Energy and Power in Electrical Circuits

When a battery is connected between the ends of a conductor, a current is established. The battery is supplying energy to the device which is connected in the circuit. Consider a circuit in which a battery of voltage V is connected to the resistor as shown in Figure 2.15.

Assume that a positive charge of dQ moves from point a to b through the battery and moves from point c to d through the resistor and back to point a. When the charge moves from point a to b, it gains potential energy dU = V.dQ and the chemical potential energy of the battery decreases by the same amount.

When this charge dQ passes through resistor it loses the potential energy dU = V.dQ due to collision with atoms in the resistor and again reaches the point a. This process occurs continuously till the battery is connected in the circuit. The rate at which the charge loses its electrical potential energy in the resistor can be calculated.

The electrical power P is the rate at which the electrical potential energy is delivered,

P = $$\frac{dU}{dt}$$ = $$\frac{(V.dQ)}{dt}$$ = V$$\frac{dQ}{dt}$$ ……….. (2.31)

Since the electric current I = $$\frac{dQ}{dt}$$, the equation (2.31) can be rewritten as

P = VI ………….. (2.32)

This expression gives the power delivered by the battery to any electrical system, where I is the current passing through it and V is the potential difference across it. The SI unit of electrical power is watt (1W = 1 J s-1). Commercially, the electrical bulbs used in houses come with the power and voltage rating of 5W-220V, 30W-220V, 60W-220V etc. (Figure 2.16)

Usually these voltage rating refers AC RMS voltages. For a given bulb, if the voltage drop across the bulb is greater than voltage rating, the bulb will fuse. Using Ohm’s law, power delivered to the resistance R is expressed in other forms

P = IV = I(IR) = I2R ………….. (2.33)
P = IV = $$\frac{V}{R}$$V = $$\frac{V^{2}}{R}$$ ……….. (2.34)

The total electrical energy used by any device is obtained by multiplying the power and duration of the time when it is ON. If the power is in watts and the time is in seconds, the energy will be in joules. In practice, electrical energy is measured in kilowatt hour (kWh). 1 kWh is known as 1 unit of electrical energy.

(1 kWh = 1000 Wh = (1000 W) (3600 s)
= 3.6 × 106 J)

Example 2.15

A battery of voltage V is connected to 30 W bulb and 60 W bulb as shown in the figure.

(a) Identify brightest bulb
(b) which bulb has greater resistance?
(c) Suppose the two bulbs are connected in series, which bulb will glow brighter?

Solution

(a) The power delivered by the battery P = VI. Since the bulbs are connected in parallel, the voltage drop across each bulb is the same. If the voltage is kept fixed, then the power is directly proportional to current (P ∝ I). So 60 W bulb draws twice as much as current as 30 W and it will glow brighter than 30 W bulb.

(b) To calculate the resistance of the bulbs, we use the relation P = $$\frac{V^{2}}{R}$$. In both the bulbs, the voltage drop is the same. So the power is inversely proportional to the resistance or resistance is inversely proportional to the power (R ∝$$\frac{1}{P}$$). It implies that, the 30W has twice as much as resistance as 60 W bulb.

(c) When the bulbs are connected in series, the current passing through each bulb is the same. It is equivalent to two resistors connected in series. The bulb which has higher resistance has higher voltage drop. So 30W bulb will glow brighter than 60W bulb. So the higher power rating does not always imply more brightness and it depends whether bulbs are connected in series or parallel.

Example 2.16

Two electric bulbs marked 20 W – 220 V and 100 W – 220 V are connected in series to 440 V supply. Which bulb will get fused?

Solution

To check which bulb will get fused, the voltage drop across each bulb has to be calculated.

The resistance of the bulb,

For 20W-220V bulb
R1 = $$\frac{(220)^{2}}{20}$$ Ω = 2420 Ω

For 100W-220V bulb,
R2 = $$\frac{(220)^{2}}{100}$$ Ω = 484 Ω

Both the bulbs are connected in series. So same current will pass through both the bulbs. The current that passes through the circuit, I = $$\frac{V}{R_{t o t}}$$.

Rtot = (R1 + R2)
Rtot = (484 + 2420)Ω = 2904Ω
I = $$\frac{440V}{2904Ω}$$ ~ 0.151 A

The voltage drop across the 20W bulb is
V1 = IR1 = $$\frac{440}{2904}$$ × 2420 ~ 366.6 V

The voltage drop across the 100W bulb is
V2 = IR2 = $$\frac{440}{2904}$$ × 484 ~ 73.3 V

The 20 W bulb will get fused because the voltage across it is more than the voltage rating.

Physics is now simple when learning with CBSELibrary – Get all important Physics Topics with detailed explanation.

Effects of Various Types of Environmental Pollution

Atmospheric pollution is generally studied as tropospheric pollution.

Different types of atmospheric pollutions are

• Air pollution
• Water pollution
• Soil pollution

Air pollution

Any undesirable change in air which adversely affects living organisms is called air pollution. Air pollution is limited to troposphere and stratosphere. Air pollution is mainly due to the excessive discharge of undesirable foreign matter in to the atmospheric air.

Types of Air Pollutants

Air pollutants may exist in two major forms namely, gases and particulates.

Gaseous Air Pollutants

Oxides of sulphur, oxides of nitrogen, oxides of carbon, and hydrocarbons are the gaseous air pollutants.

a. Oxides of Sulphur

Sulphur dioxide and sulphur trioxide are produced by burning sulphur containing fossil fuels and roasting sulphide ores. Sulphur dioxide is a poisonous gas to both animals and plants. Sulphur dioxide causes eye irritation, coughing and respiratory diseases like asthma, bronchitis, etc.

Sulphur dioxide is oxidised into more harmful sulphur trioxide in the presence of particulate matter present in polluted air.

SO3 combines with atmospheric water vapour to form H2SO4, which comes down in the form of acid rain.

SO3 + H2O → H2SO4

Some harmful effects of acid rain will be discussed in section 15.3

b. Oxides of Nitrogen

Oxides of nitrogen are produced during high temperature combustion processes, oxidation of nitrogen in air and from the combustion of fuels (coal, diesel, petrol etc.).

NO + O3 → NO2 + O2

The oxides of nitrogen are converted into nitric acid which comes down in the form of acid rain. They also form reddish brown haze in heavy traffic. Nitrogen dioxide potentially damages plant leaves and retards photosynthesis. NO2 is a respiratory irritant and it can cause asthma and lung injury. Nitrogen dioxide is also harmful to various textile fibres and metals.

c. Oxides of Carbon

The major pollutants of oxides of carbon are carbon monoxide and carbon dioxide.

(i) Carbon Monoxide

Carbon monoxide is a poisonous gas produced as a result of incomplete combustion of coal or firewood. It is released into the air mainly by automobile exhaust.

It binds with haemoglobin and forms carboxy haemoglobin which impairs normal oxygen transport by blood and hence the oxygen carrying capacity of blood is reduced. This oxygen deficiency results in headache, dizziness, tension, Loss of consciousness, blurring of eye sight and cardiac arrest.

(ii) Carbon Dioxide

Carbon dioxide is released into the atmosphere mainly by the process of respiration, burning of fossil fuels, forest fire, decomposition of limestone in cement industry etc.

Green plants can convert CO2 gas in the atmosphere into carbohydrate and oxygen through a process called photosynthesis. The increased CO2 level in the atmosphere is responsible for global warming. It causes headache and nausea.

(d) Hydrocarbon

The compounds composed of carbon and hydrogen only are called hydrocarbons. They are mainly produced naturally (marsh gas) and also by incomplete combustion of automobile fuel. They are potential cancer causing (carcinogenic) agents. For example, polynuclear aromatic hydrocarbons (PAH) are carcinogenic, they cause irritation in eyes and mucous membranes.

Greenhouse Effect and Global Warming:

In 1987, Jean Baptiste Fourier a French mathematician and scientist coined the term“Greenhouse Effect” for trapping of heat in the atmosphere by certain gases.

The earth’s atmosphere allows most of the visible light from the Sun to pass through and reach Earth’s surface. As Earth’s surface is heated by sunlight, it radiates part of this energy back toward space as longer wavelengths (IR).

Some of the heat is trapped by CH4, CO2, CFCs and water vapour present in the atmosphere. They absorb IR radiation and effectively block a large portion of earth’s emitted radiation. The radiation thus absorbed is partly reemitted to earth’s surface. Therefore, the earth’s surface gets heated up by a phenomenon called greenhouse effect.

Thus Greenhouse effect may be defined as the heating up of the earth surface due to trapping of infrared radiations reflected by earth’s surface by CO2 layer in the atmosphere”. The heating up of earth through the greenhouse effect is called global warming.

Without the heating caused by the greenhouse effect, Earth’s average surface temperature would be only about – 18 °C (0 °F). Although the greenhouse effect is a naturally occurring phenomenon, it is intensified by the continuous emission of greenhouse gases into the atmosphere.

During the past 100 years, the amount of carbon dioxide in the atmosphere increased by roughly 30 percent and the amount of methane more than doubled. If these trends continue, the average global temperature will increase which can lead to melting of polar ice caps and flooding of low lying areas. This will increase incidence of infectious diseases like dengue, malaria etc.

Acid Rain

Rain water normally has a pH of 5.6 due to dissolution of atmospheric CO2 into it. Oxides of sulphur and nitrogen in the atmosphere may be absorbed by droplets of water that make up clouds and get chemically converted into sulphuric acid and nitric acid respectively. As a result, pH of rain water drops below the level 5.6, hence it is called acid rain.

Acid rain is a by-product of a variety of sulphur and nitrogen oxides in the atmosphere. Burning of fossil fuels (coal and oil) in power stations, furnaces and petrol, diesel in motor engines produce sulphur dioxide and nitrogen oxides. The main contributors of acid rain are SO2 and NO2. They are converted into sulphuric acid and nitric acid respectively by the reaction with oxygen and water.

