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- 1. Electricity Unit 3 1
- 2. 8.1 Electricity3.1 Introduction3.2 Electricity fundamentals 2.1 Particles and laws 2.2 Electricity generation 2.3 Electricity magnitudes3.3 Circuits 3.1 Ohm’s law. Electric Power 3.2 Elements and Symbols 3.3 Circuit associations. Calculations 3.4 Basic circuits 3.5 Electric Measurements 2
- 3. Indicador Prueba ValorComprensión del concepto de corriente eléctrica y su Ex El 2base física. Diferenciación entre corriente continua yalterna.Interpretación de esquemas de circuitos sencillos de Poster 2corriente continuaCálculos en circuitos de corriente continua basados ex el 4en la ley de OhmConocimiento de la función de cada componente Ex El 2eléctrico y de su simbologíaFormulación y resolución de problemas Ex El 2 3
- 4. 3.1 IntroductionWhat would happen if we didn’t have electricity? ? 4
- 5. 3.1 Introduction However, we have to remember that we can decrease the amount of energy that we waste everyday, helping our sustainable development. 5
- 6. 3.1 IntroductionWhat are we going to learn today?The electric current and its physic principles 6
- 7. 3.1 Introduction So, what is electricity?The concept of electricity includes all the phenomena related to electric charges 7
- 8. 3.2.1 Particles and lawsMatter can be electrically charged when the charge distribution is upsetFor example, you can change the charge distribution of a pen by rubbing it against your hair. Then you can attract some pieces of paper 8
- 9. 3.2.1 Particles and lawsMatter is formed by atoms, which contain smaller particles inside with electric charges: ProtonThe electrons and protons Electron Atom 9
- 10. 3.2.1 Particles and lawsElectrons and protons have negative and positive charges respectively .These charges create forces between them that can be attraction or repulsion forces according to the value of the charge: Equal charges: repulsion Different charges: attraction attraction repulsion repulsion 10
- 11. 3.2.1 Particles and lawsThe use of electricity needs a flow of electrons, the electric currentcThe electric current is the displacement of the electrical charges through the matter 11
- 12. 3.2.1 Particles and lawsIn solid metals such as wires, the positive charge carriers are immobile, and only the negatively charged electrons can flow. 12
- 13. 3.2.1 Particles and lawsAll the flowing charges are assumed to have positive polarity, and this flow is called conventional current. 13
- 14. 3.2.1 Particles and lawsBecause the electron carries negatives charges, the electron motion in a metal is in the direction opposite to that of conventional (or electric) current. 14
- 15. 3.2.2 Electricity generationWhat are we going to learn next?Explain how the electricity can be created. 15
- 16. 3.2.2 Electricity generationExercise 3.2.2a :Explain how the electricity can be created. Describe all elements needed in a common electric generator. Solution 16
- 17. 3.2.2 Electricity generationLong time ago… Hans Christian Oersted discovered that a electric current can disturb a compass 17
- 18. 3.2.2 Electricity generationSince the compass only is disturbed with a magnet, Volta concluded that an electric current creates a artificial magnetic Artificial magnetNatural 18magnet
- 19. 3.2.2 Electricity generationThe magnet is stronger if we use a spiral instead of a simple circuit.Therefore electric magnets are formed by several spirals connected to a battery 19
- 20. 3.2.2 Electricity generation Using Volta experiment, Mr Michael Faraday discovered that an electric current can be created when a magnet is moving close to an electric circuit or vice versa 20
- 21. 3.2.2 Electricity generationIn conclusion, if we want to create an artificial electric current, we need:Mechanical energyto move the circuit or themagnet Artificial magnet Closed circuit 21
- 22. 3.2.2 Electricity generation This is the diagram of an industrial electric generator Artificial magnetClosed circuit 22
- 23. 3.2.2 Electricity generationThe electricity is created when the mechanical energy moves the closed circuit inside the artificial magnet Electric engine video Electric generator and 23 Engine video
- 24. 3.2.2 Electricity generationWhen we have all elements together, we find that we create an alternate electric current due to the movement of the circuit inside the magnetic field. 24
- 25. 3.2.2 Exercise 3.2.2b Electricity generationExercise 3.2.2b :1. Explain the difference between the AC and the DC.2. Indicate what type of current do we use in this electrical devices: Torch, Bedroom lamp, TV, Mobil phone, Fridge.3. Draw the diagram of a mobile phone power supply.4. Find the input and output current in your mobile phone power supply. solution 25
