Electrical Machines Formulas for GATE EE Exam | Electrical Machines - Electrical Engineering (EE) PDF Download

 Page 1


TRANSFORMERS: 
? Gross cross sectional area = Area occupied by magnetic material + Insulation 
material. 
? N et cross sectional area = Area occupied by only magnetic material excluding area 
of insulation material.  
? Hence for all calculations, net cross sectional area is taken since ? (flux) majorly 
flows in magnetic material. 
? = BA
n
 
t
f
? Specific weight of t/f = 
K
We
VA
i
 
g
ra
ht
t
 
i
of
n
f
g of
t
 
? Stacking/iron factor :- (k
s
) = 
Net Cross Sectional area
Gross Cross Sectional area
Total C.S Area
? k
s
 is always less than 1 
? Gross c.s Area = A
G
 = length × breadth  
? Net c.s Area = A
n
 = k
s
 × A
G
 
? Utilization factor of transformer core = 
Effective C.S.Area
 U.F of cruciform core = 0.8 to 
0.85 
? Flux = 
mmF
Reluctance
 = = ?
m
sin ?t 
? According to faradays second law  e
1
= -N
1
d?
dt
= -N
1
d
dt
 ? ?
m
sin ?t? 
? Transformer emf equations :- 
E
1
 = 4.44 N
1
 B
max
 A
n
f ??  (1) 
E
2
 = 4.44 N
2
 B
max
 A
n
f   ??  (2) 
N
1
? Emf per turn in I
ry
 = 
E
1
 = 4.44 B
max
A
n
f
N
2
? Emf per turn in II
ry
= 
E
2
 = 4.44 B
max
 A
n
 f
? Emf per turn on both sides of the transformer is same 
E
1
N
1
N
2
= 
E
2
 
?   
E
1
E
2
= 
N
1
N
2
= 
1
k
Transformation ratio = K = 
E
2
E
1
=
Instantaneous value 
of emf in primary  
e
1
= N
1
?
m
 ?sin? ?t - 
p
?
2
?
N
2
N
1
Turns ratio = 
1
K
= N
1
: N
2
ELECRIAL MACHINES (Formula Notes)
Page 2


TRANSFORMERS: 
? Gross cross sectional area = Area occupied by magnetic material + Insulation 
material. 
? N et cross sectional area = Area occupied by only magnetic material excluding area 
of insulation material.  
? Hence for all calculations, net cross sectional area is taken since ? (flux) majorly 
flows in magnetic material. 
? = BA
n
 
t
f
? Specific weight of t/f = 
K
We
VA
i
 
g
ra
ht
t
 
i
of
n
f
g of
t
 
? Stacking/iron factor :- (k
s
) = 
Net Cross Sectional area
Gross Cross Sectional area
Total C.S Area
? k
s
 is always less than 1 
? Gross c.s Area = A
G
 = length × breadth  
? Net c.s Area = A
n
 = k
s
 × A
G
 
? Utilization factor of transformer core = 
Effective C.S.Area
 U.F of cruciform core = 0.8 to 
0.85 
? Flux = 
mmF
Reluctance
 = = ?
m
sin ?t 
? According to faradays second law  e
1
= -N
1
d?
dt
= -N
1
d
dt
 ? ?
m
sin ?t? 
? Transformer emf equations :- 
E
1
 = 4.44 N
1
 B
max
 A
n
f ??  (1) 
E
2
 = 4.44 N
2
 B
max
 A
n
f   ??  (2) 
N
1
? Emf per turn in I
ry
 = 
E
1
 = 4.44 B
max
A
n
f
N
2
? Emf per turn in II
ry
= 
E
2
 = 4.44 B
max
 A
n
 f
? Emf per turn on both sides of the transformer is same 
E
1
N
1
N
2
= 
E
2
 
