SS2

Scheme of work

Week Topic

1. Electric field/ electrical conduction through liquids (electrolysis)

- Electrolytes and non- electrolytes

- Dynamics of charged particles (ions) in electrolyte

- Examples of electrolysis (Faraday’s law of electrolysis) , application of electrolysis

2. Electric field

- Conduction of electricity through gases

- Hot cathode, thermionic emission

- The cathode value

- Application of hot cathode (thermionic emission)

- Cathode ray oscilloscope

3. Electric field

- Electric force between point charges (coulomb’s law)

- Concept of electric field / electric field intensity, electric potential

4. Electric field

- Capacitors and capacitance ( definition and arrangement of capacitors)

- Energy stored in a capacitor

- Application of

5. Magnetic field

- Concept of magnetic fields; properties of magnets, magnetic flux and flux density

- Magnetic field around a bar magnet, a straight conductor carrying current, a

solenoid

- Methods of making magnets

- Methods of demagnetization

6. Magnetic field

- Magnetic properties, iron and steel

- Magnetic screening or shielding

- Electromagnetic and application of electromagnetism

- Temporary magnets (The electric bell, telephone, earpiece etc)

7. Magnetic field

- The earth’s magnetic field

- Magnetic elements of a place ( declination and angle of dip) , horizontal component

of the earth’s magnetic field

- Bar magnetic force in the earth’s field, natural point, hammers and compass.

8. Electromagnetic field (i) Magnetic force on a charge moving in a magnetic field (ii)

Concept of electromagnetic field (iii) Interaction between magnetic field and currents in a

current – carrying wire in a magnetic field. (iv) A current carrying solenoid in a magnetic

field (v) Application of electromagnetic field (electric motor, moving coil galvanometer.

9. Electromagnetic field (i) Electromagnetic induction (ii) Faraday’s law (iii) Lenz law (iv)

Motor / generator (v) Eddy current

10. Electromagnetic (i) The transformer (ii) Power transmission (iii) The induction coil

11. Revision

12. Examination

8/5/ 19

ELECTROLYSIS

Electrolysis is defined as a process of decomposing certain liquids and solutions into their

components parts by the passage of electricity through them.

An electrolyte is a liquid which will conduct electricity and which can be decomposed into its

component parts in the process.

Ions are charged particles which exist in electrolytes and takes part in electrolysis. The ions that

move to the anode are called ANIONS, those that move to the cathode are called CATIONS

A voltameter is the vessel containing the electrolyte and the electrode.

- Organic compounds e.g kerosene, petroleum products, diesel etc are non-

electrolytes.

- Compounds which are inorganic e.g salt solutions, bases , acids etc are electrolytes.

- Voltameters are always connected in parallel because they have high resistance

- Ammeter has low resistance , therefore they are always connected in

- Electrodes are materials in the form of rods or plates through which current enter or

leaves the electrolytes.

IONIC THEORY

A Swedish scientist, Arrhenius developed the ionic theory. He considers an electrolyte as

dissociating into charged particles called IONS. The dissociation occurs irrespective of whether

or not an electrolyte . e.g ;

CuSo 4 in a solution dissociates into copper ion and tetraoxosulphate (vi) ion

CuSO 4 ________ Cu 2+

+ SO 2-

4

The charges (ions) so formed moved randomly in solution until negative and positive electrodes

are placed in the electrolyte and a battery connected to the electrode.

CONDUCTION OF ELECTRICITY THROUGH

GASES

Gases conduct electricity at low pressure and high voltage. The pressure of about 0.01mmHg and

voltage of about 1000V under these conditions, a gas in a discharge tube will glow because its

atoms are given energy by the flow of electricity through the tube.

Conduction of electricity through gases is studied using a discharge tube. It consists of a long

glass tube with metal electrodes sealed to each end of the tube.

THE DISCHARGE TUBE

The pressure of the gas in the tube can be altered through the vacuum pump connected to it.

At high voltage and low pressure, the gas in the tube breaks in to ions. The positive ions move

towards the cathode. The negative ions move towards the anode. The positive ions knock off

electrons from the metal plate of the cathode. These electrons produced at the cathode are

cathode rays.

PROPERTIES OF CATHODE RAYS

1. They are fast moving electrons.

2. They travel in a straight line

3. The cause gas and other materials to fluorescence or glow with a greenish colour.

VISIT NEW SCHOOL PHYSICS FOR MORE DETAILS

4. They are deflected by magnetic and electrical fields as shown above

5. They can ionize a gas

6. They can affect photographic plates.

7. They can produce x-rays from high density metals when they are suddenly stopped by

such metals.