- 26. 3.2.2 Electricity generationWhat are we going to learn next?Difference between the direct and the alternate current 26
- 27. 3.2.2 Electricity generationThe difference between AC and DC is :AC is an alternating current (the amount of electrons) that flows in both directions and DC is direct current that flows in only one direction AC changes its Voltage value from +A to –A (+ and – represent the direction) and the DC has a constant Voltage value 27
- 28. 3.2.2 Electricity generationBut, most electric at devices like an iPhone, need a Direct Current that keeps its magnitude and direction constant 28
- 29. 3.2.2 Electricity generationIf we apply an AC to an electrical device like a light bulb, we obtain a light pulse as you see in a bike dynamo: At home, the pulse of the AC supply is so fast that the light pulse cannot be detected by the human eyes 29
- 30. 3.2.2 Electricity generationIn order to obtain a direct current we use powers supplies that include different elements that transforms AC into DC. DC AC 30
- 31. 3.2.2 Electricity generationThese are the elements of a Power supply that that transforms AC into DC: transformer, rectifier, smoothing and regulator. 31
- 32. 3.2.2 Electricity generationThese are the elements of a Power supply that that transforms AC into DC: transformer, rectifier, smoothing and regulator. 32
- 33. 3.2.2 Electricity generationThese are the elements of a Power supply that that transforms AC into DC: transformer, rectifier, smoothing and regulator. 33
- 34. 3.3.1 CIRCUITSWhat are we going to learn next?Analysis of the direct current circuits. 34
- 35. 3.3.1 Exercise 3.3.1:Electricity magnitudesExercise 3.3.1:Define Voltage, Intensity and Resistance and its units Solution 35
- 36. 3.3.1 Electricity magnitudesTo better understand the concept of the electric current, we can think of it as a stream where the drops are the electric charges We use the water power from the drops’ movement to create energy 36
- 37. 3.3.1 Electricity magnitudesIn order to understand electricity, we have to first know the main electric magnitudes: VOLTAGE INTENSITYResistance 37
- 38. 3.3.1 Electricity magnitudesThe electrons need energy to be able to move through a material. This is the Voltage EnergyWe define the Voltage as the energy per charge unit that makes them flow through a material. This magnitude is measured in Volts (V). 38
- 39. 3.3.1 Electricity magnitudesThe higher the Voltage is, the more energy the electric charges will have to keep on moving. Higher voltageLowervoltage 39
- 40. 3.3.1 Electricity magnitudesIntensity is the amount of charges that goes through a conductor per time unit. It is measured in Ampere (A) Higher intensity Lower intensity 40
- 41. 3.3.1 Electricity magnitudesWe can see that the stream will have more strength if there is more water in the tank. It’s similar to what happened to the electricity More water pressureLesswaterpressure 41
- 42. 3.3.1 Electricity magnitudesElectric resistance is the opposition to the movement of the charges through a conductor. It is measured in Ohms (Ω) Higher Resistance Lower Resistance 42
- 43. 3.3.1 Electricity magnitudesExercise 3Which one of these lamps will give light? 43
- 44. 3.3.1 Electricity magnitudesExercise 4Will this lamp turn on? 44
- 45. 3.3.1 Electricity magnitudesThe electric current always goes through the route with less resistance, like water Here we have the resistance needed to create light No resistance 45
- 46. 3.3.2 Ohm’s law. Electric PowerWhat are we going to learn next?Magnitudes analysis using the Ohm’s Law. 46
- 47. 3.3.2 Exercise 3.3.2a Ohm’s law. ElectricPowerExercise 3.3.2a:Justify how the Intensity will be if: We have a low voltage V We have a low Resistance R Solution 47
- 48. 3.3.2 Ohm’s law. Electric PowerOhm’s law links the three electric magnitudes as is shown: V I= R V= Voltage (voltsV) I= Intensity (ampereA) R= Resistance (ohm Ω) 48
- 49. 3.3.2 Ohm’s law. Electric PowerIntensity is directly proportional to voltage:If the voltage is high, the charges will have a lot of energy. Therefore the Intensity will be high too V I= R 49
- 50. 3.3.2 Ohm’s law. Electric Power Intensity is inversely proportional to Resistance If there is a high Resistance, the intensity will be low. VThe charges will cross thematerial slowly I= R 50
- 51. Solution3.3.2 Exercise 3.3.2.bOhm’s law. Electric PowerExercise 3.3.2.b: V (V) R (Ω) I (A)Calculate the value of the intensity in these cases: 2 14 2 2 20 2 V I= 12 6 R 252 1000 64 20 51
- 52. 3.3.2 c Exercise Ohm’s law. Electric PowerWe have a fan connected to the domestic electric supply. The amperimeter indicate a 0,52A on the circuit.1. What’s is the resistance of the fan?2. What’s the power measured in KwCalculate the Intensity that flaws in a electrical engine of 1,2 Kw installed in my washing machine. What would be the Intensity and the power if we have 380V at home.Extra data: house electrical supply 230V, 50Hz Solution52
