ENVIROMENTAL PHYSICS
(i) Agriculture physics
- Influence of solar radiation on plant growth.
- Influence of wind, humidity, rainfall and air temperature on plant growth.
- Soil environmental component which influence plant growth.
(ii) Energy from the environment
Photovoltaic energy
Wind energy
Geothermal energy
Wave energy
(iii) Geophysics (Earth quakes)
Elastic rebound theory
Types of seismic waves
Propagation of seismic waves
Seismology
(iv) — Environmental pollution
Types of pollutant in the atmosphere
Transport mechanisms of atmospheric pollutant
Nuclear waste and their disposal
Effects of pollution on visibility and optical properties of materials.
INTRODUCTION

Environmental physics is an interdisciplinary subject that integrates the physics processes in
the following disciplines: the atmosphere, the biosphere, the hydrosphere, and the
geosphere.
Environmental physics can be defined as the response of living organisms to their
environment within the framework of the physics of environmental processes and issues.
It is structures within the relationship between the atmosphere, the oceans (hydrosphere),
land (lithosphere), soils and vegetation (biosphere).
It embraces the following themes:
(i) Human environment and survival physics,
(ii) Built environment
(ili) Renewable energy
(iv) Remote sensing
(v) Weather, climate and climate change, and
(vi) Environmental health.
The environment may be defined as the medium in which any entity finds itself, For
example, for a cloud its environment may be the region of the atmosphere in which it is
formed.
AGRICULTURE PHYSICS
Agriculture physics is concerned with physics environment in relation to plant growth.
(a) Influence of Radiation Environment on Plant Growth
Radiation environments. Refer to radiations present in the atmosphere, commonly
coming from the sun.
Components of solar radiation
The main components of solar radiation are:

(i) Visible light
(ii) Infrared radiation, and
(iii) Ultraviolet radiation.
HEATING EFFECT OF SOLAR RADIATION ON PLANTS
Positive effect
An optimum amount of heat on plant favours the process of photosynthesis. This enables a
plant to make its own food and hence provide its growth.
Negative effects
(i) Excessive solar radiation (ultraviolet light) on plants leads to bleaching of green pigment
(chlorophyll), This lowers the amount of food produced by photosynthesis to plant and
hence a plant may dic.
(ii) Excessive solar radiation on plants leads to excessive water loss in the form of water
vapour commonly on plant leaves (transpiration). Hence wilting (drying) of plants may
occur.
(b) Influence of Aerial Environment on Plant Growth
Aerial environments refer to the atmospheric condition resulting from a series of
processes occurring in the atmosphere. These include air temperature, wind, humidity
and rainfall.
WIND EFFECT ON PLANT GROWTH
Positive effects
(a) Wind acts as pollinating agent for some plants and hence favours plant productivity.
(b) Wind also favours evaporation of water from plant leaves and thus maintains water
balance for proper plant growth.
Negative effects

(a) Excessive wind on environments leads to plant breaking or cutting of tree
branches. This may lead to the death of plant.
(b) As the wind speed increases further, cell and Cuticular damage occurs, followed by death
of plant tissue, and a gnarled appearance becomes more apparent.
(c) At low wind speeds, the effect seems to be an increase in transpiration, which results in
water stress. This stress causes the plant to adapt by decreasing leaf area and internodes
length, while increasing root growth and stem diameter.
(d) Strong wind may also cause shade off flowers; this lowers plant productivity.
Effect of Rainfall on Plant Growth
Positive effect
An optimum amount of rainfall on plants favours its growth. Water is a raw material for the
process of photosynthesis from which plants obtain their food and hence their growth.
Negative effect
Excessive rainfall leads to water logging in soil which in turn leads to root spoil and hence
the death of plant.
Effect of Humidity on Plant Growth
Positive effect
Favourable humidity on plants help plants to conserve water for various activities and in
seeds helps the development of new leaves.
Negative effect
Low humidity results into a greater rate of transpiration and hence may result into plant
drying.
Effect of Air Temperature of Plant Growth
Positive effect

le
action.

Capillary action refers to the attraction of water into soil pores — an attraction which makes
water move in soil. Capillary action involves two types of attraction — adhesion and
cohesion.
Adhesion is the attraction of water to solid surfaces.
Cohesion is the attraction of water to itself.
Speed of water in a particular soil type depends on:
(i) How much water is in the soil, and
(ii) Porosity of the soil.
The movement of water in the solid is mainly due to gravity. The porosity gives a measure
of how much water the soil can hold and the rate at which water flows through the
soil. Large pore spaces give a faster rate and vice versa.
An experiment to study water movement in soil
An experiment to demonstrate the rate of flow of water in the soil is done using a glass
tube and sand type filled in it. Water is poured into the tube and the time taken for water
to reach the bottom of the tube in notes.
Unsaturated soil Saturated soil
Soil type Water speed Soil type Water speed
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Sandy loam Clay loam
(a) )
i. Sand soil have large pore spaces thus allows water to travel downwards through it at a
fastest rate.
ii. Clay soil can hold water as has very fine pore spaces.
iii. Loamy soil allows water movement at a medium rate.
Heat transfer in the soil
Within the soil heat is transferred by a conduction process. Since soil is poor conductor of
heat most of the heat from the atmosphere appears at the surface of the earth.
An optimum soil temperature favours plants growth but a high temperature can lead to the
rotting of plant roots.
(d) Techniques for the Improvement of the Plant Environment
Plant environment can be improved by using wind breaks, shading and mulching.
Shading

Shading is the process of obstructing plants from excessive solar radiation.
Positive Impacts of Shading
1. Prevents excessive loss of water by plants through transpiration. This enhances plant
productivity.
2. Preserve moisture in the soil and hence water supply to plant.
Mulching
Mulching is the process of covering the soil by dry leaves, grasses and or papers.
Benefits (Advantages) of Mulching
1. Improve soil moisture. Bare soil is exposed to heat, wind and compaction loses water
through evaporation and is less able to absorb irrigation or rainfall. Using mulches, the
soil has greater water retention, reduced evaporation, and reduced weeds. Mulch can also
protect trees and shrubs from drought stress and cold injury
2. Reduce soil erosion and compaction, Mulches protect soils from wind water, traffic
induced erosion and compaction that directly contribute to root stress and poor plant
health.
3. Maintenance of optimal soil temperatures. Mulches have shown to lower soil
temperatures in summer months. Extreme temperatures can kill fine plant roots which
can cause stress and root rot. Mulches protect soils from extreme temperatures, cither
cold or hot.
4. Increase soil nutrition. Mulches with relatively high nitrogen content often result in
higher yields, but low nitrogen mulches, such as straw, sawdust and bark, can also
increase soil fertility and plant nutrition.
5. Reduction of salt and pesticide contamination. In arid landscapes, evaporating water
leaves behind salt crusts. Because mulches reduce evaporation, water is left in the soil
and salts are diluted. Organic mulches can actively accelerate soil desalinization and help
degrade pesticides and other contaminants.
6. Improve plant establishment and growth. Mulches are used to enhance the
establishment of many woody and herbaceous species. Mulches improve seed
germination and seed survival, enhance root establishment, transplant survival, and
increase plant performance.

7. Reduction of disease. Mulches will reduce the splashing of rain or irrigation water,
which can carry spores of disease organisms to stems and leaves of plants. Populations
of beneficial microbes that reduce soil pathogens can be increased with
mulches. Mulches can combat disease organisms directly as well.
8. Reduction of Weeds. Using mulches for weed control is highly effective. Mulches can
reduce seed germination of many weed species and reduce light, which stresses existing
weeds.
9. Reduce pesticide use. Mulches reduce weeds, plant stress, and susceptibility to pests and
pathogens which translates to reduced use of herbicides, insecticides, and fungicides.
Mulch Problems (disadvantages of mulching)
1. i, Acidification. Some types of mulches can increase soil acidity.
2. ii, Disease. Many mulches made from diseased plant materials can be composted or treated
at temperatures that kill pathogens that can be transmitted to healthy plants,
3. iii.Pests. Many organic mulches, especially wood — based mulches, have the reputation as
being “pest magnets”.
4. iv, Weed contamination. Improperly treated crop residues and composts as well as bark
mulches are often carriers of weed seed. Mulch must be deep enough to suppress weeds and
promote healthy soils and plants. Weed control and enhanced plant performance are directly
linked to mulch depth.
v. Wind Breaks
Wind breaks are long rooted strong plants (trees) that are used to obstruct the path of
wind or to slow down the wind.
Windbreaks provide many benefits to soil, water, plants, animals and man. They are an
important part of the modern day agricultural landscape. Windbreaks come in many
different sizes and shapes to serve many different conservation purposes.
In agriculture, wind breaks protect small growing plants from strong blowing wind
Advantages of Windbreaks to Plant Environment
1. i. Control soil erosion. Windbreaks prevent wind erosion from causing loss of soil
productivity. This eliminates plant roots stresses and thus favours plant growth condition.

Solar constant
Solar constant is defined as the solar energy falling per second on a square meter placed
normal to the sun’s rays at the edge of the Earth’s atmosphere, when the Earth is at mean
distance from the sun.
Its value is about 1.35 kWm?
The amount of solar radiation received at any point on the earth’s surface depends on:
(i) The geographical location,
(ii) The season, (summer or winter)
(iii) The time of the day, the lower the sun is in the sky the greater is the atmospheric
absorption.
(iv) The altitude; the greater the height above sea level the less is the absorption by the
atmosphere, clouds and pollution
PHOTOVOLTAIC DEVICES (SOLAR CELLS)
A solar cell (PV, cells) is a PN junction device which converts solar energy directly into
electrical energy.
How it Works
PV cells are made of at least two layers of semiconductor material. One layer has a positive
charge (p — type material), the other negative (n-type material). When light enters the cell,
some of the photons from the light are absorbed by the semiconductor atoms, freeing
electrons from the cell’s negative layer to flow through an external circuit and back into the
positive layer. This flow of electrons produces electric current.

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Uses of the solar cell
1. (i)Are used to power electronics in satellite and space vehicles.
2. (ii)Are used as power supply to some calculators.
3. iii)Are used to generate electricity for home, office and industrial uses.
Series arrangement of solar cells
Solar panel (module) is a sealed, weatherproof package containing a number of
interconnected solar cells so as to increase utility of a solar cell.
When two modules are wired together in series, their voltage is doubled while the current
stays constant.
When two modules are wired in parallel, their current is doubled while the voltage stays
constant.
To achieve the desired voltage and current, modules are wired in series and parallel into
what is called a PV array.
The flexibility of the modular PV system allows designers to create solar power systems that
can meet a wide variety of electrical needs, no matter how large or small.

Cell Module
Efficiency of a photovoltaic system
The output power of a solar cell depends on:
(i) The amount of light energy from the sun falling on a solar panel (the intensity of
light).
(ii) The orientation of the solar panel. More electricity is produced if light falls
perpendicular to panels.
(iii) The surface area of the panel. Large area collects more solar energy and hence
greater electricity.
The best designed solar cell can generate 240 Wm” in bright sun light at an efficiency of
about 24%.
Advantages of photovoltaic systems
1. Solar cells can produce electricity without noise or air pollution.
2. A photovoltaic system requires no fuels to purchase.
3. Panels of photovoltaic cells are used for small — scale electricity generation in remote
areas where there is sufficient sun.
4. Net metering: This has the potential to help shave peak loads, which generally coincide

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also vary from layer to layer, the solid part showing greater speed and the liquid ones lower
speed.
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SURFACE WAVES/LONG WAVES
Surfaces waves are produced when earthquake energy reaches the Earth’s surface.
These are the slowest moving waves, but are the most destructive for structures on earth
There are two types of L- Waves:
(i) Love long waves
(ii) Rayleigh long waves
i. Love Waves
Love waves are Transverse horizontal motion, perpendicular to the direction of propagation
and generally parallel to the Earth’s surface.
They are formed by the interaction of S waves with Earth’s surface and shallow structure and
are dispersive waves. The speed at which a dispersive wave travels depends on the wave’s
period.
Characteristics of Love Waves

1. i. Love waves are transverse and restricted to horizontal movement (horizontally polarized).
2. ii. The amplitude of ground vibration caused by a Love wave decrease with depth. The rate
of amplitude decrease with depth also depends on the period/frequency.
3. iii. Loves wave are dispersive, i.e. wave velocity is dependent on frequency; low frequency —
higher velocity.
4. iv. Speed of love waves is between 2.0 and 4.4 km/s
5. v. Love waves travels within the earth’s crust only.
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LOVE WAVE
Rayleigh Waves
Rayleigh waves are vertically polarized long waves. The slowest of all the seismic wave types
and in some ways the most complicated.
Characteristics of Rayleigh Waves
1. Rayleigh waves are transverse and restricted to vertical movements (vertically polarized).

2. The amplitude of Rayleigh wave decreases with depth. The rate of amplitude decrease
with depth depends on the period/frequency
3. Rayleigh wave are dispersive, i.e. wave velocity dependent on frequency; low frequency
—high velocity
4. Speed of love waves is between 1.0 and 4.2 km/s slowest of all waves.
5. Travels within the earth’s crust only.
6. Depth of penetration of the Rayleigh waves depend frequency, with lower frequencies,
penetrating greater depth.
PROPAGATION OF SEISMIC WAVES
Like all other types of waves, seismic waves may undergo,
(i) Reflection, (ii) Refraction, (iii) Dispersion, (iv) Diffraction, (v) Attenuation.
Seismic reflection:
Seismic waves bounce (reflect) rock boundaries of different rock type (density).

