Electrical machine slide share

ELECTRICAL
MACHINE
Mohammed Waris Senan

Content
Transformer
Introduction
Laws governing working
Understanding Lenz’s Law
Ideal Transformer
EMF Equation of Transformer
Phasor diagram of ideal transformer
Practical Transformer
Equivalent Circuit diagram
Phasor diagram
Concept of referred Values
31-May-19 Mohammed Waris Senan 2

Transformer
A static device which transfer power (or energy) from one circuit
to another circuit by changing the level of voltage (and current)
without changing frequency.

A CUT SECTION
Transformer works on two LAWS

Transformer
1st Law:
Faraday law of Electromagnetic Induction
Flux linking with a coil changes an emf is induced in the coil which is
directly proportional to the rate of change of flux linkage.
dt
d
e

 λ = Flux Linkage
λ = N 𝜙
dt
d
Ne



Transformer
2nd Law:
Lenz’s Law
The direction of induced EMF is such that if it is allowed to cause
current by short-circuiting the coil, then the current so produced
opposes the cause
dt
d
Ne
dt
d
e
ei




..
Where the sign depends on
Lenz’s law and which terminal
is taken as positive

Transformer
N SN S
dt
d
Neab


abe
NS
a b

Transformer
N SN S
dt
d
Neab


abe
N S
a b

Transformer
dt
d
Neab


abe
NS N SN S
a b

Transformer
Ideal Transformer
1. No losses
 Copper losses and iron losses are zero
2. Infinite Permeability (i.e. Zero Reluctance)
3. No leakage Flux
mFlux ,
v1
e1
e2 v2
+
+
-
N1 N2
N1>N2
i1
2l1l

Transformer
)90sin(
cos
)sin(
,
sin,





tNe
tNe
dt
td
Ne
So
tLet
dt
d
Ne
m
m
m
m





Equation of induced EMF
mfNE  11 2
mfNE  22 2
For primary winding having N1 turns
For Secondary winding having N2 turns
RMS Value
m
m
m
fNE
πfω
fN
E
N
E



2
)2(
2
)2(
2




So,

Transformer
Ideal Transformer
e1
𝜙𝜙m
𝜙 , e1
t
1V
2E2V
1E
m
(Flux)

Transformer
Ideal Transformer
mFlux ,
+
+
-
N1 N2
L
O
A
D
1V 1E 2V2E
1I 2I
At Load

Transformer
Turns Ratio/Voltage Ratio/Transformation Ratio (a)
a
2
1
2
1
2
1
N
N
V
V
E
E
m
m
fNE
fNE


22
11
2
2


As,
Also, in case of Ideal Transformer
V1 = E1 and V2 = E2
a
2
1
2
1
N
N
E
E
So,

Transformer
Turns Ratio/Voltage Ratio/Transformation Ratio (a)
In an ideal transformer there is no leakage flux
Net mmf,
0ININ 2211 
0I
0,NI
NI
As,







if
μA
l
2
'
2
1
2
1
2211
II
N
N
I
ININSo,


1
2
2
1
2
1
2
1
I
I
N
N
V
V
E
E
 a
Therefore we can write,
)0( 21  ll 

Transformer
Ideal Transformer on Load (at lagging pf)
22 VE 
11 VE 
m
(Flux)
2I
a
2
1
I
I 1
2
 The primary winding is behaving
as a sink and simultaneously
secondary winding is behaving as
a source w.r.t. load.
 Ideal transformer is just
transferring only and consuming
no power in between

Transformer
Practical Transformer
So, here no load current (I0) is required to produce flux in the core and for core losses.
 No losses
 Copper losses and iron losses are present
 Infinite Permeability (i.e. Zero Reluctance)
 Definite Permeability (Reluctance is present)
 No leakage Flux Leakage
 Leakage Flux present
02
1
2
1
012211
012211
II
N
N
I
INININ
INININ



02
'
1 III 
CurrentLoadCurrent/NoExcitingI0 
Ideal Transformer

Transformer
R1jX1 jX2R2
RcjX 𝜙1V 1E 2E 2V
cII
0I
1I 0I 2
'

At No Load
01
2
'
2
II
0I0I


0I2 

Transformer
22 VE 
11 VE 
m
(Flux)
10 II 
0
I
cI Hysteretic Angle
At No Load
01
2
'
2
II
0I0I