?   
E
1
E
2
= 
N
1
N
2
= 
1
k
Transformation ratio = K = 
E
2
E
1
=
Instantaneous value 
of emf in primary  
e
1
= N
1
?
m
 ?sin? ?t - 
p
?
2
?
N
2
N
1
Turns ratio = 
1
K
= N
1
: N
2
ELECRIAL MACHINES (Formula Notes)
? For an ideal two-winding transformer with primary voltage V1 applied 
across N1 primary turns and secondary voltage V2 appearing across N2 secondary 
turns: 
V1 / V2 = N1 / N2 
? The primary current I1 and secondary current I2 are related by: 
I1 / I2 = N2 / N1 = V2 / V1 
? For an ideal step-down auto-transformer with primary voltage V1 applied 
across (N1 + N2) primary turns and secondary voltage V2 appearing 
across N2 secondary turns: 
V1 / V2 = (N1 + N2) / N2 
? The primary (input) current I1 and secondary (output) current I2 are related by: 
I1 / I2 = N2 / (N1 + N2) = V2 / V1. 
? For a single-phase transformer with rated primary voltage V1, rated primary 
current I1, rated secondary voltage V2 and rated secondary current I2, the voltampere 
rating S is: 
S = V1I1 = V2I2 
? For a balanced m-phase transformer with rated primary phase voltage V1, rated 
primary current I1, rated secondary phase voltage V2 and rated secondary current I2, 
the voltampere rating S is: 
S = mV1I1 = mV2I2 
? The primary circuit impedance Z1 referred to the secondary circuit for an ideal 
transformer with N1 primary turns and N2 secondary turns is: 
Z12 = Z1(N2 / N1)
2
 
? During operation of transformer :- 
B
m
 ? 
E
1
f f
 ? 
V
1
 
B
max
= constant ? 
V
1
f
 = constant 
Equivalent ckt of t/f under N.L condition :- 
No load /shunt branch. 
N
1
  N
2
 
E
2
 E
1
 V
1
 
I
0
  
I
w
  
R
0
   
X
0
   
I
µ
  
Page 3


TRANSFORMERS: 
? Gross cross sectional area = Area occupied by magnetic material + Insulation 
material. 
? N et cross sectional area = Area occupied by only magnetic material excluding area 
of insulation material.  
? Hence for all calculations, net cross sectional area is taken since ? (flux) majorly 
flows in magnetic material. 
? = BA
n
 
t
f
? Specific weight of t/f = 
K
We
VA
i
 
g
ra
ht
t
 
i
of
n
f
g of
t
 
? Stacking/iron factor :- (k
s
) = 
Net Cross Sectional area
Gross Cross Sectional area
Total C.S Area
? k
s
 is always less than 1 
? Gross c.s Area = A
G
 = length × breadth  
? Net c.s Area = A
n
 = k
s
 × A
G
 
? Utilization factor of transformer core = 
Effective C.S.Area
 U.F of cruciform core = 0.8 to 
0.85 
? Flux = 
mmF
Reluctance
 = = ?
m
sin ?t 
? According to faradays second law  e
1
= -N
1
d?
dt
= -N
1
d
dt
 ? ?
m
sin ?t? 
? Transformer emf equations :- 
E
1
 = 4.44 N
1
 B
max
 A
n
f ??  (1) 
E
2
 = 4.44 N
2
 B
max
 A
n
f   ??  (2) 
N
1
? Emf per turn in I
ry
 = 
E
1
 = 4.44 B
max
A
n
f
N
2
? Emf per turn in II
ry
= 
E
2
 = 4.44 B
max
 A
n
 f
? Emf per turn on both sides of the transformer is same 
E
1
N
1
N
2
= 
E
2
 
?   
E
1
E
2
= 
N
1
N
2
= 
1
k
Transformation ratio = K = 
E
2
E
1
=
Instantaneous value 
of emf in primary  
e
1
= N
1
?
m
 ?sin? ?t - 
p
?
2
?
N
2
N
1
Turns ratio = 
1
K
= N
1
: N
2
ELECRIAL MACHINES (Formula Notes)
? For an ideal two-winding transformer with primary voltage V1 applied 
across N1 primary turns and secondary voltage V2 appearing across N2 secondary 
turns: 
V1 / V2 = N1 / N2 
? The primary current I1 and secondary current I2 are related by: 
I1 / I2 = N2 / N1 = V2 / V1 
? For an ideal step-down auto-transformer with primary voltage V1 applied 
across (N1 + N2) primary turns and secondary voltage V2 appearing 
across N2 secondary turns: 
V1 / V2 = (N1 + N2) / N2 
? The primary (input) current I1 and secondary (output) current I2 are related by: 
I1 / I2 = N2 / (N1 + N2) = V2 / V1. 
? For a single-phase transformer with rated primary voltage V1, rated primary 
current I1, rated secondary voltage V2 and rated secondary current I2, the voltampere 
rating S is: 
S = V1I1 = V2I2 
? For a balanced m-phase transformer with rated primary phase voltage V1, rated 
primary current I1, rated secondary phase voltage V2 and rated secondary current I2, 
the voltampere rating S is: 
S = mV1I1 = mV2I2 
? The primary circuit impedance Z1 referred to the secondary circuit for an ideal 
transformer with N1 primary turns and N2 secondary turns is: 
Z12 = Z1(N2 / N1)
2
 