8. They will turn a light paddle which in the tube showing that they have momentum and

therefore mass and energy.

9. They can produce intense heat on objects which stop them, confirming that they are

highly energetic particles.

10. They have high penetrating power and can penetrate through metals such as aluminium

plate, steel plate and gold foil.

CATHODE RAY OSCILLOSCOPE

STATIC ELECTRICITY

Static electricity is the study of charges at rest. A charge is an atom that loses or gains electrons.

An electric field is a region or space where a system of electric charges experiences an electric

force.

COULOMB’S LAW

The electric force between two charges is governed by coulomb’s law. It states that the force of

repulsion or attraction between two point charges q1 and q2 is directly proportional to the

product of their charges and inversely proportional to the square of the distance between the two

charges.

F Q 1 Q 2 / r2

But K = 1/ 4 where Q 1 → charge 1

Q 2 → charge 2

F= Q1 Q2/ 4 r = distance between the point charges

= permittivity of free space >

K = 9.0 X 109NM

2C

-2

∑0 = 8.854 X 10 -12

Fm -1

Or

8.854 X 10 -12

C2 N

-1m

-2

Example 1

The positive charges of 12 µC and 8µC respectively are 10cm apart in vacuum. Calculate the

force between them.

SOLUTION

F= Q1 Q2/ 4 r2

= 9.0 X 109 x12 x 8 X 10

-12 / (10 x 10

-12 )

2

= 9.0 x12 x 8 x 10 9 -12

/ 100 x 10-4

= 9x12x8 x 10 9 -12

/ 100 x 10-4

= 864/ 100 x 101

= 864/ 10 = 86.4N

Example 2

If the distance between two points charge is increased by a factor of 4. The magnitude of the

electrostatic force between them will be?

Use inverse square law

F= 1/ r2

, where r=4

F= 1/4 2 = 1/16

ELECTRIC FIELD INTENSITY

Electric field intensity (E) also known as the strength of an electric field at any point is defined

as the force experienced by a unit positive charge at that point .

E = f/ Q; The unit is NC -1

E = QQ/ 4 r2

Q = Q/ 4 r2

or kQ/ r2

EXAMPLE

Calculate the magnitude of the electric field intensity, E, at a field point 3.0 m from a point

charge 5.0 NC -1

Solution

A student is at a height 4m above the ground during a thunderstorm, given that the potential

difference between the ground is 107

V, the electric field created by the storm is --------------

Solution

E= v/ r = 10 7

/ 4 = 2.5 x 10 6

NC -1

ELECTRIC POTENTIAL

Electric potential (V) at a point is defined as the work done in bringing a unit positive charge

from infinity to that point against the electrical forces of the field.

The electric potential is different from potential difference in that, the potential difference is

concerned with the circuit electric potential while electric potential is concerned about bringing a

unit charge from infinity. The unit of electric potential in the Volt; V = W/ q

Also electric intensity is related to electric potential

V= Ed but E = f/Q

= Q 1d/ 4 d 2

= Q/ 4 d

Example : Calculate the potential at a point 40cm from a charge of 3x 10 -6.

K = 9 x 10 9

NmC-2

V

= KQ/ d = 9 x 10

9 x 3x 10

-6 / 0.4 = 6.75 x10

4V

2. how much work is required to carry a charge of 3x10 -5

C from a point 50cm from a charge of

2 x 10 -4

C to a point 20cm from it.

Solution

Q2 = 2 x10 -4

C

Q 1 = 3 x 10 -5

C

V AB = Q/ 4 r = KQ/ r

Potential at A due to a charge 2x 10 -4

C

9x 10 9 x 2x 10

-4 /2 x 1/ 0.5 =

potential at B due to a charge of 2x10 – 4

C

V AB is given by V AB = 9 x 10 9 x 2 x 10

-4 ( 1/ 0.2 – 1/ 0.5)

= 54 x 10 5 V

Hence W = QV AB

= 3 x 10 -5

x 54 x 105

= 162 joules

CAPACITORS

A capacitor is an electric device that stores electric charges. It is made of two conductors (metal

plates) carrying opposite charges. Between the plates are insulators called dielectric substance

Capacitance © of a capacitor is defined as the ability of a capacitor to store charges . it is the

ratio of charge to potential difference

The unit of capacitance is the Farad. Other units are microfarad ( µF), millifarad (mf) and nano

farad ( nf) .

