Electrical Engineering (EE) Exam > Electrical Engineering (EE) Notes > Short Notes for Electrical Engineering > Short Notes: Transformers
Short Notes: Transformers | Short Notes for Electrical Engineering - Electrical Engineering (EE) PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
? ? ?
Read More
FAQs on Short Notes: Transformers - Short Notes for Electrical Engineering - Electrical Engineering (EE)
| 1. What is a transformer in electrical engineering? |  |
A transformer is an electrical device that is used to transfer electrical energy between two or more circuits through electromagnetic induction. It consists of two or more coils of wire, known as windings, that are wound around a core made of magnetic material. Transformers are essential in electrical power transmission and distribution systems as they can step up or step down the voltage levels.
| 2. How does a transformer work? | |
A transformer works on the principle of electromagnetic induction. When an alternating current (AC) flows through the primary winding, it creates a changing magnetic field around the winding. This changing magnetic field induces a voltage in the secondary winding, which is connected to the load. The ratio of the number of turns in the primary and secondary windings determines the voltage transformation ratio of the transformer.
| 3. What are the different types of transformers? | |
There are several types of transformers used in electrical engineering. Some common types include:
- Power transformers: These are used in power transmission and distribution systems to step up or step down voltages.
- Distribution transformers: These are used to step down the voltage for distribution to households and commercial buildings.
- Instrument transformers: These are used for measurement and protection purposes, such as current transformers (CTs) and voltage transformers (VTs).
- Autotransformers: These have a single winding that serves as both the primary and secondary winding, resulting in a voltage transformation ratio between the input and output.
- Isolation transformers: These provide electrical isolation between the input and output circuits, often used for safety purposes.
| 4. What are the main components of a transformer? | |
The main components of a transformer include:
- Core: The core is made of a magnetic material, such as laminated iron or steel, and provides a path for the magnetic flux.
- Windings: There are two or more windings, including the primary winding and secondary winding, which are made of insulated copper or aluminum wire. These windings are responsible for transferring the electrical energy.
- Insulation: Insulation materials, such as varnish or enamel, are used to electrically isolate the windings and prevent short circuits.
- Tap changer: Some transformers have a tap changer that allows adjusting the turns ratio to compensate for variations in the input or output voltage.
- Cooling system: Transformers may have cooling systems, such as oil or air cooling, to dissipate the heat generated during operation.
| 5. What are some common problems with transformers? | |
Some common problems with transformers include:
- Overheating: Transformers can overheat due to excessive loading, poor cooling, or insulation degradation. This can lead to insulation failure and reduced efficiency.
- Winding faults: Short circuits or open circuits in the windings can cause a loss of voltage or current, resulting in decreased performance.
- Insulation breakdown: Over time, the insulation in a transformer can degrade, leading to breakdowns and potential electrical faults.
- Voltage regulation issues: Transformers should provide a stable output voltage, but variations in input voltage or load can cause voltage regulation problems.
- Transformer noise: Transformers can produce audible noise due to magnetostriction, vibration, or loose components. Excessive noise can indicate a potential issue.
Related Exams
About this Document
|
2.1K Views |
|
4.71/5 Rating |
|
Dec 22, 2024 Last updated |
Document Description: Short Notes: Transformers for Electrical Engineering (EE) 2024 is part of Short Notes for Electrical Engineering preparation.
The notes and questions for Short Notes: Transformers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Short Notes: Transformers covers topics
like and Short Notes: Transformers Example, for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Short Notes: Transformers.
Introduction of Short Notes: Transformers in English is available as part of our Short Notes for Electrical Engineering
for Electrical Engineering (EE) & Short Notes: Transformers in Hindi for Short Notes for Electrical Engineering course.
Download more important topics related with notes, lectures and mock test series for Electrical Engineering (EE)
Exam by signing up for free. Electrical Engineering (EE): Short Notes: Transformers | Short Notes for Electrical Engineering - Electrical Engineering (EE)
Description
Full syllabus notes, lecture & questions for Short Notes: Transformers | Short Notes for Electrical Engineering - Electrical Engineering (EE) - Electrical Engineering (EE) | Plus excerises question with solution to help you revise complete syllabus for Short Notes for Electrical Engineering | Best notes, free PDF download
Information about Short Notes: Transformers
In this doc you can find the meaning of Short Notes: Transformers defined & explained in the simplest way possible. Besides explaining types of
Short Notes: Transformers theory, EduRev gives you an ample number of questions to practice Short Notes: Transformers tests, examples and also practice Electrical Engineering (EE)
tests
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Important questions
,mock tests for examination
,Short Notes: Transformers | Short Notes for Electrical Engineering - Electrical Engineering (EE)
,ppt
,Sample Paper
,Short Notes: Transformers | Short Notes for Electrical Engineering - Electrical Engineering (EE)
,study material
,video lectures
,Viva Questions
,Exam
,MCQs
,Free
,Short Notes: Transformers | Short Notes for Electrical Engineering - Electrical Engineering (EE)
,Previous Year Questions with Solutions
,Objective type Questions
,Summary
,Extra Questions
,Semester Notes
,practice quizzes
,shortcuts and tricks
,past year papers
,
Additional Information about Short Notes: Transformers for Electrical Engineering (EE) Preparation
Short Notes: Transformers Free PDF Download
The Short Notes: Transformers is an invaluable resource that delves deep into the core of the Electrical Engineering (EE) exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Short Notes: Transformers now and kickstart your journey towards success in the Electrical Engineering (EE) exam.
Importance of Short Notes: Transformers
The importance of Short Notes: Transformers cannot be overstated, especially for Electrical Engineering (EE) aspirants.
This document holds the key to success in the Electrical Engineering (EE) exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Short Notes: Transformers
Short Notes: Transformers Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Short Notes: Transformers.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Short Notes: Transformers Notes on EduRev are your ultimate resource for success.
Short Notes: Transformers Electrical Engineering (EE) Questions
The "Short Notes: Transformers Electrical Engineering (EE) Questions" guide is a valuable resource for all aspiring students preparing for the
Electrical Engineering (EE) exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Short Notes: Transformers on the App
Students of Electrical Engineering (EE) can study Short Notes: Transformers alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Short Notes: Transformers,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Short Notes: Transformers is prepared as per the latest Electrical Engineering (EE) syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup