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Transformer

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Transformer

A transformer is a device that changes the voltage in an alternating current (AC) circuit. It works based on electromagnetic induction, where the primary coil creates a magnetic field that induces a voltage in the secondary coil. Transformers are mainly used to either increase (step up) or decrease (step down) voltage, making them vital for power distribution systems. They are key components in power grids, household appliances, and many industries, helping to transmit and distribute electrical energy efficiently.

1.0Definition of Transformer

It is an electrical device that changes alternating current from one voltage level to another. When it raises the input voltage, it's called a step-up transformer, and when it lowers the input voltage, it's known as a step-down transformer.

2.0Principle of Transformer

It runs on the principle of mutual induction, which means that when a changing current flows through one of the two inductively coupled coils, it generates an induced electromotive force (emf) in the other coil.

3.0Construction of Transformer

A transformer primarily consists of two coils made of insulated copper wire, each having a different number of turns, wound around a common soft iron core.
The coil to which electrical energy is supplied is referred to as the primary coil, while the coil from which energy is extracted or output is called the secondary coil.
A laminated core is used to reduce energy losses from eddy currents. The high permeability of soft iron keeps most of the magnetic flux within the core, ensuring it passes through the secondary coil and preventing stray currents in surrounding conductors, thus minimizing power loss.

There are typically two types of configurations used for winding the primary and secondary coils in a transformer.

1.Shell Type:

In this type of transformer, the primary and secondary coils are wound on top of each other on the same limb of the iron core, with the coils being extensively surrounded by the iron core.
It consists of primary and secondary copper coils. The effective resistance between the primary and secondary coils is infinite because the electrical circuit between them is open. $(R_{p s} = \infty)$
It consists of primary and secondary copper coils. The effective resistance between the primary and secondary coils is infinite because the electrical circuit between them is open.

2.Core Type

In core-type transformers, the primary and secondary coils are wound on different limbs of the core, enabling the core to be mostly enclosed by the coils. Many modern transformers are designed as closed-core types.
Both Cu coils are tightly wound over a bulk metal piece of high magnetic permeability (eg. soft iron) called core. Both coils are electrically insulated to the core but the core part magnetically coupled to both the coils.

4.0Working of Transformer

As a.c flows through the primary coil, it generates a changing magnetic flux in the core, inducing an electromotive force (emf) in the secondary coil and a self-induced emf in the primary coil.
If there is no magnetic flux leakage, the flux linked with each turn of the primary coil will be the same as that linked with each turn of the secondary coil.
It regulates A.C. voltage and transfers the electrical power without change in frequency of input supply. (The alternating current changes itself.)

5.0Types of Transformers

Feature	Step-Up Transformer (S.U.T.) $N_{S} > N_{P}$	Step-Down Transformer (S.D.T) $N_{S} < N_{P}$
Function	Increases voltage	Decreases voltage
Number of Turns (Primary)	Fewer turns in the primary coil	Greater turns in the primary coil
Number of Turns (Secondary)	More turns in the secondary coil	Fewer turns in the secondary coil
Voltage (Primary)	Low voltage	High voltage
Voltage (Secondary)	High voltage	Low voltage
Current (Primary)	Low current	High current
Current (Secondary	High current	Low current
Usage	Power transmission over long distances	Household appliances, low-power applications
Example	Power plants, electrical grids	Home electrical systems, chargers, small appliances

S.U.T converts low voltage, high current into high voltage, low current
S.D.T. converts high voltage, low current into low voltage, high current.
Power transmission is carried out always at "High voltage, low current" so that voltage drop and power losses are minimum in the transmission line.

$Voltage Drop = I_{L} R_{L}$

$(I_{L} = line Current, R_{L} = Total Line Resistance)$

$I_{L} = \frac{Power to be transmitted}{Line Voltage}$

$Power Losses = I_{L}^{2} R_{L}$

Sending power always at high voltage & low current (By. S.U.T.) and receiving power always at low voltage & high current (By S.D.T.)
High voltage coils have more turns and are always made of thin wire and high current coils having less number of turns and always made of thick wires.