- 53. 3.3.2 D Exercise Ohm’s law. Electric PowerA) Calculate the Intensity that flaws in a electrical engine of 1,2 Kw installed in my washing machine.B) B)What would be the Intensity and the power if we have 380V at home.Extra data: house electrical supply 230V, 50Hz Solution 53
- 54. 3.3.2 Ohm’s law. Electric PowerPower is defined as the work consumed per time unit, and it’s measured in watt W W P= t 54
- 55. 3.3.2 Ohm’s law. Electric Power W W VQV= ⇒ W = VQ Q P= = = VI t t QI= tIf we apply the Ohm’s law, we obtain a new formula that explains the power of an electrical device P = VI 55
- 56. 3.3.2 Ohm’s law. Electric PowerFrom this expression we can derivate in other two using the Ohm’s LawP = VI P= I R 2V = IR 2 V VI= P= R R 56
- 57. 3.3.2 Ohm’s law. Electric PowerThere is a bulb lamp of 100w connected to 220V electric grid. Calculate:Intensity throw the lamp.Power of the lamp.Calculate the Kwh that the electricity meter will show after 20 hours.Indicate the price of the electricity consumed if the price is 0,90€/Kwh 57
- 58. 3.3.2 Ohm’s law. Electric PowerThere is a bulb lamp of 100w connected to 220V electric grid. Calculate:Intensity throw the lamp. P = 100W P = VI P 100 V = 220V I= = = 0,45A V 220 58
- 59. 3.3.2 Ohm’s law. Electric PowerThere is a bulb lamp of 100w connected to 220V electric grid. Calculate:Power of the lamp.As is says in the wording it is 100WP=100WCalculate the Kwh that the electricity meter will show after 20 hours.P=100W=0,1kWPKWh=0,1x20h=2Kwh 59
- 60. 3.3.2 Ohm’s law. Electric PowerIndicate the price of the electricity consumed if the price is 0,90€/Kwh Price Cost =Energy consumed × Energy € Cost =Kwh × Kwh 0,9€ Cost =2Kwh × Cost = 1,8€ Kwh 60
- 61. 3.3.2 Ohm’s law. Electric PowerElectrical companies measure our electrical consume in Kwh with a electricity meter situated in the basement of a building.PowerConsumed = P • h 61
- 62. 3.3.3 Elements and SymbolsWhat are we going to learn next?Electric components and their applications. 62
- 63. 3.3.3 Elements and symbolsAn electric circuit is a group of elements that allows us to control electric current through some sort of material 63
- 64. 3.3.3 Elements and symbolsThe essential elements in a circuit are:Generator: it creates an electric current supplying voltage to the circuit. They can be: Batteries: they supply electric current but only for a short time. Power supplies: they give a constant and continuous electric current. 64
- 65. 3.3.3 Elements and symbolsThe essential elements in a circuit are:Control elements: we can manipulate the electricity through the circuit. Switch: they keep the ON or OFF positions. For example, the switch lights at the bathroom Push button: The On position only works while you are pressing the button. For example the door bell. Diverter switch: it is used to switch a light on or off from different points in the same room, as you have in your bedroom 65
- 66. 3.3.3 Elements and symbolsControl elements: Change Direction diverter switch: We use this element to invert the direction of the electric current through a component. 66
- 67. 3.3.3 Elements and symbolsControl elements: Relay: It is an electrically operated switch. Thanks to that we can control a circuit from a remote control applying just electricity An electrical magnet attracts the switch that closes the circuit 67
- 68. 3.3.3 Elements and symbolsSafityIt is used to control high electrical volt devices, because we don’t touch the main circuit 68
- 69. 3.3.3 Elements and symbolsReceptors: they are the elements that transform the electric energy into other ones that are more interesting for us. For example:Incandescent lights: when the electric current goes through the lamp filament it gets really hot and starts emitting light.Engine: the electricity creates a magnetic field that moves the metal elements of the engineResistance: we use it to decrease the intensity of a circuit 69
- 70. 3.3.3 Elements and symbolsConductor: all the elements have to be connected to a material that transmits the electric charges.The circuit has to be CLOSED in order to allow the electricity to circulate around it from the positive to the negative pole. 70
- 71. 3.3.3 Elements and symbolsProtection elements: they keep all the circuit elements safe from high voltage rises that can destroy the receptors. Fuse: the first one will blow, cutting the circuit in case of a voltage rise. They are easily replaced Circuit breaker: they are used in new electrical installations, at home or at factories. If there is a voltage rise, you don’t have to replace them, only 71 reload them.
