Transcript Principle of Transformer Action

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 1max cos t  N 1max sin  t  
dt
2

 N 1max
e1   N 1
E 1max
 

e1  E 1max sin  t  
2

E 1max
N 1max
E 1RMS 

 2 fN 1max  4.443fN 1max
2
2
d
 

e 2  N 2
  N 2max cos t  N 2max sin  t  
dt
2

E 2 max
N 2max
E 2 RMS 


2
2
E1
N1

 2 f max
E2
N2
2 fN 2max  4.443fN 2max
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  536.9 0
I1'
V1
250 0
0



50


36
.
9
 50(0.8- j0.6)
0
Z T 536.9
I1' N1 50  1
I2 

 5A
N2
10
Secondary terminal voltage = I2ZL
V1
250 0
2
2
I2 


5
380

230
 5  444  2220V
Z T 536.9 0
V1 2500 0
0
IC 


0
.
5

0
RC 50000
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