6.0Key Points of Transformer

It cannot operate with a DC supply. When a battery is connected to the primary side, the output across the secondary side remains zero, meaning the transformer does not function.
It cannot be considered an 'amplifier' since it does not provide power gain like a transistor.
Since the transformer has no moving parts, there are no mechanical losses, making its efficiency higher compared to that of generators and motors.

7.0Ideal Transformer $(η = 100%)$

8.0

No Flux Leakage:-

$ϕ_{s} = ϕ_{p} = \frac{- d ϕ _{s}}{d t} = \frac{- d ϕ _{p}}{d t}$

$e_{s} = e_{p} = e$ (induced emf per turn of each coil is also same)

total induced emf for secondary, $E_{s} = N_{s} e$

total induced emf for primary, $E_{p} = N_{p} e$

$\frac{E _{s}}{E _{p}} = \frac{N _{s}}{N _{p}} = n or p$ ……….(1)

$n \to Turn Ratio$

$p \to Transformation Ratio$

(b) No load condition :-

$V_{p} = E_{p} and E_{s} = V_{s}$

$\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$ ………….(2)

from(1) and (2)

$\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}} = n or p$ ………..(3)

(c) No power loss :-

$P_{o u t} = P_{in}$

$V_{S} I_{S} = V_{P} I_{P}$

$\frac{V _{S}}{V _{P}} = \frac{I _{P}}{I _{S}}$ …………………….(4)

From equation (3) and (4)

$\frac{V _{S}}{V _{P}} = \frac{I _{P}}{I _{S}} = \frac{N _{s}}{N _{p}} = n or p$

Sp. Note : Generally transformers deals in ideal condition i.e. $P_{in} = P_{o u t}$ , if other information is not given.

9.0Real Transformer

Some power is always lost due to flux leakage, hysteresis, eddy currents, and heating of coils.Hence $P_{o u t} < P_{in}$ always
Efficiency of a Transformer $η = \frac{P _{o u t}}{P _{in}} = \frac{V _{S} I _{S}}{V _{P} I _{P}} \times 100$

10.0Losses In Transformer

Copper or joule heating losses :-

Where : These losses occur in both coils of the shell part.

Reason : Due to the heating effect of current $(H = I^{2} Rt)$ .

Remedy : To minimise these losses, high current coil is always made up with thick wire and for removal of produced heat, circulation of mineral oil should be used.

Flux leakage losses :-

Where : These losses occur in between both coils of the shell part.

Cause : Due to air gap between both the coils.

Remedy : To minimise these losses both coils are tightly wound over a common soft iron core (high magnetic permeability) so a closed path of magnetic field

lines formed itself within the core and tries to make coupling factor $K \to 1$

Iron losses :-

Where : These losses occur in the core part.

Types:(a) Hysteresis Loss (b) Eddy Current Loss

(a) Hysteresis Loss

Cause : Transformer core always present in the effect of alternating magnetic field $(B = B_{0} S inω t)$ so it will be magnetised & demagnetised with very high frequency (f=50Hz) .During its demagnetization a part of magnetic energy is left inside the core part in the form of residual magnetic field. Finally, this residual energy wasted as heat.

Remedy : To minimise these losses material of transformer core should be such that it can be easily magnetised & demagnetised. For this purpose, magnetic soft materials should be used.

Example- Soft Iron (Low Retentivity and Low Coercivity)

(b) Eddy Current Loss

It is a group of induced currents which are produced when metal bodies placed in a time varying magnetic field or they move in an external magnetic field in such a way that flux through them changes with respect to time.

11.0Advantages And Disadvantages of Transformer

Advantages of Transformers

1.Transformers are highly efficient

2.They can easily step up or step down voltages

3.Transformers have no moving parts.

Disadvantages of Transformers

1. Transformers only work with alternating current (AC)

2.Heavy, and difficult to install.

3.Under high load or inefficient operation, transformers can overheat

12.0Significant Points of Transformer

These currents are produced only in closed paths within the entire volume and on the surface of the metal body. Therefore their measurement is impossible.
The circulation plane of these currents is always perpendicular to the external magnetic field direction.
Generally resistance of metal bodies is low so the magnitude of these currents is very high.
These currents can heat up the metal body and some time body will melt out (Application :Induction furnace)
Due to these induced currents a strong eddy force (or torque) acts on the metal body which always opposes the translatory (or rotatory) motion of the metal body, according to Lenz law. The transformer core is always present in the effect of alternating magnetic field \left(B=B_0Sin\omega t\right). Due to this eddy currents are produced in its volume, so a part of magnetic energy of the core is wasted as heat.