- 72. 3.3.3 Elements and symbolsElectric symbols are used to represent electric circuits with drawings that replace the real circuit elements. 72
- 73. 3.3.3 Elements and symbolsExercise: Draw the following circuit using electric symbols 73
- 74. 3.3.3 Elements and symbolsExercise: Draw the following circuit using electric symbols 74
- 75. 3.3.3 Elements and symbolsExercise: Draw the following circuit using electric symbols 75
- 76. 3.3.3 Elements and symbols Generator Battery Battery association Conductors: When two conductors are crossed without any contact we indicate it with a curve 76
- 77. 3.3.3 Elements and symbols Control elements Push button Switch Diverter switch 77
- 78. 3.3.3 Elements and symbols Protection elements Fuse Receptors: Lamp Resistances: they have two symbols Engines 78
- 79. 3.3.4 Circuit associationsThe behaviour of electrical elements depends on how they connect to each other.There are three possible configurations :SeriesParallelMixed 79
- 80. 3.3.4 Circuit associations SERIES circuitThe series circuit connects the electrical elements one behind the other.In this way, there is only one connection point between elements 1 & 2 are connected only by A 2 & 3 are connected only by B 80
- 81. 3.3.4 Circuit associations PARALLEL associationIn this association, all the elements are connected by two points.So, 1, 2 & 3 are connected by A and B 81
- 82. 3.3.4 Circuit associations MIXED associationMixed association has elements associated in parallel and in series.1, 2 & 3 are in parallel and all of them are in series with 4 82
- 83. 3.3.4 Circuit associationsBut what happens to receptors when they are connected in series or parallel associations?Series and parallel association change the valueof intensity and voltage through receptors. 83
- 84. 3.3.4 Circuit associations Voltage Series ParallelVoltage is distributed Voltage is the same in allbetween elements, that is elements, so all thethe reason why they have lamps have the sameless energy for each energy and the light islamp, so the light is higher.lower in each lamp. 84
- 85. Intensity Series ParallelAll the lamps are in line, All lamps are separated soso they create a high they create a lowresistance, that is the resistance, that is thereason why the intensity reason why the intensityis lower but the battery through the lamps iswill have a longer life. higher, but the battery will have a shorter life 85
- 86. Circuit Series ParallelIf there is any cut along If there is any cut alongthe conductor, the the conductor, theelectric current will not electric current can gobe able to go from the through any other waypositive to the negativepole. CUT 86
- 87. 3.3.4 Circuit associationsIf there is any cut in series association, the water can’t go further. But in parallel association there will be no problem. cut 87
- 88. 3.3.4 Circuit associations ExercisesWhich of these elements are associated with series, parallel or mixed circuits? Name the connection points with letters. 88
- 89. 3.3.4 Circuit associations ExercisesWhich of these elements are associated with series, parallel or mixed circuits. Name the connection points with letters 89
- 90. 3.3.4 Circuit associations ExercisesWhich of these elements are associated with series, parallel or mixed circuits?. Name the connection points with letters. 90
- 91. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt 91 Solution
- 92. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt 92
- 93. 3.3.4 Circuit associations. Calculations SERIES circuitIn this association we find that:VT=V1+ V2+V3 RT=R1+R2+R3IT=I1= I2=I3 ET=E1+ E2+E3 ET RT VT ET=VT IT 93
- 94. 3.3.4 Circuit associations. Calculations VT I1 ET RT V1 IT I2 I3 V2 V3 94
- 95. 3.3.4 Circuit associations. Calculations SERIES circuit INTENSITY= EQUAL IT=I1=I2=I3 I1 ET RT VT ET=VT IT IT I I2 I3 T 95
- 96. 3.3.4 Circuit associations. Calculations SERIES circuit VOLTAGE= DIVED VT=V1+ V2+V3 V1 ET RT VTT V ET=VT IT IT I V2 V3 T 96
- 97. 3.3.4 Series Exercise Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 97
- 98. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance R1= 0,5 Ω R2= 6,3 Ω R3= 6/5 Ω RT= 1º I1= 2º I2= 2º I3= 2º IT= 2º VT= 220V V1= 3º V2= 4º 5º V 3= 98
- 99. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance RT= 8 Ω R1= 0,5 Ω R2= 6,3 Ω R3= 6/5 Ω IT= 27,5 A I1= I 2= I3= 1º VT= 220V V1= V 2= V3= R t = R1 + R2 + R3 2º IT = VT 1 6 RT R t = + 6,3 + 2 5 220 IT = R t = 8Ω 8 I T = 27,5Ω 99
- 100. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance RT= 8 Ω R1= 0,5 Ω R2= 6,3 Ω R3= 6/5 Ω IT= 27,5 A I1= 27,5 A I2= 27,5 A I3= 27,5 A VT= 220V V1= V 2= V3= Series ⇒I t = I1 = I2 = I3 = 27,5A 100
