ELECTROMAGNETISM

ELECTROMAGNETISM

Electromagnetism is the effect produced by the interaction of an electric current with a magnetic field.

OR

Electromagnetism is the production of a magnetic field by current in a conductor

MAGNETIC FIELD DUE TO A CURRENT –CARRYING CONDUCTOR

If a current of few amperes is passes through a wire a weak magnetic field is produced. 

NOTE:

• The strength of the magnetic field produced increases as the current passing through                 conductor increases. (Strength of the magnetic field depends on the magnitude of the                  electric current)
• The effect of the magnetic field can be detected by iron fillings when the current through            the conductor reaches 20A. 
• The direction of the magnetic field can be determined by applying two rules; Right-hand            Grip Rule and Maxwell’s cork screw rule

RIGHT-HAND GRIP RULE

The Right-Hand Grip Rule states that;

“Imagine the wire carrying the current is gripped by the right hand with the thumb pointing in the direction of the convectional current, the fingers will curl around the wire pointing in the direction of the magnetic field”

MAXWELL’S RIGHT-HAND CORK SCREW RULE

The Maxwell’s Right-Hand Cork Screw Rule states that;

“If a right-hand screw advances in the direction of the current, then the direction of rotation of the screw represents the direction of the magnetic field due to the current”

DIRECTION OF THE MAGNETIC FIELD DUE TO CURRENT-CARRYING CONDUCTOR

FLEMING’S LEFT HAND RULE

The Fleming’s left hand rule states that;

“If the first three fingers of the left hand are held mutually at right angles to each other, such that the middle finger points in the direction of the current, and the forefinger in the direction of magnetic field lines, then the thumb points in the direction of motion of the current-carrying conductor”

The Fleming’s left hand rule determines the direction of the force acting on a current-carrying conductor in a magnetic field. The Fleming’s left hand rule also known as Motor Rule

FORCE ON A CURRENT CARRYING CONDUCTOR IN A MAGNETIC FIELD

When a current carrying conductor is placed in a magnetic field, it experiences a force. This force causes conductor to move when suspended freely.

NOTE:

The magnitude of the force on the current-carrying conductor depends on;

• Magnetic field strength: Force on the current-carrying conductor is proportional to                       strength of the magnetic field   ∴ Fα B

• Magnitude of current: Force on the current-carrying conductor is proportional to the                   current through conductor.      ∴ Fα I

• Conductor Length: Force on the current-carrying conductor is proportional to the length             of  the conductor    ∴ Fα l

• Angle between magnetic field and conductor: The magnitude of the force is proportional            to the component of the field that is at right angles to the conductor.

• The force is at a maximum when the conductor is at right angles to the field,

• The force is zero when the conductor is parallel to the field

FORCE DUE TO PARALLEL CONDUCTORS CARRYING CURRENT

When electric current passes through parallel wires, a force is developed between them. The force developed can be attractive or repulsive depending on the direction of the current through the wires.

When the current flow in the same direction

Two parallel wires that carry current in the same direction attract each other

When the current flows in the same direction, the magnetic field between the conductors (wires) cancel out while outside magnetic field add up. This causes weak magnetic field between the conductors than on the outside, hence the resultant force pushes the conductors towards each other.

When the current flow in the opposite direction

Two parallel wires that carry current in the opposite direction repel each other

When the current flows in the opposite direction, the magnetic field between the conductors (wires) add up while outside magnetic field cancel out. This causes strong magnetic field between the conductors than on the outside, hence the resultant force is towards the outside of each conductor.

NOTE:

The Magnitude of the force between parallel wires carrying current inversely proportional to the distance between wires

The shorter the distance of separation the stronger the force of attraction or repulsion

CLASS ACTIVITY

1. Under what conditions does the charge affect a magnet?

2. The figure below shows two parallel wires. What would be the effect of passing

    (a)  Upward current through both wires.

    (b)  An upward current through wire A and a download current through wire B?

3.  (a)   An electric current flows in a copper wire from north to south direction. A compass 

                needle is placed below this wire. What is your observation and inference?

            (b)   Sketch the magnetic lines of force due to a current through a straight conductor.

4.  Two straight parallel placed conductors carrying current in the same direction tends to                        contract. Why?

5.  On what factor does the strength of the magnetic field due to a straight conductor carrying                   current depend?

