Types Of Manufacturing

The term “manufacturing” covers a broad spectrum of activities. Metal working industries, process industries like chemical plants, oil refineries, food processing industries, electronic industries making microelectronic components, printed circuit boards, computers and entertainment electronic products etc. are examples of manufacturing industries. Manufacturing involves fabrication, assembly and testing in a majority of situations. However, in process industries operations are of a different nature.

Manufacturing industries can be grouped into four categories:

1. Continuous Process Industries
In this type of industry, the production process generally follows a specific sequence. These industries can be easily automated and computers are widely used for process monitoring, control and optimization. Oil refineries, chemical plants, food processing industries, etc are examples of continuous process industries.

2. Mass Production Industries
Industries manufacturing fasteners (nuts, bolts etc.), integrated chips, automobiles, entertainment electronic products, bicycles, bearings etc. which are all mass produced can be classified as mass production industries. Production lines are specially designed and optimized to ensure automatic and cost effective operation. Automation can be either fixed type or flexible.

3. Batch Production (Discrete Manufacturing)
The largest percentage of manufacturing industries can be classified as batch production industries. The distinguishing features of this type of manufacture are the small to medium size of the batch, and varieties of such products to be taken up in a single shop. Due to the variety of components handled, work centres should have broader specifications. Another important fact is that small batch size involves loss of production time associated with product changeover.

As mentioned earlier, integration of computer in process industries for production automation, process monitoring and control and optimization is relatively easy. In the case of mass production and batch production computer integration faces a number of problems as there are a large number of support activities which are to be tied together. These are discussed in detail later in this chapter.

Automation of manufacture has been implemented using different techniques since the turn of the 20th Century. Fixed automation is the first type to emerge. Single spindle automatic lathe, multi spindle automatic lathe and transfer lines are examples of fixed automation. Fixed automation using mechanical, electrical, pneumatic and hydraulic systems is widely used in automobile manufacturing. This type of automation has a severe limitation - these are designed for a particular product and any product change will require extensive modifications to the automation system.

The concept of programmable automation was introduced later. These were electrically controlled systems and programs were stored in punched cards and punched tapes. Typical examples of programmable automation are:

i. Electrical programme controlled milling machines
ii. Hydraulically operated Automatic lathes with programmable control drum
iii. Sequencing machines with punched card control /plug board control

Development of digital computers, microelectronics and microprocessors significantly altered the automation scenario during 1950-1990. Machine control systems are now designed around microprocessors and microelectronics is part and parcel of industrial drives and control. The significant advances in miniaturization through integration of large number of components into small integrated chips and the consequent improvement in reliability and performance have increased the popularity of microelectronics. This has resulted in the availability of high performance desktop computing machines as well as file servers which can be used for industrial control with the help of application software packages.

Introduction of Computer Integrated Manufacturing

Computer Integrated Manufacturing (CIM) encompasses the entire range of product development and manufacturing activities with all the functions being carried out with the help of dedicated software packages. The data required for various functions are passed from one application software to another in a seamless manner. For example, the product data is created during design. This data has to be transferred from the modeling software to manufacturing software without any loss of data. CIM uses a common database wherever feasible and communication technologies to integrate design, manufacturing and associated business functions that combine the automated segments of a factory or
a manufacturing facility. CIM reduces the human component of manufacturing and thereby relieves the process of its slow, expensive and error-prone component. CIM stands for a holistic and methodological approach to the activities of the manufacturing enterprise in order to achieve vast improvement in its performance.

This methodological approach is applied to all activities from the design of the product to customer support in an integrated way, using various methods, means and techniques in order to achieve production improvement, cost reduction, fulfillment of scheduled delivery dates, quality improvement and total flexibility in the manufacturing system. CIM requires all those associated with a company to involve totally in the process of product development and manufacture. In such a holistic approach, economic, social and human aspects have the same importance as technical aspects.