- 101. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance RT= 8 Ω R1= 0,5 Ω R2= 6,3 Ω R3= 6/5 Ω IT= 27,5 A I1= 27,5 A I2= 27,5 A I3= 27,5 A VT= 220V V1= 13,75V V2= 173,25V V3= 33V Series ⇒V t= V1 + V2 + V3 = 220V V1 = I1R1 = 0,5 • 27,5 = 13,75V V2 = I2 R2 = 6,3 • 27,5 = 173,25V V3 = I3 R3 = 1,2 • 27,5 = 33V Series ⇒V t= 13,75 +173,25 + 33 = 220V 101
- 102. 3.3.4 Circuit associations PARALLEL associationVT=V1= V2=V3 1 1 1 1 = + + RT R1 R2 R3IT=I1+ I2+I3 ET=E1+ E2+E3 102
- 103. 3.3.4 Circuit associations VT=V123=V1= V2=V3 IT=I123=I1+ I2+I3 103
- 104. 3.3.4 Circuit associations PARALLEL association INTENSITY= DIVED IT=I1+I2+I3 I1 I2 IT I3 104
- 105. 3.3.4 Circuit associations PARALLEL association INTENSITY= DIVED IT=I1+I2+I3 I1 I2 IT I3 105
- 106. 3.3.4 Circuit associations PARALLEL association VOLTAGE: IQUAL VT=V1= V2=V3 V1 V2 V3 VT 106
- 107. 3.3.4 Circuit associations PARALLEL association V1 R T= R1= 6 Ω R2= 18 Ω R3= 4 Ω V2 IT= I1= I2= I3= V3 VT= 20V V1= V2= V 3= VT RT= R1= 6 Ω R2= 18 Ω R3= 4 Ω IT= I1= I 2= I3= VT= 220V V1= V 2= V 3= 107
- 108. 3.3.4 Circuit associations PARALLEL association V1 R T= R1= 6 Ω R2= 18 Ω R3= 4 Ω V2 IT= I1= I2= I3= V3 VT= 20V V1= V2= V 3= VT R1= 6 Ω R2= 18 Ω R3= 4 Ω RT= 1º IT= 2º I1= 3º I 2= 4º I3= 5º VT= 220V V1= 20V V2= 20V V 3= 20V 108
- 109. 3.3.4 Circuit associations PARALLEL EXERCISE RT=2,11Ω R1= 6 Ω R2= 18 Ω R3= 4 Ω IT= I1= I2= I3= VT= 20V V1= V2= V3= V1 V2 V3 VT 109
- 110. 3.3.4 Circuit associations PARALLEL EXERCISE RT=2,11Ω R1= 6 Ω R2= 18 Ω R3= 4 Ω IT= I1= I2= I3= VT= 20V V1= V2= V3= V1 V2 V3 VT 110
- 111. 3.3.4 Circuit associations PARALLEL EXERCISE RT=2,11Ω R1= 6 Ω R2= 18 Ω R3= 4 Ω IT=9,48A I1= I2= I3= VT= 20V V1= V2= V3= V1 VT IT = RT V2 20 IT = V3 2,11 VT IT = 9,48A 111
- 112. 3.3.4 Circuit associations VOLTAGE: IQUAL VT=V1= V2=V3 VT=20 RT=2,11Ω R1= 6 Ω R2= 18 Ω R3= 4 Ω IT=9,48A I1= I2= I3= V1 VT= 20V V1=20V V2=20V V3=20V V2 V3 VT 112
- 113. 3.3.4 Circuit associations PARALLEL EXERCISE RT=2,11Ω R1= 6 Ω R2= 18 Ω R3= 4 Ω IT=9,48A I1=3,33A I2=1,11A I3=5A VT= 20V V1=20V V2=20V V3=20V V1 V1 20 I1 = = = 3,33A R1 6 V2 V2 20 I2 = = = 1,11A R2 18 V3 V3 20 VT I3 = = = 5A 113 R3 4
- 114. 3.3.4 Circuit associations MIXED association We have to convert the parallel associations into series in order to obtain Vt, It and Rt. T 1º3º 2º Vt, It and RtI1, I2 , I3 and V4, I4 and V123 V123 114
- 115. 3.3.4 Circuit associations MIXED associationIT=I123=I4 series associationI123=I1+ I2+I3 parallel association 115
- 116. 3.3.4 Circuit associations MIXED associationIT=I123=I4 series associationI123=I1+ I2+I3 parallel association I1 I2 I1 116
- 117. 3.3.4 Circuit associations MIXED associationI123=I1+ I2+I3, but they don’t have to be equal, it depends on the R in each path. I1 I2 I1 117
- 118. 3.3.4 Circuit associations VT=V123+V4 series association V123=V1= V2=V3 parallel association V1 VT=V123+V4 V2 V3 118
- 119. 3.3.4 Circuit associations VT=V123+V4 series association IT=I123=I4 series association V123=V1= V2=V3 parallel association I123=I1+ I2+I3 parallel association 119
- 120. 3.3.4 Circuit associations VT=V123+V4 series association V123=V1= V2=V3 parallel association I123=I1+ I2+I3 parallel association 3º 1ºI1, I2 , I3 and 2º Vt, It V4, I4 and and Rt V1 V123 IT=I123=I4 series association 120
- 121. 3.3.4 Ex 1 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance R1= 6 Ω R2= 18 Ω R3= 3/2 Ω RT= 1º IT= 2º I1= 5º I 2= 6º I3= 2º VT= 20V V1= 4º V 2= 4º V 3= 3º 121
- 122. 3.3.4 Circuit associations. Calculations 1º RT= 6 Ω IT= VT= 20V 122
- 123. RT= 6 Ω R1= 6 Ω R2= 18 Ω R3= 3/2 ΩIT= 10/3 A I1= I2= I3=VT= 20V V1= V 2= V 2= 2º VT 20 IT = = RT 6 10 IT = A 3 123
- 124. RT= 6 Ω R1= 6 Ω R2= 18 Ω R3= 3/2 ΩIT= 10/3 A I1= I2= I3= 10/3 AVT= 20V V1= V 2= V 3= 2º Mixed Mixed IT = I1 + I2 = I1− 2 = I3 10 /3A = I1 + I2 = I1− 2 = 10 /3 I2 I3 I1 I1-2 I3 124
- 125. RT= 6 Ω R1= 6 Ω R2= 18 Ω R3= 3/2 ΩIT= 10/3 A I1= I2= I3= 10/3 AVT= 20V V1= V 2= V3= 5V Mixed VT = V1−2 + V3 3º V1−2 = V1 = V2 10 10 3 I3 = IT = A ⇒ V3 = I3 R3 = • = 5V 3 3 2 125
- 126. RT= 6 Ω R1= 6 Ω R2= 18 Ω R3= 3/2 ΩIT= 10/3 A I 1= I2= I3= 10/3 AVT= 20V V1= 15V V2= 15V V3= 5V 4ºOne way Other way 10 9Mixed : V1−2 = I1−2R1−2 = • = 15V 3 2VT = V1−2 + V3 V1−2 = V1 = V2 = 15VV1−2 = VT1 − V3 V1 = 15V V2 = 15VV1−2 = 20 − 5 = 15V Two ways to calculate V1 and V2 126
- 127. RT= 6 Ω R1= 6 Ω R2= 18 Ω R3= 3/2 ΩIT= 10/3 A I1= 2,5 A I2= 5/6 A I3= 10/3 AVT= 20V V1= 15v V2= 15V V 2= 5 V 5º 6º V1 15 5 I1 = = = ⇒ I1 = 2,5 A R1 6 2 V2 15 5 5 I2 = = = ⇒ I1 = A R2 18 6 6 127
- 128. 3.3.4 Circuit associations VT=V123+V4 series association V123=V1= V2=V3 parallel association I123=I1+ I2+I3 parallel association 3º 1ºI1, I2 , I3 and 2º Vt, It V4, I4 and and Rt V1 V123 IT=I123=I4 series association 128