Seismic refraction:
Waves change velocity and direct (refract) when they enter a medium of different density it
the one they just passed through.
Seismic Dispersion:
surface waves are dispersive which means that different periods travel at different
velocities. The effects of dispersion become more noticeable with increasing distance
because the long travel distance spreads the energy out (it disperses to energy).
SEISMIC WAVE PATHS
By comparing the data recorded by many stations all over the world the nature, speed and
the paths of the seismic waves can be determined. This information can be used to tell us
about the earth’s interior such as density sand state in each layer.
L— Waves travel within the Earth’s crust only
P and S waves travel through the carth in a curve path. The waves are refracted because their
speeds a constantly changing with depth due to continue increase in density. Waves are
also strongly refracted the Mantle — Core boundary.
Surface waves travels through the Farth crust only
Shadow zone is the region on the Earth’s surface where no S or P waves are present.
This lies between 105° and 140°. Only surface waves may be detected in this region.
Shadow zone occurs because:
(i) P~ Waves are strongly refracted at the liquid outer core.
(ii) S— Waves can’t travel through the liquid outer core.
Seismic waves can also be used to locate the discontinuities in the earth’s crust. A
change in density or crack would affect the propagation of the waves.
This alteration in the wave’s path or speed would indicate the discontinuity.

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The fact that S waves do not travel through the core provides evidence for the existence
of a liquid layer beneath the rocky mantle
The change in the velocity of P waves at crust - Mantle boundary reveals the presence
of Mohorovicic discontinuity
P waves passing through the inner core show increased velocity suggesting that the
inner core is solid.
Both P and S — Waves slow down when they reach the asthenosphere. Because of
this scientists know that the asthenosphere is partially liquid
MEASUREMENTS OF EARTHQUAKES

i. Seismology is the scientific study of earthquakes (seismic waves) and artificially produced
vibrations in the earth.Seismograph is a sensitive instrument that is used to record
earthquakes and seismic waves (i.e. ground movements).
ji, Seismogram is the record of ground movement drawn by a seismograph.
The arrival of seismic waves at a station
surface waves
minute mark P s
_
Seismograph consists of a heavy weight suspended from a frame fixed into the ground. When
the earth vibrates the frame moves but the heavy weight remains stationary due to great
inertia. A pen attached to weight plots the earth’s movements on a chart recorder to produce
a seismogram. To obtain a complete record of the earthquake measurements must be taken
in all three planes (x, y and 2).

SIZE OF AN EARTHQUAKE
The size of an earthquake can be measured in terms of its intensity (Mercalli/Wood Neumann
scale) or its magnitude (Richter scale).
Mercalli Intensity Scale
The Mercalli scale measures the intensity of how people and structures are affected by the
seismic event. In essence, it measures damage. It is much more subjective and uses numbers
ranging from 1 (no damage) to 12 (total destruction).
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ISOSEISMAL LINES
Intensity distribution maps can be drawn up showing the intensities of an earthquake over

a region. The earthquake is most intense at the epicenter and decreases with distance.

Isoseismal lines are line joining points of equal intensity.
Richter magnitude scale
The magnitude of an earthquake is measured in terms of energy released by an
earthquake. This is determined from the amplitude of the seismic wave recorded on a
seismogram 100 km from the epicenter. The magnitude is equal to the logarithm of the
amplitude. Therefore each successive number represents a tenfold (x10) increase in the
ground motion. The Richter scale starts at 0 but has no upper limit. -However 8 represent
an earthquake that causes total destruction within the region.
Magnitude Amount of explosives (TNT) needed to release the equivalent energy, in tons
a
Intensity of an earthquake is a measure of its strength based on the changes it causes to the
landscape.
EARTHQUAKE PREDICTIONS (WARNINGS)
Forecasting (predicting) earthquakes is very difficult, although there are a number of warning
signs which occur before an earthquake happens.
(i) Change in the velocity of p — waves
(ii) Electrical resistivity of the rocks decreases.
(iii) An increase in radon, emission (radon is an inert gas, radon is found to increase in soil
and water samples).
(iv) Increase in fore shock (small tumors that occur just before an earthquake).
(v) Local variations in the magnetic field
(vi) Animals begin to behave strange

(vii) Water levels rise or fall in wells few days before earthquake.

(viii) Increase in temperature of the area few months before the occurrence of an
earthquake
PRECAUTIONS
Some of the world’s populations are living in regions where there is a high risk of an
earthquake. Most of these regions lie along fault lines. However a few precautions can be
taken to reduce the damage caused.
(a) Build structures that can withstand the forces of an earthquake. One method is to include
shock absorbers into the buildings foundations.
(b) Scientific rescarch has shown that pumping water out of the carth reduces the stress in the
crust hence preventing an earthquake. However this technique is very expensive.
() Stay away from tall buildings or structures during an carthquake if you are outside on
occurrence,
(d) Ifyou are inside a house, stay in a safe place where things will not fall on you
EARTHQUAKE HAZARDS
Earthquake give rises to a number of hazards which pose a great risk to human life, animals,
property and the environment at large. The following are some hazards:
1. Landslides and avalanches: The shaking caused by an earthquake can cause unstable
hillsides, mountain slops’ and cliffs to move downwards creating landslides. Earthquakes
can also trigger avalanches on snow slopes
2. Tsunamis: If an earthquake occurs under the sea or ocean, the shock waves disturb the
water. The ocean floor can rise or fall causing the water to rise and fall too. This
movement creates huge water waves called tsunamis that travel across the ocean.
3. Collapsing building: Buildings or structures may collapse during a_ strong
earthquake, The collapse of the building may kill people.
4. Fire outbreak: Earthquakes can cause gas or oil pipes to break and or the collapse of
electricity lines. This may set up fire.

5. Backward rivers: Tilting ground due to earthquakes can make rivers change their course.

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Example 11: Necta 1994 P;
(a) (i) Name the lowest layer of the atmosphere and the lowest layer of the ionosphere.
(ii) State the importance of each of these layers.
(b) What is the ozone layer?
Answers
(a)(i) The lowest layer of the atmosphere is troposphere and the lowest layer of the
ionosphere is called the D — layer.
(ii) The t troposphere supports life
The D layer is important for communication purposes as it reflects radio waves
(b) The ozone layer is within the stratosphere. In the ozone layer molecular oxygen (O2) is
dissociated into atomic oxygen (O) which is then reformed into ozone (Os)
The ozone so formed absorbs ultra violet radiation thus protecting plants and shielding
people from skin cancer and eye cataracts.
Example 12: Necta 1994 P,
(a) Illustrate the component of the earth’s magnetic field at a given point P in the earth’s
atmosphere by a suitable diagram.
(b) Using a tangent galvanometer, explain how you could determine the earth's magnetic
field.
Answers

Example 13: Necta 1995 P;
(a) (i) which region of the solid earth includes the e earth’s centre?
(ii) On which region of the solid earth do the continent rests directly?
{iii Which region of the ionosphere has the highest electron density?
(b) Briefly explain how earthquake can be detected
Answers
(a) (i) inner core (ii) crust (iii) F—region
(b) Detection of earthquake is done by recording or measuring the seismic waves generated
by the earthquakes. These waves are recorded by instrument called seismograph.
Example 14: Necta 1995 P,
(a) Draw a well labeled diagram which shows the interior structure of the earth. Indicate also
which part of the interior are in solid form and which are in liquid form.
(b) Name and distinguish the type of waves that are produced by an earthquake.
(c) Briefly describe the three ways in which signal form ground based transmitter can reach
the receiver.
Answers
(a) There are four types of seismic waves:
Body waves — divided into P and S- waves
Surface waves — divided into love and Rayleigh
(b) A telecommunication problem.
Ground wave, sky wave and space waves
Example 15: Necta 1998 P;
(a) State any three magnetic components of the earth’s magnetic field

(b) The horizontal and vertical components of the earth’s magnetic field at a certain location
are; 2.73 x 10% and 2.1 x 10°T respectively. Determine the earth’s magnetic field at
the location and its angle of inclination @
Solution
(a) Components of the earth magnetic field are:
Vertical component (which point vertically downward)
Horizontal component which comprise If:
Eastly component (towards geographic north pole)
Northly component (towards magnetic north pole)
(b) Be =3-44x10" TandO =37.66°
Example 16: Necta 1998 P; B
(a) What is the origin of the earth’s magnetic field?
(b) The following diagram shows the main layers forming the interior of the earth name the
layers indicated by letters A to G.

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c
Answers
(a) Refer notes
(b) A = Farth’s surface, B = Crust, C = Moho discontinuity, D = Gutenberg discontinuity, F =
outer core, F = Mantle and G = inner core.
Example 17: Necta 1998 ?2B
(a) Explain the following terms; Earthquake, Earthquake focus, Epicenter and body waves.
(b) List down three (3) sources of earthquakes,
(c) (i) Define ionosphere
(ii) Mention the ionosphere layers that exist during the day time

(iii) Give the reason for better reception of radio waves for high frequency signal of
night than during day time.
(a) Explain briefly three different types of radio waves traveling from a transmitting station to
a receiving antenna.
Answers
(a) Refer notes
(b) Refer notes
(c) (i) During the day time all the layers D,E,F,, and F2 — layers exists.
(ii) Refer Necta 1992 (b)
(d) Ground (surface wave)
Space wave
Sky waves) (refer telecommunication notes)
Example 18: nectar 2000 P;
(a) With reference to an earthquake on a certain point of the carth explain the terms ‘focus’
and ‘Epicenter’
(b) What is importance of the following layer of the atmosphere?
(i) The lowest layer
(ii) The ionosphere
(c) (i) Describe two ways by which seismic waves may be produced.
{ii) Describe briefly the meaning and application of “seismic prospecting”.
Answers
(a) Refer notes
{b) (i) Importance of troposphere is supports life on earth

(ii) lonosphere enhances communication over long distances.
(c) (i) Describe any two causes of earth quake
(ii) Seismic prospecting is an artificial production of seismic waves purposely for
searching underground fuels and oils or gases
Example 19: Necta 2001 P;
(a) (i) Define the terms “angle of declination” as used in the specification of the
earth’s magnetic field at a point
(ii) The horizontal component of the earth’s magnetic field at a location was found to
be 26.0 ME while the angle of inclination was 59°- Find the magnitude of the
field and the vertical component of the field at the location
(b) (i) Define an earthquake
(ii) Distinguish between P and S waves. What factors influence their velocities?
Answers
(a) (i) Refer notes
(ii) Be =50-48 pTBy = 4527 pT
(b) The velocities of P and S waves are influenced by;
Density, ? of the media
Shear modulus, 7 of the media, and
Bulk modulus, B of the media.
Example 20: Necta 2002 P1
(a) (i) What is the importance of ionosphere to mankind?
(ii) Explain why transmission of radio waves is better at night than at day time.
(b) (i) What is an earthquake?

(ii) Explain briefly any four (4) causes of earthquake
Example 21: Necta 2003 P,
(a) Explain the following:
{i) Earthquake (ii) Earthquake focus (iii) The epicenter.
(b) List down three sources of earthquake
(c) (i) Define the ionosphere
(ii) State the ionosphere layer that exists during day time.
(iii) Give the reason for better waves reception for light frequencies signal at night than
during the day time
Example 22: Necta 2004 P;
(a) (i) Explain the terms epicenter and focus as applied to earthquake.
(ii) State any four (4) indications that may predict the occurrence of an earthquake.
(iii) State and explain two variations of the earth magnetic field.
(iv) State one necessary precaution to be taken to people living in a region with a high risk
of occurrence of earthquakes.
(b) Explain the following
(i) Solar wind (ii) Magnetopause (iii) lonosphere.
Example 23: Necta 2005 P;
{a) Define the following terms
(i) Epicentral distance (ii) Body wave (iii) Seismograph
(b) (i) explain the meaning of reflection seismology state its application
(ii) Show how the magnetic field within the atmosphere is generated?

Seismic station
Focus (close to the earth surface)
Distance travelled by the waves (distance between focus and seismic station) is
d= RY2 =64x10° x./2 =9.05x10°m
Time taken by P —- waves to arrive at the station is
a 9.05 x 10°
8 = So = gage = 1131.25sec = 189min
Time taken by the waves to arrive at the station is
<= 10° _ 1508.33 = 25.11
= =a = jsec = 25.1min
The time interval between the arrival of the two waves is t = t)—t; = 25.1 = 18.9 = 6.2 minutes.
Example 31: Necta 2012 P,
(a) (i) What do you understand by the word environmental physics?