ELECTROMAGNETS

Electromagnet is a core of magnetic material (as iron) surrounded by a coil of wire through which an electric current is passed to magnetize the core temporary.

SOLENOID

Solenoid s a long coil of wire through which a current is passed to generate magnetic field

The strength of the magnetic field increases by inserting soft iron core in the solenoid.

APPLICATIONS OF ELECTROMAGNETS

Electromagnets are wide applied in;

• Industries, where industry electromagnets are used for picking magnetic materials                      (metallic) by attraction.

• Electrical devices, as mostly applied in electric bells and telephone receivers

ELECTRIC BELL

The electric bell is a small electric device that converts electric energy into mechanical energy in the form of sound.

Mode of action of electric bell

The step by step process of working of electric bell is described below

• The switch (K) is pressed and current flows through the electric circuit.

• The electromagnet (E) is powered and generates a magnetic field that attracts the iron                 strip    armature (A) towards it.

• The striker (S) strikes the gong (bell) (B).

• When the striking arm strikes the gong, the contact is broken and current stops flowing             through the circuit.

• This causes the electromagnet to lose its magnetic field.

• The connected spring arm (T) returns the striker to its original rest position.

• The contact is restored and current flows through the circuit. (provided the main switch is           still pressed)

• The process is repeating from the beginning.

THE TELEPHONE RECEIVER (EARPIECE)

Telephone receiver is an electric device used for converting electrical energy into sound energy.

When you speak to a friend through telephone, your sound will be converted into electric current by the mouthpiece of the telephone. The current produced is varying current and the frequency of the current will be the same as the frequency of your sound. The current will be sent to the earpiece of the telephone of your friend.

Mode of action of the telephone receiver (earpiece)

The step by step process of working of electric bell is described below

• When the current passes through the solenoid, the iron core become magnetized and                   magnetic field is produced

• The strength of the magnetic field changes according to the varying current
• When the current is high, the magnetic field become stronger, and

• When the current is low, the magnetic field become weaker

• The soft-iron diaphragm is pulled by the electromagnet (magnetized iron core) and                     vibrates at the frequency of the varying current.

• The air around the diaphragm vibrates and produces sound wave.

• The frequency of the sound produced in the telephone earpiece will be the same as your             sound

• This sound is what listens by your friend.

THE MOVING COIL GALVANOMETER

A galvanometer is an instrument used to detect the presence of an electric current in a circuit.

OR

A galvanometer is a device used to detect the flow of current in an electrical circuit

Working principle of the moving coil galvanometer

• The coils together with the soft iron core are suspended at Axle (E) such that they are free          to rotate about a vertical axis.
• The action of the hair spring is to provide restoring couple or force to the coil as it rotates.
• When the current flow through the coil, a force due to current carrying conductor placed            in a magnetic field develops.
• The forces (direction indicated by Fleming’s Left Hand Rule) on the two vertical sides of           the coil give rise to a couple.
• The coil rotates moving a pointer which is attached to the coil.
• If the current is switched off the coil is brought back by the action of hair springs.

SENSITIVITY OF THE GALVANOMETER

The sensitivity of a moving galvanometer is the deflection of the pointer in divisions per unit current through the coil. The unit of Sensitivity is divisions per ampere (div/A).

The sensitivity of the moving galvanometer depends on;

• Strength of the restoring hair spring;

If the spring is less powerful, the coil turns easily and faster than when they are powerful

• Strength of the magnetic flux;

If the magnet is strong, the force on the coil is also large, which give rise to large deflection

• Number of turns of the coil;

If the coil has many turns, the force on the coil is large.

• Area of the coil;

If the area of the coil is large, the deflection becomes high.

CLASS ACTIVITY

1. List down some applications of electromagnets

2. (a) What is solenoid?

           (b) Indicate the magnetic field lines due to a solenoid.

           (c) What is the core material used in a solenoid to obtain an electromagnet?

3. (a) What rule would you use to find the force acting on a current carrying conductor    

                 placed in a magnetic field? 

            (b) Name any two devices that work on this principle.

4. Draw a labelled diagram of an electric bell and explain its mode of action.

5. How does telephone receiver work?

6. Describe the structure of galvanometer.

7. Give three ways in which the sensitivity of the moving coil galvanometer could be increased

CONVERSION OF GALVANOMETER INTO AN AMMETER

Ammeter is an instrument used to measure electric current in amperes

A galvanometer whose scale is graduated to measure l current in milliamperes is called milliameter. 

The capacity of the current a galvanometer can withstand is small, this is because the resistance of the coil small.

To convert a moving coil galvanometer into an ammeter a resistor of low resistance (shunt) is connected in parallel to the coil of the galvanometer.

When a large current is passed, a small current flows through galvanometer coil (G) and the rest of the current flows through the shunt (S)

Suppose;

Galvanometer coil resistance = G

Shunt resistance = S

Current through galvanometer = Ig

Current through shunt = Is

Total current = I

The potential difference across the galvanometer coil = the potential difference across the shunt

Vg=Vs

Ig G=Is S

But total current (I) =Ig +Is

Hence; Is=I-Ig

S(I-Ig)=Ig G

S=(Ig G)/((I-I(g)) )

Therefore the shunt can be determined by

Example 1

Suppose the galvanometer coil resistance, G = 10Ω and the full-scale deflection current, Ig = 15mA. If it is be converted so that it gives a full-scale deflection current, I= 1.5A then what will be the required shunt?

Solution

Data given

Galvanometer coil resistance, G = 10Ω

Current through galvanometer coil, Ig = 15mA

Total current, I = 1.5A

Shunt, S= ?

S=(I_g G)/((I-I_(g)) )

=0.015x10/((1.5-0.015)

=0.150/1.488

= 0.101Ω

The required shunt is 0.101Ω

Example 2

A milliammeter has a resistance of 5Ω and a full-scale deflection of 10mA. The meter is to be adapted to measure current up to 1A. Calculate the shunt resistance required.

Solution

Data given

Galvanometer coil resistance, G = 5Ω

Current through galvanometer coil, Ig = 10mA

Total current, I = 1A

Shunt, S= ?

S=(I_g G)/((I-I_(g)) )

=0.01x5/((1-0.01)

=0.05/0.99

= 0.051Ω

The required shunt is 0.051Ω

CONAVERSION OF GALVANOMETER INTO VOLATAMETER

Voltmeter is an instrument used to measure potential difference between the two ends of a current carrying conductor.

To convert a moving coil galvanometer into a voltmeter a resistor of high resistance (multiplier) is connected in series to the coil of the galvanometer.

Suppose;

Galvanometer coil resistance = G

Multiplier resistance = R

Potential difference across galvanometer = Vg

Potential difference across multiplier = VR

Total Potential difference = V

The current through the galvanometer coil = the current through across the shunt

Ig=IR=I 

Vg/G=  VR/R

But total potential difference, V= Vg + VR

V=IG+IR

IR=V-IG

R=V/I-IG/I

=V/I-G

R=(V-IG)/I

Therefore the shunt can be determined by

Example 1

If the resistance of the galvanometer coil, G = 10Ω, current flowing through the coil, Ig = 15mA and the instrument is to be converted to measure a full-scale deflection potential difference of 3V, what will be the resistance of the multiplier required?

Solution

Data given

Resistance of the galvanometer coil, G = 10Ω

Current through the coil, I = 15mA

Potential difference, V = 3V

The multiplier, R =?

R=(V-IG)/I

=(3-0.015 x 10)/0.015

=2.85/0.015

=190Ω

The required multiplier is 190Ω

Example 2

A milliammeter has a resistance of 5Ω and a full-scale deflection of 10mA. The meter is to be adapted to measure p.d up to 10V. Calculate the multiplier resistance required.

Solution

Data given

Resistance of the galvanometer coil, G = 5Ω

Current through the coil, I = 10mA

Potential difference, V = 10V

The multiplier, R =?

R=(V-IG)/I

=(10-0.01 x 5)/0.01

=(10-0.05)/0.01

=9.95/0.01

=995Ω

The required multiplier is 995Ω

A SIMPLE DIRECT CURRENT (D.C) ELECTRIC MOTOR

DC Motor is an electrical device that converts electrical energy into mechanical energy.

Working principle of the DC Motor

• The permanent magnet provides a constant magnetic field.
• When the current flows through the coil (armature), the force exerted as a result of the                  interaction of magnetic field and electric current
• The force produced on the two sides of the coil is such that the coil (armature) starts to                 move (rotates) in the direction of force as described by Fleming’s Left hand rule.
• The armature is connected to a DC power source through a pair of commutator rings.
• When the current flows through the coil an electromagnetic force is produced on it, so the          coil (armature) will start to rotate.