2SO2 + O2 + 2H2O → 2H2SO4
4NO2 + O2 + 2H2O → 4HNO3

Harmful effects of acid rain:

Some harmful effects are discussed below.

(i) Acid rain causes extensive damage to buildings and structural materials of marbles. This attack on marble is termed as Stone leprosy.

CaCO3 + H2SO4 → CaSO4 + H2O + CO2

(ii) Acid rain affects plants and animal life in aquatic ecosystem.

(iii) It is harmful for agriculture, trees and plants as it dissolves and removes the nutrients needed for their growth.

(iv) It corrodes water pipes resulting in the leaching of heavy metals such as iron, lead and copper into drinking water which have toxic effects.

(v) It causes respiratory ailment in humans and animals.

Particulate matter (Particulate Pollutants)

Particulate pollutants are small solid particles and liquid droplets suspended in air. Many of particulate pollutants are hazardous. Examples: dust, pollen, smoke, soot and liquid droplets (aerosols) etc,.

They are blown into the atmosphere by volcanic eruption, blowing of dust, incomplete combustion of fossil fuels induces soot. Combustion of high ash fossil fuels creates fly ash and finishing of metals throws metallic particles into the atmosphere.

Types of Particulates:

Particulate in the atmosphere may be of two types, viable or non-viable.

a. Viable Particulates

The viable particulates are the small size living organisms such as bacteria, fungi, moulds, algae, etc. which are dispersed in air. Some of the fungi cause allergy in human beings and diseases in plants.

b. Non-Viable Particulates

The non- viable particulates are small solid particles and liquid droplets suspended in air. They help in the transportation of viable particles. There are four types of nonviable particulates in the atmosphere. They are classified according to their nature and size as follows

(i) Smoke

Smoke particulate consists of solid particles (or) mixture of solid and liquid particles formed by combustion of organic matter. For example, cigarette smoke, oil smoke, smokes from burning of fossil fuel, garbage and dry leaves.

(ii) Dust

Dust composed of fine solid particles produced during crushing and grinding of solid materials. For example, sand from sand blasting, saw dust from wood works, cement dust from cement factories and fly ash from power generating units.

(iii) Mists

They are formed by particles of spray liquids and condensation of vapours in air. For example, sulphuric acid mist, herbicides and insecticides sprays can form mists.

(iv) Fumes

Fumes are obtained by condensation of vapours released during sublimation, distillation, boiling and calcination and by several other chemical reactions. For example, organic solvents, metals and metallic oxides form fume particles.

Health effects of particulate pollutants:

(i) Dust, mist, fumes,etc., are air borne particles which are dangerous for human health. Particulate pollutants bigger than 5 microns are likely to settle in the nasal passage whereas particles of about 10 micron enters the lungs easily and causes scaring or firosis of lung lining.

They irritate the lungs and causes cancer and asthma. This disease is also called pneumoconiosis. Coal miners may suffer from black lung disease. Textile workers may suffer from white lung disease.

(ii) Lead particulates affect children’s brain, interferes maturation of RBCs and even cause cancer.

(iii) Particulates in the atmosphere reduce visibility by scattering and absorption of sunlight. It is dangerous for aircraft and motor vehicles

(iv) Particulates provide nuclei for cloud formation and increase fog and rain.

(v) Particulates deposit on plant leaves and hinder the intake of CO2 from the air and affect photosynthesis.

Techniques to reduce particulate pollutants

The particulates from air can be removed by using electrostatic precipitators, gravity settling chambers, and wet scrubbers or by cyclone collectors. These techniques are based on washing away or settling of the particulates.

Smog

Smog is a combination of smoke and fog which forms droplets that remain suspended in the air.

Smog is a chemical mixture of gases that forms a brownish yellow haze over urban cities. Smog mainly consists of ground level ozone, oxides of nitrogen, volatile organic compounds, SO2, acidic aerosols and gases, and particulate matter.

There are two types of smog. One is Classical smog caused by coal smoke and fog, second one is photo chemical smog caused by photo chemical oxidants. They are discussed below in detail.

(i) Classical smog or London smog

Classical smog was first observed in London in December 1952 and hence it is also known as London smog. It consists of coal smoke and fog.

It occurs in cool humid climate. This atmospheric smog found in many large cities. The chemical composition is the mixture of SO2, SO3 and humidity. It generally occurs in the morning and becomes worse when the sun rises. This is mainly due to the induced oxidation of SO2 to SO3, which reacts with water yielding sulphuric acid aerosol.

Chemically it is reducing in nature because of high concentration of SO2 and so it is also called as reducing smog.

Effects of classical smog:

• Smog is primarily responsible for acid rain.
• Smog results in poor visibility and it affects air and road transport.
• It also causes bronchial irritation

(ii) Photo chemical smog or Los Angel Smog

Photo Chemical smog was first observed in Los Angels in 1950. It occurs in warm, dry and sunny climate. This type of smog is formed by the combination of smoke, dust and fog with air pollutants like oxides of nitrogen and hydrocarbons in the presence of sunlight.

It forms when the sun shines and becomes worse in the afternoon. Chemically it is oxidizing in nature because of high concentration of oxidizing agents NO2 and O3, so it is also called as oxidizing smog.

Photo chemical smog is formed through sequence of following reactions.

N2 + O2 → 2NO
2NO + O2 → 2NO2

(O) + O2 → O3
O3 + NO → NO2 + O2

NO and O3 are strong oxidizing agent and can react with unburnt hydrocarbons in polluted air to form formaldehyde, acrolein and peroxy acetyl nitrate (PAN).

Effcts of photo chemical smog

The three main components of photo chemical smog are nitrogen oxide, ozone and oxidised hydro carbon like formaldehyde (HCHO), Acrolein (CH2=CH-CHO), peroxy acetyl nitrate (PAN).

Photochemical smog causes irritation to eyes, skin and lungs, increase in chances of asthma.

High concentrations of ozone and NO can cause nose and throat irritation, chest pain, uncomfortable in breathing, etc.

PAN is toxic to plants, attacks younger leaves and cause bronzing and glazing of their surfaces. It causes corrosion of metals stones, building materials and painted surfaces.

Control of Photo Chemical Smog

The formation of photochemical smog can be suppressed by preventing the release of nitrogen oxides and hydrocarbons into the atmosphere from the motor vehicles by using catalytic convertors in engines. Plantation of certain trees like Pinus, Pyrus, Querus Vitus and juniparus can metabolise nitrogen oxide.

Find free online Chemistry Topics covering a broad range of concepts from research institutes around the world.

Ohm’s Law Definition – Formula and its Current Electricity

The ohm’s law can be derived from the equation J = σE. Consider a segment of wire of length l and cross sectional area A as shown in Figure 2.7

When a potential difference V is applied across the wire, a net electric field is created in the wire which constitutes the current in the wire. For simplicity, we assume that the electric field is uniform in the entire length of the wire, then the potential difference (voltage V) can be written as

V = El

As we know, the magnitude of current density
J = σE = σ $$\frac{V}{l}$$ ………… (2.14)

But J = $$\frac{I}{A}$$, so we write the equation (2.14) as
$$\frac{I}{A}$$ = σ $$\frac{V}{l}$$.

By rearranging the above equation, we get
V = I($$\frac{1}{σA}$$) ………….. (2.15)

The quantity $$\frac{1}{σA}$$ is called resistance of the conductor and it is denoted as R. Note that the resistance is directly proportional to the length of the conductor and inversely proportional to area of cross section.

Therefore, the macroscopic form of ohm’s law can be stated as

V = IR …………… (2.16)

From the above equation, the resistance is the ratio of potential difference across the given conductor to the current passing through the conductor.

R = $$\frac{V}{I}$$ ………………. (2.17)

The SI unit of resistance is ohm (Ω). From the equation (2.16), we infer that the graph between current versus voltage is straight line with a slope equal to the inverse of resistance R of the conductor. It is shown in the Figure 2.8 (a).

Materials for which the current versus voltage graph is a straight line through the origin, are said to obey Ohm’s law and their behaviour is said to be ohmic as shown in Figure 2.8(a). Materials or devices that do not follow Ohm’s law are said to be nonohmic. These materials have more complex relationships between voltage and current. A plot of I versus V for a non-ohmic material is non-linear and they do not have a constant resistance (Figure 2.8(b)).

Example 2.5

A potential difference across 24 Ω resistor is 12 V. What is the current through the resistor?

Solution

V = 12 V and R = 24 Ω
Current, I = ?
From Ohm’s law, I = $$\frac{V}{R}$$ = $$\frac{12}{24}$$ = 0.5A

Resistivity

In the previous section, we have seen that the resistance R of any conductor is given by

R = $$\frac{1}{σA}$$ …………… (2.18)

where σ is called the conductivity of the material and it depends only on the type of the material used and not on its dimension. The resistivity of a material is equal to the reciprocal of its conductivity.

ρ = $$\frac{1}{σ}$$ ……….. (2.19)

Now we can rewrite equation (2.18) using equation (2.19)

R = ρ$$\frac{l}{A}$$ …………. (2.20)

The resistance of a material is directly proportional to the length of the conductor and inversely proportional to the area of cross section of the conductor. The proportionality constant ρ is called the resistivity of the material.

If l = 1 m and A = 1 m2, then the resistance R = ρ. In other words, the electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit length having unit area of cross section. The SI unit of ρ is ohm-metre (Ω m).

Based on the resistivity, materials are classified as conductors, insulators and semiconductors. The conductors have lowest resistivity, insulators have highest resistivity and semiconductors have resistivity greater than conductors but less than insulators. The typical resistivity values of some conductors, insulators and semiconductors are given in the Table 2.1

Example 2.6

The resistance of a wire is 20 Ω. What will be new resistance, if it is stretched uniformly 8 times its original length?

Solution

R1 = 20 Ω, R2 = ?
Let the original length of the wire (l1) be l.
New length, l2 = 8l1 (i.e;) l2 = 8l
Original resistance, R1 = ρ$$\frac{l_{1}}{A_{1}}$$
New resistance R2 = ρ$$\frac{l_{2}}{A_{2}}$$ = $$\frac{\rho(8 l)}{A_{2}}$$
Though the wire is stretched, its volume remains unchanged.
Initial volume = Final volume
A1l1 = A2l2, A1l = A2(8l)
$$\frac{A_{1}}{A_{2}}$$ = $$\frac{8l}{l}$$ = 8
By dividing equation for R2 by equation for R1, we get

Substituting the value of $$\frac{A_{1}}{A_{2}}$$, we get
$$\frac{R_{2}}{R_{1}}$$ = 8 × 8 = 64
R2 = 64 × 20 = 1280 Ω

Hence, stretching the length of the wire has increased its resistance

Example 2.7

Consider a rectangular block of metal of height A, width B and length C as shown in the figure.

If a potential difference of V is applied between the two faces A and B of the block (figure (a)), the current IAB is observed. Find the current that flows if the same potential difference V is applied between the two faces B and C of the block (figure (b)). Give your answers in terms of IAB.

Solution

In the first case, the resistance of the block

RAB = ρ$$\frac{length}{Area}$$ = ρ$$\frac{C}{AB}$$
The current IAB = $$\frac{V}{R_{A B}}$$ = $$\frac{V}{ρ}$$.$$\frac{AB}{C}$$ ………. (1)

In the second case, the resistance of the block RBC = ρ$$\frac{A}{BC}$$
The current IBC = $$\frac{V}{R_{B C}}$$ = $$\frac{V}{\rho}$$. $$\frac{BC}{A$$ ………… (2)

To express IBC interms of IAB, we multiply and divide equation (2) by AC, we get IBC = $$\frac{V}{ρ}$$. $$\frac{BC}{A}$$$$\frac{AC}{AC}$$ = ($$\frac{V}{ρ}$$ $$\frac{AB}{C}$$).$$\frac{C^{2}}{A^{2}}$$ = $$\frac{C^{2}}{A^{2}}$$.IAB Since C > A, the current IBC > IAB

Resistors in Series and Parallel

An electric circuit may contain a number of resistors which can be connected in different ways. For each type of circuit, we can calculate the equivalent resistance produced by a group of individual resistors.

Resistors in Series

When two or more resistors are connected end to end, they are said to be in series. The resistors could be simple resistors or bulbs or heating elements or other devices. Figure 2.9 (a) shows three resistors R1, R2 and R3 connected in series.

The amount of charge passing through resistor R1 must also pass through resistors R2 and R3 since the charges cannot accumulate anywhere in the circuit.

Due to this reason, the current I passing through all the three resistors is the same. According to Ohm’s law, if same current pass through different resistors of different values, then the potential difference across each resistor must be different.

If V1, V2 and V3 be the potential differences (voltage) across each of the resistors R1, R2 and R3 respectively, then we can write V1 = IR1, V2 = IR2 and V3 = IR3. But the supply voltage V must be equal to the sum of voltages (potential differences) across each resistor.

V = V1 + V2 + V3 = IR1 + IR2 + IR3 ……… (2.21)
V = I (R1 + R2 + R3)
V = IRS ………… (2.22)

where Rs is the equivalent resistance.

RS = R1 + R2 + R3 ………….. (2.23)

When several resistors are connected in series, the total or equivalent resistance is the sum of the individual resistances as shown in the Figure 2.9 (b).

Note:
The value of equivalent resistance in series connection will be greater than each individual resistance.

Example 2.8

Calculate the equivalent resistance for the circuit which is connected to 24 V battery and also find the potential difference across each resistors in the circuit.

Solution

Since the resistors are connected in series, the effective resistance in the circuit
= 4 Ω + 6 Ω = 10 Ω
current I in the circuit = $$\frac{V}{R_{e q}}$$ = $$\frac{24}{10}$$ = 2.4A
Voltage across 4Ω resistor
V1 = IR1 = 2.4A × 4Ω = 9.6V
Voltage across 6Ω resistor
V2 = IR2 = 2.4A × 6Ω = 14.4V

Resistors in Parallel

Resistors are in parallel when they are connected across the same potential difference as shown in Figure 2.10 (a).

In this case, the total current I that leaves the battery is split into three separate components. Let I1, I2 and I3 be the current through the resistors R1, R2 and R3 respectively. Due to the conservation of charge, total current in the circuit I is equal to sum of the currents through each of the three resistors.

I = I1 + I2 + I3 ……….. (2.24)

Since the voltage across each resistor is the same, applying Ohm’s law to each resistor, we have

Substituting these values in equation (2.24), we get

Here Rp is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocals of resistance of the individual resistors is equal to the reciprocal of the effective resistance of the combination as shown in the Figure 2.10 (b).

Note:
The value of equivalent resistance in parallel connection will be lesser than each individual resistance.

House hold appliances are always connected in parallel so that even if one is switched off, the other devices could function properly.

Example 2.9

Calculate the equivalent resistance in the following circuit and also find the values of current I, I1 and I2 in the given circuit.

Solution

Since the resistances are connected in parallel, the equivalent resistance in the circuit is

The resistors are connected in parallel, the potential difference (voltage) across them is the same.

The current I is the sum of the currents in the two branches. Then,
I = I1 + I2 = 6A + 4A = 10 A

Example 2.10

Two resistors when connected in series and parallel, their equivalent resistances are 15 Ω and $$\frac{56}{15}$$Ω respectively. Find the values of the resistances.

Solution

Rs = R1 + R2 = 15 Ω ……….. (1)
Rp = $$\frac{R_{1} R_{2}}{R_{1}+R_{2}}$$ = $$\frac{56}{15}$$Ω
∴ R1R2 = 56
R2 – $$\frac{56}{R_{1}}$$ Ω ………….. (2)
From equation (1) substituting for R1 + R2 in equation (2)
$$\frac{R_{1} R_{2}}{15}$$ = $$\frac{56}{15}$$Ω
∴ R1R2 = 56
R2 = $$\frac{56}{R_{1}}$$Ω …………… (3)
Substituting for R2 in equation (1) from equation (3)
R1 + $$\frac{56}{R_{1}}$$ = 15
Then, $$\frac{R_{1}^{2}+56}{R_{1}}$$ = 15
R12 + 56 = 15R1
R12 – 15R1 + 56 = 0

The above equation can be solved using factorisation

R1 = 8 Ω (or) R1 = 7 Ω
If R1 = 8 Ω
Substituting in equation (1)
8 + R2 = 15
R2 = 15 – 8 = 7 Ω
If R2 = 7 Ω i.e; (when R1 = 8 Ω ; R2 = 7 Ω)
If R1 = 7 Ω
Substituting in equation (1)
7 + R2 = 15
R2 = 8 Ω, i.e; (when R1 = 7 Ω; R2 = 8 Ω)

Example 2.11

Calculate the equivalent resistance between A and B in the given circuit.

Solution

In all the sections, the resistors are connected in parallel.

Equivalent resistance is given by

R = Rp1 + Rp2 + Rp3
R = 1Ω + 2Ω + 3Ω = 6Ω
The circuit becomes

Equivalent resistance between A and B is

Example 2.12

Five resistors are connected in the configuration as shown in the figure. Calculate the equivalent resistance between the points a and b.

Solution

Case (a)

To find the equivalent resistance between the points a and b, we assume that a current is entering the junction at a. Since all the resistances in the outside loop are the same (1Ω), the current in the branches ac and ad must be equal. Hence the points C and D are at the same potential and no current through 5 Ω. It implies that the 5 Ω has no role in determining the equivalent resistance and it can be removed. So the circuit is simplified as shown in the figure

The equivalent resistance of the circuit between a and b is Req = 1 Ω

Colour code for Carbon resistors

Carbon resistors consists of a ceramic core, on which a thin layer of crystalline carbon is deposited as shown in Figure 2.11. These resistors are inexpensive, stable and compact in size. Colour rings are used to indicate the value of the resistance according to the rules given in the Table 2.2.

Three coloured rings are used to indicate the values of a resistor: the first two rings are significant figures of resistances, the third ring indicates the decimal multiplier after them. The fourth colour, silver or gold,

shows the tolerance of the resistor at 10% or 5% as shown in the Figure 2.12. If there is no fourth ring, the tolerance is 20%.

For the resistor shown in Figure 2.12, the first digit = 5 (green), the second digit = 6 (blue), decimal multiplier = 103 (orange) and tolerance = 5% (gold). The value of resistance = 56 × 103 Ω or 56 kΩ with the tolerance value 5%.

Temperature dependence of resistivity

The resistivity of a material is dependent on temperature. It is experimentally found that for a wide range of temperatures, the resistivity of a conductor increases with increase in temperature according to the expression, ρr = ρ0[1 + α(T – T0)] ……….. (2.27)

where ρT is the resistivity of a conductor at T0C, ρ0 is the resistivity of the conductor at some reference temperature T0 (usually at 20°C) and α is the temperature coefficient of resistivity. It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity at T0.

From the equation (2.27), we can write

where ∆ρ = ρτ – ρ0 is change in resistivity for a change in temperature ∆T = T – T0. Its unit is per oC.

α of Conductors

For conductors α is positive. If the temperature of a conductor increases, the average kinetic energy of electrons in the conductor increases. This results in more frequent collisions and hence the resistivity increases. The graph of the equation (2.27) is shown in Figure 2.13.

Even though, the resistivity of conductors like metals varies linearly for wide range of temperatures, there also exists a nonlinear region at very low temperatures. The resistivity approaches some finite value as the temperature approaches absolute zero as shown in Figure 2.13(b).

Using the equation ρ = R $$\frac{A}{l}$$ in equation (2.27), we get the expression for the resistance of a conductor at temperature T°C as

Rr = R0[1 + α(T – T0)] ……………. (2.28)

The temperature coefficient of resistivity can also be obtained from the equation (2.28),

where ∆R = RT – R0 is change in resistance during the change in temperature ∆T = T – T0

α of Semiconductors

For semiconductors, the resistivity decreases with increase in temperature. As the temperature increases, more electrons will be liberated from their atoms (Refer unit 9 for conduction in semi conductors)

Hence the current increases and therefore the resistivity decreases as shown in Figure 2.14. A semiconductor with a negative temperature coefficient of resistivity is called a thermistor.

The typical values of temperature coefficients of various materials are given in table 2.3.

We can understand the temperature dependence of resistivity in the following way. In section 2.1.3, we have shown that the electrical conductivity, σ = $$\frac{n e^{2} \tau}{m}$$. As the resistivity is inverse of σ, it can be written as ρ = $$\frac{m}{n e^{2} \tau}$$ …………. (2.30)

The resistivity of materials is

(i) inversely proportional to the number density (n) of the electrons
(ii) inversely proportional to the average time between the collisions (τ).

In metals, if the temperature increases, the average time between the collision (τ) decreases and n is independent of temperature. In semiconductors when temperature increases, n increases and τ decreases, but increase in n is dominant than decreasing τ, so that overall resistivity decreases.

Example 2.3

If the resistance of coil is 3 Ω at 20° C and α = 0.004/°C then determine its resistance at 100 °C.

Solution

R0 = 3 Ω, T = 100° C, T0 = 20°C
α = 0.004/°C, RT = ?
RT = R0(1 + α(T – T0))
R100 = 3(1 + 0.004 × 80)
R100 = 3.96 Ω

Example 2.4

Resistance of a material at 20°C and 40°C are 45 Ω and 85 Ω respectively. Find its temperature coefficient of resistivity.

Solution

T0 = 20°C, T = 40°C, R0 = 45 Ω, R = 85 Ω
α = $$\frac{1}{R_{\circ}}$$ $$\frac{∆R}{∆T}$$
α = $$\frac{1}{45}$$($$\frac{85-45}{40-20}$$) = $$\frac{1}{45}$$(2)
α = 0.044 per°C

Physics is now simple when learning with CBSELibrary – Get all important Physics Topics with detailed explanation.

Electric Current Definition – Formula and its Flow of Current

Matter is made up of atoms. Each atom consists of a positively charged nucleus with negatively charged electrons moving around the nucleus. Atoms in metals have one or more electrons which are loosely bound to the nucleus. These electrons are called free electrons and can be easily detached from the atoms.

The substances which have an abundance of these free electrons are called conductors. These free electrons move randomly throughout the conductor at a given temperature.

In general due to this random motion, there is no net transfer of charges from one end of the conductor to other end and hence no current in the conductor. When a potential difference is applied by the battery across the ends of the conductor, the free electrons drift towards the positive terminal of the battery, producing a net electric current. This is easily understandable from the analogy given in the Figure 2.1.

In the XI Volume 2, unit 6, we studied, that the mass move from higher gravitational potential to lower gravitational potential. Likewise, positive charge flows from region of higher electric potential to region of lower electric potential and negative charge flows
from region of lower electric potential to region of higher electric potential. So battery or electric cell simply creates potential difference across the conductor.

The electric current in a conductor is defined as the rate of flow of charges through a given cross-sectional area A. It is shown in the Figure 2.2

If a net charge Q passes through any cross section of a conductor in time t, then the current is defined as I = $$\frac{Q}{t}$$. But charge flow is not always constant. Hence current can more generally be defined as

Iavg = $$\frac{∆Q}{∆t}$$ …………. (2.1)

Where ∆Q is the amount of charge that passes through the conductor at any cross section during the time interval ∆t. If the rate at which charge flows changes with time, the current also changes. The instantaneous current I is defined as the limit of the average current, as ∆t → 0

The SI unit of current is the ampere (A)
1A = $$\frac{1C}{1s}$$

That is, 1A of current is equivalent to 1 coulomb of charge passing through a perpendicular cross section in a conductor in one second. The electric current is a scalar quantity.

Example 2.1

Compute the current in the wire if a charge of 120 C is flowing through a copper wire in 1 minute.

Solution

The current (rate of flow of charge) in the wire is

I = $$\frac{Q}{t}$$ = $$\frac{120}{60}$$ = 2A

Conventional Current

In an electric circuit, arrow heads are used to indicate the direction of flow of current. By convention, this flow in the circuit should be from the positive terminal of the battery to the negative terminal. This current is called the conventional current or simply current and is in the direction in which a positive test charge would move.

In typical circuits the charges that flow are actually electrons, from the negative terminal of the battery to the positive terminal. As a result, the flow of electrons and the direction of conventional current point in opposite direction as shown in Figure 2.3. Mathematically, a transfer of positive charge is the same as a transfer of negative charge in the opposite direction.

Drift Velocity

In a conductor the charge carriers are free electrons. These electrons move freely through the conductor and collide repeatedly with the positive ions. If there is no electric field, the electrons move in random directions, and hence their velocities are also randomly oriented. On an average, the number of electrons travelling in any direction will be equal to the number of electrons travelling in the opposite direction. As a result, there is no net flow of electrons in any direction and hence there will not be any current.

Suppose a potential difference is set across the conductor by connecting a battery, an electric field $$\vec{E}$$ is created in the conductor. This electric field exerts a force on the electrons, producing a current.

The electric field accelerates the electrons, while ions scatter the electrons and change their direction of motion. Thus, we see zigzag motion of electrons. In addition to the zigzag motion due to the collisions, the electrons move slowly along the conductor in a direction opposite to that of $$\vec{E}$$ as shown in the Figure 2.4.

This velocity is called drift velocity $$\vec{v}_{d}$$. The drift velocity is the average velocity acquired by the electrons inside the conductor when it is subjected to an electric field. The average time between two successive collisions is called the mean free time denoted by τ. The acceleration $$\vec {a}$$ experienced by the electron in an electric field $$\vec {E}$$ is given by

$$\vec {a}$$ = $$\frac{-e \vec{E}}{m}$$ (since $$\vec {F}$$ = – e$$\vec {E}$$) ………….. (2.3)

The drift velocity $$\vec{v}_{d}$$ is given by

Here µ = $$\frac{eτ}{m}$$ is the mobility of the electron and it is defined as the magnitude of the drift velocity per unit electric field.

µ = $$\frac{\left|\vec{v}_{d}\right|}{|\vec{E}|}$$ ………… (2.6)

The SI unit of mobility is m2V-1s-1.

Example 2.2

If an electric field of magnitude 570 N C-1 is applied in the copper wire, find the acceleration experienced by the electron.

Solution

E = 570 N C-1, e = 1.6 × 10-19C,
m = 9.11 × 10-31 kg and a = ?
F = ma = eE

Misconception

(i) There is a common misconception that the battery is the source of electrons. It is not true. When a battery is connected across the given wire, the electrons in the closed circuit resulting the current. Battery sets the potential difference (electrical energy) due to which these electrons in the conducting wire flow in a particular direction. The resulting electrical energy is used by electric bulb, electric fan etc. Similarly the electricity board is supplying the electrical energy to our home.

(ii) We often use the phrases like ‘charging the battery in my mobile’ and ‘my mobile phone battery has no charge’ etc. These sentences are not correct.

When we say ‘battery has no charge’, it means, that the battery has lost ability to provide energy or provide potential difference to the electrons in the circuit. When we say ‘mobile is charging’, it implies that the battery is receiving energy from AC power supply and not electrons.

Microscopic model of current

Consider a conductor with area of cross section A and let an electric field $$\vec{E}$$ be applied to it from right to left. Suppose there are n electrons per unit volume in the conductor and assume that all the electrons move with the same drift velocity $$\vec{v}_{d}$$ as shown in Figure 2.5

The drift velocity of the electrons = υd

If the electrons move through a distance dx within a small interval of dt, then
υd = $$\frac{dx}{dt}$$; dx = vddt …………. (2.7)

Since A is the area of cross section of the conductor, the electrons available in the volume of length dx is
= volume × number of electrons per unit volume
= A dx × n ………….. (2.8)

Substituting for dx from equation (2.7) in (2.8)
= (A υddt) n
Total charge in the volume element dQ = (charge) × (number of electrons in the volume element)

dQ = (e)(Aυddt)n
Hence the current I = $$\frac{dQ}{dt}$$
I = ne Aυd ………….. (2.9)

Current Density (J)

The current density (J) is defined as the current per unit area of cross section of the conductor.

J = $$\frac{I}{A}$$

The S.I unit of current density is $$\frac{\mathrm{A}}{\mathrm{m}^{2}}$$ (or) A m-2
J = $$\frac{n e A v_{d}}{A}$$ (form equation 2.9)
J = neυd …………… (2.10)

The above expression is valid only when the direction of the current is perpendicular to the area A. In general, the current density is a vector quantity and it is given by

$$\vec {J}$$ = ne$$\vec {υ}$$d
Substituting $$\overline{v_{d}}$$ from equation (2.4)

But conventionally, we take the direction of (conventional) current density as the direction of electric field. So the above equation becomes

$$\vec {J}$$ = σ$$\vec {E}$$ ………….. (2.12)

where σ = $$\frac{n e^{2} \tau}{m}$$ is called conductivity. The equation (2.12) is called microscopic form of ohm’s law.

The inverse of conductivity is called resistivity (ρ) [Refer section 2.2.1].

ρ = $$\frac{1}{σ}$$ = $$\frac{m}{n e^{2} \tau}$$ ……….. (2.13)

Example 2.3

A copper wire of cross-sectional area 0.5 mm2 carries a current of 0.2 A. If the free electron density of copper is 8.4 × 1028 m-3 then compute the drift velocity of free electrons.

Solution

The relation between drift velocity of electrons and current in a wire of crosssectional area A is

Example 2.4

Determine the number of electrons flowing per second through a conductor, when a current of 32 A flows through it.

Solution

I = 32 A , t = 1 s
Charge of an electron, e = 1.6 × 10-19 C
The number of electrons flowing per second, n = ?

Physics is now simple when learning with CBSELibrary – Get all important Physics Topics with detailed explanation.

Introduction of Haloalkens and Haloarenes – Definition, Classification, Uses

In the previous unit we learnt about the chemistry of hydrocarbons. In this unit us learn about organic compounds containing halogens. When one or more hydrogen atoms of aliphatic or aromatic hydrocarbons are replaced by the corresponding number of halogens like fluorine, chlorine, bromine or iodine, the resultant compounds are called either haloalkanes or halo arenes. They serve as starting materials for many organic synthesis.

Halogen substituted organic compounds are widely spread in nature and find application in our day to day life as well as in industry. Certain compounds like chloramphenicol produced by soil microbes are used in the treatment of typhoid; chloroquine is used in the treatment of malaria, halothane is used as an anesthetic, and halogenated solvents like trichloroethylene are used for cleaning electronic equipments.

Haloalkanes are hydrocarbons containing aliphatic alkane with one or more hydrogen atoms replaced by halogens. Haloarenes are hydrocarbons containing aromatic alkane with one or more hydrogen atoms replaced by halogens.

Classification of Haloalkanes

The haloalkanes, also known as alkyl halides, are a group of chemical compounds comprised of an alkane with one or more hydrogens replaced by a halogen atom (fluorine, chlorine, bromine, or iodine). The classification is determined by the number of carbons bonded to the carbon bearing the halide.

Name of Haloalkanes and Haloarenes

IUPAC name – The IUPAC (International Union of Pure and Applied Chemistry) naming system is the standard naming system that chemists generally use.

Reaction of Haloalkanes and Haloarenes

Haloarenes are more stable than the haloalkane that’s why haloarenes are less reactive than the haloalkane. The reasons for being stable are dependent on mainly three factors that are polarity of carbon-halogen bond, hybrid state of the carbon atom in the haloarenes and the presence of resonance.

Halide Functional Group

An acyl halide (also known as an acid halide) is a chemical compound derived from an oxoacid by replacing a hydroxyl group with a halide group. If the acid is a carboxylic acid, the compound contains a – COX functional group, which consists of a carbonyl group singly bonded to a halogen atom.

Functional Group of Haloalkenes

Alkyl halides [haloalkanes] consist of an alkyl group attached to a halogen: F, Cl, Br, I. Chloro, bromo and iodo alkyl halides are often susceptible to elimination and/or nucleophilic substitution reactions. Aldehydes have a hydrogen and an alkyl (or aromatic) group attached to a carbonyl function.

Find free online Chemistry Topics covering a broad range of concepts from research institutes around the world.

Orbital Overlap | Definition, Examples, Diagrams

When atoms combines to form a covalent molecule, the atomic orbitals of the combining atoms overlap to form a covalent bond. The bond pair of electrons will occupy the overlapped region of the orbitals. Depending upon the nature of overlap we can classify the covalent bonding between the two atoms as sigma (σ) and pi (π) bonds.

Sigma and Pi bonds

When two atomic orbitals overlap linearly along the axis, the resultant bond is called a sigma (σ) bond. T is overlap is also called ‘head-on overlap’ or ‘axial overlap’. Overlap involves an s orbital (s-s and s-p overlaps) will always result in a sigma bond as the s orbital is spherical.

Overlap between two p orbitals along the molecular axis will also result in sigma bond formation. When we consider x-axis as molecular axis, the px-px overlap will result in σ-bond.

When two atomic orbitals overlaps sideways, the resultant covalent bond is called a pi(π)bond. When we consider x-axis as molecular axis, the py-py and pz-pz overlaps will result in the formation of a π-bond.

Following examples will be useful to understand the overlap:

Formation of hydrogen (H2) Molecule

Electronic configuration of hydrogen atom is 1s1.

During the formation of H2 molecule, the 1s orbitals of two hydrogen atoms containing one unpaired electron with opposite spin overlap with each other along the internuclear axis. This overlap is called s-s overlap. Such axial overlap results in the formation of a σ-covalent bond.

Formation of fluorine molecule (F2):

Valence shell electronic configuration of fluorine atom:
2s2 2px2, 2py2, 2pz1

When the half filled pz orbitals of two fluorine overlaps along the z-axis, a σ-covalent bond is formed between them.

Formation of HF molecule:

Electronic configuration of hydrogen atom is 1s1
Valence shell electronic configuration of flourine atom: 2s2 2px2, 2py2, 2pz1
When half filled 1s orbital of hydrogen linearly overlaps with a half filled 2pz orbital flourine, as σ-covalent bond is formed between hydrogen and flourine.

Formation of oxygen molecule (O2):

Valence shell electronic configuration of oxygen atom:
2s2 2px2, 2py1, 2pz1

When the half filled pz orbitals of two oxygen overlaps along the z-axis (considering molecular axis as z axis), a σ-covalent bond is formed between them. Other two half filled py orbitals of two oxygen atoms overlap laterally (sideways) to form a π-covalent bond between the oxygen atoms.

Thus, in oxygen molecule, two oxygen atoms are connected by two covalent bonds (double bond). The other two pair of electrons present in the 2s and 2px orbital do not involve in bonding and remains as lone pairs on the respective oxygen.

Find free online Chemistry Topics covering a broad range of concepts from research institutes around the world.

Aromatic Hydrocarbons – Formula, Definition, Structure, Properties

Take a moment and think of substances that have a strong fragrance. What kind of things come to your mind?

Perfume, Vanila or cinnamon? They smell differently, they have something in common. These substances are made of aromatic compounds [Greek: AromaPleasant smelling]. However, some compounds are chemically aromatic but do not have distinct smell. The aromatic hydrocarbons are classified depending upon number of rings present in it.

(i) Monocyclic aromatic hydrocarbon (MAH) (Ex) Benzene (C6H6) and Toluene (C7H8)

(ii) Polycyclic aromatic hydrocarbon (PAH) (Ex) Naphthalene (C10H8) and Anthracene (C14H10).

Nomenclature and Isomerism

We have already discussed about nomenclature of aromatic hydrocarbons in Unit: 11. The first member of aromatic hydrocarbon is benzene (C6H6) represented by a regular hexagon with a circle inscribed in it.

Since, all the six hydrogen atom in benzene are equivalent, it can give only one monosubstituted compound (Ex) methyl benzene (C6H5-CH3) which named as toluene.

When di substitution occurs either by a similar monovalent atom or two different atoms or groups in benzene, then three different position isomers are possible. Their relative positions are indicated as ortho (1, 2), meta (1, 3) and para (1, 4). For example, consider dimethyl benzene which is named as xylene.

Aromaticity

Huckel proposed that aromaticity is a function of electronic structure. A compound may be aromatic, if it obeys the following rules

• The molecule must be co-planar
• Complete delocalization of π electron in the ring
• Presence of (4n+2) π electrons in the ring where n is an integer (n=0, 1, 2….)

This is known as Huckel’s rule

Some of the examples for the Huckel rule

Structure of benzene:

1. Molecular formula

Elemental Analysis and molecular weight determination have proved that the molecular formula of benzene is C6H6. This indicates that benzene is a highly unsaturated compound.

2. Straight chain structure not possible:

Benzene could be constructed as a straight chain or ring compound but it not feasible since it does not show the properties
of alkenes or alkynes. for example, it did not decolourise bromine in carbon tetrachloride or acidified KMnO4. It did not react with water in the presence of acid.

3. Evidence of cyclic structure:

I. Substitution of Benzene:

Benzene reacts with bromine in the presence of AlCl3 to form mono bromo benzene.

Formation of only one monobromo compound indicates that all the six hydrogen atoms in benzene were identical. This is possible only if it has a cyclic structure of six carbons each containing one hydrogen.

II. Addition of Hydrogen:

Benzene can add on to three moles of hydrogen in the presence of nickel catalyst to give cyclohexane.

This confirms cyclic structure of benzene and the presence of three carbon-carbon double bond.

4. Kekule’s Structure of Benzene:

In 1865, August Kekule suggested that benzene consists of a cyclic planar structure of six carbon with alternate single and double bonds. There were two objections:

(i) Benzene forms only one ortho disubstituted products whereas the Kekule’s structure predicts two o-di
substituted products as shown below.

(ii) Kekule’s structure failed to explain why benzene with three double bonds did not give addition reactions like other alkenes. To overcome this objection, Kekule suggested that benzene was mixture of two forms (1 and 2)which are in rapid equilibrium.

5. Resonance description of benzene:

The phenomenon in which two or more structures can be written for a substance which has identical position of atoms is called resonance. The actual structure of the molecule is said to be resonance hybrid of various possible alternative structures. In benzene, Kekule’s structures I & II represented the resonance structure, and structure III is the resonance hybrid of structure I & II.

The structures 1 and 2 exist only in theory. The actual structure of benzene is the hybrid of two hypothetical resonance structures.

6. Spectrosscopic Measurments:

Spectroscopic measurements show that benzene is planar and all of its carboncarbon bonds are of equal length 1.40A°. This value lies between carbon-carbon single bond length 1.54A° and carboncarbon double bond length 1.34A°.

7. Molecular Orbital Structure:

The structure of benzene is best described in terms of the molecular orbital theory. All the six carbon atoms of benzene are sp2 hybridized. Six sp2 hybrid orbitals of carbon linearly overlap with six 1s orbitals of hydrogen atoms to form six C – H sigma bonds. Overlap between the remaining sp2 hybrid orbitals of carbon forms six C-C sigma bonds.

All the σ bonds in benzene lie in one plane with bond angle 120°. Each carbon atom in benzene possess an un hybridized p-orbital containing one electron. The lateral overlap of their p-orbital produces 3 π- bond. The six electrons of the p-orbitals cover all the six carbon atoms and are said to be delocalised.

Due to delocalization, strong π-bond is formed which makes the molecule stable. Hence unlike alkenes and alkynes benzene undergoes substitution reactions rather addition reactions under normal conditions.

Representation of Benzene:

Hence, there are three ways in which benzene can be represented.

Benzene and its Homologous Series

Benzene and its homologous series are colorless liquids with pleasant odour. They are lighter than water and insoluble in it. Their vapours are highly flammable, and volatile and toxic in nature.

Sources of aromatic compound:

• Benzene and other aromatic compound are obtained from coal tar and petroleum
• It can also be prepared in laboratory using some simple aliphatic compounds

1. Preparation of Benzene

(i) industrial preparation of benzene from coal tar:

Coal tar is a viscous liquid obtained by the pyrolysis of coal. During fractional distillation, coal tar is heated and distills away its volatile compounds namely benzene, toluene, xylene in the temperature range of 350 to 443 K. These vapours are collected at the upper part of the fractionating column (Table 13.5.)

(ii) From Acetylene

Acetylene on passing through a red – hot tube trimerises to give benzene. We have already studied this concept in polymerization of alkynes.

(iii) Laboratory Methods Of Preparing Benzene And Toluene

(a) Decarboxylaation Of Aromatic Acid. When sodium benzoate in heated with sodalime, benzene vapours distil over.

(b) Preparation Of Benzene From Phenol

When phenol vapours are passed over zinc dust, then it is reduced to benzene.

(c) Wurtz – Fitting Reaction:

When a solution of bromo benzene and iodo methane in dry ether is treated with metallic sodium, toluene is formed.

(d) Friedel Craft’s Reaction:

When benzene is treated with methyl chloride in the presence of anhydrous aluminium chloride, toluene is formed.

Physical Properties:

1. Benzene is a colourless liquid, insoluble in water and solution alcohol, ether and chloroform.
2. It burns with luminous sooty flme in contrast to alkanes and alkenes which usually burn with bluish flame.
3. Their vapours are highly toxic which on inhalation produce loss of consciousness.

Chemical Properties:

1. Benzene contains delocalized π-electrons which make the ring to act as an electron rich centre. So electrophilic substitution reaction occurs in benzene.
2. Benzene ring is stabilized by delocalized π electrons. Though it is highly stable, it undergoes addition and oxidation reaction under specific conditions.

1. Electrophilic Substitution Reaction

(a) Nitration:

When benzene is heated at 330K with a nitrating mixture (Con. HNO3 + Con. H2SO4), nitro benzene is formed by replacing one hydrogen atom by nitronium ion NO+2 (electrophile)

Concentrated H2SO4 is added to produce nitronium ion NO+2

(b) Halogenation:

Benzene reacts with halogens (X2=Cl2, Br2,) in the presence of Lewis acid such as FeCl3, FeBr3 or AlCl3 and give corresponding halo benzene. In the absence of catalyst, Fluorine reacts vigoursly with benzene even in the absence of catalyst. However iodine is very inactive even in the presence of catalyst.

(c) Sulphonation:

Benzene reacts with fuming sulphuric acid (Con H24 + SO3) and gives benzene sulphonic acid. The electrophile SO3 is a molecule. Although it does not have positive charge, it is a strong electrophile. This is because the octet of electron around the sulphur atom is not reached. The reaction is reversible and desulphonation occurs readily in aqueous medium.

(d) Friedel Craft’s Alkylation: (Methylation)

When benzene is treated with an alkyl halide in the presence of only AlCl3, alkyl benzene is formed.

(e) Friedel Craft’s Acylation: Acetylation

When benzene is treated with acetyl chloride in the presence of AlCl3, acyl benzene is formed

(f) Electrophilic Subistution Reactions: Mechanism

Benzene undergoes electrophilic substitution reaction because it is an electron-rich system due to delocalised π electron. So it is easily attacked by electrophilies and gives substituted products.

Mechanism:

Step: 1
Formulation of the electrophile

E – Nu + Catalyst → E+ + Nu-Catalyst

Step: 2

The electrophile attacks the aromatic ring to form a carbocation intermediate which is stabilized by resonance.

Step: 3

Loss of proton gives the substitution product.

(ii) Addition Reaction:

a. Hydrogenation of benzene:

Benzene reacts with hydrogen in the presence of Platinum or Palladium to yield Cyclohexane. This is known as hydrogenation.

b. Chlorination of Benzene:

Benzene reacts with three molecules of Cl2 in the presence of sun light or UV light to yield Benzene Hexa Chloride (BHC) C6H6Cl6. This is known as gammaxane or Lindane which is a powerful insecticide.

(iii) Oxidation:
a. Vapour – phase oxidation:-

Although benzene is very stable to strong oxidizing agents, it quickly undergoes vapour phase oxidation by passing its vapour mixed with oxygen over V2O5 at 773k. The ring breaks to give maleic anhydride.

b. Birch Reduction:

Benzene can be reduced to 1, 4-cyclohexadiene by treatment with Na or Li in a mixture of liquid ammonia and alcohol. It is the convenient method to prepare cyclic dienes.

Directive Influence of a Functional Group in Monosubtituted Benzene:

When mono substituted benzene undergoes an electrophilic substitution reaction, the rate of the reaction and the site of attack of the incoming electrophile depends on the functional group already attached to it. Some groups increase the reactivity of benzene ring and are known as activating groups. While others which decrease the reactivity are known as deactivating groups.

We further divide these groups into two categories depending on the way they influence the orientation of attack by the incoming groups. These which increases electron density at ‘ortho’ and ‘para’ position are known as orthopara directors while those which increase electron density at ‘meta’ position is known as meta-directors. Some examples of directive influence of functional groups in mono-substituted benzene are explained below.

Ortho and Para Directing Groups

All the activating groups are ‘orthopara’ directors. Example – OH, – NH2, – NHR, – NHCOCH3, – OCH3 – CH3 – C2H5 etc. Let us consider the directive inflences of phenolic (-OH) group. Phenol is the resonance hybrid of following structures.

In these resonance structures, the (-) charge residue is present on ortho and para position of ring structure. It is quite evident that the lone pair of electron on the atom which is attached to the ring involves in resonance and makes the ring more electron rich than benzene. The electron density at ortho and parapositions increases as compared to the meta position. Therefore phenolic group activates the benzene ring for electrophilic attack at ‘ortho’ and ‘para positions and hence -OH group is an orthopara director and activator.

In aryl halides, the strong -I effect of the halogens (electron withdrawing tendency) decreases the electron density of benzene ring, thereby deactivating for electrophilic attack. However the presence of lone pair on halogens involved in the resonance with pi electrons of benzene ring, increases electron density at ortho and para positions. Hence the halogen group is an ortho-para director and deactivator.

META DIRECTING GROUPS

Generally all deactivating groups are meta-directors. For example -NO2, -CN, -CHO, -COR, -COOH, -COOR, -SO3H etc. Let us consider the directive influence of aldehyde (-CHO) group. Benz aldehyde is the resonance hybrid of following structures.

In these resonance structures, the (+) charge residues is present on the ring structure. It is quite evident that resonance delocalizes the positive charge on the atoms of the ring, making the ring less electron rich than benzene. Here overall density of benzene ring decreases due to -I effect of -CHO group there by deactivating the benzene for electrophilic attack. However resonating structure shows that electron density is more in meta position. Compared to o & p-position. Hence -CHO group is a meta-director and deactivator.

Carcinogenity and Toxicity

Benzene and polycyclic aromatic hydrocarbons (PAH) are ubiquitous environmental pollutants generated during incomplete combustion of coal oil, petrol and wood. Some PAH originate from open burning, natural seepage of petroleum and coal deposits and volcanic activities. They are toxic, mutagenic and carcinogenic. It has hematological. immunological and neurological effect on humans. They are radiomimetic and prolonged exposure leads to genetic damage. Some of the examples of PAH are.

“L” shaped polynuclear hydrocarbons are much more toxic & carcinogenic

Found in cigarette smoke Found in tobacco and cigarette and charcoal boiled food

FLOWCHART AND REACTION SUMMARY OF HYDROCARBON

Find free online Chemistry Topics covering a broad range of concepts from research institutes around the world.

Alkynes – Formula, Definition, Structure, Properties

Alkynes are unsaturated hydrocarbons that contain carbon-carbon triple bonds in their molecules. Their general formula is CnH2n-2. The first member of alkyne series is Ethyne popularly known as acetylene. Oxyacetylene torch is used in welding.

Nomencluture of alkynes:

Let us write the IUPAC name for the below mentioned alkynes by applying the general rules of nomenclature that we already discussed in unit No.11

Preparation of alkynes from alkenes:

This process involves two steps:

(i) Halogenation of alkenes to form vicinal dihalides
(ii) Dehalogenation of vicinal dihalides to form alkynes.

Preparation of alkene from gem dihalides:

A compound containing two halogen atoms on the same carbon atom is called gem dihalide (Latin word ‘Gemini’ means twins). On heating with alcoholic KOH, gem dihalides give alkynes.

3. Preparation of alkynes from electrolysis of salts of unsaturated dicarboxylic acids. (Kolbe’s electrolytic method)

Electrolysis of sodium or potassium salt of maleic or fumaric acid yields alkynes.

4. Industrial preparation of ethyne:

Ethyne can be manufactured in large scale by action of calcium carbide with water.

Calcium carbide required for this reaction is prepared by heating quick lime and coke in an electric furance at 3273 K

Physical properties of alkynes:

1. The first three members are gases, next eight are liquids and the higher alkynes are solids. They are all colourless and odourless except acetylene which has garlic odour.

2. They are slightly soluble in water but dissolve readily in organic solvents like benzene, acetone and ethyl alcohol.

Chemical properities of alkynes

Terminal Alkynes are acidic in nature. It undergoes polymerization and addition reaction.

1. Acidic nature of alkynes:

An alkyne shows acidic nature only if it contains terminal hydrogen. This can be explained by considering sp hybrid orbitals of carbon atom in alkynes.