? During operation of transformer :- 
B
m
 ? 
E
1
f f
 ? 
V
1
 
B
max
= constant ? 
V
1
f
 = constant 
Equivalent ckt of t/f under N.L condition :- 
No load /shunt branch. 
N
1
  N
2
 
E
2
 E
1
 V
1
 
I
0
  
I
w
  
R
0
   
X
0
   
I
µ
  
? No load current = I
0
= I
µ
+ I
w
= I
0
 ? -?
0
 
  I
w
= I
0
cos ?
0
 
   I
µ
= I
0
sin ?
0
 
? No load power = v
1
I
0
cos ?
0
= v
1
I
w
= Iron losses. 
R
0
= 
v
1
I
w
1
; X
0
= 
v
1
I
µ
? I
w
= 
No load power
V
1
Transferring from ???????? ????????
 to ???? ????????
:- 
I
 
2
 2
 R
2
= I
 
1
2
 R
  
2
1
  
R
  
2
1
= R
2
 ?
I
2
?
2
K
2
I
1
       = 
R
2
  
? 
R
  
2
1
= 
K
R
2
2
From ???? ????????
 to ???????? ????????
 :- 
I
 
1
 2
R
1
= I
 
2
 2
. R
 
1
1
 
R
 
1
1
= 
I
 
1
 2
I
 
2
 2
. R
1
  
R
 
1
 1
= R
1
. K
2
 
? Total resistance ref to primary = R
1
+ R
  
2
1
 
R
01
= R
1
+ R
2
/k
2
 
? Total resistance ref to secondary = R
2
+ R
 
1
 1
 
R
02
= R
2
+ k
2
R
1
 
? Total Cu loss = I
 
1
2
R
01
 
         Or 
      I
 
2
 2
 R
02
 
R  R
2
1
1
Per unit resistance drops :-  
E
1
? P.U primary resistance drop = 
I
1
R
1
 
E
2
? P.U secondary resistance drop = 
I
2
 R
2
 
E
1
? Total P.U resistance drop ref to I
ry
 = 
I
1
R
01
 
E
2
? Total P.U resistance drop ref to II
ry
 = 
I
2
 R
02
 
? The P.U resistance drops on both sides of the t/f is same 
I
1
 R
01
=
I
2
R
02
  
E
1
E
2
Losses present in transformer :- 
1. Copper losses
2. Iron losses
3. Stray load losses
4. Dielectric losses
major losses 
Iron parts   
minor losses 
t/f windings  
t/f core 
cu parts  
insulating materials. 
Page 4


TRANSFORMERS: 
? Gross cross sectional area = Area occupied by magnetic material + Insulation 
material. 
? N et cross sectional area = Area occupied by only magnetic material excluding area 
of insulation material.  
? Hence for all calculations, net cross sectional area is taken since ? (flux) majorly 
flows in magnetic material. 
? = BA
n
 
t
f
? Specific weight of t/f = 
K
We
VA
i
 
g
ra
ht
t
 
i
of
n
f
g of
t
 
? Stacking/iron factor :- (k
s
) = 
Net Cross Sectional area
Gross Cross Sectional area
Total C.S Area
? k
s
 is always less than 1 
? Gross c.s Area = A
G
 = length × breadth  
? Net c.s Area = A
n
 = k
s
 × A
G
 
