Short Notes: Transformers | Short Notes for Electrical Engineering - Electrical Engineering (EE)

 Page 1


 
 
 
 
 
Impact of dimensions on various parameters of Transformer 
KVA Rating ? (Core Dimension)
4
 
Voltage Rating ? (Core Dimension)
2 
Current Rating ? (Core Dimension)
2
 
No-Load Current ? Core Dimension 
Core Loss ? Core Volume 
Induced EMF in a Transformer 
?
?
?
?
??
??
11
22
1 1 m
2 2 m
d
EN
dt
d
EN
dt
E (rms) 4.44fN
E (rms) 4.44fN
 
? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively. 
? F is the flux in the transformer and F m is maximum value of flux. 
? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary 
should be such that primary and secondary flux should oppose each other. 
? Also, primary current enters the positive terminal of primary winding as primary absorbs 
power and secondary current leaves the positive terminal of secondary winding as 
secondary delivers power and this way we can mark emf polarities. 
 
Exact equivalent circuit 
 
 
 
 
 
 
 
 
Page 2


 
 
 
 
 
Impact of dimensions on various parameters of Transformer 
KVA Rating ? (Core Dimension)
4
 
Voltage Rating ? (Core Dimension)
2 
Current Rating ? (Core Dimension)
2
 
No-Load Current ? Core Dimension 
Core Loss ? Core Volume 
Induced EMF in a Transformer 
?
?
?
?
??
??
11
22
1 1 m
2 2 m
d
EN
dt
d
EN
dt
E (rms) 4.44fN
E (rms) 4.44fN
 
? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively. 
? F is the flux in the transformer and F m is maximum value of flux. 
? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary 
should be such that primary and secondary flux should oppose each other. 
? Also, primary current enters the positive terminal of primary winding as primary absorbs 
power and secondary current leaves the positive terminal of secondary winding as 
secondary delivers power and this way we can mark emf polarities. 
 
Exact equivalent circuit 
 
 
 
 
 
 
 
 
 
 
 
 
Exact equivalent circuit w.r.t. primary 
 
 
 
 
 
2 2 2
1 1 1
2 2 2 2 L L
2 2 2
N N N
R = R ;  X = X ;  Z = Z
N N N
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
 ;                
? Approximately Equivalent Circuit 
 
 
 
 
 
01 1 2
R = R R
?
? 
01 1 2
X = X X
?
? 
Tests Conducted on a Transformer 
(i) Open Circuit Test 
o Conducted on LV side keeping HV side open circuited 
o Equivalent Circuit 
 
 
 
 
 
 
Page 3


 
 
 
 
 
Impact of dimensions on various parameters of Transformer 
KVA Rating ? (Core Dimension)
4
 
Voltage Rating ? (Core Dimension)
2 
Current Rating ? (Core Dimension)
2
 
No-Load Current ? Core Dimension 
Core Loss ? Core Volume 
Induced EMF in a Transformer 
?
?
?
?
??
??
11
22
1 1 m
2 2 m
d
EN
dt
d
EN
dt
E (rms) 4.44fN
E (rms) 4.44fN
 
? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively. 
? F is the flux in the transformer and F m is maximum value of flux. 
? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary 
should be such that primary and secondary flux should oppose each other. 
? Also, primary current enters the positive terminal of primary winding as primary absorbs 
power and secondary current leaves the positive terminal of secondary winding as 
secondary delivers power and this way we can mark emf polarities. 
 
Exact equivalent circuit 
 
 
 
 
 
 
 
 
 
 
 
 
Exact equivalent circuit w.r.t. primary 
 
 
 
 
 
2 2 2
1 1 1
2 2 2 2 L L
2 2 2
N N N
R = R ;  X = X ;  Z = Z
N N N
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
 ;                
? Approximately Equivalent Circuit 
 
 
 
 
 
01 1 2
R = R R
?
? 
01 1 2
X = X X
?
? 
Tests Conducted on a Transformer 
(i) Open Circuit Test 
o Conducted on LV side keeping HV side open circuited 
o Equivalent Circuit 
 
 
 
 
 
 
 
