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EEE205

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EEE205

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  1. EEE205 Yrd. Doç. Dr. Mehmet Ali Aktaş

  2. UNITS ASSOCIATED WITH BASICELECTRICAL QUANTITIES • The system of units used in engineering and science isthe (International systemof units), usually abbreviated asSI units.

  3. UNITS ASSOCIATED WITH BASICELECTRICAL QUANTITIES • Derived SI units use combinations of basic units andthere are many of them. Two examples are: • Velocity – metres per second (m/s) • Acceleration – metres persecondsquared(m/s2)

  4. UNITS ASSOCIATED WITH BASICELECTRICAL QUANTITIES • SI unitsmay bemade larger or smaller by using prefixeswhich denote multiplication or divisionby a particularamount. The six most common multiples, with theirmeaning, arelistedbelow:

  5. UNITS ASSOCIATED WITH BASICELECTRICAL QUANTITIES • CHARGE • The unit of charge is the coulomb (C) where onecoulomb is one ampere second (1 coulomb = 6.24×1018 electrons). The coulomb is defined as the quantityof electricity which flows past a given point in an electriccircuit when a current of one ampere is maintainedforonesecond. charge, in coulombsQ=I.t I is the current in amperes, t is the time insec.

  6. UNITS ASSOCIATED WITH BASICELECTRICAL QUANTITIES • Problem 1. If a current of 5A flows for 2minutes,find the quantity of electricity transferred. • Quantity of electricity Q = I t coulombs • I = 5 A, t = 2×60 = 120 s • HenceQ= 5×120 = 600C

  7. UNITS ASSOCIATED WITH BASICELECTRICAL QUANTITIES • Force • The unit of force is the newton (N) where one newtonis one kilogram metre per second squared. The newtonis defined as the force which, when applied to a mass ofone kilogram, gives it an acceleration of one metre persecond squared. force, in newtonsF=ma • where m is the mass in kilograms and a is the accelerationin metres per second squared.

  8. UNITS ASSOCIATED WITH BASICELECTRICAL QUANTITIES • Problem 2. A mass of 5000g is accelerated at2m/s2 by a force. Determine the force needed. • Force = mass×acceleration • = 5kg×2m/s2= 10 kgm/s2 = 10N.

  9. UNITS ASSOCIATED WITH BASICELECTRICAL QUANTITIES • Problem 3. Find the force acting verticallydownwards on a mass of 200 g attached to a wire. Mass=200g=0.2 kg and acceleration due to gravity,g=9.81m/s2 (Gravitationalforce, or weight, is mg, where g = 9.81m/s2.) • Force actingdownwards = weight • =mass×acceleration • = 0.2kg×9.81m/s2 • = 1.962N

  10. UNITS ASSOCIATED WITH BASICELECTRICAL QUANTITIES • Work The unit of work or energy is the joule (J) where onejoule is one newton metre. The joule is defined as thework done or energy transferred when a force of onenewton is exerted through a distance of onemetre in thedirection of theforce. work done on a body, in joules, W=Fs where F is the force in newtons and s is the distance inmetres moved by the body in the direction of the force.Energy is the capacity for doing work.

  11. UNITS ASSOCIATED WITH BASICELECTRICAL QUANTITIES • Power The unit of power is thewatt (W) where onewattis onejoule per second. Power is defined as the rate of doingworkortransferringenergy. power, in watts, P= W / t where W is the work done or energy transferred, injoules, and t is the time, in seconds. energy, in joules, W=Pt

  12. UNITS ASSOCIATED WITH BASICELECTRICAL QUANTITIES • Problem 4. A portablemachinerequires a forceof 200N to move it. How much work is done if themachine is moved 20m and what average power isutilized if the movement takes 25s? Work done = force×distance = 200N×20m =4 000Nm or 4 kJ Power = work done /time taken =4000J / 25s =160J/s= 160W

  13. LESSON 2 • Resistance • Resistors • DifferentTypes Of Resistor • ResistanceAndResistivity • ResistorColorCode • LetterAndDigitCodeForResistors

  14. Resistance • What is Resistance? • Resistance is anything that slow or stops movement or keeps movement from happening. • Resistance in an electrical circuit is the limiting or opposition to current flow.