? Utilization factor of transformer core = 
Effective C.S.Area
 U.F of cruciform core = 0.8 to 
0.85 
? Flux = 
mmF
Reluctance
 = = ?
m
sin ?t 
? According to faradays second law  e
1
= -N
1
d?
dt
= -N
1
d
dt
 ? ?
m
sin ?t? 
? Transformer emf equations :- 
E
1
 = 4.44 N
1
 B
max
 A
n
f ??  (1) 
E
2
 = 4.44 N
2
 B
max
 A
n
f   ??  (2) 
N
1
? Emf per turn in I
ry
 = 
E
1
 = 4.44 B
max
A
n
f
N
2
? Emf per turn in II
ry
= 
E
2
 = 4.44 B
max
 A
n
 f
? Emf per turn on both sides of the transformer is same 
E
1
N
1
N
2
= 
E
2
 
?   
E
1
E
2
= 
N
1
N
2
= 
1
k
Transformation ratio = K = 
E
2
E
1
=
Instantaneous value 
of emf in primary  
e
1
= N
1
?
m
 ?sin? ?t - 
p
?
2
?
N
2
N
1
Turns ratio = 
1
K
= N
1
: N
2
ELECRIAL MACHINES (Formula Notes)
? For an ideal two-winding transformer with primary voltage V1 applied 
across N1 primary turns and secondary voltage V2 appearing across N2 secondary 
turns: 
V1 / V2 = N1 / N2 
? The primary current I1 and secondary current I2 are related by: 
I1 / I2 = N2 / N1 = V2 / V1 
? For an ideal step-down auto-transformer with primary voltage V1 applied 
across (N1 + N2) primary turns and secondary voltage V2 appearing 
across N2 secondary turns: 
V1 / V2 = (N1 + N2) / N2 
? The primary (input) current I1 and secondary (output) current I2 are related by: 
I1 / I2 = N2 / (N1 + N2) = V2 / V1. 
? For a single-phase transformer with rated primary voltage V1, rated primary 
current I1, rated secondary voltage V2 and rated secondary current I2, the voltampere 
rating S is: 
S = V1I1 = V2I2 
? For a balanced m-phase transformer with rated primary phase voltage V1, rated 
primary current I1, rated secondary phase voltage V2 and rated secondary current I2, 
the voltampere rating S is: 
S = mV1I1 = mV2I2 
? The primary circuit impedance Z1 referred to the secondary circuit for an ideal 
transformer with N1 primary turns and N2 secondary turns is: 
Z12 = Z1(N2 / N1)
2
 
? During operation of transformer :- 
B
m
 ? 
E
1
f f
 ? 
V
1
 
B
max
= constant ? 
V
1
f
 = constant 
Equivalent ckt of t/f under N.L condition :- 
No load /shunt branch. 
N
1
  N
2
 
E
2
 E
1
 V
1
 
I
0
  
I
w
  
R
0
   
X
0
   
I
µ
  
? No load current = I
0
= I
µ
+ I
w
= I
0
 ? -?
0
 
  I
w
= I
0
cos ?
0
 
   I
µ
= I
0
sin ?
0
 
? No load power = v
1
I
0
cos ?
0
= v
1
I
w
= Iron losses. 
R
0
= 
v
1
I
w
1
; X
0
= 
v
1
I
µ
? I
w
= 
No load power
V
1
Transferring from ???????? ????????
 to ???? ????????
:- 
I
 
2
 2
 R
2
= I
 
1
2
 R
  
2
1
  
R
  
2
1
= R
2
 ?
I
2
?
2
K
2
I
1
       = 
R
2
  
? 
R
  
2
1
= 
K
R
2
2
From ???? ????????
 to ???????? ????????
 :- 
I
 
1
 2
R
1
= I
 
2
 2
. R
 
1
1
 
R
 
1
1
= 
I
 
1
 2
I
 
2
 2
. R
1
  
R
 
1
 1
= R
1
. K
2
 
? Total resistance ref to primary = R
1
+ R
  
2
1
 
R
01
= R
1
+ R
2
/k
2
 
? Total resistance ref to secondary = R
2
+ R
 
1
 1
 
R
02
= R
2
+ k
2
R
1
 
? Total Cu loss = I
 
1
2
R
01
 
         Or 
      I
 
2
 2
 R
02
 
R  R
2
1
1
Per unit resistance drops :-  
E
1
? P.U primary resistance drop = 
I
1
R
1
 