(ii) Briefly explain three effects of seismic waves.
(b) (i) Mention three types of environmental pollution
(ii) Explain on the following climatic factors which influence plant growth: Temperature,
Relative humidity and wind.
Example 32: Necta 2013 P;
(a) (i) The main interior of the earth core is believed to be in molten form. What seismic
evidence supports this belief?
(ii) Explain why the small ozone layer on the top of the stratosphere is crucial for human
survival
(b) Electrical properties of the atmosphere are significantly exhibited in the ionosphere.
(i) What is the layer composed of and what you think is the origin of such constituents
(ii) Mentioned two uses of the ionosphere
(c) Briefly explain why long distance radio broadcasts make use of short wave
Answers
(a) (i) When P and S seismic waves are sent from one side of earth to the other, only P waves
can be detected on the other side. The fact that S waves do not travel through the core
provides evidence for the existence of a liquid core.
(ii) Ozone absorbs harmful radiation from the sun. The Ozone projects plant and shield
people from skin cancer and eye cataracts.
(b) (i) The layer is composed of free electrons and positive ions. The ionosphere is created by
atoms absorbing UV radiation, gamma and x-rays.
{ii) Uses of the ionosphere
lonosphere supports radio communication over long distances
Particles in the ionosphere absorbs U.V radiation gamma and X-rays, thus protecting
people from harmful effects of these radiations
(c) Refer telecommunication notes.

Example 33: Necta 2013 P,
(a) Briefly explain on the following types of environmental pollution:
(i) Thermal pollution
(ii) Water pollution
(b) Describe the soil temperature with regard to agriculture, physics which causes lower crop
growth at a particular area
Answers
(b) High soil temperature causes the crop roots to rot, this leads to insufficient water supply
to plant leaves and hence lower the growth of crop.
Lower soil temperature inactivates soil organisms. Decomposition of organic matter is
lowered and hence the supply of nutrients to crop which in turn lead to lower crop
growth.
TRY YOURSELF
(a) (i) What are auroras?
(ii) Define the homosphere
(b) (i) What are the factors which contribute toward volcanic eruptions?
(ii) What are the effects of volcanic eruptions?
(iii) What are lahars?
Lahars are rapidly flowing mixtures of rock debris and water that originate on the slopes of a
volcano. They are also referred to as volcanic mudflows or debris flow. Volcanic eruptions
may directly trigger one of more lahars by quickly melting snow and on a volcano or eject
water from a crater lake. The form in a variety of at always including through intense rainfall
‘on loose volcano rock deposits and as a consequence of debris of debris avalanches
ENVIRONMENTAL POLLUTION
Pollution is the addition of unwanted materials or pollutants into the environment.

Pollutant is any substance that does not belong in the natural system and disrupts the natural
balance.
Type of Environmental pollution
(a) Air pollution (atmospheric pollution)
(b) Water pollution (hydrosphere pollution)
(c) Land (soil) pollution
(a) Noise pollution
(¢) Thermal pollution
ATMOSPHERIC (AIR) POLLUTION
ag ________poauurigat
This is a form of environmental pollution caused by the release of gaseous materials and dust
particles in the atmosphere. The main pollutants found in the air we breathe include,
particulate matter, lead, ground-level ozone, heavy metals, sulphur dioxide, benzene, carbon
monoxide and nitrogen dioxide
Causes of Air Pollution
Man made causes:
(i) Clearing (deforestation) and burning of vegetation. This releases carbon dioxide in the
atmosphere and dust particles which may be carried by wind on bare land
(ii) Burning of fuels: This releases green house gases in the atmosphere. Fuels are burnt in
cars, power stations and industries.
(iii) Construction activities, like road, building, ete construction, can add dust particles in the
atmosphere.
(iv) Automobile exhausts. Car, trains, etc burns fuels as they move his releases pollutant
gases in the atmosphere.
(v) Smokes from industries also pollute the atmosphere.
(vi) Agriculture activities. The use of pesticide/insecticides pollutes the air.

Spherical cloud
+o of positive charge
-° ae
© a)
+ O+ (—) —)
o +
+
© Electron
‘The number of electrons is such that their negative charge is equal to the positive charge of the
atom, This atom is electrically neutral
This model was called Thomson's plum pudding model because the negatively charge electrons
(the plums) were embedded in a sphere of uniform positive charge (the pudding).
Drawbacks of this Model
1 It could not provide stability to the atom it is because the positive and negative charges are
stationary and will be drawn towards each other, thus destroying the individual negative and
positive charges.
2. Itcould not explain the presence of discrete spectral lines emitted by hydrogen and other
atoms,
RUTHER FORD’S MODEL OF ATOM
‘The salient features of this model are
(i)Every atom consist of a tiny central core, called the nucleus which contains all the atom’s
positive charge and most of its mass (99.9%).
(ii) The radius of the nucleus is of the orde of 10-15m and that of the atom is of the order 10-1°m.
Therefore nucleus occupies only an extremely small portion of the size of atom
(iii) The electrons occupy the space outside the nucleus. Since an atom is electrically
neutral the positive charge on the nucleus is equal to the negative charge on electrons
surrounding thenucleus.
(iv) Electrons are not stationary but revolve around the nucleus in various circular orbits as do
the planets around the sun.
In this way Rutherford provided stability to the atom. It is because the centripetal force required
by the electrons for revolution is provided by the electrostatic force of attraction between
electrons and the nucleus.

m
Ze
ge 4megr?
e= charge on electron
z=total number of protons in the nucleus
m=mass of the electron
=distance of electron from the nucleus
v= linear velocity of the electron
Force of attraction between electron and the nucleus is
Zeje
4negr
F. Ze?
elle
Frege
where Ze is a nuclear charge
The centripetal force required to keep the electron moving in circular path is
mv?
pom
r
Since the atom is stable Fe = Fe

mv? __Ze?
r+ 4mepr?
Ze?
es
Kinetic energy of electron
KE = 5mv?
From equation (1)
Ze?
my? = —_
4neqr
Ze?
2K.E =——
4negr
Ze?
tae
m= Greer
Ze?
ORE = Great?
Ze?
ME = Seer?
Potential energy of electron
(Ze)(-e)
PE cor
—Ze?
PE= Fe
‘Total energy of electron

E=KE+PE
a Ze? + —Ze?
~ Bxeor \4neor,
—Ze?
al
Bneor
The total energy of electron in the orbit is negative hence the electron is bound to the positive nucleus
For hydrogen Atom
For hydrogen atom z= 1. Therefore K. E and P.E OF electron in hydrogen atom are
e
KE = Sregr
-e
PE=——
Gregor
The total energy of electrons hydrogen atom is
E=K.E+P.E
e e
pe.
Bregr  4megr
a
B=
8negr
Limitations of Rutherford’s model of atom
1. According to Maxwell’s theory of electromagnetism a charge that is accelerating radiates
energy as electromagnetic waves
‘The electron moving around the nucleus is under constant accelerating radiates energy as.
electromagnetic waves.

- Due to this continuous loss of energy the electrons in Rutherford’s model were bound to
spiral towards the nucleus and fall into it when all of their rotational energy were radiated
- Hence Rutherford’s atomic model cannot be stable while in actual practice, an atom is stable
This shows that Rutherford’s model is not correct
1. During inward spiraling the electron’s angular frequency continuously increases
- _ Asresult electrons will radiate electromagnetic waves of all frequency i.e. the spectrum of
these waves will be continuous in nature because these are continuous loss of energy.
- But this is contrary to observation experiments shows that an atom emits line spectra and each
line corresponds to a particular frequency or wavelength.
Rutherford’s model failed to account for the stability of the atom. It was also unable to explain the
emission of line spectra.
BOHR’S MODEL OF ATOM
According to Bohr's atomic model, the revolving electrons in the atom do not emit radiations under all
conditions. They do so under certain conditions as expalined by him in his model.
BASIC POSTULATES OF BOHR’S MODEL OF ATOM
1. The electrons revolve around the nucleus of the atom in circular orbits. The centripetal force
required by electrons for revolution is provided by the electrostatic force of attraction between
the electrons and the nucleus.
2. Anelectron can revolve only in those circular orbits in which its angular momentum is an
integral multiple of 2/2
mvr= nh
“On
h= Plank’s constant.
Radius of orbit r
From,
_ ah
mT On

nh
me
2umv
Since n is a whole number only certain value of r is allowed.
‘Thus according to Bohr, an electron can revolve only in certain orbits of definite radii not in all these are
called stable orbits (stationary orbit)
According to this postulate the angular momentum of the electron does not have continuous range i.e.
the angular momentum of the revolving electron is quantized.
While revolving in stable or stationary orbits the electrons do not radiate energy inspite of
their acceleration towards the centre of the orbit.
- For this reason these permitted orbits are called stable or stationary orbits.
e= charge on electron
m= mass of electron
r= radius of the n™ orbit
vn= velocity of electron in the n™ orbit
Z= number of positive charge (protons)
Positive charge on nucleus Ze
RADIUS OF BOHR’S STATIONARY ORBITS
As the centripetal force is provided by the electrostatic force of attraction between the nucleus and
electron.
mv,?__1 (Zee
Ta | 4%" Ta?
1 Ze?
Grey ty 10)
According to Bohr

h
Man = 5 @
Consider equation
mu? = 2-282
in tee” Tm
=——.Ze?
mata = Ge 2e
Take equation (ii) square it
h
Mata = N5— Gi)
nth?
mn he? =
Take equation (iii) “equation (i)
MVaTE nth? 4néotn
MV; = 4m? Ze?
qh? \n?
T= — _
nme?) Z
Itis clear that ’" © n2, radii of the stationary orbits are in ratio 12: 22:32“ clearly the stationary
orbits are not equally spaced.
For hydrogen atom
For hydrogen atom z= 1, so that equation become
Eqh?
t= (5) in?
ame’
zon?
Now =me* = 0.53 x10%m

m=m,
e=electroniccharge T= (0.5310 metres
Forn=1
¥ = (053 x 10-9) x (12)
1m = 053 x 107m
4 =053A
Forn=2
1» = (0.53 x 107°) x (22)
rm = 212 x 107?
ry =2.12A
Forn=3
75 = (0.53 x 10-9) x (3?)
3 =4.77x 107m
Thus the radii of the first, second and third stationary orbits of hydrogen atom are 0.53 A, 2.12 A
and 4.77A respectively.
2. VELOCITY OF ELECTRON IN BOHR’S STATIONARY ORBIT
From equation below, we have
mvgt,= 2
te On
—
» = Damn,
Putting the value of "® into that equation

__nh_ nmZe?
Y= Damn, “2eonth:
_ Ze?
n= Deonh
wt
Um =
itis clearthat "7 in other words, electrons move at a lower speed in higher orbits and vice versa.
For hydrogen atom
2-1
Then
n= Jeon
3. FREQUENCY OF ELECTRON IN STATIONARY ORBIT
The number of revolution completed per second by the electron in a stationary orbit around the nucleus
Velocity of electron in the 8" orbit
Tn =TaWn
Uy = Umrah
=<
In = 2a,
Ze? 1
In= 2eonh 2nr_
__ Ze?
n= 4negnhr,
For hydrogen atom

z-1
Then,
Ze?
In = Sregnr,
eqn,
Frequency of electron in the first orbit of hydrogen atom is n=1, 1:=0.53x10™ ,,
e@
h= ake
egnhr,,
get —#
"Amey 1XhXH
(1.6 x 10-19)?
=9x 10° x 6 x10?
fi = 9% 10" x TE 62x 10° x O53 x10
fy = 657 x 10 r.p.s
Electron in first orbit of hydrogen atom will have a frequency of 6.57x 10"Srevolutions per second.
4. TOTAL ENERGY OF ELECTRON IN STATIONARY ORBIT
The total energy E. of the electron in the n® orbit is the sum of kinetic and potential energy in the n®
orbit.
- ‘The K.E of electron in the n" orbit is
K.E,= 1, mv,?
xe.<Z#
vn Brom
The potential energy of electron in the n™ orbit is

pe, = 2 OC)
Breo tm
-ze?
PEn= Faon
Total energy of electron in the n" orbit is
E,=PE,+KE,
Ze? Ze?
By = —- +
4%, BME OTm
g.- met (2
2 Beg2h? \n?
But
me*
ae -10
Begepa = 217 10
n
Eq = 21.7 x 1079
w
21.7 10719 Z?
En =—T6x10-% “ni
13.6
E,=-—> eV
n
Thus as n increases i.e. electron moves to higher orbit, the total energy of the electron increases i.e.
total energy becomes less negative.
For hydrogen atom 2=1
-13.6
E,=—a ev

Thus the total energy of electron in a stationary orbit is negative which means that the electron is bound
to the nucleus and it is not free to leave the atom.
We can find the total energy of electron in the various orbits of hydrogen atoms as under.
13.6
Ey=— = -13.6ev First orbitn=1
‘aoe
13.6
Ey =—“Fy = —13.6eV Second orbit n=2
13.6
Ey =~ = —13.6eV third orbit n-3
The total energy of electron increases i.e. becomes less negative as the electron goes to higher orbits
When nee E, =0 and the electron becomes free
Ground state/ normal state
This is the state of atom when the entire electrons in it occupies their lowest energy levels as required
by their n and / values.
The energy of an atom is least i.e. largest negative value when n=1 i.e. when electron revolves in the
first orbit.
The energy of hydrogen atom in the ground state is 13.6eV.
Excited state
This is the state of an atom when electrons in an atom occupy energy levels higher than those permitted
by the values of n and / values.
At room temperature most of the hydrogen atoms are in the ground state
If hydrogen atom absorbs energy i.e. due to rise in temperature it may be promoted to one of the higher
orbits (i.e. n=2, 3, 4.....)
The atom is said to be in the excited state.