0IofcomponentgMagnetisinI 
0c IofcomponentlossCoreI 
Phasor diagram of transformer at NO LOAD

Transformer
1V
R1jX1 jX2R2
RcjX 𝜙 1E 2E 2V
cII
0I
1I 2
'
I 2I
L
O
A
D
222222 XIRIVE j111111 XIRIEV j
c0 III  
At Load

Transformer
222222 XIRIVE j
111111 XIRIEV j
c0 III  
1I
m
2E
1E
0I
I
cI
2I
2
'
I
2V
22RI
22XIj
11RI
11XIj1V
Phasor diagram of transformer at LOAD (cos 𝜙 lag)

Transformer
Referred Value
1V 1E 2E 2V
1I 2I
L
O
A
D
loadZ
a
1
2
2
1
I
I
V
V
As,
12
1
2 IIand
a
V
V a
2
2
I
V
Z load
1
1
I
V
Z
a
aload 
1
1
2
I
V1
Z 
a
load
)Z
I
V
(
Z
Z 1
1
1
2
1
 
a
load
ZZ 2
1 loada
ZZAlso, 2'
loadload a
Secondary load impedance
referred to Primary side

Transformer
222222 XIRIVE j
222222 XIRIVE jaaaa 
2
22
2
22
2
'
2
'
1 X)
I
(R)
I
(VEE a
a
ja
a

22
'
22
'
2
'
2
'
1
''
XIRIVEE j
Equivalent Circuit
Referred to primary

Transformer
jX’2R’2
2
'
1 EE  2
'
V
R1jX1
RcjX 𝜙1V
cII
0I
1I 2
'
I
Equivalent Circuit
2
'
I

Question (1):
The parameters of equivalent circuits of 150 kVA, 2400/240 V transformer are
R1 = 0.2 Ω X1 = 0.45 Ω
R2 = 2 mΩ X2 = 4.5 mΩ
Rc = 10 kΩ X 𝜙 = 1.55 kΩ
Using the circuit referred to primary, determine the primary input voltage, input current
and input pf of transformer operating at rated load with 0.8 pf lag.
Transformer
Solution :
Equivalent Circuit

Transformer
Solution :
Equivalent Circuit

Transformer
jX’2R’2
2
'
1 EE  2
'
V
R1jX1
RcjX 𝜙1V
cII
2
'
I
Approximate Equivalent Circuit
2
'
I
0I
R1ejX1e
2
'
V
X1e = X1 + X’2R1e = R1 + R’2
Where,
and

Transformer
2
'
V
R1ejX1e
RcjX 𝜙1V
cII
2
'
I
Equivalent Circuit
0I
2
'
V
Z1e = jX1e + R1e
1V
2
'
I

Transformer
Equivalent Circuit
2
'
V
Z1e = jX1e + R1e
1V
2
'
I
Referred to primary
2V
Z2e = R2e + jX2e
1
'
V
21
'
II 
Referred to Secondary
X2e = X’1 + X2R2e = R’1+ R2
Where, and

Transformer
Phasor Diagram
2V
Z2e = R2e + jX2e
1
'
V
21
'
II 
Referred to Secondary
2e22e221
'
XIRIVV j
1
'
V
2V
2I

2e2RI
2e2XIj
2e2ZI
L
O
A
D

►Open Circuit Test and Short Circuit Test are carried out on a transformer to
predict its performance without actually loading it.
►Open Circuit Test (OC Test) is carried out at rated frequency and rated
voltage to determine the core loss.
►OC test is usually carried out with the instruments placed on the low voltage (lv)
side while the high voltage (hv) side is kept open circuited.
►This is done because it is easier to arrange rated voltage supply at the low
voltage level, the instruments are also cheaper in cost and it is safer to work at lv
side.
►Since 𝜙0 is very low, so a low pf wattmeter should be used.
Transformer
Open Circuit Test (OC Test)

Transformer
A
V
Rated
frequency,
Adjustable
voltage
supply
Open
Circuitlv hv
I0
V0
Poc
Connection Diagram