Remedy : To minimise these losses the transformer core should be laminated. With the help of the lamination process, the circulation path of eddy current is greatly reduced & net resistance of the system is greatly increased. So, these currents become feeble.

13.0Solved Examples On Transformer

Q-1.In a transformer, the AC voltage is increased from 220 V to 2200 V. If the secondary coil has 2000 turns, how many turns will be in the primary coil?

Solution:

$\frac{V _{p}}{V _{s}} = \frac{N _{p}}{N _{s}} \Rightarrow N_{p} = \frac{220}{2200} \times 2000 = 200$

Q-2.An ideal transformer is connected to a 240 V AC mains, and the output voltage is 24 V. The transformer is then used to power a 24V, 24W bulb. Calculate the current flowing through the primary coil of the transformer.

Solution:Current in primary coil $I_{p} = \frac{V _{S} I _{S}}{V _{P}} = \frac{24}{240} = 0.1 A$

Q-3.A bulb (100 W, 110 V) is operated using a transformer by supplying 220 V, 0.5 A, find efficiency of transformer.

Solution:

$P_{in p u t} = P_{o u tp u t}$

$Efficiency (η) = \frac{P _{o u tp u t}}{P _{in p u t}} \times 100$

$Efficiency (η) = \frac{100}{( 220 ) ( 0.5 )} \times 100 = 90.90%$

Q-4.A transformer has primary and secondary coils with 50 and 1500 turns, respectively. The magnetic flux coupled with the primary coil is given by $Φ = (2 + 4 t) Wb$ , where t represents time. What is the output voltage across the secondary coil?

Solution:

$N_{s} = 100 N_{p} = 50 Φ = (2 + 4 t) Wb$

$\frac{d Φ}{d t} = 4 \Rightarrow e = ν_{i} = 4$

$\frac{V _{0}}{V _{i}} = \frac{N _{s}}{N _{p}} \Rightarrow \frac{V _{0}}{4} = \frac{1500}{50} \Rightarrow V_{0} = 120 V$

So, output voltage across the secondary coil is 120 V.

Q-5.A transformer has an input voltage of 2500 volts and an output current of 80 amperes. The transformer ratio of the primary coil to the secondary coil is 20:1. Assuming the transformer operates at 100% efficiency, what is the voltage in the secondary coil?

Solution:

$V_{I n p u t} = 2500 \Rightarrow I_{O u tp u t} = 80$

$η = 100% \Rightarrow \frac{N _{P}}{N _{S}} = \frac{20}{1}$

$P_{I n p u t} = P_{O u tp u t}$

$V_{P} I_{P} = V_{S} I_{S} (\frac{V _{S}}{V _{P}} = \frac{N _{S}}{N _{P}} = \frac{I _{P}}{I _{S}})$

$\frac{V _{S}}{2500} = \frac{1}{20} \Rightarrow V_{S} = 125 V$

So, output voltage across the secondary coil is 125 V.

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I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Transformer

A transformer is a device that changes the voltage in an alternating current (AC) circuit. It works based on electromagnetic induction, where the primary coil creates a magnetic field that induces a voltage in the secondary coil. Transformers are mainly used to either increase (step up) or decrease (step down) voltage, making them vital for power distribution systems. They are key components in power grids, household appliances, and many industries, helping to transmit and distribute electrical energy efficiently.

1.0Definition of Transformer

It is an electrical device that changes alternating current from one voltage level to another. When it raises the input voltage, it's called a step-up transformer, and when it lowers the input voltage, it's known as a step-down transformer.

2.0Principle of Transformer

It runs on the principle of mutual induction, which means that when a changing current flows through one of the two inductively coupled coils, it generates an induced electromotive force (emf) in the other coil.

