1. 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

2. Principle of Transformer Action

3. 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

4. Principle of Transformer Action

5. 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

6. 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

9. Ideal Transformer on an inductive load

10. The exciting current leads the flux by hysteretic angle,

11. Transformer on LOAD

19. 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

20. 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

21. 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- j30+0.5- j1= 51∟-37.4° b) Primary current I 1 = 51A Primary p.f = cosθ 1 = cos37.4° = 0.794 lagging (c) Load p.f cosθ 2 = 380°/ (380 2 +230 2 )= 0.855 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 = 0.5 2 *0.2 =500 Watts Power Input = V 1 I 1 cosθ 1 = 250*51*0.794 = 10123.5