Transformer
R1jX1
RcjX 𝜙
ocV
cII
0I
ocII0 
Open
Circuit
Equivalent Circuit
Since, I0 is very small, therefore the power measured may be considered to represent only
core loss as the primary copper loss corresponding to such a small current would be
negligible.
The voltage drop across the primary impedance is extremely low and can be neglected.
Poc

Transformer
RcjX 𝜙
ocV
cII
0I
ocII0 
Open
Circuit
Equivalent Circuit
Yoc
Rc=
(𝑽𝒐𝒄) 𝟐
𝑷 𝒐𝒄
𝝓 𝟎 = 𝒄𝒐𝒔−𝟏 (
𝑷 𝒐𝒄
𝑽 𝒐𝒄
𝑰 𝒐𝒄
)
X 𝝓=
𝑹 𝒄
𝒕𝒂𝒏𝝓 𝟎
Quickest
Poc

Transformer
RcjX 𝜙
ocV
cII
0I
ocII0 
Open
Circuit
Equivalent Circuit
Yoc
Yoc =
𝑰 𝒐𝒄
𝑽 𝒐𝒄
∠ − 𝒄𝒐𝒔−𝟏(
𝑷 𝒐𝒄
𝑽 𝒐𝒄
𝑰 𝒐𝒄
)
= Gc - jB 𝝓
R 𝒄 =
𝟏
Gc
X 𝝓 =
𝟏
B 𝝓
Short-Cut
Poc

Transformer
Question (2):
The instruments connected at lv side, during open circuit test, the readings of for 10
kVA, 450V/120 V, 50 Hz transformer are 120 V, 4.2 A & 80 W. Find the resistances
and reactance of equivalent exciting circuit referred to hv side.
a =
450
120
=
15
4
Given:
Poc = 80 W
Voc = 120 V
I0 = 4.2 A
Solution :
Rc(lv) =
𝑉 𝑜𝑐
2
𝑃 𝑜𝑐
=
1202
80
= 180 Ω
R’
c(hv) =
15
4
2
x 180 = 2.53 kΩ
𝜙0 = cos-1(
𝑃 𝑜𝑐
𝑉 𝑜𝑐
×𝐼 𝑜
)
= cos-1(
80
120×4.2
)
= 80.87°
Now,
X 𝜙(lv) =
180
tan(80.87°)
= 28.93 Ω
X’
𝜙(hv) =
15
4
2
× 28.93 = 406.83 Ω

►Short Circuit Test (SC Test) is carried out at rated current to determine the
full load copper loss. Since, resistance is not affected too much by the frequency at
lower frequencies, the test may be carried out at a frequency even slightly different from
rated frequency.
►SC test is usually carried out with the instruments placed on the high voltage
(hv) side while the low voltage (lv) side is short circuited by thick wire (low
resistance wire).
►This is done because rated current on hv side is lower and therefore cost saving
would be obtained on the instruments.
►Since, the voltage required to circulate full load current at short circuit is only 5-
10% of rated voltage, the core loss under short circuit test is ignored.
Transformer
Short Circuit Test (SC Test)

Transformer
V
Adjustable
voltage
supply
Short
Circuit
hv lv
Isc
Vsc
Psc
Connection Diagram
A

Transformer
R1jX1
RcjX 𝜙
scV
cII
0I
scI
Short
Circuit
Equivalent Circuit
jX’
2 R’
2
Since, the voltage required to circulate full load current at short circuit is only 5-10% of
rated voltage so, the exciting current under reduced voltage is extremely small and can be
neglected.
fl2I
)Vof10%(5 rated
jX1e R1eflsc 2II 

Transformer
scV
Short
Circuit
Equivalent Circuit
jXeq Reqflsc 2II 
Zeq Zeq =
𝑽 𝒔𝒄
𝑰 𝒔𝒄
∠ 𝐜𝐨𝐬−𝟏(
𝑷 𝒔𝒄
𝑽 𝒔𝒄
×𝑰𝒔𝒄
)
= Req + jXeq