WAVE LENGTH OF EMITTED RADIATION.
When an electron jumps from a higher orbit (n2) to the lower orbit (m1) the energy difference
between the two orbits is released because the energy of electron in the higher orbit is more than
in the lower orbit. Consider two orbits having principle quantum numbers n2 and ni where n2>ni
Then energy of electron in the two orbits is given by
mZze*
Ens =~ Bente
Beq?n* 2h
mZe*
Ens = ~ Bett ht
Beg nh
As the electron jumps from orbit nz to n, energy is released in the form of electromagnetic radiation.
Ena — Ena = hf
where
f= frequency of the emitted radiation
—mZ7e* —mZ7e*
Ind he| ~ \e,2n2 he | ~ OF
Beq2n?,h?|  [Beq?n? he
—mZ7e* — —mZ*e* af
Bey?nyh?  Bey?n, 7h
f mZ7e* [2 1 |
Beh In?, n?,
The wavelength of the emitted radiation is given by
cof

af
Ase
1_ mZe* [ 1 1 |
X” Bep?hFcln?, 22.
This equation gives the wavelength of emitted radiation.
Now,
i
3 -¥ wave number
v= mZe* [ 1 1 |
~ Begth8cln2, 22.
Wave number
These are the number of waves in a unit length,
For hydrogen atom
For hydrogen atom z=1
1_ mZe* [ 1 1 |
X~ Begthecln®, n?,
This gives the mathematical formula for the wavelength of radiation emitted by hydrogen atom when
electron jumps from outer orbit to inner orbit.
1 11
tended
pried Ca os
where
Ruis Rydberg constant. The value of Ri can be calculated as the value of e, m, h and c are known

mzZe*
Ry =
Beq?ch
_ (9.1 x 10754) x (1.6 x 10-19)*
8x (6.854 x 107)? x (3 x 10") x (6.62 x 10-*)>
Ry = 1.097m*
HOW TO CALCULATE THE RYDBERG CONSTANT USING CALCULATOR
From
Ry =e
Begtch?
m= 9.1.x 10734 kg
m=16x10-%¢
£9 = 8.854 x 10-2F m=
¢=3x 10®ms"?
h = 662x 104s
y= (9.1.x 10-5) x (1.6 x 10-*9)*
4 Bx (8.854 x 10-22)? x (3 x 10°) x (6.62 x 10-74)?
pe 9.1 x (1.6)* 10-31 x (10-79)*
#8 x (8.854) x 3 x (6.627)* (10-22)? x 108 x (10-4)>
Ry = 1.093 x 10-* x =
_ 10-™* x 108 x 10-77
10-31 x 10-76
Ry = 1.093 x 10-* x a
10m,
Ry = 1.093 x 10-4 x zal
Ry = 1.093 x 10-* x 10%
Ry = 1.093 x 10’m=*

Now saturation current is of higher value as shown in figure below
/
L
L
y
—Ve
_——____* a
Retarding potential Anode PotentialV
This is expected because the greater the intensity of incident radiation the greater is the photo electric
current
Stopping potential
‘Stopping potential is the minimum retarding potential at which photoelectric current becomes zero
Or is the potential difference when no electrons are able to reach the anode.
Itis also known as stopping voltage or cut-off potential.
It is denoted by Vp or V,
Stopping potential is a measure of the maximum kinetic energy of the photo electrons.
Since potential difference V
y — Working done(W)
Charge(Q)
For stopping Voltage Vo

W = QVo
For an electron
W =QV, =Q =e
w=ev,
ne ee
zmv? = eV,
‘At Vo, even the photo electrons having maximum kinetic energy KE mu (ie. fastest photo electrons)
cannot reach the anode A.
Therefore, the stopping potential Vo is a measure of the maximum kinetic energy K.E may of the photo
electrons.
‘eVp is the work done by the retarding force to stop the photo electron with maximum kinetic
energy and is therefore equal to K.Enax
At Vo, it is found that the photoelectric current cannot be obtained even if we increase the intensity of
radiation. It is same for different intensities |,, lp and | of incident radiation.
3. Effect of frequency of incident radiation on stopping potential.
‘We now study the relation between the frequency f of the incident radiation and the stopping potential
Vo.
For this purpose, we take the radiations of different frequencies but of the same intensity.
For one frequency say f,, of the incident radiation, we plot the graph between photoelectric current and
potential of anode A with respect to cathode C at a constant intensity of incident radiation
Keeping the intensity of incident radiation the same, we repeat the experiment for frequency f, of the
incident radiation.

The following is the resulting graph
Yo
°
5 t
,
/
/
Observations from the graphs
1. The value of stopping potential is different from radiation of different frequencies
2. The value of stopping potential is move in low higher frequency. This implies that the value of
maximum kinetic energy depend on the frequency of incident radiation.
The greater the frequency of incident radiation, the greater is the kinetic energy of emitted photo
electrons.
3. ‘The value of saturation current depends on the intensity of incident radiation but is
independent of the frequency of incident radiation
if we draw a graph between the frequency of incident radiation (f) and the stopping potential (Vo) at
constant intensity of radiation, it will be a straight line AB as shown in figure below

Ym
B
A
° 7 fo f
/
/
/
From the graph
At fo, stopping potential Vo= 0. It means that at fo, the photo electric current is just zero (i.e.
photo electrons and emitted with zero velocity) and there is no retarding potential.
Vo=0
‘This limiting frequency fo is called threshold frequency for the cathode material.
It is a minimum frequency of the incident radiation which is just sufficient to eject photo
electrons (i.e. with zero velocity) from the surface of a metal.
Stopping potential is directly proportional to the frequency of incident radiation.
Voaf
The greater the frequency of incident radiation, the higher is the stopping potential and vice versa.
Experiments show that photo electric emission is an instantaneous process.

As soon as light of suitable frequency (equal to or greater than f.) is incident on the surface of the metal,
photo electrons are emitted from the metal surface.
The time delay is less than 10° second
LAWS OF PHOTOELECTRIC EMISSION
‘The above experimental study of photoelectric effect leads to the following laws of photoelectric
emission.
i Fora given metal, there exists a certain minimum frequency of incident radiation below
which no emission of photo-electrons takes place. This cut off frequency is called threshold
frequency fo.
ii For a given metal and frequency of incident radiation (>fo) the photo electric current is
directly proportional to the intensity of incident radiation.
iii, Above the fo, the maximum kinetic energy of the emitted photo-electron is independent of the
intensity of the incident radiation but depends only upon the frequency of the incident radiation.
iv. The photoelectric emission is an instantaneous process.
The above laws of photoelectric emission cannot be explained on the basis of light or radiation. This
gave death blow to the wave theory of light or radiation.
FAILURE OF WAVE THEORY/CLASSICAL PHYSICS TO EXPLAIN PHOTOELECTRIC EFFECT
The wave theory of radiation failed to explain photoelectric effect. This will become clear from the
following discussion.
1. According to wave theory of radiation the greater the intensity of the wave the
greater the energy of the wave.
‘So wave theory does explain why the number of emitted photoelectrons increase as the intensity of
radiation is increased.
But it fails to explain the experimentally observed fact that the velocity or kinetic energy of the
emitted photoelectron is independent of the intensity of incident radiation.
‘According to the wave theory, an increasing in the intensity of radiation should increase the kinetic
energy of the emitted photoelectrons but it is contrary to the experimentally observed fact.
Il. According to wave theory, intensity of radiation is independent of it is frequency it depends upon
the amplitude of electric field vector.

Therefore, an increase in the frequency of radiation should not affect the velocity or kinetic energy
of the emitted electrons.
But it is observed experimentally that if the frequency of the incident radiation is increased, the
kinetic energy of the emitted electrons also increases.
Ill. According to the wave theory, electrons should always be emitted from a metal by radiation of any
frequency if the incident been is strong enough
However experiments show that no matter how great is the intensity of the incident radiation; no
electrons are emitted from the metallic surface if the frequency of radiation is less than a particular
value i.e. threshold frequency.
IV. According to the wave theory the energy of radiation is spread continuously over the wave fronts of
the radiation.
Therefore, a single electron in the metal will intercept only a small fraction of the wave’s energy.
Consequently considerable time should be needed for an electron to absorb enough energy from the
wave to escape the metal surface.
But experiment show that electron are emitted as soon as radiation of suitable frequency falls on
the metallic surface.
In other words photoelectric emission is instantaneous there is no delay.
The above discussion is a convincing proof of the inability of the wave theory to explain the
photoelectric effect.
EINSTEIN QUANTUM THEORY OF LIGHT
Einstein explained photoelectric effect on the basis of Planck’s quantum theory.
According to Einstein light radiation consist of tiny packets of energy called quanta.
Photon
Photon is the single quantum of light radiation which travels with the speed of light.
The energy of a photon is given by E.
E=hf

where

7. In Bohr’s model of hydrogen atom the electron circulates around nucleus ona path of
radius 0.51A at a frequency of 6.8x19°~ is rev/second calculate the magnetic field
induction at the centre of the orbit.
Solution
‘The circulating electron is equivalent to circular current loop carrying current I given by
dQ_e
fu 2 =~
|
l=ef
“19 15
1216 * 107 x 6.8 x 10
“3
l= 11% 10 A
Magnetic field at the centre due to this current is
Hol
B, =fe
contre =>
_ Gm x 1077) - (4.1 x 1079)
2 x (0.51 x 10-*°)
Beentre = 44T
8. A long straight wire carries a current of SOA. An electron moving at 10"ms is Sem
from the wire

Ve
TI
Sem —_
Find the Magnetic field acting on the electron velocity is directed
(i) Towards the wire
(ii) Parallel to the wire
(iii) Perpendicular to the directions defined by land ii
Solution
‘The magnetic field produced by current carrying long wire at a distance r
= Hol
8 2nr
pa ttX10 50
"Qn 5x10?
B=2x10“T
The field is directed downward perpendicular to the plane of the paper
(1) The velocity V1 is towards the wire. The angle between Vi and B is 90°
force on electron
F=Bovein®

F = 2x 10% x 1.6x10°x10"x Sin 90°
F=3.2x101°N
(ii) When the electron is moving is moving parallel to the wire
,angle between V2 and B is again 90ES Therefore, force is again
3.2x10"N
(iii) When the electron is moving perpendicular to the directions defined
by (i) and (ii) the angle between V and B is O
F=-O
9. A solenoid has a length of 1.23 mand inner diameter 4cm it has five layers
of windings of 850 turns each and carries a current of 5.57A. what is the magnitude of
the magnetic field at the centre of the solenoid
Solution
The magnitude of the magnetic field at the centre of a solenoid is given by
B=yponl
But
N _ 5xe50
n=7 =a = 34553
B = 41x 10” x 3455.3 x 5.557
B=24.2x10°T
10. A to void has a core ( non - ferromagnetic) of inner radius 20cm and over radius 25em
around which 1500 tums ofa wire are wound. If current in the wire is 2A
Calculate the magnetic field
(i) Inside the to void

(ii) Outside the to void
Solution
20+25
Mean radius = ———
r = 22.Sem = 22.5 x 107m
L=2ar
mean length |= 2nr
= 2m x 22.5x 107
=1413m
(i) The magnitude of the magnetic field inside the toroid is given by
B=jonl
B= 41x 1077 x Sx
Perciery
B=0.003T
(ii) The magnetic field outside the toroid is Zero. It is all inside the toroid.
11. A solenoid 1.5m long and dem in diameter possess 10 turns/cm. A current of 5A is flowing
through it. Calculate the magnetic induction
()) Inside and
(ii) At one end on the axis of the solenoid
Solution
N
p= 1 = 10turns/cm _ 10° turns/cm
(i) Inside the solenoid , the magnetic induction is given by

pateal
pa4tX 1077 x 10° x 5
patX10°T
(ii) Atthe end of the solenoid the magnetic induction is given by
patent _ 2m x 107%
2 2
B=1X10°T
12. (a) How will the magnetic field intensity at the centre of a circular loop carrying current
change, if the current through the coil is doubled and the radius of the coil is halved?
(b) A long wire first bent in to a circular coil of one tum and then into a circular
coil of smaller radius having n turns, if the same current passes in both the cases,
find the ratio of magnetic fields produced at the centers in the two cases.
(c) A and B are two concentric coils of centre O and carry currents Ia and In_as shown in figure
I
C-) ,

If the ratio of their radii is 1:2 and ratio of flux densities at O due to A and B is 1:3, find the
tp
value of 4
Solution
(a) Magnetic field at the centre of circular coil
Honl
p = Hort
2r
I
Ba-
;
KI
pa Kt
+
Br
ak’
I
Byr, _ Bat
A L
B,_h mh
Ro An
k=l
ner
I=21
:
net
= 2
By
By
B
By = 4B,
(b)Suppose r is the radius of one turn coil and the r’is the radius of n-turn coil. Then
Nn* 2nr* = 2nr
rss
N

First ease Second case
so! we be
B= B,="T
B,_ wNI_ 2r
By 2("/y) Hol
By
Br _ ye
B,
Solution
C. Magnetic field at the centre of circular coil
Hol
a
Ban
=
Bry
=
Bara _ Bate
Llp
ta _ Bata
Ty” Bate
boil
Iz 32
Ip 6
13. A helium nucleus makes a full rotation in a circle of radius 0.8m in two seconds. Find the
value of magnetic field at the centre of the circle.
Solution
The charge on helium nucleus

a: +
a=*?* 1.6 x10%
Current produced | =
1=2x16x10"
1=1.6 x10%A
Magnetic field at the centre of the circle orbit of the helium is,
= bel
B or
‘2m x10" x16x 10-19
pe a08
B= 1.256 x10"
14, A soft Iron ring has a mean diameter of 0.20m and an area of cross section 5x10“m? it is
uniformly wound with 2000turns carrying a current of 2A and the magnetic flux in the iron is 8x
10°Wb. What is the relative permeability of iron?
Solution
Length of ring
122"
1=27"x0.10m
Number of turns per unit length n
N 2000
nel 22mx0.1
If Mis the absolute permeability of iron, then magnetic flux density of iron ring is
p=Monl

p- X pose * 2
B= 2x2 x 10*Wb/
cd m?
Magnetic flux ® = BA
H ‘ +
Ex2x10*x5 x10
™
10,
= 20H
10,
8x10 =—*
Magnetic flux ® = BA
w= 82x 10-*H/M
Relative permeability of Iron yt.
_# _8xx107*
Mr ig 4X 1077
bi, = 2000
15. Two flat circular coils are made of two identical wires each of length 20cm
one coil has number of turns 4 and the other 2. Ifthe some current flows though the wire
in which will magnetic field at the centre will be greater?
Solution
For the first coil
Hor
p, = tom
1 ry

For second coil
HoNI
a, = Holl!
2 2
By mh
By mh
Length of the wire | = ny; 2nr, = n,27y
mh =m
my = nm
mn
By om my
By mz Mm
By c y
B,  \n,
z-(5)
B, \2
Ply
B,
= Therefore, magnetic field will be greater in coil with 4 turns
16. A plat circular coil of 120 turns has a radius of 18cm and carries currents of 3A. What is the
magnitude of magnetic field at a point on the axis of the coil at a distance from the centre equal
to the radius of the coil?
Solution
Number of turns n = 120
Radius of the coil r= 0.18 m
Axial distance x = 0.18m_
Current in coil 1= 3A.

Nir?
B= ——
2(r? + x2)"
B= (44x 10-7 x 120 x3 x0.18?
2(0.18? * 0.182) 3?
Be 44x 047
17. A current of 5A is flowing upward in a long vertical wire. This wire is placed in a uniform
northward magnetic field of 0.021. How much force and in which direction will this field exert
on 0.06 length of the wire?
Solution
F=BiLsing
8=0.02T
1=5A
L=0.06
e =90°
Fe 002x3 x0esing00
P0000
By Fleming's Left hand rules the force is directed towards West
18. A straight wire of mass 200g and length 1.5m carries a current of 2A. It is suspend
in mind air by a uniform horizontal magnetic field B. What is the magnitude of the
magnetic field?
solution
M=200 X 10° kg
1=2A
1= 1.5m
B=?
F=BIL

Since magnetic field strength B is the magnetic field lines passing per unit area of
the material, it is a measure of magnetic permeability of the material.
Suppose magnetic flux density in air or vacuum ise, If vacuum/air is replaced by a material,
suppose the magnetic flux density in the material becomes B
Then ratio B/e called the relative permeability“. of the material
(i) Relative permeability “> .
Isthe ratio of magnetic flux density Bin that material to
the magnetic flux density 3 that would be if the material were replaced
by vacuum/ air.
Bo
Ho= B= 1
Clearly isa pure number and its value per vacuum/air is 1
Bo
==1
By By
Relative permeability of a material may also be defined as the ratio of absolute
permeability “ of the material to absolute permeability “2 of vacuum/air.
#
wo=h=1
on Ho
(ii) Magnetizing force/ Magnetic intensity H
Is the number of ampere - turns flowing per unit length of the toroid.
‘The SI Unit of magnetizing force His Ampere —turns per meter (AT/m)
Consider a toroid with n turns per unit length carrying a current I. if
the absolute permeability of toroid material is M, then magnetic flux density B in
the material is

B=ynl
B= Hou,nl
The quantity ™ is called magnetizing force or magnetic intensity
Zz
Therefore, the ratio” ina material lis from
B= pH. B= Holly § Holy H
Bo = oll
Thus if the some magnetizing force is applied to two identical air cored and iron cored toroid, then
magnetic flux density produced inside the toroid is
Bo = Mob H
(iii) Intensity of magnetization (1) is the magnetic moment developed per unit volume of
the material.
When a magnetic material is subjected toa magnetizing force ,
the material is magnetized
Intensity of magnetization is the measure of the extent to which the material is a
magnetized and depends upon the nature of the material
> M
i=—
v
where:
M = magnetic moment developed in the material
V= volume of the material

If mis the pole strength developed,
@ is the area of X — section of the material and 21 is the magnetic length. Then
mx 21
1=———
ax 21
m
1=7
a
Hence Intensity of magnetization of a material may be defined as the pole strength developed
per unit area of cross-section of the material.
‘Thus the Slunit of lis Am! which is the same as the SI unit of H
Magnetic susceptibility ** is the ratio of intensity of magnetic on I developed in
the material to the applied magnetizing force H
, 1
Itis represented by Xm =F
The magnetic susceptibility of a material indicates how easily the material can be magnetized.
‘The unit of | is the same as that of H1 so that ~™ is a number
Since I is magnetic moment per volume *™ is also called volume susceptibility
of the material .
Consider a current carrying toroid having core material of relative permeability “
The total magnetic flux density B inthe material is given by

B= Bot Bm
Where
5. — magnetic flux density due to current in the coils.
Fm = magnetic flux density due to the material (Magnetization of the material)
Bo = Modi)
Bm = Hol . sali)
Here | is the intensity of magnetization induced in the toroid material
B=30 45x
B= ul + ul
B=y(H+1)
Now,
x= I
"OH
I=X,,H
B=yo(H+ X,,H) = uoH(1+ X,,)
B=pH = pou
* HoH = woH(1+ X,,)
Or
Equation (iii) give the relation between relative permeability (4, ) and magnetic susceptibility (Xm).
CLASSIFICATION OF MAGNETIC MATERIALS:
All materials or substances are affected by the external magnetic field. Some attain weak magnetic
properties and acquire strong magnetic properties.

Path described by particle, Q
x ox x x x xi x
x oe KOK x _
x x x Keo x x
a a a a a a ae a
RMR KK OK KR
KOK KK KK KX
X = Magnetic fields into the page
The velocity V can be resolved into two rectangular components
i) vi= V°°S9 Acting in the direction of the field
ii) V2= Vsin® acting perpendicular to the direction
The perpendicular component V2 moves the charged particle in
a circular party while the horizontal component “imoves it in the direction
ofthe magnetic field
In other words, the charged particle corers circular path as well as linear path.
Consequently the charged particle will follow a helical path,
‘The charged particle rotates in a circle at speed V2 while moving in the direction of
the field with a speed Vi
PARAMETERS OF MOTION
The perpendicular component of velocity V2 determines the parameters
of the circular motions while the horizontal component of velocity Vi decides the pitch
of helix
i) Radius of path

From
Fy = Ma
Mv,?
BQV, = ——
.
sq = MB
=o
MY,
roth
BQ
= MVsin@
r= Muang
(i) ~~ T,fandw
Since time period (1), frequency (f) and Angular frequency(«) of a
charged particle moving in a uniform magnetic field
are independent of speed V and radius (r) of the path, these values
remain the same
2nM
T=
BQ
f= 2nM
w= ™M
(iii) Pitch of helix (d)
Itis the linear distance covered by charged particle when it completes one

circular revolution Or It is the linear distance covered by charged particle during
time T
a=4r
a= vos Oy
cos@ x 2
d=V BQ
2am
d= 0 (Vcos6]
The following points may be noted about the behavior of charged particle in a Uniform magnetic field
i If acharged particle is at rest V=0 in a magnetic field, it experiences
no force
From
Fra pgysind
Frau yQX 0X sind
Fra)
(i) If a moving charged particle enters a uniform magnetic field at right angles
to the field it describes a circular path
(iii) Ifa moving charged particle enters a uniform magnetic field. Making
an angle to the direction of the field it describes a helical path

(iv) A moving charged particle in a magnetic field experience maximum
force when angle between V and Bis 90°
(v) Since magnetic force does not change the speed of a charged particle it means
that K.E of the charged particle remains constant in the magnetic field.
(vi) Since magnetic force (Fm) is perpendicular to V, it does not work. Therefore
work done by the magnetic force on the charged particle is zero
WORKED EXAMPLE
1. Anelectron and a proton moving in a circular path at 3x10° Ms" in a uniform
magnetic field of magnitude 2 x 10“T. Find the radius of the path
Solution
v2
Bey ==
r
abe
"= 'Be
9 x 1073 x (3 x 10°)
(2x 10-4) x (1.6 x 10749)
r =85x 107m
2. Anelectron anda proton moving with the same speed enter the same magnetic field region
at right angles to the direct of the field. For which of the two particles will the radius of circular path
be smaller?
Solution
From
_MV
"Be

rom
Since the mass of electron is less than that of the proton the radius of the circular path of
electron will be smaller.
3. (a) What will be the path of a charged particle moving along the direction of a
uniform magnetic field?
(b) A moving charged particle enters a magnetic
Solution
(a) When a charged particle moves along the direction of a uniform magnetic
field ,it experiences no force © 0° therefore the charged particle will
more along its original straight path
(b) Helical path since the velocity of the charged particle can
be resolved into two rectangular components one along the field,
and the other perpendicular to the field, The velocity component
perpendicular to the field causes the charged particle to more in
acircular path while the velocity component along the field cause
it to more itin the direction of the field, The combination
of these two motions course the charged particle to. move ina helical
path.
4. Abecome of a-particles and of proton of the same velocity V, entries
a uniform magnetic field at right angles to the field lines. The particle
describes circular paths. What is the ratio of radii of the two paths?
Solution
Radius of a-particle path "

EAd
en)
Of proton path "2
av 4
rr od
Take equation (i) *. equation (ii)
nm _ MW BQ,
7 BQ, MV
TG My
% QQ M2
1x4
™% 2
-
haa
n
Therefore, radius of x - particle path is twice that of proton’s path
Note
a-particle
Charge = 2e
Mass = mass of helium nucleus
5. A proton with charge ~ mass ratio of 10° CKg “is moving in a circular orbit
in a uniform magnetic field of 0.ST. calculate the frequency of revolution

d=4.37x10°M
7. (a) A particle of charge 0leef moves in acircular path of radius r in
a uniform is P = BQr
(b) An electron emitted by a heated cathode and accelerated through a potential
difference of 2.0KV enters a region of magnetic field of 0.15 determine the trajectory of
the electron if the field
(i) _ is traverse to its initial velocity
(ii) makes an angle of 30 with the initial velocity
¢)  Aproton a deuteron and an a ~ particle whose kinetic energies are same enter
perpendicularly to a uniform magnetic field. Compare the radii of their circular paths
(a) Solution
‘The magnetic force Fm provides the necessary centripetal force Fe
Fm = Fe
Bq=%
momentum
BQ=
Momentum P = BQr
(b) Solution
When an electron (e) is accelerated through a p.d of V, it acquires energy eV. IF Vis
the velocity gained by electron, then

IM(v?)
y = Me)
° 2
w
-\e*#)
Vv = J(@x(46 x 10* — 19) x (2x10"3))
‘v=9x107"
@ Force on the electron due to transverse field is
Fm = BQr ores =90°
Since magnitude of Fm is constant and Fm is perpendicular to both V and B
the electron will move in a circle of radius r. The necessary centripetal force is
provided by Fm
Mv?
Paei
.
Mv?
we
BQv
__ @x10™)x
” * G0-)x 6x10)
re
10°M
(ii) When electrons enters the magnetic field making an angle 0"ce/ = 30° with the field
M (rSin 8°)?
BQBVSin® = aoe
MVSin@
pel)
BQ

__9x107%4x Sin 30°
r * 0.A5x16x 10-9
r =05x10°M
(©) Solution
Let 1, 2 and 3 be the suffix force proton, deuteron and a - particle respectively
K E,=KE=KE;
1 MiVi = | M2V2= | MaVs
2 2 2
IfMI then My =2M and Ms = 4M
MV" = 2MV2 = 4MV?s
viz VY¥2=2y,
V2= VI
v2
Also
%
Yaz
Radius of the path r
r= Mv
8a
I=, Then @2=Q anda3=20
n=MM = MW
8Q, Ba

p= M.V2= 2M. V;
sa, sa Y?
nae
BQ3  2BQ “2
hi = My: uy, : MY,
BQ BQ BQ
ninine 12 4
NUMERICAL PROBLEMS
1. What is the radius of the path of an electron (mass 9 x 107!)Kg and charge( 1.6 x 10°C)
moving at a speed of 3 x10'm/s in a magnitude field of 6 x 10“T perpendicular to it? What its
frequency? Calculate its energy in KeV(1Ev=1.6x10"'J)
= 0.28m
& 1.7x107 Hz
E=2.53KeV
2. Anelectron after being accelerated through a p.d of 100V enters a uniform magnetic field of
0.0041 perpendicular to its direction of motion. Find the radius of the path described by the
electron
1=8.410°m
3. Ana-particle is describing a circle of radius of 0.45m in a field of magnetic Induction
1,2Wb/m”, Find its speed, frequency of rotation and kinetic energy. What potential difference
will be required which will accelerate the particle so as to this much energy to it? The mass of a-