3.0Construction of Transformer

A transformer primarily consists of two coils made of insulated copper wire, each having a different number of turns, wound around a common soft iron core.
The coil to which electrical energy is supplied is referred to as the primary coil, while the coil from which energy is extracted or output is called the secondary coil.
A laminated core is used to reduce energy losses from eddy currents. The high permeability of soft iron keeps most of the magnetic flux within the core, ensuring it passes through the secondary coil and preventing stray currents in surrounding conductors, thus minimizing power loss.

There are typically two types of configurations used for winding the primary and secondary coils in a transformer.

1.Shell Type:

In this type of transformer, the primary and secondary coils are wound on top of each other on the same limb of the iron core, with the coils being extensively surrounded by the iron core.
It consists of primary and secondary copper coils. The effective resistance between the primary and secondary coils is infinite because the electrical circuit between them is open. $(R_{p s} = \infty)$
It consists of primary and secondary copper coils. The effective resistance between the primary and secondary coils is infinite because the electrical circuit between them is open.

2.Core Type

In core-type transformers, the primary and secondary coils are wound on different limbs of the core, enabling the core to be mostly enclosed by the coils. Many modern transformers are designed as closed-core types.
Both Cu coils are tightly wound over a bulk metal piece of high magnetic permeability (eg. soft iron) called core. Both coils are electrically insulated to the core but the core part magnetically coupled to both the coils.

4.0Working of Transformer

As a.c flows through the primary coil, it generates a changing magnetic flux in the core, inducing an electromotive force (emf) in the secondary coil and a self-induced emf in the primary coil.
If there is no magnetic flux leakage, the flux linked with each turn of the primary coil will be the same as that linked with each turn of the secondary coil.
It regulates A.C. voltage and transfers the electrical power without change in frequency of input supply. (The alternating current changes itself.)

5.0Types of Transformers

Feature	Step-Up Transformer (S.U.T.) $N_{S} > N_{P}$	Step-Down Transformer (S.D.T) $N_{S} < N_{P}$
Function	Increases voltage	Decreases voltage
Number of Turns (Primary)	Fewer turns in the primary coil	Greater turns in the primary coil
Number of Turns (Secondary)	More turns in the secondary coil	Fewer turns in the secondary coil
Voltage (Primary)	Low voltage	High voltage
Voltage (Secondary)	High voltage	Low voltage
Current (Primary)	Low current	High current
Current (Secondary	High current	Low current
Usage	Power transmission over long distances	Household appliances, low-power applications
Example	Power plants, electrical grids	Home electrical systems, chargers, small appliances

S.U.T converts low voltage, high current into high voltage, low current
S.D.T. converts high voltage, low current into low voltage, high current.
Power transmission is carried out always at "High voltage, low current" so that voltage drop and power losses are minimum in the transmission line.

$Voltage Drop = I_{L} R_{L}$

$(I_{L} = line Current, R_{L} = Total Line Resistance)$

$I_{L} = \frac{Power to be transmitted}{Line Voltage}$

$Power Losses = I_{L}^{2} R_{L}$

Sending power always at high voltage & low current (By. S.U.T.) and receiving power always at low voltage & high current (By S.D.T.)
High voltage coils have more turns and are always made of thin wire and high current coils having less number of turns and always made of thick wires.

6.0Key Points of Transformer

It cannot operate with a DC supply. When a battery is connected to the primary side, the output across the secondary side remains zero, meaning the transformer does not function.
It cannot be considered an 'amplifier' since it does not provide power gain like a transistor.
Since the transformer has no moving parts, there are no mechanical losses, making its efficiency higher compared to that of generators and motors.

7.0Ideal Transformer $(η = 100%)$

8.0

No Flux Leakage:-

$ϕ_{s} = ϕ_{p} = \frac{- d ϕ _{s}}{d t} = \frac{- d ϕ _{p}}{d t}$

$e_{s} = e_{p} = e$ (induced emf per turn of each coil is also same)

total induced emf for secondary, $E_{s} = N_{s} e$

total induced emf for primary, $E_{p} = N_{p} e$

$\frac{E _{s}}{E _{p}} = \frac{N _{s}}{N _{p}} = n or p$ ……….(1)

$n \to Turn Ratio$

$p \to Transformation Ratio$

(b) No load condition :-

$V_{p} = E_{p} and E_{s} = V_{s}$

$\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}}$ ………….(2)

from(1) and (2)

