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 V1 = E1 and V2 = E2 .
Let the load at secondary is inductive load which causes the secondary current I2 to lag the secondary voltage V2 by Φ2. Primary current I1 must meet two conditions
i) Primary current must supply the no load current I0 to meet the iron losses in the transformer and to provide flux in the core.
ii) Primary current must supply a current I2′ to counteract the demagnetizing effect of secondary current I2. The magnitude of I2′ will be
N1 I2′ = N2 I2
Or I2′ = I2 N2/N1 = K I2
The phasor sum of I2′ and I0 is the total primary current I1.
I1 = I2′ + I0
Where I2′ = – K I2
I2′ is 180˚ out of phase with I2.
Phasor diagram of practical transformer on load
Figure 2 shows the phasor diagram of practical transformer on load for inductive load. Here E1 and E2 are lagging behind by mutual flux Φ by 90˚. Phasor sum of I0 and I2′ is the primary current I1. I2′ is anti phase with I2. The value of K is assumed unity so primary phasor is equal to secondary phasor.
Primary power factor = cos Φ1
Secondary power factor = cos Φ2
Primary input power = V1 I1 cos Φ1
Secondary input power = V2 I2 cos Φ2
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 R1 and X1 occurs in primary side so V1 > E1 and voltage drop R2 and X2 occurs in secondary side so V2 < E2.
Let take an inductive load which causes the secondary current I2 to lag behind the secondary voltage V2 by Φ2. Primary current I1 must follow the two requirements
i) Primary current must supply no load current I0 to meet the iron losses in the transformer and to provide flux in the core.
ii) It must supply a current I2′ to counteract the demagnetizing effect of secondary current I2. The magnitude of I2′ will be
N1 I2′ = N2 I2
Or I2′ = I2 N2/N1 = K I2
The phasor sum of I2′ and I0 is the total primary current I1.
I1 = I2′ + I0
Where I2′ = – K I2
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 E1 and E2 are lagging behind by mutual flux Φ by 90˚.
Primary power factor = cos Φ1
Load power factor = cos Φ2
Primary input power = V1 I1 cos Φ1
Secondary input power = V2 I2 cos Φ2