HOW TO IMPROVE THE EFFICIENCY OF THE D.C MOTOR

Efficiency of a DC motor can be increased by increasing -

• the number of turns in the coil
• the strength of the current
• the area of cross-section of the coil
• the strength of the radial magnetic field

CLASS ACTIVITY

1. A moving coil galvanometer has coil resistance 20Ω and can carry a maximum of 15mA.

(a)   What is the value of the shunt required to enable the galvanometer register 10A full-scale                   deflection? How will the shunt be connected?

(b)  What is the value of the multiplier required to enable the galvanometer register 10V full-                     scale deflection? How will the multiplier be connected?

2. What happens if a current carrying conductor is placed in a magnetic field?

3. When does a current carrying conductor experience a maximum force in the magnetic field?

4. What is the use of a commutator in an electric motor?

5. Describe the structure and mode of action of a simple d.c motor.

ELECTROMAGNETIC INDUCTION

FACTORS INFLUENCING MAGNITUDE OF INDUCED E.M.F

The magnitude of the induced electromotive force (e.m.f) is determined by

• The number of turns of wire in the coil.

The e.m.f is directly proportional to the number of turn in the coil, the more the number of turns the higher the e.m.f induced.

• The speed of the relative motion between the coil and the magnet;

The e.m.f is directly proportional to the speed of the coil and the magnet, the more the speed the higher the e.m.f induced

• The strength of the magnetic field.

The stronger the magnetic field the more e.m.f produced because there are more lines of force to cut.

• The medium in which the conductor is placed.

Soft iron increases the strength of magnetic field and therefore a large rate of magnetic flux linkage change would result.

MAGNETIC FLUX

Magnetic flux is the amount of number of magnetic field lines that passes through a certain area.

The SI-unit of magnetic flux is Weber (Wb)

LAWS OF ELECTROMAGNETIC INDUCTION

THE FARADAY’S LAW OF ELECTROMAGNETIC INDUCTION

Faraday’s Law of electromagnetic induction states that;

“Whenever a circuit experiences a change in magnetic flux electromotive force (e.m.f) is induced, the magnitude of electromotive force (e.m.f) induced is proportional to the rate of change of the magnetic flux linking the conductor”

THE LENZ’S LAW

The Lenz’s law states that;

“The direction of an induced e.m.f is such that it opposes the change producing it”

CASE-I: When a magnet is moving towards the coil. 

When the north pole of the magnet is approaching towards the coil, the magnetic flux linking to the coil increases

According to Faraday's law of electromagnetic induction, when there is change in flux, an e.m.f and hence current is induced in the coil and this current will create its own magnetic field. Now according to Lenz's law, this magnetic field created will oppose its own or we can say opposes the increase in flux through the coil and this is possible only if approaching coil side attains north polarity, as we know similar poles repel each other. Once we know the magnetic polarity of the coil side, we can easily determine the direction of the induced current by applying right hand rule. In this case, the current flows in anticlockwise direction

CASE-II: When a magnet is moving away from the coil

When the north pole of the magnet is moving away from the coil the magnetic flux linking to the coil decreases 

According to Faraday's law of electromagnetic induction, an e.m.f and hence current is induced in the coil and this current will create its own magnetic field. Now according to Lenz's law, this magnetic field created will oppose its own or we can say opposes the decrease in flux through the coil and this is possible only if approaching coil side attains south polarity, as we know dissimilar poles attract each other. Once we know the magnetic polarity of the coil side, we can easily determine the direction of the induced current by applying right hand rule. In this case, the current flows in clockwise direction

FLEMING’S RIGHT HAND RULE

The Fleming’s Right hand rule states that;

“The thumb, first finger and second finger of the right hand are extended mutually at right angle, the first finger pointing in the direction of the field and the thumb in the direction of the motion then the second finger will point in the direction of the induced current”

Sometimes this Law is called the Dynamo Rule

SELF- INDUCTION

Self-induction is that phenomenon in which a change in electric current in a coil produces an induced e.m.f in the coil itself.

OR

Self-induction is the induction of the e.m.f in a current carrying conductor as a result of change in the current that flows through the conductor. 