CIM also encompasses the whole lot of enabling technologies including total quality management, business process reengineering, concurrent engineering, workflow automation, enterprise resource planning and flexible manufacturing.

A distinct feature of manufacturing today is mass customization. This implies that though the products are manufactured in large quantities, products must incorporate customer-specific changes to satisfy the diverse requirements of the customers. This requires extremely high flexibility in the manufacturing system.

Manufacturing industries strive to reduce the cost of the product continuously to remain competitive in the face of global competition. In addition, there is the need to improve the quality and performance levels on a continuing basis. Another important requirement is on time delivery. In the context of global outsourcing and long supply chains cutting across several international borders, the task of continuously reducing delivery times is really an arduous task. CIM has several software tools to address the above needs.

Manufacturing engineers are required to achieve the following objectives to be competitive in a global context.

• Reduction in inventory
• Lower the cost of the product
• Reduce waste
• Improve quality
• Increase flexibility in manufacturing to achieve immediate and rapid response to:
• Product changes
• Production changes
• Process change
• Equipment change
• Change of personnel

CIM technology is an enabling technology to meet the above challenges to the manufacturing.

The advances in automation have enabled industries to develop islands of automation. Examples are flexible manufacturing cells, robotized work cells, flexible inspection cells etc. One of the objectives of CIM is to achieve the consolidation and integration of these islands of automation. This requires sharing of information among different applications or sections of a factory, accessing incompatible and heterogeneous data and devices. The ultimate objective is to meet the competition by improved customer satisfaction through reduction in cost, improvement in quality and reduction in product development time. 

CIM makes full use of the capabilities of the digital computer to improve manufacturing. Two of them are: 

i. Variable and Programmable automation
ii. Real time optimization

The computer has the capability to accomplish the above for hardware components of manufacturing (the manufacturing machinery and equipment) and software component of manufacturing (the application software, the information flow, database and so on).

The capabilities of the computer are thus exploited not only for the various bits and pieces of manufacturing activity but also for the entire system of manufacturing. Computers have the tremendous potential needed to integrate the entire manufacturing system and thereby evolve the computer integrated manufacturing system.

Voltage Divider Rule ( Electrical tearms )

Since in a series circuit, same current flows through each of the given resistors, voltage drop varies directly with its resistance. In Figure is shown a 24- V battery connected across a series combination of three resistors.

Total resistance R = R 1 + R2 + R3 = 12 Ohm

According to Voltage Divider Rule, various voltage drops are :

V1 = VR1/R = (24x2) / 12 = 4 v  ;  V2 = 8 V  ; V3 = 12 V

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Electrical Resistance in Series

When some conductors having resistances R1,R2 and R3 etc. are joined end-on-end as in Fig. 1, they are said to be connected in series. It can be proved that the equivalent resistance or total. resistance between points A and D is equal to the sum of the three individual resistances. Being a series circuit, it should be remembered that (i) current is the same through all the three conductors (ii) but voltage drop across each is different due to its different resistance and is given by Ohm's Law and (iii) sum of the three voltage drops is equal to the voltage applied across the three conductors. There is a progressive fall in potential as we go from point A to D as shown in Fig. 2.

V = V1 + V2+ V3= IR1 + IR2 + IR3   _ _ _ _ _ _ _ _ _ Ohms Law

But  V = IR

where R is the equivalent resistance of the series combination.

IR = IR1 + IR2 + IR3 or R = R1 + R2+ R3

$\frac{1}{G}=\frac{1}{G1}+\frac{1}{G2}+\frac{1}{G3}$

As seen from above, the main characteristics of a series circuit are :

1. same current flows through all parts of the circuit.
2. different resistors have their individual voltage drops.
3. voltage drops are additive.
4. applied voltage equals the sum of different voltage drops.
5. resistances are additive.
6. powers are additive.

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Ohm's law ( in terms of electrical tearms )

This law applies to electric to electric conduction through good conductors and may be stated as follows:

The ratio of potential difference (V) between any two points on a conductor to the current (I) flowing between them, is constant. provided the temperature of the conductor does not change.

In other words,  V/I = constant  or V/I = R

where R is the resistance of the conductor between the two points considered. Put in another way, it simply means that provided R is kept constant, current is directly proportional to the potential difference across the ends of a conductor. However, this linear relationship between V and I does not apply to all non-metallic conductors. For example, for silicon carbide, the relationship is given by $V=K{I}^m$ where K and m are constants and m is less than unity. It also does not apply to non-linear devices such as Zener diodes and voltage-regulator (VR) tubes.

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Example: A coil of copper wire has resistance of 0 at 20°C and is connected to a 230-V supply. By how much must the voltage be increased in order to maintain the current consant if the temperature of the coil rises to 60°C? Take the temperature coefficient of resistance of copper as 0.00428from 0°c

Solution

As seen from page Temperature Coefficient of Electrical Resistance 

$\frac{R60}{R20}=\frac{1+60\times 0.00428}{1+20\times 0.00428}$
so R60=90 x 1.2568 / 1.0856 = 104.2 Ohm

Now, current at 20°C = 230/90 =23/9 A

Since the wire resistance has become 104.20 Ohm at 60°C, the new voltage required for keeping the
current constant at its previous value = 104.2 x 23/9 = 266.3 V
So, increase in voltage required=266.3 - 230 = 36.3 V

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Example: Three resistors are connected in series across a 12-V battery. Thefirst resistor has a value of 1 Ohm, second has a voltage drop of 4 V and the third has a power dissipation of 12 W. Calculate the value of the circuit current.

Solution

Let the two unknown resistors be R2and R3and I the circuit current

$I^{2}R3 =12 and IR3=4$
So,$\mathrm{ \; R}3=\frac{3}{4}R{2}^2 Also I=\frac{4}{R2}$

Now,     I (1+R2+R3) = 12

Substituting the values of I and R3, we get

$\frac{4}{R2}\left(1+R2+\frac{3}{4}R{2}^2\right) =12 or 3R{2}^2 -8R2+4=0$


$R2=\frac{8\pm \sqrt{64-48}}{6} So R2=2\Omega or \frac{2}{3}\Omega$


$R3=\frac{3}{4}R{2}^2=\frac{3}{4}{2}^2 =3\Omega or \frac{3}{4}{\left(\frac{2}{3}\right)}^2=\frac{1}{3} \Omega$

So I = 2A or I = 6A

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Variations of Electrical Resistivity with Temperature

Not only resistance but specific resistance or resistivity of metallic conductors also increases with rise in temperature and vice versa. As seen from Fig. the resistivities of metals vary linearly with temperature over a significant range of temperature-the variation becoming non-linear both at very high and at very low temperatures. Let, for any metallic conductor.

ρ1 = resistivity at t1 °C
ρ2 = resistivity at t2 °C
m = Slope of the linear part of the curve

then ,$m=\frac{\rho 2-\rho 1}{t2-t1}$

The ratio of m/ρ1 is called the temperature coefficient of resistivity at temperature t1 °C. It may be defined as numerically equal to the fractional change in ρ1 per °C change in the temperature from t1 °C. It is almost equal to the temperature-coefficient of resistance ${\alpha }_1$ . Hence, putting ${\alpha }_1$= m/ρ1 we get

$\rho 2=\rho 1\left[1+\alpha 1(t2-t1)\right] or \rho t=\rho 0(1+\alpha 0t)$

Note: It has been found that although temperature is the most significant factor influencing the resistivity of metals. other factors like pressure and tension also affect resistivity to some extent For most metals except lithium and calcium. increase in pressure leads to decrease in resistivity. However, resistivity increases with increase in tension.

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Examples based on theory 1) Electrical Conductance and Conductivity 2) Effect of Temperature on Electrical Resistance  3) Temperature Coefficient of Electrical Resistance  4) Value of α at Different Temperatures ( In terms of...
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Example A copper conductor has its specific resistance of 1.6×10-6 ohm-cm at O°C and a resistance temperature coefficient of 1/254.5 per °C at 20°C. Find (1) the specific resistance and (2) the resistance - temperature coefficient at 60°C. 

Solution :

$\alpha 20=\frac{\alpha 0}{1+\alpha 0\times 20} or \frac{1}{254.5}=\frac{\alpha 0}{1+\alpha 0\times 20}$$$     so ,$\alpha 0=\frac{1}{234.5}per °C$
$$
(1)       ρ60 = ρ0 (1 +$\alpha$0 x 60) =  2×10-6 Ω-cm

(2)      $\alpha$60= $\alpha$0 / (1+$\alpha$0x60) = 1/294.5 per °C.

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Example: A platinum coil has a resistance of 3.146 ohm at 40°C and 3.767 ohm at 100°C. Find the resistance at O°C and the temperature-coefficient of resistance at 40°C. 

Solution:

R100 = R0 (1+100 $\mathrm{\alpha <span\; style="font-family:\; Arial,Helvetica,sans-serif;">0\; )</span>}$  _ _ _ _ _ _ _ (1)

R40 = R0 (1+ 40$\mathrm{<span\; style="font-family:\; Arial,Helvetica,sans-serif;"> \; </span>\alpha 0) \; <span\; style="font-family:\; Arial,Helvetica,sans-serif;">\_\; \_\; \_\; \_\; \_\; \_ \; \_\; (2)</span>}$

$\frac{3.767}{3.146}=\frac{1+100 \alpha 0}{1+40 \alpha 0} or \alpha 0=\frac{1}{264} per °C$

From eq (1)       3.767 = R0 (1+100x0.00379 )  so R0=2.732 ohm

and  α40 = α0 / (1+40 α0 ) = 1/304 per °C

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Example: A potential difference of 250 V is applied to afield winding at 15°C and the current is 5 A. What will be the mean temperature of the winding when current has fallen to 3.91 A. applied voltage being constant. Assume $\mathrm{\alpha 15}$ =1/254.5.

Solution:

Let R1 =winding resistance at 15°C; R2 = winding resistance at unknown mean temperature t2°c.

R1 = 250/5 = 50 ohm; R2= 250/3.91 = 63.94 ohm.

R2 = R1 [1 + α15 (t2 - t1)]   so t2=86 °C

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Example: Two coils connected in series have resistances of 600 ohm and 300 ohm with temp coeff. of 0.1% and 0.4% respectively at 20°C. Find the resistance of the combination at a tempt. of 50°C. What is the effective tempt. coeff. of combination?

Solution:

Resistance of 600 ohm resistor at 50°C is = 600 [1 + 0.001 (50 - 20)] = 618 Ohm
Similarly, resistance of 300 ohm resistor at 50°C is = 300 [1 + 0.004 (50 - 20)] = 336 Ohm

So, total resistance of combination at 50°C is = 618 + 336 =954 ohm

Let β = resistance-temperature coefficient at 20°C

Now, combination resistance at 20°C = 900 ohm
Combination resistance at 50°C = 954 ohm

954 = 900[ ] + β (50 - 20)]   so, β = 0.002

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Example: Two wires A and B are connected in series at O°C and resistance of B is 3.5 times that of A. The resistance temperature coefficient of A is 0.4% and that of the combination is 0.1%. Find the resistance temperature coefficient of B.

Solution:

A simple technique which gives quick results in such questions is illustrated by the diagram of Fig. It is seen that. RB/RA =0.003 / ( 0.001- $\alpha$)

$$or 3.5 = 0.003/(0.001- $\alpha$)$$

or a = 0.000

143 °C^-1   or 0.0143%
 

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Value of α at Different Temperatures ( In terms of Electrical )

So far we did not make any distinction between values of a at different temperatures. But it is found that value of a itself is not constant but depends on the initial temperature on which the increment in resistance is based. When the increment is based on the resistance measured at O°C, then α has the value of ${\alpha }_0$.At any other initial temperature t° C,value of α is αt, and so on. It should be remembered that, for any conductor, ${\alpha }_0$ has the maximum value.

Suppose a conductor of resistance R0 at 0 °C (point A in Figure) is heated to t °C (point B). Its resistance Rt after heating is given by

$Rt=R0 ( 1+{\alpha }_0 t )$ _ _ _ _ _ _ _ (1)

where α0 is the temperature-coefficient at 0°C.

Now, suppose that we have a conductor of resistance Rt, at temperature t°C.. Let this conductor be cooled from t°C.to 0°C.. Obviously, now the initial point is B and the final point is A. The final resistance R0 is given in tenns of the initial resistance by the following equation$$

$R0=Rt [1+\alpha t(-t)]=Rt (1-\alpha t\times t)$ _ _ _ _ _ _ _ _ (2)

From Eq. (ii) above, we have $\alpha t=\frac{Rt-R0}{Rt\times t}$

Substuting the value of Rt from Eq. (1), we get

$\alpha t=\frac{R0 ( 1 +\alpha 0 t)-R0}{R0 ( 1 +\alpha 0 t)\times t}$

$so, \; \alpha t \; =\frac{\alpha 0}{1+\alpha 0 \mathrm{t }}$         _ _ _ _ _ _ _ (3)

In general, let αt =temp. coeff. at t1°C;  α2 = tempt. coeff. at t2°C.  Then from Eq. (3) above, we get

$\; \alpha 1 \; =\frac{\alpha 0}{1+\alpha 0 \mathrm{t1 }}$ or $\frac{1}{\alpha 1}=\frac{1+\alpha 0 \mathrm{t1}}{\alpha 0}$

Similarly,

$\frac{1}{\alpha 2}=\frac{1+\alpha 0 \mathrm{t2}}{\alpha 0}$

So, 

$\frac{1}{\alpha 1}-\frac{1}{\alpha 2}=(t2-t1) or \frac{1}{\alpha 2}=\frac{1}{\alpha 1}+ ( t2-t1) or \alpha 2=\frac{1}{1/\alpha 1 + ( t2-t1 )}$

Values of α for copper at different temperatures are given in Table below

Temp in C	0	5	10	20	30	40	50
α	0.00427	0.00418	0.00409	0.00393	0.00378	0.00364	0.00352

In view of the dependence of α on the initial temperature, we may define the temperature coefficient of resistance at a given temperature as the charge in resistance per ohm per degree centrigrade change in temperaturefrom the given temperature.

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Temperature Coefficient of Electrical Resistance

Let a metallic conductor having a resistance of Ro at 0°C heated of t°C and let its resistance at this temperature be Rt. Then, considering normal ranges of temperature, it is found that the increase in resistance $∆R=Rt-R0$ depends on,

(1) directly on its initial resistance
(2) directly on the rise in temperature
(3) on the nature of the material of the conductor.

or,  $Rt-R0 \propto R\times t or Rt-R0=\alpha R0t$ _ _ _ _ _ _ _ (1)

where  (alpha) is a constant and is known as the temperature coefficient of resistance of the conductor.

Rearranging Eq. (1), we get

$\alpha =\frac{Rt-R0}{R0\times t}=\frac{∆R}{R0\times t}$

if , $R0=1\Omega , t= 1^0 C then \alpha =∆R=Rt-R0$

Hence, the temperature-coefficient of a material may be defined as : the increase in resistance per ohm original resistance per °C rise in temperature.