- 129. 3.3.4 Ex 2 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 4Ω 4Ω 12 Ω220V 8Ω RT= R1= 4 Ω R2= 4 Ω R3= 8 Ω R4= 12 Ω IT= I1= I 2= I3= I 4= sol VT= 220V V 1= V3 V4 = V 2= 129
- 130. 3.3.4 Circuit associations. CalculationsE represents the electrometric force that is the Total Voltage that provides a electric source like a battery or a power supply, 130
- 131. 3.3.4 Circuit associationsExercise 3.3.4. Try to make a cheat sheet with all the formulas needed to solve electric circuits. It has to be no bigger than 25cm2. 131
- 132. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 132 Sol Sol
- 133. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 133
- 134. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 134
- 135. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 135
- 136. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalentRt, It and Vt. Calculate the I and V in eachResistance 136
- 137. 3.3.4 Circuit associationsExercise:Which of these lamps have more lightness? 137
- 138. 3.3.4 Circuit associationsExercise:Which of these assemblies generates electricity? 138
- 139. 3.3.4 Circuit associations Exercise: What happen in this exercise if we activate: 1. 1 and 7 2. 1 and 8 3. 1,2 and 3 4. 1,2,3,4,5,6 and 7 5. 1,4,6 and 7 What do you have to activate to switch on…? 1. A and B 2. A, B and C 3. A, B , C and M 139
- 140. 3.3.4 Circuit associationsDraw the electric circuit of a car that has :1.2 front lamps2.2 Back lamps3.A forward and backward engine that stops automatically when it crash with a object 140
- 141. 3.3.4 Circuit associations Why are these assemblies no correct? 141
- 142. 3.3.6 Electric Measurements. PolimeterOhmmeter: It measures the resistance and the continuity of acircuit or component.If there is no resistance that means that there is a shortcircuitIf there is a maximum resistance that means that the cricuit is open Ω Resistance control 142
- 143. 3.3.6 Electric Measurement. Polimeter• In order to get a precise lecture we have tochoose the correct range of measurement. 4K7Ω 143
- 144. 3.3.6 Electric Measurements. PolimeterVoltimeterAplication: If we want to measure the voltage present in acomponent we have to conect the electrodes in parallel V Voltage in a component 144
- 145. 3.3.6 Electric Measurements. PolimeterIf we want to measure the voltage of batterywe have to place the control in the DirectCurrent and in the proper scale 145
- 146. 3.3.6 Electric Measurements. PolimeterAmperimeter: In order to be able to measure thecurrent we have to place the electrodes BETWEENthe circuitr or in sereies A CONTROL DE CONSUMO 146
- 147. 3.3.6 Electric Measurements. PolimeterAmperimeter: we chose the position accordingto the scale of the current. 4K7 Ω 147
- 148. 3.3.4 Circuit associations Exercise: According to this assembly, chose the correct answer: The I through the lamp is 2,3 A The voltage between the lamp is 2,3 V The resistance of the lamp is 2,3 Ω 148
- 149. 3.3.4 Circuit associations Point out in the table if the engine and lamps works for the following situations A closed A open A closed B open B closed B closedEnginelamp 1lamp 2 149
- 150. 3.2.2 Electricity generationExercise 3.2.2a :Explain how the electricity can be created. Describe all elements needed in a common electric generator.Thanks to the discoveries made by Faraday and Oesterd, we know that when a closed circuit moves inside a magnetic field an electric current is created inside the circuit.Therefore, nowadays we use an artificial magnet (made with a electric current that flows through a spiral that has a iron bar inside), closed circuit (spirals) and mechanical energy to create electricity. The closed circuit is moved thanks to the mechanical energy inside the artificial magnet. Thanks to the energy of the magnetic field created by the magnet the electrons from the circuit start to move creating the electric current. Back 150
- 151. 3.2.2 Sol Exercise 3.2.2b Electricitygeneration1. Explain the difference between the AC and the DC.The difference between AC and DC is that AC is an alternating current (the amount of electrons) that flows in both directions and DC is direct current that flows in only one direction; Also, AC changes its Voltage value from +A to –A (+ and – represent the direction) and the DC has a constant Voltage value 151
- 152. 3.2.2 Electricity generationExercise 3.2.2b :2.-Indicate what type of current do we use in this electrical devices: Torch, Bedroom lamp, TV, Mobil phone, Fridge.AC power iswhat fuelsour homes.The wiresoutside ofour houseareconnectedat two endsto AC 152
- 153. 3.2.2 Electricity generationExercise 3.2.2b :2.-Indicate what type of current do we use in this electrical devices: Torch, Bedroom lamp, TV, Mobil phone, Fridge.DC is found in batteries andsolar cells and it used for mostof our electric devices. Weneed to convert the AC intoDC in most of electric devicesusing power suplies 153