The current through the coil produces a magnetic flux which links with the coil itself. If we vary the resistance in the circuit, the current through the coil changes and the magnetic flux through the coil also change.

This change in flux indicates an e.m.f in the coil itself. Such an e.m.f is called self-induced e.m.f and the phenomenon is called self-induction.

SELF INDUCTANCE

Self-inductance of a coil is defined as the ratio of self-induced e.m.f to the rate of change of current in the coil

The SI-unit of inductance is Henry

MUTUAL INDUCTION

Mutual induction is the induction of e.m.f in a conductor as a result of change in current in another conductor.

OR

Mutual induction is the phenomenon in which a change of current in one coil causes an induced e.m.f in another coil placed near to the first coil.

The coil in which current is changed is called primary coil and the coil in which e.m.f is induced is called secondary coil.

When current is passed through the primary coil, magnetic flux is produced. This magnetic flux is also linked with the secondary coil. If the current is changed by varying the resistance in the primary circuit, the magnetic flux also change. As this changing flux is linked with the secondary coil, it induces an e.m.f in it. 

This phenomenon of inducing e.m.f in a coil by changing current in another coil is known as mutual inductance

MUTUAL INDUCTANCE

Mutual inductance is defined as the ratio of e.m.f induced in the secondary coil to the rate of change of electric current in the primary coil.

APPLICATIONS OF MUTUAL INDUCTION

Mutual induction are widely applied in

• Transformers
• The car ignition system
• Flash tubes in cameras and strobe lights
• Wireless telegraphy
• Generators and motors

SIMPLE A.C GENERATOR

Simple A.C generator is a device for converting mechanical energy into electrical energy.

It works on the converse principle of electromagnetic induction. It is also called Dynamo

Working Principle of A.C generator

The simple dynamo consists of a permanent magnet which rotates around a central shaft with a coil of wire placed next to this magnetic field.

As the magnet spines, the magnetic field around the top and bottom of the coil constantly changes between north and South Pole.

This rotational movement of the magnetic field results in alternating e.m.f being induced into the coil as defined by Faraday’s Law of electromagnetic induction

A.C Generator consists of the following parts.

• Powerful field magnet with concave poles.
• Armature:

It is a rectangular coil of large number of turns of wire wound on laminated soft-iron core of high permeability and low hysteresis loss.

• Slip rings:

The ends of the coil are joined to two separate copper rings fixed on the axle (S1 & S2).

• Carbon brushes:

Two carbon brushes remain pressed against each of the rings which form the terminals of the external circuit

EDDY CURRENTS

Eddy currents are loop circulating currents which are induced in a conductor when it is in relative motion to a magnetic field. 

NOTE:

Since eddy currents are closed loop currents, they form circular or whirling currents and produce own magnetic field

Eddy currents cause an increase in the temperature of the metals due to collisions between moving charges and the atoms of the metal, as well as the direct agitation of atoms by a magnetic field changing direction at high frequency

APPLICATIONS OF EDDY CURRENTS

• Electromagnetic braking.

Electromagnetic braking systems utilize eddy currents to interact with electromagnets as to create a stopping force for moving vehicles

• Electromagnetic cookers.

Some electric cookpops contain induction coils instead of heating coils. These set up rapidly changing magnetic fields, inducing eddy currents in the metal of the saucepan

• Electromagnetic damping.

Certain galvanometers have a fixed core made of nonmagnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly.

• Induction furnace.

Induction furnace can be used to produce high temperatures and can be utilized to prepare alloys, by melting the constituent metals

CLASS ACTIVITY

How can you convert an AC generator to a DC generator?

Briefly explain the mode of action of dynamo

(a) What are eddy currents?

(b) Give three ways that eddy currents are benefit

TRANSFORMERS

Transformer is a device that uses mutual induction between two coils to convert alternating voltage to larger or smaller alternating voltage across the other coil.

TYPES OF TRANSFORMERS

There are two types of transformers

Step-up transformers

Step-down transformers

STEP-UP TRANSFORMERS

Step-up transformers are transformers for increasing alternating voltage.

The number of turns in primary winding is smaller than in the secondary winding 

The output voltage (in secondary) is higher than input voltage (in primary)

STEP-UP TRANSFORMERS

Step-up transformers are transformers for decreasing alternating voltage.

