is the phasor sum of I_{1}’ and I_{0}. - Primary applied voltage V
_{1}is the phasor sum of V_{1}’ and the voltage drop in the primary winding.

V_{1} = V_{1}’ + voltage drop

I_{1}R_{1} is in phase with I_{1} and I_{1}X_{I} is in quadrature with I_{1}.

- The phasor difference between V
_{1}and I_{1}gives the power factor angle ϕ_{1}of the primary side of the transformer. - The power factor of the secondary side depends upon the type of load connected to the transformer.

## 11 Comments

Very good explained.. Clear all doubts, thanks..

It’s really helpful

I like it

Thanks very nice notes and very helpful

very well explained.

helped a lot.

thanks..

Thanks

Great Content.

It is very easy process

And it is very well

I have doubt why voltage drop due to resistance is in phase with current

Ohm’s Law

Thank you, this is very well explained. I however have some clarifications i would like to request for..

Current I0 lags the voltage V1’ by 90 degrees. <- is this correct or is it V1

Current I1’ is drawn equal and opposite to the current I2 <- is this correct or is it I1

thanks!! 🙂

Very very nicely explained . Thanks so much