The device which raises or lowering voltage is called transformer. Previously I have discussed about ideal transformer and practical transformer, working principal of transformer also. Transformer gives same output power which we give as input power but raise or down voltage corresponding current decrease or increase. Input and output frequency is same. We also know transformer is used only ac current.

Ideal transformer is quite different from practical transformer. Because ideal transformer has no loss other hand practical transformer has core loss, winding resistance, flux leakage. Two types of practical transformer combination can be i) practical transformer on no load and ii) practical transformer on load. Here I will discuss about practical transformer on load.

For practical transformer on load we will consider two cases

**Case 1: **when the transformer has no winding resistance and leakage flux.

**Case 2: **when the transformer has winding resistance and leakage flux

**Case 1: No winding resistance and no leakage flux**

Figure 1 shows a practical transformer on load. We assume here it has no winding resistance and no leakage flux. Another word winding resistance and leakage flux is negligible here. For this assumption V_{1} = E_{1} and V_{2} = E_{2} .

Let the load at secondary is inductive load which causes the secondary current I_{2} to lag the secondary voltage V_{2} by Φ_{2}. Primary current I_{1} must meet two conditions

i) Primary current must supply the no load current I_{0} to meet the iron losses in the transformer and to provide flux in the core.

ii) Primary current must supply a current I_{2}′ to counteract the demagnetizing effect of secondary current I_{2}. The magnitude of I_{2}′ will be

N_{1} I_{2}′ = N_{2} I_{2}

Or I_{2}′ = I_{2 }N_{2}/N_{1} = K I_{2}

The phasor sum of I_{2}′ and I_{0} is the total primary current I_{1}.

I_{1} = I_{2}′ + I_{0}

Where I_{2}′ = – K I_{2}

I_{2}′ is 180˚ out of phase with I_{2}.

**Phasor diagram of practical transformer on load**

Figure 2 shows the phasor diagram of practical transformer on load for inductive load. Here E_{1} and E_{2} are lagging behind by mutual flux Φ by 90˚. Phasor sum of I_{0} and I_{2}′ is the primary current I_{1}. I_{2}′ is anti phase with I_{2}. The value of K is assumed unity so primary phasor is equal to secondary phasor.

Primary power factor = cos Φ_{1}

Secondary power factor = cos Φ_{2}

Primary input power = V_{1 }I_{1} cos Φ_{1}

Secondary input power = V_{2 }I_{2} cos Φ_{2}

**Case 2: Transformer with resistance and leakage reactance**

Figure 3 shows a practical transformer with winding resistance and leakage resistance. This is actual condition which exists in a practical transformer. Voltage drop R_{1} and X_{1} occurs in primary side so V_{1 }> E_{1} and voltage drop R_{2} and X_{2} occurs in secondary side so V_{2} < E_{2}.

Let take an inductive load which causes the secondary current I_{2} to lag behind the secondary voltage V_{2} by Φ_{2}. Primary current I_{1} must follow the two requirements

i) Primary current must supply no load current I_{0} to meet the iron losses in the transformer and to provide flux in the core.

ii) It must supply a current I_{2}′ to counteract the demagnetizing effect of secondary current I_{2}. The magnitude of I_{2}′ will be

N_{1} I_{2}′ = N_{2} I_{2}

Or I_{2}′ = I_{2 }N_{2}/N_{1} = K I_{2}

The phasor sum of I_{2}′ and I_{0} is the total primary current I_{1}.

I_{1} = I_{2}′ + I_{0}

Where I_{2}′ = – K I_{2}

## Phasor diagram of practical transformer with resistance and reactance:

Figure 4 shows the phasor diagram of a practical transformer for the usual case of inductive load. Here E_{1} and E_{2} are lagging behind by mutual flux Φ by 90˚.

Primary power factor = cos Φ_{1}

Load power factor = cos Φ_{2}

Primary input power = V_{1 }I_{1} cos Φ_{1}

Secondary input power = V_{2 }I_{2} cos Φ_{2}