  15. Resistors • The main function of resistors in a circuit is tocontrol the flow of current and voltage drops to other components. • Forexample; iftoomuch current flows through an LED it is destroyed and will not light, so a resistor isused to limit the current but not so big as it will limit all the current. • Whena currentflows through a resistor, energy is wasted and the resistor heats up.

  16. Resistors • Thiswillonlybe noticed if the resistor is working at its maximum power rating. • The greater the currentflowing through the resistor the hotter it gets. • An important property to know about resistors is how much heat energy itcan withstand before it's damaged or causes a fire.

  17. DifferentTypes Of Resistor

  18. DifferentTypes Of Resistor • Three of themostcommonmethods of constructionsare: • Wirewoundresistors • A length of wire such as nichrome or manganinis cuttothedesired value and wound around a ceramic former priorto being lacquered for protection. • Thelargephysicalsize of thistypeof resistor is a disadvantege but it has twoadvatagessuch as highdegree of accuracy, and can have a high power rating. • Wire wound resistors are used in power circuits andmotor starters. • Metal oxideresistors • Metal-oxide film resistors are made of metal oxides such as tin oxide. • The value of resistance is controlled mainly by the thickness of the coating layer (the thicker the layer, the lower is the value of resistance). • Metal oxide resistors are used in electronic equipment. • Carbonresistors • This type of resistor is made from a mixture of carbonblack resin binder and a refractory powder thatis pressed into shape and heated in a kiln to form asolid rod of standard length and width. • Theresistivevalue is predetermined by the ratio of the mixture.Metal end connections are crimped onto the rod to actas connecting points for electrical circuitry. • This typeof resistor is small and mass-produced cheaply; it haslimited accuracy and a low power rating.Carbon resistors are used in electronic equipment.

  19. ResistanceAndResistivity • The resistance of an electrical conductor depends onfourfactors. • (a) the length of the conductor. • (b) the cross-sectional area of the conductor. • (c) thetypeof material. • (d) the temperature of the material.

  20. ResistanceAndResistivity • Resistance, R, is directly proportional to length, l, of aconductor, i.e. R∝l. • For example, if the length ofa piece ofwire is doubled, then the resistance is doubled. • Resistance, R, is inversely proportional to crosssectionalarea, a, of a conductor, i.e. R∝1/a. • Forexample, if the cross-sectional area of a piece of wire isdoubled then the resistance is halved.

  21. ResistanceAndResistivity • A constant can be takenintoaccountbased on thetype of thematerial. Theconstant of proportionalityis known as the resistivity of thematerial and is given the symbol ρ (Greek rho). resistanceR = ρl /a ohms ρ is measured in ohm metres (Ωm).

  22. ResistanceAndResistivity • Resistivity varies with temperature and some typicalvalues of resistivities measured at about roomtemperaturearegivenbelow: • Copper 1.7×10−8Ωm (or 0.017μΩm) • Aluminium 2.6×10−8Ωm (or 0.026μΩm) • Carbon (graphite) 10×10−8Ωm (0.10μΩm) • Glass1×1010Ωm (or 104μΩm) • Mica1×1013Ωm (or 107μΩm) • Note that good conductors of electricity have a lowvalueof resistivity and good insulators have a high value ofresistivity.

  23. ResistanceAndResistivity • Problem 1. The resistance of a 5m length of wireis 600Ω. Determine (a) the resistance of an 8mlength of the same wire, and (b) the length of thesame wire when the resistance is 420Ω. (a) Resistance, R, is directly proportional to length, l,i.e. R∝l. Hence, 600 Ω∝5mor 600=(k)(5), where k is the coefficient of proportionality. • Hence, k = = 120 • When the length l is 8m, then resistanceR=kl=(120)(8)=960Ω (b) When the resistance is 420Ω, 420=kl, fromwhich, • = = 3.5 m