N=50
I=1A
e=0
t= 1.5 X (7.85 X10) X50X1
t=0.589NM
(b) Since torque on the loop is independent of its shape provide area (A)
Temains the same the magnitude of the torque will remain unaltered.
4 A circular coil of 20tums and radius 10 cm is placed in a uniform
magnetic field of 0.21 normal to the coil. If current in the coil is 5A find.
(i) Total torque on the coil
(ii) Total force on the coil
(“Average force on each electron in the coil due to the magnetic field. ‘The coil is
made of copper wire of cross-sectional area 10m? and force of electron density
inthe wire is 10°%m?>
Solution
@ The toque on the coil is given by
t= BANIcos@
Since
@=0
(ii) The net force ona planar current loop ina uniform magnetic field is always zero
(iii) Magnetic force on each electron
F=BeVa

F=Be, =04
BI
pa2t
nA
0.25%5
F= 102°x10-5
F=10™N
MOVING COIL METERS
A galvanometer deflects or measures small amount of current passing through it and it gives the
direction to which that current is flowing.
In these instruments a rectangular of fine insulated copper wire is suspended in an strong
magnetic field as shown in the figure below. The field is set up between soft iron poler pierces
Ns attacked to a powerful permanent magnet.
a b
Cis
s
°
re RH
”,
» a
® by sprin
ena e
1a
Coil gO
[pJewelled Bearing
(i) Millimeter

The magnet field is radial to the core and pole pieces over the region which the coil can swing. In
this case the deflected coil always comes to rest with the plane parallel to the field in which it
then situated.
The moving coil galvanometer ha s has hair spring and jewel bearings. The coil is around in the
rigid but light aluminium frame which also comes to carries a pivot. The current is led in and out
of the springs.
Aluminum pointer P shows the deflection of coil, it is balanced by counter weight Q.
‘THEORY OF MOVING COIL GALVANOMETERS
The rectangular coil is situated in the radia field B when the current is passed into it the coil
rotate to an angle Q which depend on the length of the spring.
No matter where the coil comes to rest, the field B in which it is situated always along the plane
of the coil because the field is radial.
It is shown that the torque T of the coil is given by T= B AN L ie.
In equilibrium the deflecting torque T is equal to the opposing torque due to the elastic forces in
the springs. The opposing torque
‘Torque = CQ
Where C is the constant of the spring
Example 1

Total resistance for the new scale is
=15002000 = 30 10° 4,,|
Extra resistance required is
(30-3) 10° =27 10° a,
MAGNITUDES FOR CURRENT CARRYING CONDUCTORS,
Laws of Biot and Savant.
It state that the flux density dB= at point P due to a small element dl ofa conductor carrying
Where r is the distance from the point P to the element is the angle formed it to P.
Pp dB
ao 1
|
/
B: FLUX DENSITY = INDUCTION OR MAGNETIC INDUCTION

MAGNETIC FIELD
Equation (1) can be written as
dB=KIdlsinx
Where K is the constant of proportionality and it depends on the medium in which the
conductor is situated. Also K =
=
aHoldlsine ai
— @
The formula is
Note that fp = 47 X 10-7 Henry m? B =a
r
B AT THE CENTRE OF A NARROW CIRCULAR COIL

Suppose the coil is in air has a radius of r carries a steady current I and it is considered to consist
of current clement of length dl. Each element is at the distance r from the centre 0 and it is at the
right angles the line joining it to i.e. * 90°
dl
al at
r
[~\ ad r
al e
al o%®
2 1
Ifthe coil have N tums then the length of the wire = 2711N
— _ Holdisinx
From the equation Gi) dB = "#\°5*
Then the magnetic field at the centre is given by
= Bo pPN ap = Bot py ZN
_ iat at = (Ho
= told =SN
; B a B cy
Example
A coil of wire with 15 turns of radius 6.0cm, has a current of 3.5A flowing through it. What is
the magnetic flux density at the center of the coil?
Solution
B=38 = a = 175k x 10° Tesla
2 = 5.495 x 104 Tesla

Example
What is the magnitude of the flux density produced the center of a coil of radius Scm carrying
current of 4A in air.
Solution Formula,
Given
= = BNL amtoxexd
I=4A B on 2x5x107? 1.6m
r=5x 10%m.
N=I B=5.024 x 105 Tesla
Example
A circular coil of radius 6cm consisting of 5 turns carries a current supplied from 2v accumulator
of negligible internal resistance. If the coil has a total resistance of 24,,|. Calculate the magnetic
field induced at the centre
Solution Formula. y
=Ya2-1- wt _ dato as
re 2 Axfeso?
N=S
B= 2x 105 =3.335 x 10% Tesla
r= 6x 107m = 3.335 x 10 Tesla
uw. = 4x 107
B DUE TO A LONG STRAIGHT WIRE AT A DISTANCE d SIDE THE WIRE
Consider a very long wire YN carrying a current I. Take P to be a point outside the wire but also
this point is considered to be very near to this wire.

'
N a 8 4B
da
x
di
ypa #7
A
THE HALL EFFECT
Is the phenomenon where by e.m. for voltage is set up transversely or across a current carrying
conductor when a perpendicular magnetic field is applied
Consider a piece of conducting material in a magnetic field of flux density B
Suppose that the field is directed (perpendicularly) into the paper and that there is a current
flowing from right to left.If the material is a metal the current is carried by electrons moving
from left to right
Consider the situation of one of these electrons and suppose that it has a velocity V
The electron feels a force F which by Fleming’s left hand rule, is directed downwards .
Thus in addition to the electron flow from left to right electrons are urged away from face
Y and towards face X. Anegative charged builds up on X, leaving a positive charged on Y
so that a potential difference is established between X and Y. The buildup of charge
continues until the potential difference becomes so large that it prevents any further increase
. This maximum, potential difference is called the Hall voltage
Hall voltage

Is the potential difference created across a current carrying metal strip when the strip
is placed in a magnetic field perpendicular to the current flow in the strip.
Actually, the magnetic field does not have to be totally perpendicular to the strip
the magnetic field only needs to have a component that is perpendicular
The flow ceases when the ¢ .m freaches a particular Vi called Hall voltage
MAGNITUDE HALL VOLTAGE
Suppose Vu is the magnitude of the Hall voltage and d is the width of the slab
(the separation of x and y). Then the Electric field strength E set
up across the slab is numerical equal to the potential gradient.
Ya
E=4
let Fy be the force exerted on an electron by the P.d between X and Y.
‘Therefore when the buildup of charged on X and Y has ceased
F=Fy
BeV=ek
BV=E
“a
By= ¢
Vi BV oooccccccseneeee (i)
Where
E = The strength of the uniform electric field between X and Y due to the Hall voltage
Vu = Hall voltage
d= The separation of X and Y

w=).
=e
MUTUAL INDUCTANCE (M)
the °™S may be induced by in one circuit by changing current in another. This phenomenonis often called
mutual induction and the pairs of circuits which shows it are said to have mutual inductance
The mutual inductance m between the two circuits is defined by the following equation
emf, , , nai
Induced in B by changing = M (rate of change of current in A) ie.
=i
E, = M74
The unit of mutual inductance is Henry the same as that of self-inductance
hk M
a Ec
x 8
MUTUAL INDUCTION
since 2 =the rate of change in flux in 8 then
=i mat
Ea= at at
M= dop _ Fluxchange inB
dl, current change ind
QUANTITY OF ELECTRICITY INDUCED

Consider a close circuit of total resistance R Ohms which has a total flux linkag@ with magnetic field B. if
the flux linkage starts to change
cc
E=%
Induced °F,” ~ 4 but current
122218
R Rat
—
: |
Rg ——————qxqg8F
Flux linkage will not change at a steady rate and a current will not be constant. But throughout it changes.
Its charge is being carried round the circuit. If a time t seconds is taken to reach a new constant value the
charge carried round the circuit in that time is
Q = filde. from equation i) we have
aise? gp = 15%
Om ELE at = 5h, do
Where , is the number of linkage at t=o and %., the number of linkage time t
(_-Op) _ (®p-04)
Thus * al
Q _ Change of flux linkage
R

AC THEORY
When a battery is connected to a circuit the current flows steadily in one direction, this is called a Direct
current (d.c).
The use of Direct currents is limited to a few applications e.g charging of batteries, electroplating etc.
Most of electrical energy is generated and used in the form of alternating current due to many reasons
including,
i) Alternating voltages can be changed in value very easily by means of transformers.
ii) A.c motors are simpler in construction and cheaper than d.c motors.
ALTERNATING VOLTAGE AND CURRENT
i) Alternating voltage
‘An alternating voltage is one whose magnitude changes with time and direction reverses periodically.
The instantaneous value (i.e value at any time t) of an alternating voltage is given by,
E = Ep sinwt
where,
E = Value of the Alternating voltage at time t
£ = Maximum value of the Alternating voltage
w= Angular frequency of supply
From ® = 27f where fis the frequency of the alternating voltage, If T is the time period of alternating
voltage then
w= 2nf
2n
o=—
T

f Maximum
>
+
f \e» A criy
o
& a
4 180° 360°
> Minimum
' Maximum
The voltage varies from zero to a positive peak (*Eo) then back via zero to negative peak (-Eo)
and so on.
In time period T, the wave completely cycle.
E
=——
sinwt
ii) (ii) Alternating current
This is one whose magnitude changes with time and direction reverses periodically.
The Instantaneous value (!-© value at any time t) of sinusoidally varying alternating current is given by
1=Iysinwt
where
1 = value of alternating current at time t
lo = maximum value (Amplitude) of alternating current.
w = Angular frequency of supply.
w= 2nf
2a
o——
T
Figure below shows the waveform of alternating current.

Consider the current variation over one complete cycle.
a eS
[On
2270 ho
bs 2
Average value of J?
pottst9
IP? =S67A
R.MS value of current
VP? =V5.67
=24A
«R.M.S value of current = 2.44
A.C CIRCUIT
‘An A.C circuit is the closed path followed by alternating current.
When a sinusoidal alternating voltage is applied in a circuit, the resulting alternating current is also
sinusoidal and has the same frequency as that of applied voltage.
However there is generally a phase difference between the applied voltage and the resulting
‘As we shall see, this phase difference is introduced due to the presence of inductance (L) and
capacitance (C) in circuit.
While discussing A.C circuits our main points of interest are;
(i) Phase difference between the applied voltage and circuit current
(ii) Phasor diagram. It is the diagram representation of the phase difference between the applied
voltage and the result circuit current.
(iii) Wave diagram
(iv) Power consumed.
A.C CIRCUIT CONTAINING RESISTANCE ONLY

When an alternating voltage is applied across a pure resistance, then from electrons ‘© current flow in
one direction for the first half cycle of the supply and then flow in the opposite direction during the next
half cycle, thus constitute alternate current in the circuit.
Consider a pure resistor of resistance R connected across an alternating source of &:7”- f
E
| ER |
‘Suppose the instantaneous value of the alternating &”"-F is given by
Pm Rysinetspu aun wet
If is the circuit current at that instant, then by ohm’s law
1-£
R
1 — Zasinat
R
Ey
= sino w
The value of | will be maximum |, when
sinwt =1
Therefore equation (i) becomes
1=2sinat
From R

Since sin t= 1 then I= I.
Eo
b=
oR
1) Phase Angle
It is clear from equation (i) and (ii) the applied ©” and circuit current are in phase with each other
£-€ they pass through their zero values at the same instant and attain their peak value both positive and
negative peaks at the same instant.
This is indicated in the wave diagram shown in figure below.
The Ph@SOr diagram shown in figure below also reveals that current is in phase with the applied
voltage.
e
/
f
Ps
) Wave diagram
oe
ly Fy
(0) Phasor diagram
Hence in an a.c circuit, current through R is in phase with voltage across R.
This means that current in R varies in step with voltage across R. If voltage across R is maximum current
in Ris also maximum, if voltage across R is zero, current in R is also zero and so on.

2) Power Absorbed
In a.c circuit, voltage and current vary from instant to instant. Therefore power at any instant is equal to
the product of voltage and current at that instant.
Instantaneous power P
P=El
P = (Ey sin wt). (Iysin wt)
P = Eplol[sin? wt]
P=Eole - — cos —|
P= fale - Fel cos a |
‘Since power varies from instant to instant the average power over a complete cycle is to be considered.
This is found by integrating equation. .........(iv) with respect to time for 1 cycle and dividing by the time
of 1 cycle. The time per one cycle is T.
Average power P

r r
1f Flo , 1 [Fol
P==| — d-= | —
7! 2 z/ 2 cos 2wt dt
a a
Now
Then
fj cos 2wt dt =0
1 (TE,
pP=—| dt
el. 2
Bet,'f*
p= dt
ar i
E,l,
p= 22 1e7
acd
p = Fole
P= Ee x fo
v2 v2
P = Evmslrms
Therefore, average power absorbed by a resistor in an a.c circuit is equal to the product of virtual
voltage ( Ens) across it and virtual current ( lms) through it.
Obviously, this power is supplied by the source of alternating &™-f.
Since
E,
ne |
R
— Eims
ealind
Or
P= [2asR

‘WORKED EXAMPLES
1, An a.c circuit consists of a pure resistance of 100 and is connected across an a.c supply of 230V, 50
Hz.
Calculate
i) Circuit current
ii) Power dissipated and
iii) Equations for voltage and current
Solution
Erms=230v R=100 f=SOHZ
i) Circuit current
ba oe
aia R 10
Ips = 234
ii) Power dissipated P
P= Enmslrms
= 230x 23
+ P=5290W
iii)Equations for voltage and current
Eo = V2 Enns
=V2x230

Eo =325.27V
Ip = V2XI pms
=2x23
a rf '0=32.52A
w=2"f
227 x 50
w=3145
« The equations of voltage and current
E = 325.27 sin 314t and | = 3.52 sin 314t
2. Ina pure resistive circuit, the instantaneous voltage and current are given
E=250sin 314t
1=10 sin 314t
Determine
i) Peak power
ii) Average power
Solution
Ina pure resistive circuit
i) Peak power = Zo/o
=250x 10
+ Peak power = 2500W

ii) Average power P-
Foxy
Pz 2
p= 2
= P=1250W
3. Calculate the resistance and peak current in a 1000 W hair dryer connected to 120V, 60Hz
‘supply. What happens if it is connected to 240V line?
Solution
Enms= 120V,
P=1000W
Ins =?
P=Enms lms
P
a
aa ae
Ipms= 833A
Peak current lo
1b=V2 Inms
=2x 8.33
“ Ig = 118A
Resistance of hair dryer R
E,
R=
Trms
R 120
~ 8.33
~ R=144Q2

When connected to 240V line, the average power delivered would be
— Ffms
— R
P 240?
~ 144
+ P=4000W
This would undoubtedly melt the heating element or the coils of the motor.
4. Avoltage E = 60sin 314t is applied across a 20 © resistor. What will;
i) An a.c ammeter
ii) Ordinary moving coil ammeter in series with resistor read?
Solution
i) E=60sin 314
An a.c ammeter will read the r.m.s value.
He
Ey. iva ~ *24V
ty = = 4 = 2124
v"R 20 ~*~
Therefore a.c a meter will read 2.12A
ii) An ordinary moving coil ammeter will read average value of alternating current. Since the average
value of a.c over one cycle is zero, this meter will record zero reading.
5. What is the peak value of an alternating current which produces three times the heat per second as a
direct current of 2A in a resistor R?
Solution
Heat per second by 2A

Ao
ealR
H=27R
#
' 4k
Three times heat per second
HL
T= 3X4R
#
32 =12R
If lis the r.m.s value of the a.c heat per second in R
ca
= LR
aar=!"vR
17212
VP, = Vi2
Peak value is given by, fo = V2 1,
Ip = V2 x Vi2
«Peak value = 24 = 4.9A
6. An a.c voltage of 4V peak (maximum) is connected to a 1000 resistor R
a) What is the phase of the current and voltage?
b) Calculate the current in R in mA.
c) What is the power in R in mW
Solution

a) The current and voltage are in phase

b) Current is R
Ey = V2 X Ey
E,p=4V
5, 23
=e
fai = 2 v2
Ey=2.83V
From,
= ©
ly =
2.83
ta
v™ 100
= 0.0283A
Or
ly = 28mA
c) Power inR
pale
P = 0.028" x 100
P=0.078W
P=78mW
A.C CIRCUIT CONTAINING INDUCTANCE,

When alternating voltage is applied across a pure inductive coil a back © -F (E) is induced in the coil
due to its self inductance.
a
e--ta
The negative sign indicates that induced e. mf opposes the change in current.
In order to maintain the flow of current the applied voltage must be equal and opposite to induced
voltage at every instant.
Consider a pure inductor of inductance L connected across an alternating source of @:™-f
L
Ey
n
E
E=E,sinwt
Suppose the instantaneous value of the alternating e. m. fis given by
Be Ea sin Wt enn (8)
dl
If Lis the current in the circuit and a is the rate of change of current at that instant, then e. m. f
induced in Lis given by
E L a
© dt
As applied voltage is equal and opposite to induce e.m_f at every instant
dt
E=-(12)

a o
Bliss vena
From, Equation (i) and Equation (ii)
inwt = L&
FE, Sinwt = b=
dt = © Sinwtde
Integrating both sides we get
1 Ey
i) dl= [ —sin wtdt
0 ob
Fo?
1= 2] sin wede
L 0
, Ey [ cos oy
L @ Io
E
1= 2 [-cos wt]
oL
Be, m
I= sr lsia (wt = =|
The value of | will be maximum I, when, Sin(Wt — 77/2) = 1
Then,
L=2
0 ek
poe?
Substituting the value of a2 =/o in equation (iii)
&
1=tsin (wt- 77/2)
ee al a OE)

1) Phase Angle
It is clear from equation (i) and (iv) that circuit current lags behind the applied voltage by (772) radians
or 90°.
This fact is also indicated in the wave diagram.
The phase diagram in figure below also reveals the fact that /v lags behind 2v by 902
By
a / |
=
on Wy
(a) Phasordiagram (b) wave diagram
Hence in an a.c circuit current through L lags behind the voltage across L by 902
This means that when voltage across L is zero, current through L is maximum and vice versa.
From
=:
E=Ler
a a
Now, @¢ is maximum when circuit current is zero and @¢ is zero when circuit current is
maximum
INDUCTIVE REACTANCE
Inductive reactance is the opposition in which an inductor offers to current flow. It is denoted by X,
Inductance not only causes the current to lag behind the voltage but it also limits the magnitude of
current in the circuit.

We have seen above that,
5a
Ip = wok
&
wl=%
Clearly the opposition of inductance to current flow is wl. This quantity wl is called inductive reactance
XL of the inductor.
Inductive reactance X,
X= eo
Or
X, = 2nfL
Or
Ey
ee
Le Tp
a) From
y
Kah
But £o = V2Ey andi,=V24v
Then,
X= %:
y
Zw
=x
Ket hy
b) For d.c
f=0
so that,
xe 2mfl

X22TXOXL
X, =0
Therefore a pure inductance offers zero opposition to d.c
Xi of
) X= 27 FL
Therefore,the greater f.the greater is X, and vice versa.
4) We can show that the units of X: are that of ohm
2 By vote
Xi=wl=2eK Henry = #° ~ amp/see and
volt
——=ohm
X,= ampere
Ill) Average Power consumed
£=Fosinwt
1=£osin(ut2) = locosut
Instantaneous power P
P=El
P= (Fosin wt) (cos wt)
P= -Fo lo sin wt cos wt
— he
P= 2 Sin2ut
Average power P is equal to average of power over one cycle.

Lf g he
P=T/0 2 Sin 2wtdt
P=0
Hence average power absorbed by pure inductor is zero
During one quarter cycle of alternating source of e.m_f. energy is stored in the magnetic field of the
inductor this energy is supplied by the source.
During the next quarter cycle the stored energy is returned to the source. For this reason average
power absorbed by a pure inductor over a complete cycle is zero.
NUMERICAL EXAMPLES
1. A pure inductive coil allows a current of 10A to flow from a voltage of 230V and frequency 60Hz
supply.
Find
i) Inductive reactance
ii) Inductance of the coil
iii) Power consumed
Write down equations for voltage and current:
Solution
Ey =230V.
Iv=10A
f=S0Hz
i) Inductive reactance X,
ke = B= 230
ii) Inductance of the coil L

From
Xe =20fL
L=2t=2x50
inf On
L=0.073H
iii) Power absorbed = 0
Also
Fox 230xV2 Tox 10x V2
Eo=325.27V To= 14.148
w=2t X50
w=314
Since in pure inductive circuit current lags behind the applied voltage by 7 radians. The equation for
voltage and current are,
E£=325 .27sin314t, | = 14.14 sin (314t)
2. Calculate the frequency at which the inductive reactance of 0.7H inductor is 2200
Solution
XL = 2200
L=0.7H
f=?
X, = 2nfL
« f=50Hz

3. A coil has self inductance of 1.4H. The current through the coil varies sinusoidally with amplitude of
2A and frequency 50 Hz
Calculate
i) Potential difference across the coil
li) r.m.s value of P.d across the coil.
Solution
T= Ipsinwt
(i) P.d across the coil
a
E=Lae
Uy sinwt)
E=L ae
aU, sin 2ft)
E=_ ae
=o cost,
E=L02"F cosa ft
p=2X 14 2X 2 x 50 X Cos 20x 50t
E = 880 cos 100 7
ii) r.m.s value of potential different across the coil
4S
Ey a2
=
Eyat
Ey = 622.2V

4, How much inductance should be connected to 200V, 50 Hz a.c supply so that a maximum current of
0.94 flows through it?
Solution
E,=200v
1,=0.9A
f=S0HZ
Peak value of voltage Fo
Eoz V2E,
Ez V2 200
Inductive reactance L
a
X= b
X= VEx200
ro
X= 314.270
Inductance L,
aL
L=2nf
318.27
L= 2x50
L=1H
5. An Inductor of 2H and negligible resistance is connected to 12V, SOHz supply. Find the circuit current,
what current flows when the inductance is changed to 6H?
Solution
* For the First case Xi

x= anf
X= 20 X50X2
X= 6280
Circuit current
x
ye
= &
Ty = 0.0198
* For the second case Xi’
x= 20fL
X¢=27 X50 X 6
X= 1884.0
Circuit current
w=
y=
x,
Iyz tees
1v= 0.00638
A.C CIRCUIT CONTAINING CAPACITANCE ONLY
When an alternating voltage is applied to a capacitor, the capacitor is charged first in one direction and
then in the opposite direction.
The result is that electrons move to and fro round the circuit connecting the plates, thus constituting
alternating current.

Consider a capacitor of capacitance C connected across an alternating source of e.m. f .
c
_ Ve _
!
Suppose the instantaneous value of the alternating e.m.f E is given by
= Fbsin wt (i)
If lis the current in the circuit and Q is the charge on the capacitor at this instant, then the Potential
difference across the capacitor Vc
Q
#12
ey
At every instant the applied e.m.f E must be equal to the potential difference across the capacitor.
este
|
E,Sin wt ==
a=CFosinwt
From,
l=ee
a(CESin wt)

1= cwFocos wt
1= Cw Zosin (wt 7/2)
Eo
1= ~sinwt + 2/2
Y
we
The value of lp will be maximum lo when sin (wt+7"/2) = 1
h= a
°
Toc
Substituting the value of le in equation (i)
T=Ipsin(w + 1/2)
1) Phase Angle
It is clear from equation (1) and (ii) that circuit current leads the applied voltage by n/2 radians or 90S.
This fact is also indicated in the wave diagram.
W
ey
(2) Phasor Diagram (b) Wave Diagram
It also reveals that ly leads E, by 90°, hence in a.c circuit current in capacity leads the voltage.
This means that when voltage across capacitor is zero, current in capacitor is maximum and vice versa.
When P.d across capacitor is maximum £0, the capacitor is fully charged, i.e circuit current is zero.
Since the rate at which a sinusoidally varying p.d falls is greater as it reaches zero, the current has its
maximum value when P.d across capacitor is zero. Hence, current and voltage are out of phase by 90°.

CAPACITIVE RESISTANCE
Capacitive resistance is the opposition which a capacitor offers to current flow. Itis denoted by Xc.
Capacitance not only causes the voltage to lag behind the current but it also limits the magnitude of
current in the circuit.
We have seen above that
ss
oz Moe
fo
z= oC
Then
Bifa. Bs
Tp we
Clearly, the opposition offered by capacitance to current flow is 1/wC
The quantity 1/wC is called capacity reactance Xc of the capacitor
Ec
xc=—
Ip
Or
xC = —
wc
Xewill be in 9 if fis in Hz and Cin Farad
2) From
fo
hz
Then

2. Acoil has an inductance of 1H
a) At what frequency will it have a reactance of 31420?
b) What should be the capacitance of a condenser which has the same reactance at that frequency?
Solution
a)L=1H
X= 31420
f=?
x= 2mfL
fs Int =2exa:
Jf 500H2
b) Xc= 3142, f=500 Hz
c=?
a
Xc= 2876
=
c= Bfke
=
C=iefke
-6
c=0.11% 10
c=o011eF
3. aSOHF capacitor is connected to a 230V, SOHz supply. Determine
i) The maximum charge on the capacitor

ii) The maximum energy stored in the capacitor
Solution
The charge and energy in capacitor will be maximum when p.d across the capacitor is maximum.
i) Maximum charge on the capacitor
a=cFo
a=cEyv2
= (50x20) x (230xv2)
-3
Q=16.26x10%¢
ii) Maximum energy stored in the capacitor U
u=2cE?
os
U=1/2x (50x20) x(230x V2»
U=2.65)
4. The Instantaneous current in a pure inductance of SH is given be
1= 10sin (314t-™/2)amperes
A capacitor is connected in parallel with the inductor. What should be the capacitance of the capacitor
to receive the same amount of energy as inductance at the same terminal voltage?
Solution
The current flowing through pure inductor is
1=10 Sin (314t— 7/2)
To= 108
w= 314s?

Maximum energy stored in the inductor
glgre
W= SLI
Mex 5S 107
U= 250) (i)
Now
Foz wtlo
=314x5x10
Fo=15700v
Max energy stored in the capacitor of capacitance C
a
Uc 22CE,?
=x 5 x 157007
Equate the equation (1) and (ii)
i 2
23% CX 157007).
250x2
C= is700*
-6
C=2.03x 10"f
c=2.034F
‘A.C CIRCUIT CONTAINING R AND LIN SERIES

Consider a resistor of resistance R ohms connected in series with pure inductor of L Henry.
R L
a
VR ————>+—_- VL
lv
Ev
Let
Ev= r.ms value of applied alternating e.m.f
Jozr,m.s value of the circuit current
Vn = 4vR when Vris in phase with /v
vi=/y X1 where “i leads /v by 902
Taking current as the reference phasor, the phasor diagram of the circuit can be drawn as shown in
figure.
Ev | Vi
a)
° z lv
The voltage drop Vsis in phases with current and is represented in magnitude and direction by the
phase OA.
The voltage drop V. leads the current by 902 and is represented in magnitude and direction by the phase
AB.