From Eq. (1), we find that $Rt=R0(1+\alpha t)$

Temperature Coefficient of Electrical Resistance

It should be remembered that the above equation holds good for both rise as well as fall in temperature.  As temperature of a conductor is decreased. its resistance is also decreased. In Fig. 1 is shown the temperature/resistance graph for copper and is practically a straight line. If this line is extended backwards, it would cut the temperature axis at a point where temperature is - 234.5°C (a number quite easy to remember). It means that theoretically, the resistance of copper conductor will become zero at this point though as shown by solid line, in practice, the curve departs from a straight line at very low temperatures. From the two similar triangles of Fig. 1 it is seen that :

$\frac{Rt}{R0}=\frac{t+234.5}{234.5}=\left(1+\frac{t}{234.5}\right)$

$Rt=R0 \left(1+\frac{t}{234.5}\right) or Rt=R0(1+\alpha t)$ Where α=1/234.5 for copper.

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Effect of Temperature on Electrical Resistance

(1) to increase the resistance of pure metals. The increase is large and fairly regular for normal ranges of temperature. The temperature/resistance graph is a straight line. As would be presently clarified, metals have a  temperature co-efficient at resistance.

(2) to increase the resistance of alloys, though in their case, the increase is relatively small and irregular. For some high-resistance alloys like Eureka (60% Cu and 40% Ni) and manganin, the increase in resistance is (or can assume) negligible over a considerable range of temperature.'

(3) to decrease the resistance of electrolytes, insulators (such as paper, rubber, glass, mica etc.) and partial conductors such as carbon. Hence, insulators are said to possess a negative temperature-coefficient of resistance. .

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Electrical Conductance and Conductivity

Conductance (G) is reciprocal of resistance. Whereas resistance of a conductor measures the opposition which it offers to the flow of current, the conductance measures the inducement which it offers to its flow.

From page Laws of Electrical Resistance,  R=ρlA   or  $G=\frac{1}{\rho }\times \frac{A}{l}=\frac{\sigma A}{l}$

a is called the conductivity or specific conductance of a conductor. The unit of conductance is siemens (S). Earlier,this unit was called mho.

It is seen from the above equation that the conductivity of a material is given by

$\sigma =G\frac{l}{A}=\frac{G siemens\times l meter}{A mete{r}^2}=G\frac{l}{A} siemens/meter$

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Laws of Electrical Resistance

The resistance R offered by a conductor depends on the following factors:

(1) It varies directly as its length, l.
(2) It varies inversely as the cross-section A of the conductor.
(3) It depends on the nature of the material.
(4) It also depends on the temperature of the conductor.

Neglecting the last factor for the time being, we can say that.

R∝lA or R=ρlA  _ _ _ _ _ _ _ _ (i)

where p is a constant depending on the nature of the material of the conductor and is known as its specific resistance or resistivity.

If in Eq. (i), we put , l = 1 meter and A = 1 ${\mathrm{meter}}^{}$ then R=ρ (fig 2)

Hence, specific resistance of a material may be defined as the resistance between the opposite faces of a metre cube of that material.

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The Unit of Resistance ( in tearms of Electrical )

The practical unit of resistance is ohm.A conductor is said to have a resistance of one ohm if it permits one ampere current to flow through it when one volt is impressed across its terminals.

For insulators whose resistances are very high, a much bigger unit is used i.e. megaohm = ${\mathrm{10}}^6$ ohm (the prefix 'mega' or mego meaning a million) or kilohm = ${\mathrm{10}}^3$ ohm (kilo means thousand). In the case of very small resistances, smaller units like milli-ohm = ${\mathrm{10}}^{-3}$ ohm or microhm = ${\mathrm{10}}^{-6}$ ohm are used. The symbol for ohm is Ω.

Prefix	Meaning	Contraction	Equal
Mega Kilo Centi Mili Micro	One Million One Thousand One Hundredth One Thousandth One Millionth	M Ω k Ω - m Ω μ Ω${\mathrm{10}}^6$	${\mathrm{10}}^6$ ${\mathrm{10}}^3$ - $$${\mathrm{10}}^{-3}$ $$${\mathrm{10}}^{-6}$

Table : Multiples and sub-multiples of Ohm

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Elecrtical Resistance or Electricity Resistance

It may be defmed as the property of a substance due to which it opposes (or restricts) the flow of electricity (i.e., electrons) through it.

