BASIC ELECTRICAL IDEAS AND UNITS.pdf

BASIC ELECTRICAL IDEAS AND UNITS

Electron Theory of Electricity  All matter is composed of atoms which are

made up of fundamental subatomic particle called protons, neutrons, and electrons.  Each atom represents a sort of microscopic solar system in which the nucleus contains protons and around which the electrons revolve in definite orbits.

Orbital Electrons  All negatively charged electrons revolve

about the positive nucleus in definite orbits  Basically, there is a force of attraction between the positive nucleus and each negative electron  This force is counterbalanced by one that is determined by the orbital motion of the electron 

Energy of the revolving electron  The energy required to displace the electron from the

nucleus so that it may revolve at some fixed radius from the atomic center  The energy represented by its motion around the nucleus, and  The energy that results from its spin on its own axis

 As the atoms become increasingly complex, the

  



positive charge of the nucleus is strengthened by acquiring additional protons Electrons rises proportionately to provide a structure that is electrically neutral Neutrons are also added to the nucleus but have no effect upon the atomic charge Protons and neutrons are bunched together in a central core Electrons are presumed to revolve in orbits or shells around the nucleus

Electron Shell and Orbits  Electron orbit the nucleus of an atom at certain

distance from the nucleus  Electrons near the nucleus have less energy than those in more distant orbits  Energy levels - orbit from the nucleus corresponds to a certain energy level - orbits are grouped into shells (energy band) - each shell has a fixed maximum number of electrons

- the shells are designated K, L, M, N and so on - with K being the closest to the nucleus

Valence Electrons and Conductivity in Solids  the outermost shell s known as the valence shell and

electrons in this shell are called valence electrons  Solid materials may be classified as conductors, insulators, and semiconductors  Classification depends upon the number of valence electrons -Conductor : material that easily conducts electric current. Valence electron4 ex. Phosphorus - Semiconductor: material that is between conductors and insulators in its ability to conduct electrical current valence electron = 4 ex. Germanium, silicon and carbon

Copper atom

Semiconductor material

Electric Charge and Electric Current  Electric charge unit – coulombs

 For each negatively charged electron it is 1.59 x1019 coulomb  Or one coulomb of electric charge is 6.24 x10 e 18

 When one coulomb of electric charge passes a given

point every second the electric current is said to be one ampere  One coulomb per second is one ampere

Q I  t

Where: I = current, Ampere Q = charge, coulombs t = time, sec, during which electrons move  If the current is constant, charge is transferred at a constant rate Q= It  For non-uniform current, the transferred charge will vary with current changes, q=it

 Where: q = geometrically an area

i = plotted along y-axis t = plotted along x-axis

i (ampere)

Variations of current with time

t ( seconds)

 The shaded area represents the number of coulombs

q xtransferred in t x sec

Example:  The current in a conductor changes uniformly from

zero to 2 amp in 3sec, remains steady at 2 amp for 6 sec, and then drops uniformly to 1.5 amp in 8 sec. Calculate the total amount of charge transferred in the elapsed time of 17 sec.  Between t= 1ms and t = 14ms, 8µC of charge pass though a wire. What is the current?

Electromotive Force (EMF)  Also known as electric pressure  Commonly called voltage  Unit is volt ( V)

 When an emf is applied to the ends of a conductor it

is proper to refer to the existence of a potential difference between such ends  Several methods are employed to develop an emf : - combining certain kinds of metals and chemicals into a device (battery) - building a machine which generates voltage when conductors are rotated near magnets.