The percentage of s-character of sp hybrid orbital (50%) is more than sp2 hybrid orbital of alkene (33%) and sp3 hybrid orbital of alkane (25%). Because of this, Carbon becomes more electronegative facilitating donation of H+ ion to bases. So hydrogen attached to triply bonded carbon atoms is acidic.

2. Addition reactions of alkynes

(i) Addition of hydrogen

(ii) Addition Of Halogens:

(iii) Addition Of Hydrogen Halides:

Reaction of hydrogen halides to symmetrical alkynes is electrophilic addition reaction. This reaction also follows Markovnikof ’s rule.

Addition of HBr to unsymmetrical alkene follows Markownikoff ’s rule.

(iv) Addition of Water:

Alkynes undergo hydration on warming with mercuric sulphate and dilute H2SO4 at 333K to form carbonyl compounds.

3. Ozonolysis:

Ozone adds to carbon-carbon triple bond of alkynes to form ozonides. The ozonides are hydrolyzed by water to form carbonyl compounds. The hydrogen peroxide (H2O2) formed in the reaction may oxidise the carbonyl compound to carboxylic acid.

4. Polymerisation:

Alkyne undergoes two types of polymerisation reaction

(i) Linear Polymerisation:

Ethyne forms linear polymer, when passed into a solution of cuprous chloride and ammonium chloride.

(ii) Cyclic Polymerisation:

Ethyne undergoes cyclic polymerization on passing through red hot iron tube. Three molecules of ethynepolymerises to
benzene.

Uses of Alkynes

1. Acetylene is used in oxy acetylene torch used for welding and cutting metals.
2. It is used for manufacture of PVC, polyvinyl acetate, polyvinyl ether, orlon and neoprene rubbers.

Find free online Chemistry Topics covering a broad range of concepts from research institutes around the world.

Alkanes – Formula, Definition, Structure, Properties

Alkanes are saturated hydrocarbons represented by the general formula CnH2n+2 where ‘n’ is the number of carbon atoms in the molecule. Methane CH4, is the first member of alkane family. The successive members are ethane C2H6, propane C3H8, butane C4H10, pentane C5H12 and so on. It is evident that each member differs from its proceeding or succeeding member by a – CH2 group.

Nomenclature and Isomerism:

We have already discussed the nomenclature of organic compound in Unit:11. Let us understand the nomenclature and isomerism in few examples. The first three members methane CH4, ethane C2H6 and propane C3H8 have only one structure.

However, higher members can have more than one structure leading to constitutional isomers (differ in connectivity) or structural isomers. For example, an alkane with molecular formula C4H10 can have two structures. They are n-butane and iso-butane. In n-butane, all the four carbon atoms are arranged in a continuous chain. The ‘n’ in n-butane stand for ‘normal’ and means that the carbon chain is unbranched. The second isomer iso-butane has a branched carbon chain. The word iso indicates it is an isomer of butane.

Though both the structures have same molecular formula but their carbon chains differ leading to chain isomerism. Let us understand the chain isomerism by writing the isomers of pentane C5H12

Solution:

IUPAC name for some branched alkanes

Let us write the IUPAC name for the below mentioned alkanes by applying the general rules of nomenclature that we already discussed in unit No.11

How to draw structural formula for given IUPAC name:

After you learn the rules for naming alkanes, it is relatively easy to reverse the procedure and translate the name of an alkane into a structural formula. The example below show how this is done.

Let us draw the structural formula for

(a) 3-ethyl-2,3-dimethyl pentane

Solution:

Step 1:

The parent hydrocarbon is pentane. Draw the chain of fie carbon atoms and number it.

Step 2:

Complete the carbon skeleton by attaching the alkyl group as they are specified in the name. An ethyl group is attached to carbon 3 and two methyl groups are attached to carbon 2 and 3.

Step 3:

Add hydrogen atoms to the carbon skeleton so that each carbon atoms has four bonds

Preparation of Alkanes:

Alkanes are not laboratory curiosities but they are extremely important naturally occurring compounds. Natural gas and
petroleum (crude oil) are the most important natural sources. However, it can be prepared by the following methods.

1. Preparation of alkanes from catalytic reduction of unsaturated hydrocarbons:

When a mixture hydrogen gas with alkene or alkyne gas is passed over a catalysts such as platinum or palladium at room temperature, an alkane is produced. This process of addition of H2 to unsaturated compounds is known as hydrogenation. The above process can be catalysed by nickel at 298K. This reaction is known as Sabatier Sendersens reaction

For example:

2. Preparation of alkanes from carboxylic acids:

(i) Decarboxylation of sodium salt of carboxylic acid

When a mixture of sodium salt of carboxylic acid and soda lime (sodium hydroxide + calcium oxide) is heated, alkane is formed. The alkane formed has one carbon atom less than carboxylic acid. This process of eliminating carboxylic group
is known as decarboxylation.

For example:

(ii) Kolbe’s Electrolytic Method

When sodium or potassium salt of carboxylic acid is electrolyzed, a higher alkane is formed. The decarboxylative dimerization of two carboxylic acid occurs. This method is suitable for preparing symmetrical alkanes(R-R).

3. Preparation of alkanes using alkyl halides (or) halo alkanes

(i) By reduction with nascent hydrogen

Except alkyl florides, other alkyl halides can be converted to alkanes by reduction with nascent hydrogen. The hydrogen for reduction may be obtained by using any of the following reducing agents:
Zn+HCl, Zn+CH3COOH, Zn-Cu couple in ethanol, LiAlH4 etc.,

For Example:

(ii) Wurtz Reaction

When a solution of halo alkanes in dry ether is treated with sodium metal, higher alkanes are produced. This reaction is used to prepare higher alkanes with even number of carbon atoms.

For Example:

(iii) Corey – House Mechanism

An alkyl halide and lithium di alkyl copper are reacted to give higher alkane.

For Example:

4. Preparation of Alkanes from Grignard Reagents

Halo alkanes reacts with magnesium in the presence of dry ethers to give alkyl magnesium halide which is known as Grignard reagents. Here the alkyl group is directly attached to the magnesium metal make it to behave as carbanion. So, any compound with easily replaceable hydrogen reacts with Grignard reagent to give corresponding alkanes.

For Example:

Physical Properties:

1. Boiling Point and Physical State

The boiling point of continuous chain alkanes increases with increases in length of carbon chain roughly about 30°C for every added carbon atom to the chain. Being non polar, alkanes have weak Vanderwal’s force which depends upon molecular surface area and hence increases with increase molecular size.

We observe that with same number of carbon atoms, straight chain isomers have higher boiling point compared to branch chain isomers. The boiling point decreases with increase in branching as the molecule becomes compact and the area of the contact decreases.

2. Solubility and Density

Water molecules are polar and alkanes are non-polar. The insolubility of alkanes in water makes them good water repellent for metals which protects the metal surface from corrosion. Because of their lower density than water, they form two layers and occupy top layer. The density difference between alkanes and water explains why oil spills in aqueous environment spread so quickly.

Conformations of Alkane:

Each carbon in alkanes is sp3 hybridized and the four groups or atoms around the carbon are tetrahedrally bonded. In alkanes having two or more carbons, there exists free rotation about C-C single bond. Such rotation leaves all the groups or atoms bonded to each carbon into an infinite number of readily interconvertible three dimensional arrangements. Such readily interconvertible three dimensional arrangement of a molecule is called conformations.

(i) Conformations of Ethane:

The two tetrahedral methyl groups can rotate about the carbon – carbon bond axis yielding several arrangements called conformers. The extreme conformations are staggered and eclipsed conformation. There can be number of other arrangements between staggered and eclipsed forms and their arrangements are known as skew forms.

Eclipsed Conformation

In this conformation, the hydrogen’s of one carbon are directly behind those of the other. The repulsion between the atoms is maximum and it is the least stable conformer.

Staggered Conformation:

In this conformation, the hydrogens of both the carbon atoms are far apart from each other. The repulsion between the atoms is minimum and it is the most stable conformer.

Skew Conformation:

The infinite numbers of possible intermediate conformations between the two extreme conformations are referred as skew conformations.

The stabilities of various conformations of ethane are
Staggered > Skew > Eclipsed

The potential energy difference between the staggered and eclipsed conformation of ethane is around 12.5 KJmol-1. The various conformations can be represented by new man projection formula.

Newman Projection formula for Ethane

Conformations of n-Butane:

n-Butane may be considered as a derivative of ethane, as one hydrogen on each carbon is replaced by a methyl group

Eclipsed Conformation:

In this conformation, the distance between the two methyl group is minimum. So there is maximum repulsion between them and it is the least stable conformer.

Anti or Staggered Form

In this conformation, the distance between the two methyl groups is maximum and so there is minimum repulsion between them. And it is the most stable conformer. The following potential energy diagram shows the relative stabilities of various conformers of n-butane.

Chemical Properties:

Alkanes are quite unreactive towards most reagents. However under favourable conditions, alkanes undergo the following type of reaction.

Combustion:

A combustion reaction is a chemical reaction between a substances and oxygen with evolution of heat and light (usually as a flame). In the presence of sufficient oxygen, alkanes undergoes combustion when ignited and produces carbondioxide and water.

For Example:

CH4 + 2O2 → CO2 + 2H2O ∆H° = – 890.4kJ

When alkanes burn in insufficient supply of oxygen, they form carbonmonoxide and carbon black.

Halogenation:

Halogenation reaction is the chemical reaction between an alkane and halogen in which one or more hydrogen atoms are substituted by the halogens.

Chlorination and Bromination are two widely used halogenation reactions. Fluorination is too quick and iodination is too slow. Methane reacts with chlorine in the presence of light or when heated as follows.

Mechanism:

The reaction proceeds through the free radical chain mechanism. This mechanism is characterized by three steps initiation, propagation and termination.

(i) CHAIN INITITATION:

The chain is initiated by UV light leading to homolytic fission of chlorine molecules into free radicals (chlorine atoms).

Here we choose Cl-Cl bond for fision because C-C & C-H bonds are stronger than Cl-Cl.