- 154. 3.2.2 Electricity generationExercise 3.2.2b :2.-Indicate what type of current do we use in this electrical devices: Torch, Bedroom lamp, TV, Mobil phone, Fridge. AC DC TV TORCH Fridge Phone Bedroom lamp 154
- 155. 3. Draw the diagram of a mobile phone power supply. DC AC Back 155
- 156. 3.2.2 Electricity generation Exercise 3.2.2b : 4.Find the input and output current in your mobile phone power supply.Input; 100-240 VOutput: 5,3 V 156
- 157. 3.3.1 Exercise 3.3.1:Solution BAck Electricity magnitudesExercise 3.3.1:Define Voltage, Intensity and Resistance and its units Simbol Name Definition Unit Voltage Is the energy per V Voltage charge unit that makes them Volts (V) flow through a material Intensity is the amount of Ampere I Intensity charges that goes through a (A) conductor per time unit Electric resistance is the opposition to the movement of R Resistance ohm (Ω) the charges through a 157 conductor.
- 158. 3.3.2 a Exercise Ohm’s law. Electric PowerIntensity is directly proportional to voltage:If the voltage is Low, the charges will have less energy to move. Therefore the Intensity will be low too V I= R 158
- 159. exercise 3.3.2 a Exercise Ohm’s law. Electric Power Intensity is inversely proportional to Resistance If there is Low Resistance, the intensity will be High because there is less opostion to the movement of electrons. VThe charges will cross thematerial slowly I= R 159
- 160. exercise3.3.2 Exercise 3.3.2.bOhm’s law. Electric PowerExercise 3.3.2.b: V (V) R (Ω) I (A)Calculate the value of the intensity in these cases: 2 14 0,14A 2 1Ω 2 20 2 10A V I= 72 V 12 6 R 252 1000 64 20 3,94 A 50A160
- 161. 3.32 b Solution Ohm’s law Calculations with Ohm’s law 5º Exercise SolutionV (V) R (Ω) I (A) V 2 2 2 I= I= I = 1A R 2 V 2 I= I= I = 0,5A R 4 2 4 161
- 162. 6.4 Ohm’s lawCalculations with Ohm’s law Solution V (V) R (Ω) I (A) V = IR V = 4⋅2 2 4 V = 8V V R = I 10 10 5 R = 5 R = Ω 2 162
- 163. exercise6.4 Ohm’s lawCalculations with Ohm’s law Solution V (V) R (Ω) I (A) V I= R 5 10 5 I= 10 I = 0,5 AV = IRV = 1000 ⋅ 20 20 1000V = 20000VV = 20kv 163
- 164. 3.3.2 c Solution Ohm’s law. Electric PowerWe have a fan connected to the domestic electric supply. The amperimeter indicate a 0,52A on the circuit. Extra data: house electrical supply 230V, 50Hz1º Text analysis:I= 0,52A ; V=230V; R= ? ; P=?1. What’s is the resistance of the fan? 164
- 165. 3.3.2 c Solution Ohm’s law. Electric PowerWe have a fan connected to the domestic electric supply. The amperimeter indicate a 0,52A on the circuit. Extra data: house electrical supply 230V, 50Hz1º Text analysis:I= 0,52A ; V=230V; I= ? ; P=?2º- What’s the power measured in Kw P = VI P = 230 • 0,52 = 119,6W = 0,12kw P = 0,12kw or you could aslo do V V 2 230 2 P = V× = = = 0,12kw 165 R R 442,3
- 166. 3.3.2 c Solution Ohm’s law. Electric PowerCalculate the Intensity that flaws in a electrical engine of 1,2 Kw installed in my washing machine.Extra data: house electrical supply 230V, 50HzText Analysis:R= ?; V=230V; I= ?; P=1,2kWP = VI P 1200P = VI ⇒ I = = = 5,22A V 230I = 5,22A 166
- 167. 3.3.2 c Solution Ohm’s law. Electric PowerCalculate the Intensity that flaws in a electrical engine of 1,2 Kw installed in my washing machine. What would be the Intensity and the power if we have 380V at home.Text Analysis:R= ?; V=380V; I= ?; P=1,2kWP = VI P 1200P = VI ⇒ I = = = 3,15A V 380I = 3,15A Back 167
- 168. 3.3.4 Circuit associations SolutionsParallel: 1, 2 and 3 Series: 1 and 2 areare joined together joined by A; 2 and 3by A and B are joined by B 168
- 169. 3.3.4 Circuit associations SolutionsMixed:Series: 1 and 2 are joined to ASeries: 2 and the 4 & 3 association are joined by BParallel: 4 and 3 are joined by B and C 169
- 170. 3.3.4 Circuit associations Solutions: red-series Blue: parallelSERIES: MIXED: 1 and 2 are joined by A. Series:1 is joined to 2 and 3 by A2 & 3 are joined by B Parallel: 2 and 3 are joined by A and3 & 4 are joined by C B Series: 2 and 3 are joined to 4 by C 170
- 171. 3.3.4 Circuit associations. Calculations 1 1 1 = + RT R1 R2R t = R1 + R2 + R3 1 1 1 = + = 7 + 1 RT 4 28 28 28 1 6R t = + 6,3 + 1 2 = 2 5 RT 7Rt = 8 RT = 7 2 RT = 3,5Ω 171
- 172. 3.3.4 Circuit associations. CalculationsRT = R1− 2 + R3 RT = R1− 2 + R3 1 1 1 9 3 = + RT = +R1− 2 R1 R2 2 2 1 1 1 3 1 12 = + = + RT =R1− 2 6 18 18 18 2 1 2 = RT = 6ΩR1− 2 9 9R1− 2 = Ω 172 2
- 173. 3.3.4 Circuit associations. Calculations 1 1 1 1 1 1 = + = + Rt R1− 2 R3 Rt R1− 2 R3 1 1 1 R1− 2 = R1 + R2 = + Rt 25 40 1 13 R1− 2 = 13 + 12 = Rt 200 200 R1− 2 = 25Ω RT = Ω 13 173
- 174. 3.3.4 Circuit associations. Calculations 174