E
2
? P.U secondary resistance drop = 
I
2
 R
2
 
E
1
? Total P.U resistance drop ref to I
ry
 = 
I
1
R
01
 
E
2
? Total P.U resistance drop ref to II
ry
 = 
I
2
 R
02
 
? The P.U resistance drops on both sides of the t/f is same 
I
1
 R
01
=
I
2
R
02
  
E
1
E
2
Losses present in transformer :- 
1. Copper losses
2. Iron losses
3. Stray load losses
4. Dielectric losses
major losses 
Iron parts   
minor losses 
t/f windings  
t/f core 
cu parts  
insulating materials. 
1. Cu losses in t/f:
    Total Cu loss = I
 
1
 2
R
1
+ I
 
2
 2
R
1
 
         = I
 
1
2
R
01
 
E
1
        = I
 
2
 2
 R
02
  
? Rated current on I
ry
= 
VA rating of t/f
 
E
2
Similarly current on II
ry
 = 
VA rating of t/f
  
VA rating of t/f
? Cu losses ? I
 
1
 2
 or I
 
2
 2
. Hence there are called as variable losses. 
? P.U Full load Cu loss = 
FL Cu loss in watts
 
= 
I
 
1
 2
R
01
E
1
I
1
? If VA rating of t/f is taken as base then  P.U Cu loss ? I
 
1
 2
 as remaining terms are constant. 
? P.U Cu loss at x of FL = x
2
× PU FL Cu loss 
? 
P. U resistance drop ref to I
ry
or 
P. U resistance ref to I
ry
 
? = 
I
1
R
01
E
1
I
1
× 
I
1
  
= 
I
 
1
 2
R
01
E
1
I
1
? P.U Resistance drop = P.U FL cu loss 
% FL Cu loss = % R = % Resistance drop. 
Iron (or) Core losses in t/f :- 
1. Hysteresis loss :
Steinmetz formula :- 
Where  
? = stienmetz coefficient  
B
max
 = max. flux density in transformer core.  
f = frequency of magnetic reversal = supply freq. 
v = volume of core material  
x = Hysteresis coeff (or) stienmetz exponent  
   = 1.6 (Si or CRGo steel) 
2. Eddycurrent loss:
Eddy current loss ,(W
e
) ? R
ce
 × I
 
e
 2
  
As area decreases in laminated core resistance increases as a result conductivity decreases. 
W
e
= K. B
  
ma
2
x
 f
2
. t
2
 
Constant 
Supply freq 
thickness of laminations. 
 Area under one hysteresis loop. 
 
x
?  B
  
max
   .  f  .  v W
h
 = 
(it is a function of s )
Page 5


TRANSFORMERS: 
? Gross cross sectional area = Area occupied by magnetic material + Insulation 
material. 
? N et cross sectional area = Area occupied by only magnetic material excluding area 
of insulation material.  
? Hence for all calculations, net cross sectional area is taken since ? (flux) majorly 
flows in magnetic material. 
? = BA
n
 
t
f
? Specific weight of t/f = 
K
We
VA
i
 
g
ra
ht
t
 
i
of
n
f
g of
t
 
? Stacking/iron factor :- (k
s
) = 
Net Cross Sectional area
Gross Cross Sectional area
Total C.S Area
? k
s
 is always less than 1 
? Gross c.s Area = A
G
 = length × breadth  
? Net c.s Area = A
n
 = k
s
 × A
G
 
? Utilization factor of transformer core = 
Effective C.S.Area
 U.F of cruciform core = 0.8 to 
0.85 
? Flux = 
mmF
Reluctance
 = = ?
m
sin ?t 
? According to faradays second law  e
1
= -N
1
d?
dt
= -N
1
d
dt
 ? ?
m
sin ?t? 
? Transformer emf equations :- 
E
1
 = 4.44 N
1
 B
max
 A
n
f ??  (1) 
E
2
 = 4.44 N
2
 B
max
 A
n
f   ??  (2) 
N
1
? Emf per turn in I
ry
 = 
E
1
 = 4.44 B
max
A
n
f
N
2
? Emf per turn in II
ry
= 
E
2
 = 4.44 B
max
 A
n
 f
? Emf per turn on both sides of the transformer is same 
E
1
N
1
N
2
= 
E
2
 