Electrical Resistance and Resistivity





 

Brief History One of the fundamental relationships of circuit theory is that between voltage, current and resistance This relationship and the properties of resistance were investigated by the German physicist Georg Simon Ohm Ohm found that current depended on both voltage and resistance. From his investigation he was able to define the resistance of a wire and show that the current was

inversely proportional to this resistance  Resistance of Conductors  resistance of a material depends upon several factors: - type of material - length of the conductor - cross-sectional area - temperature

 the

resistance of a conductor is dependent upon the type of material  the resistance of a metallic conductor is directly proportional to the length of the conductor  the resistance of a metallic conductor is inversely proportional to the crosssectional area of the conductor

 Factors

governing the resistance of a conductor at a given temperature may be expressed mathematically:

R  Where:

l A



ρ = resistivity l = length A = cross-sectional area

 Notes: ρ

is the constant of proportionality called resistivity o Resistivity has a unit of Ω-m if the length is in meter and area is in meter square, and a unit of CM-ohms/ ft if the length is in feet and the area is in CM  Since most conductor are circular, 2 2 cross-sectional area d  d   A   r2      4 2

 units of cross-sectional area of a conductor:

- square meter - sq ft. - Circular Mil (CM) - Square-Mil (sq.mil)

Circular-Mil (CM)  A wire that has a diameter of 1 mil has an area of 1

circular mil (CM)  d = 1 mil d2 A A A

 4



4

4 (1mil ) 2 sq.mil

1CM 

 4

sq.mil

If d = 0.001 inch A A

 4



4

(0.001inch) 2 x106 in 2

1CM 

 4

x106 in 2

Square-Mil  Unit of cross-section of small conductor whose side

is equal to one mil s= 1 mil

As

2

A  (1mil ) 2 4 1sq.mil  CM 

 If s = 0.001 inch

A  (0.001inch) 2 A  1x106 in 2  For conversion purposes CM 

4



x no. of sq mils

sq. mils 

 4

x no.of CM

For a wire with a diameter of N mils (N = any positive number) A

d2 4



 N2

substituting A

 N2 4

4 4



sq. mils

CM  1 sq mil, we have

sq.mils  (

 N2 4

)(

4



CM )  N 2CM

 Since d = N, the area in circular mils is simply equal

to the diameter in mils square, that is,

ACM  (dmils )

2

Volume to Resistance Since the volume of the body is V=LA

from R 

l A

V if L  A  V A R A V R   2 A





;

V If A= L L R  V L 2

L R   V

Resistivity of Common Elements and Alloy @ 200C Elements/ Alloy

Resistivity (Ω-CM/ft)

Copper, annealed

10.37

Aluminum

17

Tungsten

33

Zinc

36

Nickel

47

Manganin

265

Nichrome

600

Examples  Most homes use solid copper wire having a diameter

of 1.63 mm to provide electrical distribution to outlets and light sockets. Determine the resistance of a 75 meters solid copper wire having the above diameter.  Calculate the resistance of the following conductor at 200C (a) material: copper with length 1000ft and area of 3, 200CM (b) material: aluminum with length 4 miles and diameter of 162 mils

 A kilometer of wire having a diameter of 11.77mm

and a resistance of 0.031Ω is drawn so that its diameter is 5mm. What does its resistance become?  A copper wire whose diameter is 0.162 in has a resistance of 0.4Ω. If the wire drawn through a series of dies until its diameter is reduced to 0.032 in. What is the resistance of the lengthened conductor? Assume that the resistivity remain unchanged.

Seatwork  A copper wire of unknown length has a diameter of

0.25 in. and a resistance of 0.28 ohm. By several successive passes through drawing dies the diameter of the wire is reduced to 0.05 in. Assuming that the resistivity of the copper remains unchanged in the drawing process, calculate the resistance of the reduced wire.  Calculate the resistance of 1km long cable, composed of 19 strands of similar copper conductors, each strand being 1.32 mm in diameter. Allow 5% increase in length for the lay or twist of each strand in complete cable. Resistivity of copper maybe taken as

1.723 x 10-8Ω-m.  A piece of silver wire has a resistance of 1 ohm. What will be the resistance of manganin wire of one-third of the length and one third the diameter, if the specific resistance of manganin is 30 times that of silver.

