PHY2049: Chapter 31

Transformers

Purpose: to change alternating (AC) voltage to a bigger (or smaller) value

tNV B

pp

tNV B

ss

p

pB

N

V

t

input AC voltagein the primaryproduces a flux

changing fluxin secondaryinduces emf

p

sps NN

VV

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

1 1 1 max 1 max

1max 1 max

1 1max

1max 1 max1 1 max 1 max

2 2 2 max 2 max

2max 2 max2

2

cos sin2

sin2

2 4.4432 2

cos sin2

22 2

RMS

RMS

f

de N N t N t

dt

E N

e E t

E NE fN fN

de N N t N t

dt

E NE

2 max 2 max

1 1max

2 2

4.443

2

fN fN

E Nf

E N

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

sspp VIVI

s

p

s

p

p

s

N

N

V

V

II

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?

p

sps NN

VV VV 4513301240

120

step-up transformer

sspp VIVI p

ssp VV

II AVV

A 456120451

15 .

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, computethe secondary terminal voltageprimary current and power factorPower 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) (N1 / N2)2

• = (380+j230) (250/2500)2

• = 3.8+j2.3• Total impedance in the primary

Secondary terminal voltage = I2ZL

the secondary terminal voltageprimary current and power factorPower output and efficiency

AN

N

Z

V

T

510

150II

j0.6)-50(0.89.36509.365

025I

9.365 j34 j2.33.8 j0.70.2

2

1'1

2

00

01'

1

0

00

01

C

220

01

2

05.00500

0250I

2220444523038059.365

025I

C

T

R

V

VZ

V

Im= V1/jXm = 250∟0°/250∟90° =1∟-90° =0-j1

I'e = Ic + Im = 0.5+ (0-j1) = 0.5-j1I'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