The capacitance of a capacitor depends on;

- Size

- Area

- Distance between the plates

- Dielectric substance

- Shape

There are different types of capacitors

- Paper capacitors

- Electrolysis capacitors

- Air capacitors

- Mica capacitors

Which serve for different purposes.

EXAMPLE

A capacitor charge 5x 10 -8

C has a potential 100V. What is the capacitance of the

capacitor?

C= Q/V = 5 x 10 -8

/ 100 = 5 x 10 -8-2

= 5 x 10 -10

F or 5 x 10 -4

µ F

CAPACITORS IN SERIES AND IN PARALLEL

For Capacitors in series, the charge, Q, stored by each capacitor is the same.

V 1 = Q/ C1 ; V = V 1 + V2

V2 = Q/ C 2 Q/C = Q/C 1 + Q/C2 + Q/C3

V3 = Q/C3 1/CI = 1/C 1+ 1/ C2 + 1/C3

In parallel,

The voltages are equal, therefore , Q= CV , CV = C1 V + C2V + C3V

CV = VCC1 + C2 + C3 )

C= C 1 + C 2 + C 3

Example

Two capacitors of capacitances 5 µF and 3µ F are connected in series and the resulting

combination is connected across 60V. What is

a. The equivalent capacitance of the combination

b. The total charge on the combination

Solution ( Series )

a.

c =15/8 =1.875 µF

b.

1.875 x 10-6

x 600

=1.125 x10-3

Coulomb

Find the equivalent capacitances of the above .

Solution

1st find the value for the parallel

3 + 2 = 5 µF

Energy store in a capacitor

USES OF CAPACITOR

1. Tuning in radio system.

2. Storing of large quantities of charges

3. Smoothening rectified current from power supply.

4. In separating alternating current from direct current.

5. It is used to control current in a.c

6. Filtering ( tone control of an audio system)

MAGNETIC FIELD

The magnetic field is defined as a region around a magnetic in which the influence of the magnet

can be felt or detected.

The area around a magnet in which it can attract or repel objects or in which a magnetic force

can be detected is called the magnetic field of the magnet. Although this field cannot be seen, its

presence can be demonstrated and mapped out using a compass needle. A magnetic field is an

example of a force field. It is a vector quantity. A magnet is a dipole i.e it has two poles:- North

and South poles.

Magnetic lines of force-: They are imaginary lines along which a free north pole would tend to

move if placed in the field. A line of force may also be considered as a line such that the tangent

to it at any point gives the direction of the field at that point. These magnetic lines of force do not

really exist; hence we defined them above as imaginary lines. They are concepts we use to

describe different magnetic phenomena.

Magnetic lines of force can easily be defined as lines along which a free N-pole would tend to

move.

PROPERTIES OF MAGNETIC FIELD

1. Magnetic lines of force cannot cross each other because they move only in one direction.

2. Magnetic lines of force do not have two possible directions at any point of intersection.

3. The magnetic field in a bar magnet is non- uniform.

4. Lines of force are straight, uniformly spread and parallel in a uniform field.

5. Lines of force normally curve.

6. The direction of the field varies from point to point.

7. Lines of force are continuous in any region with free charges.

8. Lines of force do not begin or end in the space surrounding a charge.

9. The direction of the magnetic field at a point is given by the tangent to the line of force at

one point.

MAGNET FLUX

This refers to the number of magnetic lines of force in a given magnetic field. It is also a vector

quantity as is measured in weber (Wb)

MAGNETIC FLUX DENSITY

This is defined as the magnetic flux per unit area in a given magnetic field. It is denoted by B,

measured in wb/ m2

or Tesla (T)

B = magnetic flux / area = Q/A, Q= BA

PATTERNS OF MAGNETIC FIELD

According to the law of magnetism like poles repel and unlike poles attract.

HOW TO MAKE MAGNETS

1. Electric method

2. Single touch

3. Double touch

4. Hammering at angle of dip

ELECTRICAL METHOD -: This is the best and cheapest method. A magnetic material e.g steel

is placed in a solenoid and connected to D.C supply. After a period of time the material is

removed from the solenoid in North South direction.

SINGLE TOUCH : A soft – iron bar is placed on the bench , a magnet is stroked along its

length with one end. The magnet is kept in an inclined position

DOUBLE TOUCH

The process is the same as in single touch but in this case, each half of the specimen is stroked

repeatedly in opposite direction by the opposite poles S and N of two bar magnets. The stroking

is done simultaneously outwards from the centre of the specimen, not from its end.