?   
E
1
E
2
= 
N
1
N
2
= 
1
k
Transformation ratio = K = 
E
2
E
1
=
Instantaneous value 
of emf in primary  
e
1
= N
1
?
m
 ?sin? ?t - 
p
?
2
?
N
2
N
1
Turns ratio = 
1
K
= N
1
: N
2
ELECRIAL MACHINES (Formula Notes)
? For an ideal two-winding transformer with primary voltage V1 applied 
across N1 primary turns and secondary voltage V2 appearing across N2 secondary 
turns: 
V1 / V2 = N1 / N2 
? The primary current I1 and secondary current I2 are related by: 
I1 / I2 = N2 / N1 = V2 / V1 
? For an ideal step-down auto-transformer with primary voltage V1 applied 
across (N1 + N2) primary turns and secondary voltage V2 appearing 
across N2 secondary turns: 
V1 / V2 = (N1 + N2) / N2 
? The primary (input) current I1 and secondary (output) current I2 are related by: 
I1 / I2 = N2 / (N1 + N2) = V2 / V1. 
? For a single-phase transformer with rated primary voltage V1, rated primary 
current I1, rated secondary voltage V2 and rated secondary current I2, the voltampere 
rating S is: 
S = V1I1 = V2I2 
? For a balanced m-phase transformer with rated primary phase voltage V1, rated 
primary current I1, rated secondary phase voltage V2 and rated secondary current I2, 
the voltampere rating S is: 
S = mV1I1 = mV2I2 
? The primary circuit impedance Z1 referred to the secondary circuit for an ideal 
transformer with N1 primary turns and N2 secondary turns is: 
Z12 = Z1(N2 / N1)
2
 
? During operation of transformer :- 
B
m
 ? 
E
1
f f
 ? 
V
1
 
B
max
= constant ? 
V
1
f
 = constant 
Equivalent ckt of t/f under N.L condition :- 
No load /shunt branch. 
N
1
  N
2
 
E
2
 E
1
 V
1
 
I
0
  
I
w
  
R
0
   
X
0
   
I
µ
  
? No load current = I
0
= I
µ
+ I
w
= I
0
 ? -?
0
 
  I
w
= I
0
cos ?
0
 
   I
µ
= I
0
sin ?
0
 
? No load power = v
1
I
0
cos ?
0
= v
1
I
w
= Iron losses. 
R
0
= 
v
1
I
w
1
; X
0
= 
v
1
I
µ
? I
w
= 
No load power
V
1
Transferring from ???????? ????????
 to ???? ????????
:- 
I
 
2
 2
 R
2
= I
 
1
2
 R
  
2
1
  
R
  
2
1
= R
2
 ?
I
2
?
2
K
2
I
1
       = 
R
2
  
? 
R
  
2
1
= 
K
R
2
2
From ???? ????????
 to ???????? ????????
 :- 
I
 
1
 2
R
1
= I
 
2
 2
. R
 
1
1
 
R
 
1
1
= 
I
 
1
 2
I
 
2
 2
. R
1
  
R
 
1
 1
= R
1
. K
2
 
? Total resistance ref to primary = R
1
+ R
  
2
1
 
R
01
= R
1
+ R
2
/k
2
 
? Total resistance ref to secondary = R
2
+ R
 
1
 1
 
R
02
= R
2
+ k
2
R
1
 
? Total Cu loss = I
 
1
2
R
01
 
         Or 
      I
 
2
 2
 R
02
 
R  R
2
1
1
Per unit resistance drops :-  
E
1
? P.U primary resistance drop = 
I
1
R
1
 
E
2
? P.U secondary resistance drop = 
I
2
 R
2
 
E
1
? Total P.U resistance drop ref to I
ry
 = 
I
1
R
01
 
E
2
? Total P.U resistance drop ref to II
ry
 = 
I
2
 R
02
 
? The P.U resistance drops on both sides of the t/f is same 
I
1
 R
01
=
I
2
R
02
  