The applied voltage ©v is the phasor sums of these two voltage drops
Ey72Vq 4 Vi?
Ey (IyR)? + (lyX,)*
By fir (A? +4)
Eye lyVR? + X,?
E
y= —__
VR? + XE
1) Phase Angle
It is clear from the phasor diagram that circuit current /v lags behind the applied voltage Fv by @2.
Therefore we arrive at a very important conclusion that in an inductive circuit current lags behind the
voltage.
NUMERICAL EXAMPLE
1, Three impedance are connected in series across a 200V, 50Hz supply. The first impedance is a
104,,| resistor and the second is a coil or 15 4,,| inductive reactance and 5 4,,| resistance while
the third consists of a 15 4,,| resistor in series with a 254,,| capacitor
Calculate
i) Circuit current
il) Circuit phase angle
iii) Circuit power factor
iv) Power consumed

Solution
i) Total circuit Resistance
R=10+5+15
R=304,)
Total Circuit reactance
KEK - Xe
X=15-25
X=-104,| (capacitive)
Circuit impedance Z
z= 307+ (—10)?
2=31.64,!
Circuit current lv
W=Ev
Zz
v= 200
31.6
W=6.33A

ji) Circuit phase Angle
XX
tang = ———
R
pan 2-055
tan®= -—= -0:
mt 30
© = tan7*(—0.33)
= -18.26°
= 18.26° lead
iii) Circuit power factor
Power factor = COS 8
= COS 18.262
Power factor = 0. 949,
iv) Power consumed P
P=EvlvCos @
P = (200 x 6.33) x0. 949
P=1201.4W
alt
P=i’R
2. A230V, SOH2 supply is applied to a coil of 0.06 H inductance and 2.5 resistance connected in series
with 6.8 uF capacitor

Calculate
i. Circuit impedance
ii. Circuit current
iii, Phase angle between Ey and Iv
iv. Power factor
v. Power consumed
Solution
i) Inductive reactance X,
X= 20fL
X= 21k 50 x 0.60
X= 18.850
Capacitive reactance Xe
= 1
C" 2nft
x, 1
¢ 2n X50 X68 x 10-6
Xc = 4680
Circuit Impedance
Z= JR? + (X, — Xe)?
Z= (25)? + 4885 — 468)?

2=449.209
ii) Circuit current lv
=
I, z
= 230
W=0.152A
ili) Phase angle between Ey and ly
X,- X,
Tano = ——
R
tan o = 18:85 468
se
tan © = —179.66
0 = tan™?(—179.66)
0=-89.7 lead
iv) Power factor
° R
Cos® = :
2.5
cos aang =a at
© = 92
Cos @ = 0.0056

v) Power consumed P
P=EvisCos®
P = 230.x 0.512 x 0.0056
P=0.66W
3. Aresistance R, and inductance L = 0.01H and a capacitance C are connected in series. When an
alternating voltage E = 400Sin (3000t- 20°) is applied to the series combination, the current flowing is
10v2 Sin (3000t - 650). Find the value of R and C
Solution
The circuit current lags behind the applied voltage by @
@=65°-20
6=45°
This implies that the circuit is inductive i.e.
Xi > Xe
The net circuit reactance X
X=X Xe
Now
X=wl
X: = 3000 x 0.01
X%=300

Also
rromane== = 45"
Tan 45= z
1-2
R
Circuit Impedance Z
za Fe = 400
T,  10V2
Z=28.309
Now
P=R+xX?
Z=R?+R?
P=2R’
Z=Rv2
R=200
Now
X =X Xe
20 = 30=Xe

Xc= 100
From
x, 1
© 2nfc
c=—
2nfX,
c=———_
3000 x 10
C=33.3x10°F
4. Aseries RLC circuit is connected to an a.c (220V, 50 H) as shown in the figure below
| |
100Q
220 V, 50 Hz WY
c @)
om
If the reading of the three volt meter V; V 2 and V; are 65V, 415V and 204V respectively.
Calculate

i, The currentin the circuit
ji, The value of inductor L
ili. The value of capacitor C
Solution
Here voltmeters are considered ideal i.e. having infinite resistance.
Therefore, it is a series RLC circuit
i) Circuit current Iv
W=0.65A
ii) Inductive reactance X.
Vv; 204
X,= L ry
X= 318.850
Inductance L
L= XL - 313.85
2nf 20x50
L=1H
ili) Capacitive reactance X<

Xc= 638. 460
Capacitance C
c-_3
2nfXe
¢=—______
2m X 50 X 638.46
C=5x10°F
5. Acoil of resistance 80 and inductance 0.03H is connected to an a.c supply of 240V, 50 Hz.
Calculate
i) The current the power and power factor.
ii) The value of a capacitance which when connected in series with the above coil causes no change in
the value of current and power taken from the supply
Solution
i) Reactance of the coil X:
X, = 2nfl
X, = 2m x 50 x 0.03
X=9.420
Impedance of the coil Z
Z = R+X,
= /8? +9.42?

2= 12.460
Circuit Current ly
=”
ly =
1-24
v "12.46
W=19.42A
Power consumed
P= HR
P= (19.42)? x8
P= 3017 W
Power factor Cos ©
cosp==
os@= =
Z
- 8
o3@=——
Oo” 12.46
Cos © = 0.65 lag
ii) To maintain the same current and power, the impedance of the circuit should remain unchanged.
Thus the value of capacitance in the series circuit should be such so as to cause the current to lead by
the same angles as it previously lagged.
This can be achieved if the series capacitor has a capacitive reactance equal to twice the inductive
reactance.
Xe= 2K,
Xc=2x9.42

L
b OO OINS
Iv
The circuit will be in resonance when the circuit power factor is unity.
This means that wattles component of the circuit current should be zero.
hI
‘The resonance in a parallel a .c circuit can be achieved by changing the supply frequency because X:
and Xcare frequency dependent .
Ata certain frequency called resonant frequency fa, lc become equal to , and resonance occur.
At resonance
le=h
From
YW=Ve
I, = Ip
ae
XOX
X, = Xe

2nf,L = ———
fol =>, fC
f. 1
°  2nvlc
fo will be in Hz if Lis in Henry and Cis in farad
EFFECTS OF PARALLEL RESONANCE
1) The circuit power factor becomes unity. This implies that the circuit act as a resistor
2) The impedance (resistive) of the circuit becomes maximum.
3) The circuit current is minimum the small current |v flowing in the circuit is only the amount needed to
supply the resistance losses.
RESONANCE CURVE
This is the curve between the circuit current and the supply frequency.
Figure below shows the resonance curve of a parallel a.c circuit.
I
Ey
a re
w= 7
'
'
'
i
fo #
The circuit current ly is minimum at parallel resonance.
‘As the frequency changes from resonance, the circuit current increases rapidly.

This action can be explained as follows. For frequencies other than the resonance, the reactive currents
(i.and Ic) in the two branches of the circuit are not equal.
The resultant reactive current must be supplied by the a.c source
As the difference of the reactive currents in the two branches increase with the amount of deviation
from the resonant frequency, the circuits current will also increase.
“ADVANTAGE OF A.C OVER D.C
(i Alternating voltages can be stepped up or stepped down efficiently by a transformer.
This permits the transmission of electric power at high voltages to achieve economy and
distribute the power at utilization voltages.
(ii) A.C motors are cheaper and simpler in construction than the d.c motors.
(iil) A. Ccan be easily converted into d.c by rectifiers.
(iv) Alternating current can be controlled with a choke coil without any appreciable loss of electrical
energy.
(v) The switch gear (e.g. switches, circuit breakers) for a.c system is cheaper than the d.c system.
DISADVANTAGE OF A.C OVER D.C
{i) For the same voltage (same value of voltage), a.c is more dangerous than d.c.
(ii) The shock of a.c is attractive whereas that of d.cis repulsive.
(iii) A.C cannot be used for some processes e.g. electroplating, charging of batteries etc.
(iv) A.Cis transmitted more over the surface of the conductor than from inside. This is called skin
effect.
To avoid skin effect, a.c is transmitted over several fine insulated wires instead of a single thick wire.
WORKED EXAMPLES.
1. A coil of resistance 1009 and inductance 100j1H is connected in series with a 100 pF capacitor. The
circuit is connected to a 10V variable frequency source.
au

Gm GE EO Fan
( ( © Silicon atom
g . Free electron
y. 0 9.
P-SEMICONDUCTORS
A-P- Semiconductor is made by adding a trivalent atom (an acceptor) such as B or on to pure
semi conductor such as germanium. Since there is a production of large number of holes
(positive charges) the impure semiconductor is called P- Semiconductor.
Cae a 0 a pe
( () ie) Germanium atom
4 f.) fe} Holes
) 0) = (7
Ca (<a> LD
Z
P-N=JUNCTION/DIODE:

This is formed when P and N semiconductors are melted to form a junction between them
ys N
0 0 0 LM/
es
fo} [o} ° or
‘d
ba
° ° 1o}
‘A
P-N junction
The marrow region at the P-n junction which contains the negative and positive charge is called
depletion layer.
A barrier dip is a p.d which oppose more diffusion of charges across the junction.
This is produced when the flow of +ve and -ve Charges ceases
P—N JUNCTION AS A RECTIFIER:
FORWARD BIAS.
Is said to be forward biased when its P- semiconductor is connected to the +ve terminals of the
battery and its N- Semi conductors is connected to the -ve terminal at the battery.
In this case electrons and holes flow across the P-n junction. This happen because the +ve pole of
the battery repel the +ve charge and —Ve pole rel the -ve charges.

P N
b oe li
or H \Ey Pe)
Co - +e
or yt ~~
Or OF ||! +e <0
+ H -
REVERSE BIAS.
A-P-N junction is said to be reverse biased when its P. Semiconductor is connected to the
negative pole junction of a battery and N. Semiconductor is connected to the +ve [p:e pf the
battery in this case only a very small a current flows.
P N
+0 +0! 1o> O
+0 <0} |O> Or
+0 +0! |Or Or
= | +
P-N JUNCTION AS RECTIFIERS

Direction of traverse Direction of traverse
| _K-— 4 .
pd=+E p.d=-E
The choice of the direction of the traverse depend on yourself
Qa - :
In fig given below E=2.0 volts E, = 4.0 volts
M=10Q2 1=2.0Q2  713=5Q.
What is the current I flowing in the circuit
n n
Cc 8 ‘aD
Ley
Solution
E, and E are opposing each other but since £2 > E, then Ez must be controlling the direction of
current.
Applying Kirchhofi"s laws in the clockwise direction we have
In, + Ey — Ep + 1% +17, =0
In, + Itz + Ig = Ez — Ey

10, +m +73) = E,- Ey
pe 22 Bs
~ (+n +7)
Qa
Find the current in each resistor in the circuit shown below:
A
Lov 7 1s. C al
m2 42
32 a ee I
F hep a)
Take the loop A B E F in the clockwise direction
—15+1,+91, — 21, + 10 + (-3h,) =0
10Ly — Sly = § = Dy = Ty = Devens vn owe (A)
Take the loop BCDE in clockwise direction.
25-41-91, + 15-1, =0
But! = 1, +1,
41 =4(I, +h)
+ 25 — 4h, — 41, - 91, +15—1, =0
—141, — 41, = —40 divide by -2 both side
Ty + Dy = 20 oe eer ve (2)
To solve the equations simultaneously.
71, + 2I, = 20
2h —Ib=1
_ ith
=

:
7 (#) +21, = 20
7+7Iz + 21, = 40
111, = 33
a-yA73
1+3
==
a 2 2
1,=2A and I, = 3A
Qn.
Determine the magnitude of 4s @7d J

Therefore IP = [pRy ou. .-.(1) similarly 1Q = [pS os uo ue (2)
Dividing (1) by (2) gives
1Q=15
1P=hRp
at
Qs
Now since R a C then we have
Pa ADandQaDCso
P= a
Q. be
Qn
A wire of length 1.1 meter and Radius 7*10~°m is connected across the right gap of the metre bridge when a
resistance of 45 Q is introduced in the left gap from a resistance box connected across it- The balance point is
obtained 0,6m from left side. Calculate the specific resistance (resistivity) of the material of the wire
Solution
The value of resistance Formula of resistivity
Soe = BA
06 04 Pry
45x04 _ «— 30x 3.14 x(7.0 x 10~*)
x6 = 300 od 7
p = 4.2x 10-70m
Qn.
A 2K2 and 3kQ resistor are connected in series combination is connected across a 100v supply of negligible
internal resistance as shown in the figure below:

3k = 100V
Yo R; R
(i) What is the output voltage Vout of the load resistance Rx =30k2
(ii) What current resistance through the load resistor R2_ in (i) above.
121.15
Solution
The resistance = 2kn + <2 = 52
Vv _ 10x13
Current = 2 == 21.15
Voltage V = = x 21.5]
= 57.6V