$\frac{V _{s}}{V _{p}} = \frac{N _{s}}{N _{p}} = n or p$ ………..(3)

(c) No power loss :-

$P_{o u t} = P_{in}$

$V_{S} I_{S} = V_{P} I_{P}$

$\frac{V _{S}}{V _{P}} = \frac{I _{P}}{I _{S}}$ …………………….(4)

From equation (3) and (4)

$\frac{V _{S}}{V _{P}} = \frac{I _{P}}{I _{S}} = \frac{N _{s}}{N _{p}} = n or p$

Sp. Note : Generally transformers deals in ideal condition i.e. $P_{in} = P_{o u t}$ , if other information is not given.

9.0Real Transformer

Some power is always lost due to flux leakage, hysteresis, eddy currents, and heating of coils.Hence $P_{o u t} < P_{in}$ always
Efficiency of a Transformer $η = \frac{P _{o u t}}{P _{in}} = \frac{V _{S} I _{S}}{V _{P} I _{P}} \times 100$

10.0Losses In Transformer

Copper or joule heating losses :-

Where : These losses occur in both coils of the shell part.

Reason : Due to the heating effect of current $(H = I^{2} Rt)$ .

Remedy : To minimise these losses, high current coil is always made up with thick wire and for removal of produced heat, circulation of mineral oil should be used.

Flux leakage losses :-

Where : These losses occur in between both coils of the shell part.

Cause : Due to air gap between both the coils.

Remedy : To minimise these losses both coils are tightly wound over a common soft iron core (high magnetic permeability) so a closed path of magnetic field

lines formed itself within the core and tries to make coupling factor $K \to 1$

Iron losses :-

Where : These losses occur in the core part.

Types:(a) Hysteresis Loss (b) Eddy Current Loss

(a) Hysteresis Loss

Cause : Transformer core always present in the effect of alternating magnetic field $(B = B_{0} S inω t)$ so it will be magnetised & demagnetised with very high frequency (f=50Hz) .During its demagnetization a part of magnetic energy is left inside the core part in the form of residual magnetic field. Finally, this residual energy wasted as heat.

Remedy : To minimise these losses material of transformer core should be such that it can be easily magnetised & demagnetised. For this purpose, magnetic soft materials should be used.

Example- Soft Iron (Low Retentivity and Low Coercivity)

(b) Eddy Current Loss

It is a group of induced currents which are produced when metal bodies placed in a time varying magnetic field or they move in an external magnetic field in such a way that flux through them changes with respect to time.

11.0Advantages And Disadvantages of Transformer

Advantages of Transformers

1.Transformers are highly efficient

2.They can easily step up or step down voltages

3.Transformers have no moving parts.

Disadvantages of Transformers

1. Transformers only work with alternating current (AC)

2.Heavy, and difficult to install.

3.Under high load or inefficient operation, transformers can overheat

12.0Significant Points of Transformer

These currents are produced only in closed paths within the entire volume and on the surface of the metal body. Therefore their measurement is impossible.
The circulation plane of these currents is always perpendicular to the external magnetic field direction.
Generally resistance of metal bodies is low so the magnitude of these currents is very high.
These currents can heat up the metal body and some time body will melt out (Application :Induction furnace)
Due to these induced currents a strong eddy force (or torque) acts on the metal body which always opposes the translatory (or rotatory) motion of the metal body, according to Lenz law. The transformer core is always present in the effect of alternating magnetic field \left(B=B_0Sin\omega t\right). Due to this eddy currents are produced in its volume, so a part of magnetic energy of the core is wasted as heat.

Remedy : To minimise these losses the transformer core should be laminated. With the help of the lamination process, the circulation path of eddy current is greatly reduced & net resistance of the system is greatly increased. So, these currents become feeble.

13.0Solved Examples On Transformer

Q-1.In a transformer, the AC voltage is increased from 220 V to 2200 V. If the secondary coil has 2000 turns, how many turns will be in the primary coil?

Solution:

$\frac{V _{p}}{V _{s}} = \frac{N _{p}}{N _{s}} \Rightarrow N_{p} = \frac{220}{2200} \times 2000 = 200$

Q-2.An ideal transformer is connected to a 240 V AC mains, and the output voltage is 24 V. The transformer is then used to power a 24V, 24W bulb. Calculate the current flowing through the primary coil of the transformer.