  24. ResistanceAndResistivity • Problem 2. A piece of wire of cross-sectionalarea 2mm2 has a resistance of 300Ω. Find • (a) theresistance of a wire of the same length and materialif the cross-sectional area is 5mm2, • (b) thecross-sectional area of a wire of the same lengthand material of resistance 750. • Resistance R is inversely proportional to cross-sectionalarea, a, i.e. R∝l/a • Hence 300 Ω ∝ mm2 or 300=(k) () • from which, the coefficient of proportionality, • k=300×2=600 • (a) When the cross-sectional area a=5mm2then • Ω • (b) When the resistance is 750Ωthen • Fromwhich • cross-sectionalarea,

  25. ResistanceAndResistivity • Problem 3. A wire of length 8m andcross-sectional area 3mm2 has a resistance of0.16 Ω. If the wire is drawn out until itscross-sectional area is 1mm2, determine theresistanceof thewire. • Resistance R is directly proportional to length l, andinversely proportional to the cross-sectional area, a, i.e. • where k is the coefficient ofproportionality. • Since R=0.16, =8 and a=3, then 0.16=(k)(8/3),fromwhichk=0.16×3/8=0.06 • If the cross-sectional area is reduced to 1/3 of itsoriginal area then the length must be tripled to 3×8,i.e. 24m • New resistance

  26. ResistanceAndResistivity • Problem 4. Calculate the resistance of a 2kmlength of aluminium overhead power cable if thecross-sectional area of the cable is 100mm2. Takethe resistivity of aluminium to be 0.03×10−6 Ω m. • and . =0.6

  27. ResistanceAndResistivity • Problem 5. Calculate the cross-sectional area, in, of a piece of copper wire, 40m in length andhaving a resistance of 0.25. Take the resistivity ofcopperas . • ResistanceR = ρ/a hencecross-sectionalarea

  28. ResistanceAndResistivity • Problem 6. The resistance of 1.5km of wire ofcross-sectional area is 150 Ω. Determinethe resistivity of the wire.

  29. ResistanceAndResistivity • Problem 7. Determine the resistance of 1200mof copper cable having a diameter of 12mm if theresistivity of copper is • Cross-sectionalarea of cable,

  30. TemperatureCoefficient OfResistance • As the temperature of a material increases,most conductors increase in resistance, insulatorsdecrease in resistance, whilst the resistance of somespecial alloys remain almost constant. • The temperature coefficient of resistance of amaterialis the increase in the resistance of a 1Ωresistorof that material when it is subjected to a rise of temperature of 1◦C. • The symbol used for the temperaturecoefficient of resistance is α (Greek alpha). • Forexample, ifsome copper wire of resistance 1Ωis heated through1◦C and its resistance is then measured as 1.0043Ωthenα=0.0043Ω/Ω◦C forcopper. • Theunitsareusuallyexpressed only as ‘per ◦C’, i.e. α=0.0043/◦C forcopper. • If the 1Ωresistor of copper is heatedthrough 100◦C then the resistance at 100◦C would be 1+100×0.0043=1.43Ω.

  31. TemperatureCoefficient OfResistance • If the resistance of a material at 0◦C is known theresistance at any other temperature can be determinedfrom: where R0 = resistance at 0◦C R= resistance at temperature ◦C α0= temperature coefficient of resistance at 0◦C

  32. TemperatureCoefficient OfResistance • Problem 1. A coil of copper wire has a resistanceof 100Ωwhen its temperature is 0◦C. Determine itsresistance at 70◦C if the temperature coefficient ofresistance of copper at 0◦C is 0.0043/◦C. • Resistance. Henceresistanceat 100◦C,

  33. TemperatureCoefficient OfResistance • Problem 2. An aluminiumcable has a resistanceof 27Ωat a temperature of 35◦C. Determine itsresistance at 0◦C. Take the temperature coefficientof resistance at 0◦C to be 0.0038/◦C. • Resistance at Hence resistance at 0◦C, Ω

  34. TemperatureCoefficient OfResistance • Problem 3. A carbonresistor has a resistanceof 1kΩat 0◦C. Determine its resistance at 80◦C.Assume that the temperature coefficient ofresistance for carbon at 0◦C is −0.0005/◦C. • Resistance at temperature i.e.