Metals (as a class), acids and salts solutions are good conductors of electricity. Amongst pure metals,silver, copper and aluminium are very good conductors in the given order. This, as discussed earlier, is due to the presence of a large number of free or loosely-attached electrons in their atoms. Thesevagrant electrons assume a directed motion on the application of an electric potential difference. These electrons while flowing pass through the molecules or the atoms of the conductor, collide and other atoms and electrons, thereby producing heat.

Those substances which offer relatively greater difficulty or hindrance to the passage of these electrons are said to be relatively poor conductors of electricity like bakelite, mica, glass, rubber, p.v.c. (polyvinyl chloride) and dry wood etc. Amongst good insulators can be included fibrous substances such as paper and cotton when dry, mineral oils free from acids and water, ceramics like hard porcelain and asbestos and many other plastics besides p.v.c. It is helpful to remember that electric friction is similar to friction in Mechanics.

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The Idea of Electric Potential

In Fig. 1 is shown a simple voltaic cell. It consists of copper plate (known as anode) and a zinc rod (I.e. cathode) immersed in dilute sulphuric acid (H2SO4) contained in a suitable vessel. The chemical action taking place within the cell causes the electrons to be removed from Cu plate and to be deposited on the zinc rod at the same time. This transfer of electrons is accomplished through the agency of the diluted H2SO4 which is known as the electrolyte. The result is that zinc rod becomes negative due to the deposition of electrons on it and the Cu plate becomes positive due to the removal of electrons from it. The large number of electrons collected on the zinc rod is being attracted by anode but is prevented from returning to it by the force set up by the chemical action within the cell. But if the two electrodes are joined by a wire externally. then electrons rush to the anode thereby equalizing the charges of the two electrodes. However. due to the continuity of chemical action. a continuous difference in the number of electrons on the two electrodes is maintained which keeps up a continuous flow of current through the external circuit. The action of an electric cell is similar to that of a water pump which. while working. maintains a continuous flow of water i.e. water current through the pipe (Fig. 2).

 

 

It should be particularly noted that the direction of electronic current is from zinc to copper in the external circuit. However, the direction of conventional current (which is given by the direction of flow of positive charge) is from Cu to zinc. In the present case, there is no flow of positive charge as such from one electrode to another. But we can look upon the arrival of electrons on copper plate (with subsequent decrease in its positive charge) as equivalent to an actual departure of positive charge from it.

When zinc is negatively charged, it is said to be at negative potential with respect to the electrolyte, whereas anode is said to be at positive potential relative to the electrolyte. Between themselves, Cu plateis assumed to be at a higher potential than the zinc rod. The difference in potential is continuously maintainedby the chemical action going on in the cell which supplies energy to establish this potential difference.

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Charge Velocity and Velocity of Field Propagation

The speed with which charge drifts in a conductor is called the velocity of charge. As seen from page Electron Drift Velocity . its value is quite low, typically fraction of a metre per second.

However. the speed with which the effect of e.m.f. is experienced at all parts of the conductor resulting in the flow of current is called the velocit.v of propagation of electri8calfield. It is independent of current and voltage and has high but constant value of nearly  $3\times {10}^{8}m/s^{\mathrm{}}$.

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Example

Find the velocity of charge leading to 1 A current which flows in a copper conductor of cross-section 1 ${\mathrm{cm}}^2$ and length 10 km. Free electron density of copper $\mathrm{= \; 8.5}\times {10}^{28}{}^{\mathrm{}}$per $m^3$. How long will it take the electric charge to travel from one end of the conductor to the other.