Transformer
Question (3):
The instruments connected at hv side, during short circuit test, the readings for 50 kVA,
2400 V/240 V, 50 Hz transformer are 48 V, 20.8 A & 617 W. Find leakage impedance,
effective resistance and leakage reactance referred to lv side.
a =
2400
240
= 10
Given:
Psc = 617 W
Vsc = 48 V
Isc = 20.8 A
Solution :
Zeq(hv) =
𝑉 𝑠𝑐
𝐼 𝑠𝑐
∠ cos−1
(
𝑃 𝑠𝑐
𝑉 𝑠𝑐
×𝐼𝑠𝑐
)
=
48
20.8
∠ 𝑐𝑜𝑠−1(
617
48×20.8
)
= 1.426 + j1.814 Ω
Zeq(lv) =
1
102 (1.426 + j1.814) Ω
∴ Zeq(lv) = 14.26 + j18.14 mΩ
Req(lv) = 14.26 mΩ
Xeq(lv) = 18.14 mΩ
So,

Transformer
Per Unit (pu) Values
Rpu =
Req
Z𝑏𝑎𝑠𝑒
Rpu =
Req
𝑉 𝑟𝑎𝑡𝑒𝑑
𝐼 𝑟𝑎𝑡𝑒𝑑
Rpu =
𝑅 𝑒𝑞
×𝐼 𝑟𝑎𝑡𝑒𝑑
=
𝐹𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑒 𝑑𝑟𝑜𝑝
𝑅𝑎𝑡𝑒𝑑 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
Rpu = Full load resistive drop in pu
Rpu =
𝑅 𝑒𝑞
×
𝑰 𝒓𝒂𝒕𝒆𝒅
Rpu =
𝐹𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 𝑐𝑜𝑝𝑝𝑒𝑟 𝑙𝑜𝑠𝑠
𝑆 𝑏𝑎𝑠𝑒
= Full load copper loss in pu
Rpu =
2
×𝑅 𝑒𝑞
∴
⟹
⟹
⟹
⟹
We can write,
Also,

Transformer
Xpu =
Xeq
Zbase
Xpu =
Xeq
Xpu =
𝑋 𝑒𝑞
=
𝐹𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑒 𝑑𝑟𝑜𝑝
𝑅𝑎𝑡𝑒𝑑 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
Xpu = Full load reactive drop in pu
Xpu =
𝑋 𝑒𝑞
×
Xpu =
𝐹𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑒 𝑙𝑜𝑠𝑠
= Full load reactive loss in pu
Xpu =
2
×𝑋 𝑒𝑞
∴
⟹
⟹
⟹
⟹
Similarly,
And,

Transformer
Zpu = Full load impedance drop in pu
Zpu = Full load apparent power loss in pu
Similarly,
Tentative Values:
Rpu ≃ 0.01 pu (1%)
Xpu ≃ 0.10 pu (10%)
Rc(pu) ≃ 100 pu (104%)
X 𝜙(pu) ≃ 25 pu (2500%)
∵ Pc =
𝑉2
𝑅 𝑐
= 0.01 (Very low) =
12
𝑅 𝑐
⇒ Rc = 100
∵ V1 = I 𝜙 ∙ X 𝜙= 0.04 ∙ X 𝜙 = 1 ⇒ X 𝜙 = 25

Transformer
Question (4):
A 10 kVA, 2000/200 V, 50 Hz, single phase transformer working at rated voltage at no-load
takes an input of 125 W at 0.15 pf. Its percentage leakage impedance, based on its own kVA
is (0.5 + j1) pu. If transformer delivers 10 kW at 200 V at 0.8 pf lag on its lv side, then
determine the input power (in pu) and input pf.
Solution :

Transformer
Solution :

Transformer
Voltage Regulation
It is defined as the rise in output (secondary voltage) express as a fraction of full
load rated voltage when full load at a specified pf is reduced to zero keeping the
primary input voltage constant.
Voltage Regulation =
𝑁𝑜 𝐿𝑜𝑎𝑑 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 − 𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑 𝑉𝑜𝑙𝑡𝑎𝑔𝑒
𝐹𝑢𝑙𝑙 𝐿𝑜𝑎𝑑 𝑉𝑜𝑙𝑡𝑎𝑔𝑒
Where full load voltage is Rated Voltage
inV V
I jXeq Req
Zeq
𝑉𝑖𝑛
− 𝑉
𝑉
pu
=
𝑉𝑖𝑛
𝑉
– 1 pu
= Vin(pu) – 1 pu