HAMMERING METHOD

When a soft iron is hammered at an angle of dip after sometime it acquires magnetic property.

DEMAGNETIZATION

It is a process where a magnet is made to lose its magnetism. The processes of demagnetization

are a) Mechanical method (b) Heating and (c) electrical method

A. MECHANICAL METHOD: It is a process of hammering them hard when they are

pointing in East Direction.

B. HEATING METHOD : When magnets are heated strongly they lose their magnetism

C. ELECTRICAL METHOD: The magnet is placed in a solenoid and ac current is passed

through them.

Neutral points are points where the magnetic forces are not felt because of repulsion.

MAGNETIC PROPERTIES OF IRON AND STEEL

Iron can easily be magnetized faster than steel hence iron is used for temporary magnet while

steel is used for permanent magnets because it takes time to be magnetized and cannot easily

demagnetize. Hence iron is used to make electromagnet. When magnetism is required for short

time. Electromagnets are used in separating iron from non- iron, electric telephone or ear piece

for lifting loads in cranes.

ELECTROMAGNET

Any time a conductor carries current it has a magnetic field around it. There is a law that guides

it called right hand grip rule.

If a straight wire is grasped with the right hand so that the thumb points in the direction of the

current, the fingers are curled indicating the direction of the magnetic field.

MAGNETIC FIELD AROUND A SOLENOID

A solenoid is a long cylindrical coil of wire whose turns are usually close together. Where lines

of force come from it referred to as the North

Magnetic shield

This is used to protect sensitive instrument from the influence of external magnetic field.

External fields moves axially and leave the centre alone where the sensitive instrument is kept.

Force in magnetic field

F= qVBsin θ for a charged particle

F = BiLsin θ for a conductor

If θ = 90 0

Maximum force

If θ is 0 0

= minimum force

F= qVB sin θ

Where θ is the angle between the charges or rod and the lines of flux.

EXAMPLE: Find the magnetic force experienced by an electron projected into a magnetic field

of flux density 10 Tesla with a velocity of 5 x 10 7

m is and in a direction

i. 90 0

ii. 60 0

and

iii. coil parallel to the magnetic field

assume charge on electron Q = 1.0 x 10 -19

C.

Solution

i) F = qVBsin θ

= 1.6 x 10 -19

x 5 x 10 7 x 10 x sin 90

0

= 8 x 10

-11 N

ii) Where θ = 00 0

F = qVB sin 60 0

= 1.6 x 10 -19

x 5 x 107 x10 x sin 60

= 6.9 x 10 -11

N

iii) Θ = 0

F = qVBsin θ = qVB sin θ = 0N

EARTH MAGNETIC FIELD

The angle of declination is the angle between the magnetic north and geographical north.

Angle of dip is the angle between the earth magnetic field and the horizontal.

The angle of dip increases from 00 at the equator to about 90

0 at the earth surface.

If a mariner compass gives a bearing of N20 0

W at a place where the declination is 15 0

west of

North, then the true bearing is ( 20 + 15) = 300

ELECTROMAGNET

Electromagnet is a device which is made of a solenoid around a soft iron core.

It is used for several purposes:

1. It is used in electric bell.

2. It is used in telephone earpiece.

A solenoid is placed around a soft iron, a circuit set up with a cell, key and rheostat. When

current flows into the circuit the soft iron magnifies the magnetic field and behaves like a

magnet.

USES

1) It is used to produce intense magnetic field in electromagnetic motor, generator.

2) It is used to separate iron from mixtures containing non-magnetic substances.

3) They are used in the construction of electric bell and telephone ear piece.

4) It is used for lifting and transporting heavy pieces of iron.

WORKING PRINCIPLE IN DOOR BELL

When the bell push is pressed, the circuit is completed and current begins to flow. The soft iron

ore becomes magnetized and the electromagnet attracts the soft iron armature attached to a

spring. The clapper then strikes the gongs. The circuit is broken as soon as the spring moves

along within the soft iron. The electromagnet then losses its magnetism,The armature then flies

back to remake contact once again. The circuit B then reestablished and current flows again. The

sequence of event is repeated.

TELEPHONE EAR PIECE

The telephone ear piece is a device which converts the varying electric energy from the

microphone into varying speech coils.