Solution

                i = neAv  or v = i / neA

     so,      v=18.5×1028×(1.5×1019)×(1×10-4) =7.35×10-7 m/s =0.735 μm/s

Time taken by the charge to travel conductor length of 10 km is,

t=distancevelocity=10×1037.35×10-7=1.36×1010 second

Now. 1 year = 365 x 24 x 3600 = 31,536.000 s

           so, t =1.36×1010/31,53,6000 = <b>431</b>years

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Electron Drift Velocity

Suppose that in a conductor, the number of free electrons available per${\mathrm{<span\; style="font-family:\; "Times\; New\; Roman","serif";"> \; </span>m}}^3$ of the conductor material is n and let their axial drift velocity be v metres/second. In time dt, distance travelled would be v * dt. If A.is area of cross-section of the conductor, then the volume is vAdt and the number of electrons contained in this volume is vA dt. Obviously, all these electrons will cross the conductor cross-section in time dt. If e is the charge of each electron, then total charge which crosses the section in time dt is dq = nAev dt.

Since current is the rate of flow of charge, it is given as

$\mathrm{ \; i}=\frac{dq}{dt}=\frac{nAev dt}{dt}$   so, i=nAev

$$

Current density $j=\raisebox{1ex}{$i$}\!\left/ \!\raisebox{-1ex}{$A$}\right.=ne*v ampere/mete{r}^2$$\mathrm{}$

Assuming a normal current density  $\mathrm{J\; =\; 1.55}\times {10}^{6}A/m^2$;">, n = ${10}^{29}A/m^2$ for a copper conductor and e $\mathrm{=\; 1.6}\times {10}^{-19}$ coulomb, we get,

$\mathrm{ \; \; J\; =\; 1.55}\times {10}^6=$$\mathrm{<span\; style="font-family:\; inherit;"><math\; xmlns="http://www.w3.org/1998/Math/MathML"><mo></mo><mo><span\; style="font-family:\; inherit;">*</span></mo></math></span>}$
$\mathrm{J\; =\; <span\; style="font-family:\; inherit;">9.7</span>}\times {10}^{-5}\mathrm{<span\; style="font-family:\; inherit;">m</span>}/$s= 0.58 cm/min

It is seen that contrary to the common but mistaken view, the electron drift velocity is rather very slow and is independent of the current flowing and the area of the conductor.

Note :-   Current density i.e., the current per unit area. is a vector quantity. It is denoted by the symbol$\overrightarrow{\mathrm{ \; J}}$.

Therefore. in vector notation, the relationship between current I and is:

$\mathrm{ \; I}=\overrightarrow{J }*\overrightarrow{a }$;">     [ where $\overrightarrow{a }$ is a vector notation for area a ]$\stackrel{}{}\stackrel{}{}$
For extending the scope of the above relationship. so that it becomes applicable for area of any shape, we write:

$\mathrm{<span\; style="font-family:\; "Times\; New\; Roman","serif";"> \; </span>i}=\int \overrightarrow{J.}d \overrightarrow{a }$  

The magnitude of the current density can. therefore be written as J*α  

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Example 1

A conductor material has a free-electron density of ${\mathrm{10}}^{24}$ electrons per ${\mathrm{meter}}^3$. When a voltage is applied, a constant drift velocity of  $\mathrm{1.5}\times {10}^{-2}$metre/second is attained by the electrons. If the cross-sectional area of the material is 1 ${\mathrm{cm}}^2$, calculate the magnitude of the current. Electronic charge is $\mathrm{1.6}\times {10}^{-19}{}^{\mathrm{}}$coulomb.

Solution:

The magnitude of the current is i = nAve amperes

Here         n =  ${\mathrm{10}}^{24}$ ;  A = ${\mathrm{1cm}}^2$ = ${\mathrm{10}}^{-4}$${\mathrm{ \; m}}^2$
                 e =$\mathrm{ \; 1.6}\times {10}^{-19}C{}^{\mathrm{}}$  
                 v =$\mathrm{ \; 1.5}\times {10}^{-2}{}^{\mathrm{}}$m/s

         Ans  i = 0.24 A

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