PHY2049: Chapter 31 Transformers Purpose: to change alternating (AC) voltage to a bigger (or smaller) value input AC voltage in the primary produces a flux changing flux in secondary induces emf

Principle of Transformer Action

Principal of Transformer Action – Principle of electromagnetic induction. Ideal tωo ωinding transformer – ωinding resistances are negligible – Fluxes confined to magnetic core – Core lose negligible – Core has constant permeability V 1 I 1 MMF = N 1 I e Core flux φ folloωs, I e very closely. I e & φ sinusoidal φ =φ max sinωt

Principle of Transformer Action

Transformers Nothing comes for free, however! – Increase in voltage comes at the cost of current. – Output power cannot exceed input power! – power in = power out

Transformers: Sample Problem A transformer has 330 primary turns and 1240 secondary turns. The input voltage is 120 V and the output current is 15.0 A. What is the output voltage and input current? step-up transformer

Ideal Transformer on an inductive load

The exciting current leads the flux by hysteretic angle,

Transformer on LOAD

Equivalent circuit referred to the LT side of a 250/2500 single phase transformer is shown in fig. The load impedance connected to HT is 380+j230Ω. For a primary voltage of 250V, compute  the secondary terminal voltage  primary current and power factor  Power output and efficiency

Equivalent circuit referred to the LT side of a 250/2500 single phase transformer is shown in fig. The load impedance connected to HT is 380+j230Ω. For a primary voltage of 250V, compute Z' L = (380+j230) (N 1 / N 2) 2 = (380+j230) (250/2500) 2 = 3.8+j2.3 Total impedance in the primary Secondary terminal voltage = I 2 Z L  the secondary terminal voltage  primary current and power factor  Power output and efficiency

Im= V 1 /jXm = 250∟0°/250∟90° =1∟-90° =0-j1 I'e = Ic + Im = 0.5+ (0-j1) = 0.5-j1 I' 1 = I' 1 +I‘e = 40- j j1= 51∟-37.4° b) Primary current I 1 = 51A Primary p.f = cosθ 1 = cos37.4° = lagging (c) Load p.f cosθ 2 = 380°/ ( )= Power Output = V 2 I 2 cosθ 2 = 2220*5*0.855 = 9500 Watts Power Output = I' 1 2 R L = 50 2 *3.8 = 9500 Watt Core Loss,P C= v 1 2 / R C = Ic 2 R C = *0.2 =500 Watts Power Input = V 1 I 1 cosθ 1 = 250*51*0.794 =