WORKING PRINCIPLE

When a caller speaks into the microphone at the other end of the communication line, the

varying sound energy from his speech is converted into a varying speech current which is

transmitted to the ear piece at the receiver’s end. The variation in the electric current due to

changes in the speech energy causes a variation in the magnetism of the electromagnet. This

varying magnetism of the electromagnet causes vibrations of the diaphragm placed in front of it.

Vibrations of the diaphragms create sound waves of the same frequency in the air as the speech

current hence the speaker at the other end is heard.

ELECTROMAGNETIC FIELD

Electromagnetic field is a field representing the joint interaction of electric and magnetic forces.

Electromagnetic field is considered to have its own objective existence in space just like any

other field.

F = q( E + V x B)

VxB = VBsin θ

Interactions between magnetic fields and current

a) Force on a current carrying conductor in a magnetic field.

A conductor carrying an electric current experience a mechanical force when placed in a

magnetic field. When two metals rails are fixed on each side of a powerful horse shoe

magnet. A copper rod is placed across the rails. when we pass current through the metal rails.

It is observed that the copper rod rolls along the rails, towards the right.

Factors affecting the mechanical motion felt:

1. The magnitude of the magnetic field.

2. The current flowing through the metal rail.

3. The angle through which the copper rod is placed with the metal rail. Maximum force is felt

at 90 0

. No force is felt at 00

to the metal rail or parallel to the metal rail.

The direction of force on a current carrying conductor placed perpendicular to magnetic field

given Flemming’s left- hand rule which is stated as follows:

If the thumb, fore –finger and middle finger are held mutually at right angles to one another with

the fore finger pointing in the direction of magnetic field , and the second finger in the direction

of current , then the thumb will point in the direction of motion or force producing motion.

An electric motor that converts electrical energy to mechanical energy it is used in electric car,

car engine. When current from the source flows through the split ring commutator and then flows

into a loop of wire. As electric current flows through the loop there is a turning effect on the soft

iron armature which causes it to turn due to magnetic and electric field. The carbon brush is fixed

while the split ring commutator rotates along with the soft iron armature and touches the carbon

brush.

PRACTICAL DIRECT CURRENT ELECTRIC MOTORS

Practical electric motors have:

1. Strong magnetic fields produced by magnets.

2. Several coils of wire wound in the slots of a laminated soft iron – cylinder.

3. High current: - all these add up to produce a powerful rotation.

Energy loses in an electric motor can arise from:-

a. Workdone against friction in the bearings and comutator.

b. Eddy current formed in soft- iron cylinder. This is reduced by laminating the cylinder.

c. Heat losses (or I 2

R) in the windings due to resistance.

The soft armature increases the magnetic field.

GALVANOMETER: This is a device that measures small current or small voltage.

Parts of a galvanometer.

1. A light rectangular vertical coil

2. Two curved pole pieces hands of a horse slide magnet.

3. Two spiral non- magnet control springs.

WORKING PRINCIPLE

The current is to be measured is passed into the galvanometer from one terminal through one

spring, leaves through the other, when current flows through the coil, forces act on the arms AB

and CD as shown. The two forces constitute a couple whose torque tends to rotate the coil in a

clockwise direction across the radial magnetic field provided by the magnet and the soft iron

core. The movement of the coil is opposed by the spiral springs and the coil will come to rest in a

position where the couple done to the current is exactly balanced by the control couple due to the

springs. The angle through which the coil rotates is proportional to the size of the current. The

scale reading will therefore give a measure of the current.

To increase the sensitivity of the galvanometer

i. The magnetic field is made stronger.

ii. The number of turns in the rectangular coil is increased.

iii. The area of coil is increased. Q

ELECTROMAGNETIC INDUCTION

Electromagnetic induction

Advantages of moving coil galvanometer

1. It is used to measure small amount of current

2. The high sensitivity enables very small currents to be measured.

3. No extra measures need to be taken to shield from stray magnetic fields.

4. By fitting on a suitable resistance, it can be used to measure potential difference on the

larger currents.

DISADVANTAGE

It cannot measure alternating current because rapid changing of direction of the current.

ELECTROMAGNETIC INDUCTION

Electromagnetic induction is the production of electric current or voltage in a conductor

whenever there is a relative motion between the conductor and a magnetic field. (or a magnet)

Induced EMF in a straight conductor

Fleming’s right hand rule

The direction of induced EMF is given by Fleming’s Right Hand Rule

If the thumb, fore-finger and middle finger of the right hand are mutually at right angle to each

other, with the force- finger pointing in the direction of the field and the thumb pointing in the

direction of the motion. Then the middle finger will point in the induced E.m.f or current I.

