# Phasor diagram of practical transformer on Load

A practical or actual transformer will have winding resistance and leakage reactances both on the primary and secondary sides. When the secondary of the transformer is connected to the load terminals, the transformer is said to be on load. The load can be either resistive, inductive, or capacitive. In this section, let us learn the phasor diagram of practical transformer on load condition.

Let us consider a practical transformer as shown below, which has primary and secondary windings of resistances R_{1} and R_{2} and reactances X_{1} and X_{2} respectively.

In the above circuit, V_{1} and I_{1} are the primary voltage and current respectively, E_{1} is the induced emf in the primary winding and N_{1} is the number of turns in the primary winding. Similarly, N_{2} is the number of turns in the secondary winding, E_{2} is the induced emf in the secondary winding, V_{2} is the terminal voltage and I_{2} is the load current in the secondary winding.

The impedance of primary winding and secondary winding is given by,

Before drawing the phasor diagram of the transformer on load, it is important to learn the operation of transformer on load condition.

## Steps to draw the phasor diagram

The below diagram shows the phasor diagram of the transformer on resistive load. Follow the below procedure to draw the diagram.

- Consider the flux Φ as the reference phasor.
- The EMF induced in the transformer E
_{1}is drawn lagging the flux(Φ) by 90^{0}as OA. - The counter-induced emf(-E
_{1}) in the primary winding is drawn equal and opposite to E_{1}. - Due to winding resistance and leakage reactance, the secondary current I
_{2}lags behind the induced emf E_{2}and is drawn as OI_{2}. - Since the load is resistive, secondary voltage V
_{2}is drawn in phase with I_{2}as OC. - The voltage drops due to winding resistance(I
_{2}R_{2}) in the secondary is drawn parallel to the secondary current phasor I_{2}from V_{2}phasor as CD = I_{2}R_{2}. - Similarly, the voltage drop due to reactance(I
_{2}X_{2}) in the secondary winding is drawn perpendicular to the current phasor I_{2}from D as DB = I_{2}X_{2}. - Now, the phasor sum of terminal voltage V
_{2}, voltage drop due to secondary resistance I_{2}R_{2}, and voltage drop due to secondary reactance I_{2}X_{2}is equal to induced emf E_{2}in the secondary winding. Its magnitude is expressed as, - The current in the primary winding has two components: no-load current(I
_{0}) and load component of the primary current(I_{2}‘). - Draw the no-load current I
_{0}by lagging -E1 by no-load angle Φ_{0}. - Since I
_{2}‘ is the load component of the primary current, it is drawn equal and opposite to I_{2}. - The phasor sum of I
_{0}and I_{2}‘ gives the primary current I_{1}. - The voltage drops due to winding resistance(I
_{1}R_{1}) in the primary is drawn parallel to the primary current phasor I_{1}from -E_{1}phasor as EF = I_{1}R_{1}. - Similarly, the voltage drop due to reactance(I
_{1}X_{1}) in the primary winding is drawn perpendicular to the current phasor I_{1}from D as FG = I_{1}X_{1}. - Now, the phasor sum of counter-induced emf -E
_{1}, voltage drop due to primary resistance I_{1}R_{1}, and voltage drop due to primary reactance I_{1}X_{1}gives the applied voltage V_{1}to the primary winding. Its magnitude is expressed as, - The phase angle Φ
_{1}between V_{1}and I_{1}gives the power factor angle of the transformer.

The same procedure is repeated for inductive load and capacitive load. However, in the case of inductive load, secondary voltage V_{2} is drawn leading the secondary current I_{2} by an angle Φ_{2} as OC. In the case of capacitive load, secondary voltage V_{2} is drawn lagging the secondary current I_{2} by an angle Φ_{2} as OC.

From the phasor diagrams, the following equations can be derived.

**(a) For pure resistive load,**

Using the transformation ratio,

**(b) For resistive-inductive load,**

**(b) For resistive-capacitive load,**

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