(ii) PROPAGATION:

It proceeds as follows,

(a) Chlorine free radial attacks the methane molecule and breaks the C-H bond resulting in the generation of methyl free radical.

(b) The methyl free radical thus obtained attacks the second molecule of chlorine to give chloromethane (CH3Cl) and a chlorine free radical as follows.

(c) This chlorine free radical then cycles back to step

(a) and both step (a) and (b) are repeated many times and thus chain of reaction is set up.

(iii) Chain Termination:

After sometimes, the reactions stops due to consumption of reactant and the chain is terminated by the combination of free radicals.

3. Aromatisation

Alkanes with six to ten carbon atoms are converted into homologous of benzene at high temperature and in the presence of catalyst. This process is known as aromatization. It occurs by simultaneous cyclisation followed by dehydrogenation of alkanes.

n-Hexane passed over Cr2O3 supported on alumina at 873 K gives benzene.

4. Reaction With Steam:

Methane reacts with steam at 1273K in the presence of Nickel and decomposes to form carbon monoxide and hydrogen gas.

Production of H2 gas from methane is known as steam reforming process and it is a well-established industrial process for the production of H2 gas from hydrocarbons.

5. Pyrolysis

Pyrolysis is defined as the thermal decomposition of organic compound into smaller fragments in the absence of air through the application of heat. ‘Pyro’ means ‘fie’ and ‘lysis’ means ‘separating’. Pyrolysis of alkanes also named as cracking. In the absence of air, when alkane vapours are passed through red-hot metal it breaks down into simpler hydrocarbons.

The products depends upon the nature of alkane, temperature, pressure and presence or absence of catalyst. The ease of cracking in alkanes increases with increase in molecular weight and branching in alkanes. Cracking plays an important role in petroleum industry.

6. Isomerisation:

Isomerisation is a chemical process by which a compound is transformed into any of its isomeric forms. Normal alkanes can be converted into branched alkanes in the presence of AlCl3 and HCl at 298 k.

This process is of great industrial importance. The quality of gasoline is improved by isomerising its components.

Uses

The exothermic nature of alkane combustion reaction explains the extensive use of alkanes as fuels. Methane present in natural gas is used in home heating. Mixture of propane and butane are known as LPG gas which is used for domestic cooking purpose. GASOLINE is a complex mixture of many hydrocarbons used as a fuel for internalcombustion engines.

Carbon black is used in the manufacture of ink, printer ink and black pigments. It is also used as fillers.

Find free online Chemistry Topics covering a broad range of concepts from research institutes around the world.

Distribution of Charges In a Conductor and Action at Points

Distribution of charges in a conductor

Consider two conducting spheres A and B of radii r1 and r2 respectively connected to each other by a thin conducting wire as shown in the Figure 1.60. The distance between the spheres is much greater than the radii of either spheres.

If a charge Q is introduced into any one of the spheres, this charge Q is redistributed into both the spheres such that the electrostatic potential is same in both the spheres. They are now uniformly charged and attain electrostatic equilibrium.

Let q1 be the charge residing on the surface of sphere A and q2 is the charge residing on the surface of sphere B such that Q = q1 + q2. The charges are distributed only on the surface and there is no net charge inside the conductor.

The electrostatic potential at the surface of the sphere A is given by

VA = $$\frac{1}{4 \pi \epsilon} \frac{q_{1}}{r_{1}}$$ ……….. (1.110)

The electrostatic potential at the surface of the sphere B is given by

VB = $$\frac{1}{4 \pi \epsilon_{0}} \frac{q_{2}}{r_{2}}$$ ………… (1.111)

The surface of the conductor is an equipotential. Since the spheres are connected by the conducting wire, the surfaces of both the spheres together form an equipotential surface. This implies that

VA = VB
or $$\frac{q_{1}}{r_{1}}=\frac{q_{2}}{r_{2}}$$ ……….. (1.112)

Let the charge density on the surface of sphere A be σ1 and that on the surface of sphere B be σ2. This implies that q1 = 4πr12σ1 and q2 = 4πr22. Substituting these values into equation (1.112), we get

σ1r1 = σ2r2 ……….. (1.113)

from which we conclude that

σr = constant ……………. (1.114)

Thus the surface charge density σ is inversely proportional to the radius of the sphere. For a smaller radius, the charge density will be larger and vice versa.

Example 1.23

Two conducting spheres of radius r1 = 8 cm and r2 = 2 cm are separated by a distance much larger than 8 cm and are connected by a thin conducting wire as shown in the figure. A total charge of Q = +100 nC is placed on one of the spheres. After a fraction of a second, the charge Q is redistributed and both the spheres attain electrostatic equilibrium.

(a) Calculate the charge and surface charge density on each sphere.
(b) Calculate the potential at the surface of each sphere.

Solution

(a) The electrostatic potential on the surface of the sphere A is VA = $$\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1}}{r_{1}}$$

The electrostatic potential on the surface of the sphere B is VB = $$\frac{1}{4 \pi \epsilon} \frac{q_{2}}{r_{2}}$$

Since VA = VB. We have

But from the conservation of total charge, Q = q1 + q2, we get q1 = Q – q2. By substituting this in the above equation,

Q – q2 = ($$\frac{r_{1}}{r_{2}}$$)q2

so that q2 = Q($$\frac{r_{2}}{r_{1}+r_{2}}$$)

Therefore,

q2 = 100 × 10-9 × ($$\frac{2}{10}$$) = 20 nC

and q1 = Q – q2 = 80 nC

The electric charge density on sphere A is σ1 = $$\frac{q_{1}}{4 \pi r_{1}^{2}}$$
The electric charge density on sphere B is σ2 = $$\frac{q_{2}}{4 \pi r_{2}^{2}}$$

Therefore,

Note that the surface charge density is greater on the smaller sphere compared to the larger sphere (σ2 ~ 4σ1) which confirms the result $$\frac{\sigma_{1}}{\sigma_{2}}=\frac{r_{2}}{r_{1}}$$

The potential on both spheres is the same. So we can calculate the potential on any one of the spheres

Action of points or Corona discharge

Consider a charged conductor of irregular shape as shown in Figure 1.61 (a).

We know that smaller the radius of curvature, the larger is the charge density. The end of the conductor which has larger curvature (smaller radius) has a large charge accumulation as shown in Figure 1.61 (b).

As a result, the electric field near this edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge. This reduces the total charge of the conductor near the sharp edge. This is called action of points or corona discharge.

Lightning Arrester or lightning Conductor

This is a device used to protect tall buildings from lightning strikes. It works on the principle of action at points or corona discharge. This device consists of a long thick copper rod passing from top of the building to the ground. The upper end of the rod has a sharp spike or a sharp needle as shown in Figure 1.62 (a) and (b).

The lower end of the rod is connected to copper plate which is buried deep into the ground. When a negatively charged cloud is passing above the building, it induces a positive charge on the spike. Since the induced charge density on thin sharp spike is large, it results in a corona discharge.

This positive charge ionizes the surrounding air which in turn neutralizes the negative charge in the cloud. The negative charge pushed to the spikes passes through the copper rod and is safely diverted to the Earth. The lightning arrester does not stop the lightning; rather it diverts the lightning to the ground safely.

Van de Graaff Generator

In the year 1929, Robert Van de Graaff designed a machine which produces a large amount of electrostatic potential difference, up to several million volts (107 V). This Van de Graff generator works on the principle of electrostatic induction and action at points.

A large hollow spherical conductor is fixed on the insulating stand as shown in Figure 1.63. A pulley B is mounted at the centre of the hollow sphere and another pulley C is fixed at the bottom. A belt made up of insulating materials like silk or rubber runs over both pulleys. The pulley C is driven continuously by the electric motor. Two comb shaped metallic conductors E and D are fixed near the pulleys.

The comb D is maintained at a positive potential of 104 V by a power supply. The upper comb E is connected to the inner side of the hollow metal sphere.

Due to the high electric field near comb D, air between the belt and comb D gets ionized by the action of points. The positive charges are pushed towards the belt and negative charges are attracted towards the comb D. The positive charges stick to the belt and move up. When the positive charges on the belt reach the point near the comb E, the comb E acquires negative charge and the sphere acquires positive charge due to electrostatic induction.

As a result, the positive charges are pushed away from the comb E and they reach the outer surface of the sphere. Since the sphere is a conductor, the positive charges are distributed uniformly on the outer surface of the hollow sphere. At the same time, the negative charges nullify the positive charges in the belt due to corona discharge before it passes over the pulley.

When the belt descends, it has almost no net charge. At the bottom, it again gains a large positive charge. The belt goes up and delivers the positive charges to the outer surface of the sphere. This process continues until the outer surface produces the potential difference of the order of 107 which is the limiting value.

We cannot store charges beyond this limit since the extra charge starts leaking to the surroundings due to ionization of air. The leakage of charges can be reduced by enclosing the machine in a gas filled steel chamber at very high pressure.

The high voltage produced in this Van de Graaff generator is used to accelerate positive ions (protons and deuterons) for nuclear disintegrations and other applications.

Example 1.24

Dielectric strength of air is 3 × 106 V m-1. Suppose the radius of a hollow sphere in the Van de Graff generator is R = 0.5 m, calculate the maximum potential difference created by this Van de Graaff generator.

Solution

The electric field on the surface of the sphere is given by (by Gauss law)
E = $$\frac{1}{4 \pi \epsilon_{\circ}} \frac{Q}{R^{2}}$$

The potential on the surface of the hollow metallic sphere is given by
V = $$\frac{1}{4 \pi \epsilon_{0}}$$ $$\frac{Q}{R}$$ = ER

Since Vmax = EmaxR
Here Emax = 3 × 106Vm1. So the maximum potential difference created is given by

Vmax = 3 × 106 × 0.5
= 1.5 × 106V (or) 1.5 million volt

Physics is now simple when learning with CBSELibrary – Get all important Physics Topics with detailed explanation.