The number of turns in primary winding is larger than in the secondary winding

The output voltage (in secondary) is less than input voltage (in primary)

TRANSFORMER EQUATION

The transformer equation describes the relationship between the number of turns in the coil, voltage and current in primary and secondary windings

Suppose the number of turns in primary coils = N_p

The number of turns in secondary coils = N_s

The current in primary coils = I_p

The current in secondary coils = I_s

The voltage in primary coils (input voltage) = V_p

The voltage in secondary coils (output voltage) = V_s

For an ideal transformer, 

The power input = the power output

V_P I_p= V_s I_s

V_p/V_s =I_s/I_p 

Also the e.m.f induced is proportional to the number of turns in the coils

Induced e.m.f ∝Number of turn

V∝N

V=k.N

k=V/N

V_p/N_p =V_s/N_s 

V_p N_s=V_s N_p

V_p/V_s =N_p/N_s 

Hence the general transformer equation is given by;

V_p/V_s =I_s/I_p =N_p/N_s 

NOTE:

Ideal transformer is the transformer that has the efficiency of 100%. The ideal transformer has negligible power loss

TRANSFORMER EFFICIENCY

The efficiency of a transformer is the ratio of power in secondary coils to the power in primary coils expressed as a percentage.

OR

The efficiency of a transformer is the ratio of power output to the power input expressed as a percentage.

Efficiency=(Power output)/(Power input) x100%

Efficiency=P_s/P_p  x100%

Example 1

A transformer is used to step-down 240V main supply to 12V for laboratory use. If the primary coil has 600 turns, determine the number of turns in the secondary coil.

Solution

Data given

Voltage in primary coils 〖(V〗_p) = 240V

Voltage in secondary coils 〖(V〗_s) = 12V

Number of turns in primary 〖(N〗_p) = 600

Number of turns in secondary 〖(N〗_s) =?

V_p/V_s =N_p/N_s 

240/12=600/N_s 

N_s=12x600/240

=7200/240

=30    

 There are 30 turns in secondary coils

Example 2

A current of 0.6A is passed through a step-up transformer with a primary coil of 200 turns. A current of 0.1A is obtained in the secondary coil. Determine 

The number of turns in the secondary coil

The voltage across the secondary coil, if the primary coil is connected to a 240V mains.

Solution

Data given

Voltage in primary coils 〖(V〗_p) = 240V

Current in primary coils 〖(I〗_p) = 0.6A

Current in secondary coils 〖(I〗_s) = 0.1A

Number of turns in primary 〖(N〗_p) = 200

Number of turns in secondary 〖(N〗_s) =?

I_s/I_p =N_p/N_s 

I_s/I_p =N_p/N_s 

0.1/0.6=200/N_s 

N_s=(0.6 x 200)/0.1

=120/0.1

=1200

There are 1200 turns in secondary coils

Voltage in secondary coils 〖(V〗_s) =?

V_p/V_s =N_p/N_s 

240/V_s =200/1200

V_s=(240 x 1200)/200

=1440V   

The secondary voltage = 1440V

Example 3

A step-up transformer ha 10000 turns in the secondary coil and 100 turns through the primary coil. An alternating current of 5.0A flows in the primary circuit when connected to a 12.0V a.c supply.

Calculate the voltage across the secondary coil.

If the transformer has an efficiency of 90%, what is the current in the secondary coil?

Solution

Data given

Voltage in primary coils 〖(V〗_p) = 12.0V

Current in primary coils 〖(I〗_p) = 5.0A

Number of turns in primary 〖(N〗_p) = 100

Number of turns in secondary 〖(N〗_s) = 10000

Efficiency = 90%

Voltage in secondary 〖(V〗_s) =?

V_p/V_s =N_p/N_s 

12/V_s =100/10000

V_s=(12 x 10000)/100

=1200

The secondary voltage = 1200V

Voltage in secondary 〖(V〗_s) =?

Efficiency=P_s/P_p  x100%

Efficiency=(V_s I_s)/(V_p I_p ) x100%

90%=(1200 I_s)/(12 x 5) x100%

I_s=(12 x 5 x 90%)/(1200 x 100%)

=0.045 A

The current in secondary coil = 0.045 A

POWER LOSSES IN TRANSFORMERS

Most of practical transformers do not allow 100% efficiency due to power losses. These losses may be contributed by

Copper losses (winding resistance).