Temperature-Resistance Effect  As Temperature increases, more electrons will

escape their orbits, causing additional collision within the conductor.  Any increase in the number of collision translates into a relative increase or decrease in resistance.  For most conducting materials, the increase in the number of collisions translate into a relatively linear increase in resistance, as shown in Figure 3-6.  The rate at which the resistance of a material changes with the variation on temperature is called Temperature Coefficient (α) of the material.

 Any material for which resistance increases as

temperature increases is to have a positive temp. coefficient (+α)  For semiconductor materials, as the temperature increases the number of charge electron increases, resulting in more current.  Therefore, an increase in temp. results in a decrease in resistance.  Semiconductors have negative temp. coefficient (α)

 Referring to fig 3-6, applying similar triangle we

obtain

R2 R1  t2  T t1  T  This expression may be rewritten to solve for the

resistance, R2 at any temp t2 as follows

t2  T R2  R1 t1  T

 Derived formula of R2 in terms of α

R2  R1 1  1 (t2  t1 ) where:1  temperature coefficient of t1

Examples:  The tungsten filament in an incandescent lamp has a

resistance of 9.8Ω at a room temp of 200 C and a resistance of 132Ω at normal operating temp. Using the temp coefficient formula for resistance calculate the temperature of the heated filament.  A platinum coil has a resistance of 3.146Ω at 400 C and 3.767Ω at 1000 C. Find the resistance at 00 C and the temp coefficient at 400 C.  Two coils connected in series have a resistance of 600Ω and 300Ω with temp coefficient of 0.1% and 0.4% respectively at 200 C. Find the resistance of the

combination at a temp of 500 C. What is the effective temp coefficient of the combination.

Seatwork  An aluminum wire has a resistance of 20Ω at room

temp. (200 C). Calculate the resistance of the same wire at temp of -40o C, 100o C and 200o C.  Tungsten wire used as filament in incandescent light bulbs. Current in the wire causes the wire to reach extremely high temp. Determine the temp. of the filament of a 100W light bulb if the resistance at room temp. is measured to be 11.7Ω and when the light is on, the resistance is determined to be 144Ω  Calculate the temp coefficient of resistance of aluminum at 20 C. Using the value obtained, determine the resistance of an aluminum conductor

at 620 C if its resistance at 20 C is 7.5 Ω.  The resistance of electric device is 46 Ω at 250 C. If the temp coefficient of resistance of the material is 0.00454 at 200 C, determine the temp. of the device when its resistance is 92Ω.

Rules in Sizing a Conductor Every change of three gage number changes the circular-mil area and resistance in the ratio of 2 to 1 or 1 to 2, depending upon the direction of the change. 2. Every change of 10 gage numbers changes the circular-mil area and resistance in the ratio of 10 to 1 or 1 to 10, depending upon the direction of the change 3. Every change of one gage number changes the circular-mil area and resistance in the ratio of 1 ¼ to 1 or 1 to 1 1/4 , depending upon the direction of the change 1.

4. A No. 10 wire may be assumed, for practical purposes, to have a diameter of 100 mils, an area of 10, 000 cir mils, and a resistance of 1 ohm per 1,000ft. 5. A No.5 wire has a weight of 100lb per 1,000ft; moreover, for every change of three gage numbers the weight is halved or doubled , depending upon the direction of the change 6. A No.15 wire has 100ft per lb (very nearly); moreover, for every change of three gage numbers the number of feet per pounds is doubled or halved,

depending upon the direction of the change. 7. Every change of 10 gage numbers changes the pounds per 1,000ft and the feet per pound in the ratio of 10 to 1, depending upon the direction of the change

WIRE TABLE OF A COPPER WIRE

CONTINUATION…

Example  Without consulting the wire table, determine the

following data for a No.17 copper wire: (a) circular mils; (b) resistance per 1,000ft; (c) pounds per 1,000ft; (d) feet per pound