Transformer
Question (5):
A transformer ha pu impedance of 0.10 and pu resistance of 0.01. Calculate the phase
difference between output voltage and input voltage on full load. And also determine the
voltage at
(a) 0.8 pf lag (b) UPF (c) 0.8 pf lead
Voltage Regulation
Solution :

Transformer
Voltage Regulation
Solution :

Transformer
Voltage Regulation
inV
V
I
eqRI
eqXIj
eqZI
A
D C
O
B
𝜙 𝝓
𝜃 𝑒𝑞
(90˚ − 𝜃 𝑒𝑞)
Exact VRn =
𝑂𝐵 − 𝑂𝐴
𝑂𝐴
=
𝑂𝐶 − 𝑂𝐴
𝑂𝐴
δ
Vincosδ ≃ Vin (∵ 𝛿 → 𝑉𝑒𝑟𝑦 𝑆𝑚𝑎𝑙𝑙)
𝑉𝑖𝑛
− 𝑉
𝑉
pu
Approximate VRn =
𝑂𝐷 − 𝑂𝐴
𝑂𝐴
=
𝐴𝐷
𝑂𝐴
Neglecting DC for approximate VRn
Approx. VRn =
𝐼𝑍 𝑒𝑞
𝑉
cos (𝜃 𝑒𝑞 − 𝜙)
∴ VRn = Zpu cos (𝜽 𝒆𝒒 − 𝝓)
E

Transformer
Voltage Regulation
= Zpu (cos𝜃 𝑒𝑞 . cos𝜙 + sin𝜃 𝑒𝑞 . sin𝜙)
= (Zpu cos𝜃 𝑒𝑞) . cos𝜙 + (Zpu sin𝜃 𝑒𝑞) . sin𝜙
∴ VRn = Rpu cos𝜙 + Rpu sin𝜙
Approximate VRn = Zpu cos 𝜃 𝑒𝑞 − 𝜙

Transformer
Voltage Regulation
Maximum Voltage Regulation
VRn = Zpu cos 𝜃 𝑒𝑞 − 𝜙
When, 𝜙 = 𝜃 𝑒𝑞 (lagging)
And,
pf = cos 𝜙 = cos 𝜃 𝑒𝑞 =
𝑅 𝑒𝑞
𝑍 𝑒𝑞
So, VRn
(max) = Zpu
inV
V
I
eqRI
eqXIj
eqZI
A
BO
B
𝜙 𝝓
δ
E
𝜃 𝑒𝑞
δ = 0
C
I
eqRI
𝝓= 𝜃 𝑒𝑞

Transformer
Voltage Regulation
Zero Voltage Regulation
VRn = Zpu cos 𝜃 𝑒𝑞 − 𝜙 When, 𝜃 𝑒𝑞 - 𝜙 = 90˚ (here pf is lagging)
So, VRn = 0
But, {as 𝜃 𝑒𝑞 < 90˚ (Refer Phasor diagram)}
So, zero VRn is possible only when pf is leading
For leading pf, VRn = Zpu cos 𝜃 𝑒𝑞 + 𝜙 {As, for leading pf 𝜙 becomes - 𝜙}
𝜃 𝑒𝑞 - 𝜙 can never be equal to 90˚
So, for zero VRn 𝜃 𝑒𝑞 + 𝜙 = 90˚ ⇒ 𝜙 = 90˚ - 𝜃 𝑒𝑞
Pf (leading) = cos 𝜙 = cos (90˚- 𝜃 𝑒𝑞) = sin 𝜃 𝑒𝑞 =
𝑋 𝑒𝑞
𝑍 𝑒𝑞

Transformer
Voltage Regulation
Zero Voltage Regulation
For zero VRn 𝜃 𝑒𝑞 + 𝜙 = 90˚ ⇒ 𝜙 = 90˚ - 𝜃 𝑒𝑞
inV
V
I
eqRI
eqXIj
eqZI
A
O
C
B
𝜙
𝜃 𝑒𝑞
δ 𝜙
Here, Vincos δ = V

Transformer
Voltage Regulation
Minimum Voltage Regulation
Minimum VRn is obtained when, 𝜙 = 90˚ ⇒ pf, cos 𝜙 = zero (leading)
For leading pf, VRn = Zpu cos 𝜃 𝑒𝑞 + 𝜙
inV
V
I
eqRI
eqXIj
eqZI
AO
C B
𝜙
𝜃 𝑒𝑞
δ
Vincos δ