There are losses of power in the primary winding in the form of the heat due to the resistance of the primary coil. The winding resistance loss is given by P=I^2 R

Losses due to eddy currents (Iron losses).

The eddy currents induced in the iron cores of the transformer are part of the input energy and change into heat.

Hysteresis losses.

Work is done by the input power to alternately magnetize and demagnetize the iron core of the transformer. This process is continually repeats so long as the alternating current flows in the primary winding.

Poor magnetic flux linkage.

The poor magnetic flux linkage between primary winding and secondary winding causing a flux leakage reduces the efficiency of the transformer. Flux leakage is caused by magnetic lines radiating away from the transformer and not cutting secondary winding due to bad core design.

WAYS TO REDUCE POWER LOSSES IN TRANSFORMERS

Copper losses.

The winding resistance losses  P=I^2 R  can be minimized through reducing resistance of the winding by increasing the cross-sectional area of the conductor. That is why the windings of the transformer are made thick so as to minimize the resistances.

Losses due to eddy currents (Iron losses). 

Power losses due to eddy current can be minimized by lamination of the iron core.

Lamination is the process of splitting the solid core into thin sheets being insulated from each other to minimize power losses due to eddy currents.

Hysteresis losses.

The power losses due to hysteresis can be minimized by using soft magnetic materials such as soft iron core, steel, steel alloys and silicon steel with low hysteresis and high magnetic permeability.

Poor magnetic flux linkage.

The flux leakage can be reduced by better core design.

CLASS ACTIVITY

(a) State three factors that affect the efficiency of a transformer

(b) A transformer with 100 turns in its secondary coil supplies 8.0V to a light in a toy. The primary circuit in connected to 240V main supply. What is the number of turns in the primary coil?

Explain why the core of a transformer is;

Laminated

Made from a soft magnetic material such as iron

Abuu Talhah builds a transformer, connected to a 12V a.c power supply. He has 200 turns on the primary and 40 turns on the secondary. What is the output voltage?

(a) Briefly explain four ways contributes to the power losses in transformers

(b) Briefly explain how to minimize power losses in transformers as explained in (a) 

      Above

(a) with the aid of diagrams tabulate three differences between step-down and step-up 

            transformers

A step-up transformer connected to 20V source delivers energy at 250V. if the efficiency of the transformer is 95%, determine the current in the primary circuit when the output terminals are connected to a lamp rated 250V, 100W.

REVISION QUESTIONS

A transformer has a primary coil of 100 turns and a secondary coil of 300 turns. What will be the voltage out of the secondary coil if we apply 20V to the primary?

If we need to change 240V down to 12V, and our transformer to have 2000 turns on its primary coil, how many turns should it have on its secondary coil?

A transformer has a primary of 6000 turns and a secondary of 150 turns. The primary is connected to an a.c supply of 240V. 

A 3Ω resistor in connected across the secondary. Draw a circuit diagram to show the above arrangement.

Assuming that the transformer has an efficiency of 100%, calculate;

The voltage across the resistor

The current through the resistor

The current in the primary circuit

The power taken from the supply

The power dissipated (wasted as heat) in the resistor

A 6V 12W bulb operates from the transformer connected to a 240V a.c supply. If there are 8000 turns on the primary, 

How many turns are there on the secondary?

What it is the current in the primary?

(Assume that the transformer is 100% efficient)

(a) Briefly explain why most practical transformers have efficiency less than 100%?

(b) A transformer 600 turns in the primary circuit and 9000 turns in the secondary circuit 

      is connected to 240V mains supply. What is the current in the primary circuit?

      (Assume it has 100% efficiency)

A transformer with 400 turns in the secondary circuit and 20 turns in the primary circuit has p.d of 240V in the primary circuit. What is the p.d in the secondary circuit?

A step-down transformer has 600 turns in the primary circuit and N turns in the secondary circuit. If it is connected to 240V mains supply and the voltage in the secondary circuit is 12V, determine the value of N.

A heater rated 4W, 12V is connected to a secondary circuit of a transformer with 600 turns in the primary circuit. If the primary circuit of the transformer is connected to 240V mains, determine the number of turns in the secondary circuit and the current in the primary circuit.

A radio has a transformer with an output of 6V, and a power rating of 9W. if the primary circuit has 1200 turns and connected to 240V mains, determine;

The current in the secondary circuit.

The number of turns in the secondary circuit.