 
 
 
o Power reading = 
2
1
1 0 0
c
V
P = V I cos = 
R
? -------- (i) 
o Ammeter reading ? 
0
I = I 
o 
0
10
P
cos = 
VI
? 
o Calculate 
2
00
sin = 1 - cos ?? 
o 
2
1
1 0 0
m
V
Q = V I sin = 
X
? ------- (ii) 
Calculate 
c
R from (i) & 
m
X from (ii) 
  (ii) Short Circuit Test 
o Conducted on HV side keeping LV side short circuited 
o Equivalent Circuit 
 
o 
01
R & 
01
X are equivalent winding resistance & equivalent leakage reactor referred to 
HV side. 
o Wattmeter reading = 
2
sc 01
P = I R from this equation, we can calculate 
01
R 
o 
sc
01
sc
V
Z = 
I
   & 
22
01 01 01
X = Z R ? 
o We obtain
01
R , 
01
X & full load copper losses from this test. 
Losses on Transformers 
o Copper Loss 
  
22
Cu 1 1 2 2
P = I R I R ? 
         
22
1 01 2 02
= I R  I R ? 
 Where     
1
I = primary current 
      
2
I = secondary current 
Page 4


 
 
 
 
 
Impact of dimensions on various parameters of Transformer 
KVA Rating ? (Core Dimension)
4
 
Voltage Rating ? (Core Dimension)
2 
Current Rating ? (Core Dimension)
2
 
No-Load Current ? Core Dimension 
Core Loss ? Core Volume 
Induced EMF in a Transformer 
?
?
?
?
??
??
11
22
1 1 m
2 2 m
d
EN
dt
d
EN
dt
E (rms) 4.44fN
E (rms) 4.44fN
 
? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively. 
? F is the flux in the transformer and F m is maximum value of flux. 
? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary 
should be such that primary and secondary flux should oppose each other. 
? Also, primary current enters the positive terminal of primary winding as primary absorbs 
power and secondary current leaves the positive terminal of secondary winding as 
secondary delivers power and this way we can mark emf polarities. 
 
Exact equivalent circuit 
 
 
 
 
 
 
 
 
 
 
 
 
Exact equivalent circuit w.r.t. primary 
 
 
 
 
 
2 2 2
1 1 1
2 2 2 2 L L
2 2 2
N N N
R = R ;  X = X ;  Z = Z
N N N
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
 ;                
? Approximately Equivalent Circuit 
 
 
 
 
 
01 1 2
R = R R
?
? 
01 1 2
X = X X
?
? 
Tests Conducted on a Transformer 
(i) Open Circuit Test 
o Conducted on LV side keeping HV side open circuited 
o Equivalent Circuit 
 
 
 
 
 
 
 
 
 
 
o Power reading = 
2
1
1 0 0
c
V
P = V I cos = 
R
? -------- (i) 
o Ammeter reading ? 
0
I = I 
o 
0
10
P
cos = 
VI
? 
o Calculate 
2
00
sin = 1 - cos ?? 
o 
2
1
1 0 0
m
V
Q = V I sin = 
X
? ------- (ii) 
Calculate 
c
R from (i) & 
m
X from (ii) 
  (ii) Short Circuit Test 
o Conducted on HV side keeping LV side short circuited 
o Equivalent Circuit 
 
o 
01
R & 
01
X are equivalent winding resistance & equivalent leakage reactor referred to 
HV side. 
o Wattmeter reading = 
2
sc 01
P = I R from this equation, we can calculate 
01
R 
o 
sc
01
sc
V
Z = 
I
   & 
22
01 01 01
X = Z R ? 
o We obtain
01
R , 
01
X & full load copper losses from this test. 
Losses on Transformers 
o Copper Loss 
  
22
Cu 1 1 2 2
P = I R I R ? 
         
22
1 01 2 02
= I R  I R ? 
 Where     
1
I = primary current 
      
2
I = secondary current 
 
 
 
 
      
1
R = primary winding resistance  
      
2
R = secondary winding resistance  
   
22
12
01 1 2 02 2 1
21
NN
R = R R ; R = R R
NN
? ? ? ?
??
? ? ? ?
? ? ? ?
 
o Core Loss 
(i)  Hysteresis Loss 
             
x
n n m
P = K B f 
    X = 1.6 
    
m
B = maximum value of flux density 
    
1.6
n n m
P = K B f 
    
m
V
B  
f
? 
    V = applied voltage 
    f = frequency 
    
?
??
??
??
??
1.6
1.6 0.6
nhh
V
P = K f = K V f
f
 
                          If V is constant & f is increased, 
h
P decreases 
        (ii)  Eddy Current Loss 
    
2 2
e e m
P = K B f 
    
m
V
B  
f
? 
    
2
22
e e e
V
P = K f = K V
f
??
??
??
??
 
        Core loss = 
c e n
P = P P ? 
  