Solution:Current in primary coil $I_{p} = \frac{V _{S} I _{S}}{V _{P}} = \frac{24}{240} = 0.1 A$

Q-3.A bulb (100 W, 110 V) is operated using a transformer by supplying 220 V, 0.5 A, find efficiency of transformer.

Solution:

$P_{in p u t} = P_{o u tp u t}$

$Efficiency (η) = \frac{P _{o u tp u t}}{P _{in p u t}} \times 100$

$Efficiency (η) = \frac{100}{( 220 ) ( 0.5 )} \times 100 = 90.90%$

Q-4.A transformer has primary and secondary coils with 50 and 1500 turns, respectively. The magnetic flux coupled with the primary coil is given by $Φ = (2 + 4 t) Wb$ , where t represents time. What is the output voltage across the secondary coil?

Solution:

$N_{s} = 100 N_{p} = 50 Φ = (2 + 4 t) Wb$

$\frac{d Φ}{d t} = 4 \Rightarrow e = ν_{i} = 4$

$\frac{V _{0}}{V _{i}} = \frac{N _{s}}{N _{p}} \Rightarrow \frac{V _{0}}{4} = \frac{1500}{50} \Rightarrow V_{0} = 120 V$

So, output voltage across the secondary coil is 120 V.

Q-5.A transformer has an input voltage of 2500 volts and an output current of 80 amperes. The transformer ratio of the primary coil to the secondary coil is 20:1. Assuming the transformer operates at 100% efficiency, what is the voltage in the secondary coil?

Solution:

$V_{I n p u t} = 2500 \Rightarrow I_{O u tp u t} = 80$

$η = 100% \Rightarrow \frac{N _{P}}{N _{S}} = \frac{20}{1}$

$P_{I n p u t} = P_{O u tp u t}$

$V_{P} I_{P} = V_{S} I_{S} (\frac{V _{S}}{V _{P}} = \frac{N _{S}}{N _{P}} = \frac{I _{P}}{I _{S}})$

$\frac{V _{S}}{2500} = \frac{1}{20} \Rightarrow V_{S} = 125 V$

So, output voltage across the secondary coil is 125 V.

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Transformer

1.0Definition of Transformer

2.0Principle of Transformer

3.0Construction of Transformer

4.0Working of Transformer

5.0Types of Transformers

6.0Key Points of Transformer

7.0Ideal Transformer $(η = 100%)$

8.0

9.0Real Transformer

10.0Losses In Transformer

11.0Advantages And Disadvantages of Transformer

12.0Significant Points of Transformer

13.0Solved Examples On Transformer

Table of Contents

Join ALLEN!

Transformer

1.0Definition of Transformer

2.0Principle of Transformer

3.0Construction of Transformer

4.0Working of Transformer

5.0Types of Transformers

6.0Key Points of Transformer

7.0Ideal Transformer $(η = 100%)$

8.0

9.0Real Transformer

10.0Losses In Transformer

11.0Advantages And Disadvantages of Transformer

12.0Significant Points of Transformer

13.0Solved Examples On Transformer

Table of Contents

Join ALLEN!

Transformer

1.0Definition of Transformer

2.0Principle of Transformer

3.0Construction of Transformer

4.0Working of Transformer

5.0Types of Transformers

6.0Key Points of Transformer

7.0Ideal Transformer (η=100%)

8.0

9.0Real Transformer

10.0Losses In Transformer

11.0Advantages And Disadvantages of Transformer

12.0Significant Points of Transformer

13.0Solved Examples On Transformer

Table of Contents

Join ALLEN!

Transformer

1.0Definition of Transformer

2.0Principle of Transformer

3.0Construction of Transformer

4.0Working of Transformer

5.0Types of Transformers

6.0Key Points of Transformer

7.0Ideal Transformer (η=100%)

8.0

9.0Real Transformer

10.0Losses In Transformer

11.0Advantages And Disadvantages of Transformer

12.0Significant Points of Transformer

13.0Solved Examples On Transformer

Table of Contents

7.0Ideal Transformer $(η = 100%)$

7.0Ideal Transformer $(η = 100%)$