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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 B Vp N p t Vs N s B Vp t Np B t Ns Vs V p Np PHY2049: Chapter 31 Principle of Transformer Action 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 • • • • V1 I1 MMF = N1Ie Core flux φ folloωs, Ie very closely. Ie & φ sinusoidal φ =φmax sinωt Principle of Transformer Action 2 f d N 1max cos t N 1max sin t dt 2 N 1max e1 N 1 E 1max e1 E 1max sin t 2 E 1max N 1max E 1RMS 2 fN 1max 4.443fN 1max 2 2 d e 2 N 2 N 2max cos t N 2max sin t dt 2 E 2 max N 2max E 2 RMS 2 2 E1 N1 2 f max E2 N2 2 fN 2max 4.443fN 2max 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 I pV p I sVs I s Vp N p I p Vs N s 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? 1240 Ns 120V 451V Vs V p 330 Np I pVp I sVs step-up transformer 451V Vs 15 A 56.4 A Ip Is 120V Vp 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 the secondary terminal voltage primary current and power factor Power output and efficiency • Z'L = (380+j230) (N1 / N2)2 • = (380+j230) (250/2500)2 • = 3.8+j2.3 • Total impedance in the primary 0.2 j0.7 3.8 j2.3 4 j3 536.9 0 I1' V1 250 0 0 50 36 . 9 50(0.8- j0.6) 0 Z T 536.9 I1' N1 50 1 I2 5A N2 10 Secondary terminal voltage = I2ZL V1 250 0 2 2 I2 5 380 230 5 444 2220V Z T 536.9 0 V1 2500 0 0 IC 0 . 5 0 RC 50000 Im= V1/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 I1 = 51A Primary p.f = cosθ1 = cos37.4° = 0.794 lagging (c) Load p.f • cosθ2 = 380°/ (3802+2302 )= 0.855 • Power Output = V2I2cosθ2 = 2220*5*0.855 = 9500 Watts • Power Output = I'12RL = 502*3.8 = 9500 Watt • Core Loss ,PC= v12 / RC = Ic2 RC = 0.52*0.2 =500 Watts • Power Input = V1I1cosθ1 = 250*51*0.794 = 10123.5