  35. ResistorColorCode • In 1963, the International Electrotechnical Commission (IEC) standardized the preferrednumber series for resistors and capacitors (standard IEC 60063). • The IEC also defines how manufacturers should mark the values of resistorsand capacitors in the standard called IEC 60062. • The resistor color code is a way of showing the value of a resistor. • Instead of writingthe resistance on its body, which would often be too small to read, a color code isused. • Different colors represent the numbers 0 to 9. • For a four-band fixed resistorthe first two colored bands on thebody are the first two digits of the resistance, and the third band is the 'multiplier‘andthelastband is theTolerance. • Multiplier just means the number of zeroes to add after the first two digits.

  36. ResistorColorCode • Examples • (i) For a four-band fixed resistor (i.e. Resistancevalues with two significant figures):yellow-violet-orange-red indicates 47kwith atoleranceof ±2% • (ii) For a five-band fixed resistor (i.e. Resistancevalues with three significant figures): red-yellowwhite-orange-brown indicates 249kwith a tolerance of ± 1%

  37. ResistorColorCode

  38. ResistorColorCode • Problem 1. Determine the value and toleranceof a resistor having a colour coding of:orange-orange-silver-brown. • The first two bands, i.e. orange-orange, give 33 fromtable • The third band, silver, indicates amultiplier of 102from table, which means that the value of the resistor is 33×10−2=0.33Ω • The fourth band, i.e. brown, indicates a tolerance of ± 1% from table. Hence a colour coding oforange-orange-silver-brownrepresents a resistorof value 0.33with a tolerance of ± 1%

  39. ResistorColorCode • Problem 2. Determine the value and toleranceof a resistor having a colour coding of:brown-black-brown. • The first two bands, i.e. brown-black, give 10 from Table. • The third band, brown, indicates amultiplier of 10 from Table, which means that the value of the resistor is 10×10=100Ω • There is no fourth band colour in this case; hence, fromTable, the tolerance is ±20%. Hence a colourcodingof brown-black-brown represents a resistor of value100with a tolerance of± 20%

  40. ResistorColorCode • *Yellow, Purple, Red, Gold = 47 x 100 = 4 700 = 4.7 + 5% • *Brown, Black, Yellow, Gold = 10 x 10 000 = 100 + 5% • *Yellow, Purple, Black, Silver = 47 x 1 = 47 + 10% • *Brown, Black, Red, Red = 10 x 100 = 1 000 = 1 + 1% • *Brown, Black, Green, Gold = 10 x 100 000 = 1 000 = 1 M + 5%

  41. ResistorColorCode • Problem 3. Between what two values shoulda resistor with colour coding brown-blackbrown-silverlie? • From Table, brown-black-brown-silver indicates10×10, i.e. 100Ω, with a tolerance of ± 10% This means that the value could lie between (100−10% of 100) Ω And (100+10% of 100) Ω • i.e. brown-black-brown-silver indicates any valuebetween 90Ωand110Ω

  42. ResistorColorCode • Problem 4. Determine the colour coding for a47khaving a tolerance of ± 5%. • From Table, 47kΩ=47×103 has a colour codingof yellow-violet-orange.With a tolerance of ±5%, thefourth band will be gold. • Hence47kΩ±5%, has a colour coding of:yellow-violet-orange-gold

  43. LetterAndDigitCodeForResistors • Another way of indicating the value of resistors is theletter and digit code shown in Table. • Tolerance is indicated as follows: F=±1%, G=±2%, J =±5%, K =±10% andM= ± 20% • Forexample • R33M = 0.33Ω± 20% • 4R7K= 4.7Ω ± 10% • 390RJ= 390Ω ± 5%

  44. LetterAndDigitCodeForResistors • Problem 1. Determine the value of a resistormarkedas 6K8F. • From Table, 6K8F is equivalent to: 6.8kΩ ± 1% • Problem 2.Determine the value of a resistormarkedas 4M7M. • From Table, 4M7Mis equivalent to: 4.7MΩ ±20% • Problem 3.Determine the letter and digit codefor a resistor having a value of 68 k± 10%. • FromTable, 68kΩ ± 10% has a letter and digit code of: 68KK