Transformer
Voltage Regulation
𝜃 𝑒𝑞
𝜃 𝑒𝑞
VRn
(max)
VRn
(min)
VRn
pf
laggingleading
(90˚ − 𝜃 𝑒𝑞)
(𝜙 = 𝜃 𝑒𝑞)
90˚ (lead)

Transformer
Voltage Regulation is a figure of merit of a transformer and its LOW VALUE is desirable.
A lower VRn requires a low per unit impedance (Zpu).
VRn = Zpu cos 𝜃 𝑒𝑞 − 𝜙
Resistance is already kept low due to efficiency (η) consideration. Thus leakage reactance
must be reduced to improve VRn.
Leakage flux can be reduced by keeping the windings as much close as possible.
In Core type transformer, this is done by using concentric cylindrical windings.
In Shell type transformer, leakage flux is reduced by sandwitched winding (pancaked
winding).
Leakage reactance is less in core type as compared to shell type.
Voltage Regulation

Transformer
The leakage reactance of a hv transformer is higher because of thicker insulation
which makes the winding further apart.
In core type transformer, leakage reactance can also be reduced by increasing window
height to width ratio.
Voltage Regulation
H
W
Less
Reluctance
𝜙l Xl
H
W
More
Reluctance
𝜙l Xl

Transformer
The VRn as a figure of merit has little significant in power transformer as it operates
at full load or switched off.
pu impedance in distribution transformer is lower (Zpu ~ 0.015 pu) as compared to
power transformer (Zpu ~ 0.15 pu)
A high pu impedance has an advantage that it reduces fault MVA level (i.e. fault
current).
Voltage Regulation

Transformer
Losses in transformer
Loss
(intransformer)
Copper Loss
Pcu = I2Req
Core Loss (or Iron Loss)
Pi = Ph + Pe
Hysteresis Loss
Ph ∝ f𝑩 𝒎
𝒙
Eddy Current Loss
Pe ∝ 𝒇 𝟐
𝑩 𝒎
𝟐
𝒕 𝟐
Where, x = Steinmetz Constant
= 1.5 to 2.5
∝ Magnetic properties of core
Where, t = Thickness of insulation
= 0.35 mm for 50 Hz

Transformer
Losses in transformer
• Cold Rolled Grain Oriented (CRGO) with 3 % silicon reduces
hysteresis loss.
• More than 3 % silicon makes core brittle hence not used.
• 3 % silicon also increases resistivity, hence resistance increase
so eddy current loss reduces.
• Magnetic core is a stack of thin silicon-steel laminations.
• These laminations are insulated from one another by thin layer
of varnish in order to reduce eddy current losses.

Transformer
Separation of Hysteresis and Eddy Current Losses
Hysteresis Loss, Ph = Kh f 𝐵 𝑚
𝑥 Where, Kh = Hysteresis loss constant
Eddy Current Loss, Pe = Ke 𝑓2
𝐵 𝑚
2
𝑡2
Where, Ke = Eddy Current loss constant
Core Loss (or Iron Loss), Pi = Ph + Pe
Pi = Pe + Ph
= Ke 𝑓2 𝐵 𝑚
2 𝑡2 + Kh f 𝐵 𝑚
𝑥
Thickness of insulation (t) is constant for a transformer.
So, for a given 𝑩 𝒎
Ph ∝ f ⇒ Ph = a f
Pe ∝ 𝑓2 ⇒ Pe = b𝑓2

Transformer
𝑃𝑖
𝑓
= bf + a
So, Pi = Pe + Ph
= b𝑓2
+ a f
⇒
Pi = b𝑓2
+ a f⇒
y = mx + c
𝑃𝑖
𝑓
f
a
Extrapolation
o
So,
a – y intercept (c)
b – Slope of straight line (m)
A B
CD
E
F
From the graph, a = OF and b =
EC
DC

Transformer
Question (6):
A single phase 400 V, 50 Hz transformer has an iron loss of 5000 W at the rated condition.
When operated at 200 V, 25 Hz, the iron loss is 2000 W. Calculate hysteresis and eddy
current loss when the transformer is operated at 416 V, 52 Hz.
Solution :

Transformer
Solution :

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