Page 5


 
 
 
 
 
Impact of dimensions on various parameters of Transformer 
KVA Rating ? (Core Dimension)
4
 
Voltage Rating ? (Core Dimension)
2 
Current Rating ? (Core Dimension)
2
 
No-Load Current ? Core Dimension 
Core Loss ? Core Volume 
Induced EMF in a Transformer 
?
?
?
?
??
??
11
22
1 1 m
2 2 m
d
EN
dt
d
EN
dt
E (rms) 4.44fN
E (rms) 4.44fN
 
? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively. 
? F is the flux in the transformer and F m is maximum value of flux. 
? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary 
should be such that primary and secondary flux should oppose each other. 
? Also, primary current enters the positive terminal of primary winding as primary absorbs 
power and secondary current leaves the positive terminal of secondary winding as 
secondary delivers power and this way we can mark emf polarities. 
 
Exact equivalent circuit 
 
 
 
 
 
 
 
 
 
 
 
 
Exact equivalent circuit w.r.t. primary 
 
 
 
 
 
2 2 2
1 1 1
2 2 2 2 L L
2 2 2
N N N
R = R ;  X = X ;  Z = Z
N N N
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
 ;                
? Approximately Equivalent Circuit 
 
 
 
 
 
01 1 2
R = R R
?
? 
01 1 2
X = X X
?
? 
Tests Conducted on a Transformer 
(i) Open Circuit Test 
o Conducted on LV side keeping HV side open circuited 
o Equivalent Circuit 
 
 
 
 
 
 
 
 
 
 
o Power reading = 
2
1
1 0 0
c
V
P = V I cos = 
R
? -------- (i) 
o Ammeter reading ? 
0
I = I 
o 
0
10
P
cos = 
VI
? 
o Calculate 
2
00
sin = 1 - cos ?? 
o 
2
1
1 0 0
m
V
Q = V I sin = 
X
? ------- (ii) 
Calculate 
c
R from (i) & 
m
X from (ii) 
  (ii) Short Circuit Test 
o Conducted on HV side keeping LV side short circuited 
o Equivalent Circuit 
 
o 
01
R & 
01
X are equivalent winding resistance & equivalent leakage reactor referred to 
HV side. 
o Wattmeter reading = 
2
sc 01
P = I R from this equation, we can calculate 
01
R 
o 
sc
01
sc
V
Z = 
I
   & 
22
01 01 01
X = Z R ? 
o We obtain
01
R , 
01
X & full load copper losses from this test. 
Losses on Transformers 
o Copper Loss 
  
22
Cu 1 1 2 2
P = I R I R ? 
         
22
1 01 2 02
= I R  I R ? 
 Where     
1
I = primary current 
      
2
I = secondary current 
 
 
 
 
      
1
R = primary winding resistance  
      
2
R = secondary winding resistance  
   
22
12
01 1 2 02 2 1
21
NN
R = R R ; R = R R
NN
? ? ? ?
??
? ? ? ?
? ? ? ?
 
o Core Loss 
(i)  Hysteresis Loss 
             
x
n n m
P = K B f 
    X = 1.6 
    
m
B = maximum value of flux density 
    
1.6
n n m
P = K B f 
    
m
V
B  
f
? 
    V = applied voltage 
    f = frequency 
    
?
??
??
??
??
1.6
1.6 0.6
nhh
V
P = K f = K V f
f
 
                          If V is constant & f is increased, 
h
P decreases 
        (ii)  Eddy Current Loss 
    
2 2
e e m
P = K B f 
    
m
V
B  
f
? 
    
2
22
e e e
V
P = K f = K V
f
??
??
??
??
 
        Core loss = 
c e n
P = P P ? 
  
 
 
 
 
 Efficiency 
  
? ?
? ?
2
i Cu,FL
x KVA cos
 = 
x KVA cos P x P
?
?
? ? ?
 
  X  = % loading of Transformer 
          cos = power factor ? 
  
i
P = iron loss 
  
Cu,FL
P = Full load copper losses 
            KVA = Power rating of Transformer 
  For maximum efficiency, 
    
i
Cu,FL
P
x = 
P
  
Voltage Regulation of Transformer 
   Regulation down 
NL FL
NL
VV
 100
V
?
?? 
    Regulation up      
NL FL
FL
VV
 100
V
?
?? 
 Equivalent circuit with respect to secondary 
             K = Transformation Ratio 
2
1
N
 
N
? 
   No-load voltage  
2
 V ?         
 Full-load voltage 
2
 V
?
? 
  Approximate Voltage Regulation 
   
? ?
2 02 2 02 2
2
I R cos X sin
VR = 
V
? ? ?