E
1
E
2
Losses present in transformer :- 
1. Copper losses
2. Iron losses
3. Stray load losses
4. Dielectric losses
major losses 
Iron parts   
minor losses 
t/f windings  
t/f core 
cu parts  
insulating materials. 
1. Cu losses in t/f:
    Total Cu loss = I
 
1
 2
R
1
+ I
 
2
 2
R
1
 
         = I
 
1
2
R
01
 
E
1
        = I
 
2
 2
 R
02
  
? Rated current on I
ry
= 
VA rating of t/f
 
E
2
Similarly current on II
ry
 = 
VA rating of t/f
  
VA rating of t/f
? Cu losses ? I
 
1
 2
 or I
 
2
 2
. Hence there are called as variable losses. 
? P.U Full load Cu loss = 
FL Cu loss in watts
 
= 
I
 
1
 2
R
01
E
1
I
1
? If VA rating of t/f is taken as base then  P.U Cu loss ? I
 
1
 2
 as remaining terms are constant. 
? P.U Cu loss at x of FL = x
2
× PU FL Cu loss 
? 
P. U resistance drop ref to I
ry
or 
P. U resistance ref to I
ry
 
? = 
I
1
R
01
E
1
I
1
× 
I
1
  
= 
I
 
1
 2
R
01
E
1
I
1
? P.U Resistance drop = P.U FL cu loss 
% FL Cu loss = % R = % Resistance drop. 
Iron (or) Core losses in t/f :- 
1. Hysteresis loss :
Steinmetz formula :- 
Where  
? = stienmetz coefficient  
B
max
 = max. flux density in transformer core.  
f = frequency of magnetic reversal = supply freq. 
v = volume of core material  
x = Hysteresis coeff (or) stienmetz exponent  
   = 1.6 (Si or CRGo steel) 
2. Eddycurrent loss:
Eddy current loss ,(W
e
) ? R
ce
 × I
 
e
 2
  
As area decreases in laminated core resistance increases as a result conductivity decreases. 
W
e
= K. B
  
ma
2
x
 f
2
. t
2
 
Constant 
Supply freq 
thickness of laminations. 
 Area under one hysteresis loop. 
 
x
?  B
  
max
   .  f  .  v W
h
 = 
(it is a function of s )
 During operation of transformer :- 
B
m
 ? 
V
1
f
f
Case (i) :-  
V
1
 = constant, B
max
 = const. 
w
e
 ? f
2
 
w
e
= B f
2
  
Const. 
? w
i
= w
h
+ w
e
 
w
i
= Af + Bf
  2
 
When B
max
 = const. 
Case (ii) :-     
V
1
f
 ? constant, B
m
 ? const. 
w
e
 ? ?
V
1
f
?
2
. f
2
w
e
 ? V
1
  2
 
w
i
= w
h
+ w
e
  
w
i
= 
A V
0
 
.
1
 
6
1.6
f
+ BV
1
  2
  
VA rating of t/f
P.U iron loss :- 
? P.U iron loss = 
Iron loss in watts
 
? As VA rating is choosen as base then the P.U iron loss are also constant at all load conditions. 
To find out constant losses :-  
? W
0
 = Losses in t/f under no load condition
= Iron losses + Dielectric loss + no load primary loss (I
 
0
 2
 R
1
)
? Constant losses = W
0
- I
 
0
 2
R
1
Where ,   R
1
 = LV winding resistance.
To find out variable losses :- 
? W
sc
 = Loss in t/f under S.C condition
       = F.L Cu loss + stray load losses (Cu and Iron) + Iron losses in both wdgs 
? Variable losses = W
SC
- Iron losses corresponding to V
CC
O.C test :-  
V
1
 rated ?? W
i
 
S.C test :- 
V
SC
   ?? (W
i
)
S.C
 
W
i
  ? V
1
  2
 
W
i
(W
i
)
SC
= ?
V
1
 rated
V
SC
?
2
(W
i
)
S.C
= W
i
 × ?
V
SC
V
1  rated
?
2
? Variable losses = W
SC
- (W
i
)
SC
 ?
V
SC
V
1  rated
?
2