Laws of electromagnetic induction

1. Faraday’s law

2. 2nd

law of Faraday or Lenz law.

Faraday’s law states that whenever there is a change in the magnetic lines of force, an (Emf )

is induced , the strength of which is proportional to the rate of change of the flux linked with

the circuit.

Lenz law states that the induced EMF is such a direction as to oppose the motion or change

producing it.

EDDY CURRENT

When a solid metallic sheet swings in a magnetic field, it cuts through the magnetic flux or

field lines consequently according to Lenz’s law, E.m.f is induced in the sheet which

opposes this movement .

The current generated is Eddy current because it flows in a circular path or closed loops

within the conducting material like the Eddy current in water. Eddy current generates heat in

the material leading to wastage of energy in most electrical devices.

Some devices need Eddy current for operation e.g

1. Hot wire ammeter

2. Moving coil galvanometer

3. Moving wire ammeter.

Electrical devices using electromagnetic induction. An example of such is a dynamo.

A dynamo is a machine that converts mechanical energy into electrical energy or electrical

energy into a mechanical energy.

TYPES OF DYNAMO

1. Generator

2. AC generator

3. DC generator

2. ELECTRIC MOTOR

EMF from a generator ; E= Eosinwt

Flux Ҩ through the coil is then given by

Ҩ = ABcosϴ

If there are N number of turns in the coil the flux is given by: Ҩ = NABcosϴ

Let the coil turn with a steady angular velocity ω given by

ω = dθ/ dt

Induced e.m.f is given by

E= -dθ / dt

The above equation follows from Faraday’s and Lenz laws

Hence

( )

( )

E

But

When = 90 0

MEASUREMENT OF AC

Devices used to measure alternating current.

i. Moving iron ammeter

ii. Hotwire ammeter

iii. Moving coil galvanometer

Moving iron ammeter can be used to measure direct current and alternating current.

When a motor behaves like a generator. The EMF is reversed. This is called back EMF.

TRANSFORMER

This is an electric device for changing the size of AC voltage.

TWO TYPES OF TRANSFORMER

1. Step up transformer

2. Step down transformer

N s/Np = turn ration

Number of turning of secondary coil / number of turning of primary coil = Turns ratio

Ns /N p > 1 for step up transformation

Ns /N p < 1 for step down transformer

P s = P p P = 1V

V= E.M.F in an open current

I 0E s = I sE p

I p / I s = E s E p

I p / I s = E s / Ep

I p / I s = E s / Ep = N s / Np

MUTUAL INDUCTANCE

This is the flow of current or voltage in a coil due to an alternating current or varying current in

the neighboring coil .

Efficiency in transformer

Efficiency = power output / power input x 100

Outcomes from the secondary coil

Input comes from the primary coil

Power output/ power input = Is Ep / I p Ep

Calculations

1.Find the turns ratio in a transformer which delivers a voltage of 120 volts in a secondary coil

from a primary voltage of 100volts

Turn ratio = N s/N p

N s = No of turns in the secondary

N p = No of turns in the primary coils

E s / Ep = N s / Np

120/60 = N s / Np

Turn ratio = 2

2) A transformer has 500 turns in the primary coil and 300 turns in the secondary coil. If the

primary coil is connected to 220V mains, what voltage will be obtained from the secondary coil?

What type of transformer is this?

Solution

E s / Ep = N s / Np

E s = Ep x N s / Np

= 220 x 300/500 = 132 volts

It is a step up transformer because secondary voltage is less than primary voltage (132low) and

also turns ratio less than.

No.17 pg 388 all inclusive.

A transformer with a primary coil of 800 turns and a secondary coil of 50 turn has its primary

coil connected to a 240 V a. c mains. If the current passing through the primary coil is 0.5 A,

Calculate

i. P.d across the secondary ends.

ii. Current passing through the secondary coil assuming no power losses.

iii. Power in the secondary coil if 10% of that in the primary coil is lost.

Solution

E s = Np = 800, N s = 80 , / Ep = 240 V , Ip = 0.5A

N s / Np = E s / Ep

50/800 = E s / 240

E s = 50 x 240 / 800 = 15V

Is Es = I p Ep

Is = I p Ep/ Es = 0.5 x 240/ 15 = 8.0A