- 175. 3.3.4 Circuit associations. CalculationsR1−2R1−2 = R1 + R2R1−2 = 1000 + 500R1−2 = 1500ΩR3−4−5 R6 −7 1 1 1 1 = + + 1 1 1R 3−4−5 R3 R4 R5 = + R 6−7 R6 R7 1 1 1 1 = + + 1 1 1R 3−4−5 6000 3000 6000 = + R 6−7 2000 2000 1 1 = 1 1R 3−4−5 1500 = R 6−7 1000R 3−4−5 = 1500Ω R 6−7 = 1000Ω 175
- 176. 3.3.4 Circuit associations. CalculationsR1 _ 5 RTR1 _ 5 = R1−2 + R3−4−5 1 1 1R1 _ 5 = 1500 +1500 = + RT R1 _ 5 R6 _ 8R1 _ 5 = 3000Ω 1 1 1R6 _ 8 = + RT 3000 3000R6 _ 8 = R6−7 + R8 1 1R6 _ 8 = 1000 + 2000 = RT 1500R6 _ 8 = 3000Ω RT = 1500Ω 176
- 177. 3.3.4 Circuit associations. Calculations 177
- 178. 3.3.4 Circuit associations. CalculationsR3−4 1 1 1 = +R3−4 R3 R4 R2−3−4 1 1 1 R2−3−4 = R2 + R3−4 = +R3−4 25 100 R2−3−4 = 20 + 100 1 1 R2−3−4 = 120Ω =R3−4 20R3−4 = 20Ω 178
- 179. 3.3.4 Circuit associations. Calculations 179
- 180. 3.3.4 Circuit associations. CalculationsR2 _ 5 1 1 1 = +R2 _ 5 R2−3−4 R5 1 1 1 = +R2 _ 5 120 120 1 1 =R2 _ 5 60R2 _ 5 = 60Ω 180
- 181. 3.3.4 Circuit associations. CalculationsRT1 1 1 = +RT R1−5 R61 1 1 = +RT 160 401 1 =RT 32RT = 32Ω 181
- 182. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 182
- 183. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance RT= 80 Ω R1= 25 Ω R2= 55 Ω IT= 2,75 A I1= I2= VT= 220V V1= V2= VT IT = R t = R1 + R2 RT R t = 25 + 55 220 IT = R t = 80Ω 80 I T = 2,75 A 183
- 184. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance RT= 80 Ω R1= 25 Ω R2= 55 Ω IT= 2,75 A I1= 2,75 A I2= 2,75 A VT= 220V V1= 68,75 V V2= 151,25V Series ⇒ I t= I1 = I 2 = 2,75 A V1 = I1 R1 = 25 • 2,75 = 68,75V V2 = I 2 R2 = 55 • 2,75 = 151,25V Series ⇒V t= V1 + V2 = 220V 184
- 185. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance back 185
- 186. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance RT= 200/13 Ω IT= VT= 20V 1 1 1 = + R T R 1−2 R 3R 1−2 = R 1 + R 2 = 13 +12R 1−2 = 25Ω 1 1 1 1 1 13 = + = + = R T R 1−2 R 3 25 40 200 200RT = Ω 186 13
- 187. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance RT= 200/13 Ω IT= 1,3 A VT= 20V 20 V 20 IT = T = = 1 Mixed circuit RT 200 200 13 13 I T = I1−2 + I 3 I T = 1,3A I1−2 = I1 = I 2 VT = V1−2 = V1 + V2 = V3 187
- 188. RT=200/13 R1= 13 Ω R2= 12 Ω R3= 40 ΩIT=1,3A I1= I2= I3=0,5AVT= 20V V1= V 2= V3=20VMixed circuitI T = I1−2 + I 3 VT = V1−2 = V3 = 20V V1−2 20I1−2 = I1 = I 2 I1−2 = I1 = I 2 = = R 1−2 25VT = V1−2 = V1 + V2 = V3 I1−2 = I1 = I 2 = 0,8 A V3 20 I3 = = R 3 40 I 3 = 0,5 A 188
- 189. RT= 200/13 Ω R1= 13 Ω R2= 12 Ω R3= 40 ΩIT= 1,3 A I1= 0,8 A I2= 0,8 A I3= 0,5 AVT= 20V V1= 10,4 V V2= 9,6 V V3= 20 V V1 = I1R 1 = 0,8 •13 V1 = 10,4V V2 = I 2 R 2 = 0,8 •12 V2 = 9,6V 189
- 190. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 190
- 191. RT= R1= 3 Ω R2= 2 Ω R3= 10 ΩIT= I1= I2= I3=VT= V1= V 2= V 3= RT = R1 + R2 + R3 RT = 3 + 2 + 10 ⇒ RT = 15Ω VT = ET = E1 + E2 = 5 − 20 = −15V VT 15 IT = = = 1A RT 15 Series association ⇒ I T = I1 = I 2 = I 3 = 1A 191
- 192. RT= Ω R1= 3 Ω R2= 2 Ω R3= 10 ΩIT= 1A I1=1A I2= 1A I3= 1AVT= 15V V1= 3V V2= 2V V3=10VPT= 15W P1= 3W P2= 2W P3=10WV1 = I1 R1 = 1 ⋅ 3 = 3V P = V1 I1 = 3 ⋅1 = 3W 1V2 = I 2 R2 = 1 ⋅ 2 = 2V P2 = V2 I 2 = 2 ⋅1 = 2WV3 = I 3 R3 = 1 ⋅10 = 10V P3 = V3 I 3 = 10 ⋅1 = 10W 192
- 193. 3.3.4 sol Ex 2 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 4Ω 4Ω 12 Ω220V 8Ω RT= 1º R1= 4 Ω R2= 4 Ω R3= 8 Ω R4= 12 Ω IT= 2º I1=2º I2= 7º I3=6º I4=2º VT = V1= =3º V2= 5º V3=5º V4=4º 220V 193
- 194. 3.3.4 sol Ex 2 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 4Ω 4Ω 12 Ω220V 8Ω RT= 1º R1= 4 Ω R2= 4 Ω R3= 8 Ω R4= 12 Ω IT= 2º I1=2º I2= 7º I3=6º I4=2º VT = V1= =3º V2= 5º V3=5º V4=4º 220V 194
- 195. 3.3.4 Circuit associations. Calculations 4Ω 4Ω 12 Ω 220V 8Ω 4Ω 8/3Ω 12 Ω220V 56/3 Ω220V 195
- 196. 3.3.4 Circuit associations. Calculations 4Ω 1º RT= 6 Ω 4Ω 12 Ω IT= 8Ω VT= 220V220V R T =R1 +R 2 −3 +R 4 R T =R1 +R 2 −3 +R 4 1 1 1 8 = + RT =4 + + 12 R 2 −3 R2 R3 3 1 1 1 2 1 56 = + = + RT = ≈18,67 R 2 −3 4 8 8 8 3 1 3 = R 2 −3 8 8 196 R 2 −3 = W 3
- 197. RT= 18,67Ω R1= 4 Ω R2= 4 Ω R3= 8 Ω R4= 12 ΩIT= 11,78 A I1= I2= I3=VT= 220V V1= V2= V3= V4= VT 220 IT = = = 11,78 RT 18,67 IT = 11,78A 2º 131/8 Ω 220V 197
- 198. RT= 18,67Ω R1= 4 Ω R2= 4 Ω R3= 8 Ω R4= 12 Ω A I1= 11,78A I 2= I3=IT= 11,78 I4= 11,78 AVT= 220V V1= V 2= V2= Mixed IT = I1 = I2 + I3 = I2− 3 = I4 = 11,78A 2º 4Ω 4Ω 12 Ω 220V 8Ω 198
- 199. RT= 18,67Ω R1= 4 Ω R2= 4 Ω R3= 8 Ω R4= 12 Ω I2= I 3=IT= 11,78 A I1= 11,78A I4= 11,78 AVT= 220V V 2= V 3= V1= 47,12V V4= 141,36VMixedVT = V1−2 + V3 3ºV1−2 = V1 = V2 I1 = IT = 11,78A ⇒ V1 = I1R1 = 11,78 • 4 = 47,12V I4 = IT = 11,78A ⇒ V4 = I4 R4 = 11,78 • 12 = 141,36V 199
- 200. RT= 18,67Ω R1 = 4 Ω R2= 4 Ω R3= 8 Ω R4= 12 ΩIT= 11,78 A I1= 11,78A I2= I3= I4= 11,78 A 4ºVT= 220V V1= 47,12V V2=31,52 V3=31,52 V4= 141,36V V VOne way Other way 8Mixed : V2-3 = I 2-3R 2-3 = 11,78 • = 31,52V 3VT = V1 +V2-3 + V3 V2-3 = V2 = V3 = 31,52VV2-3 = VT − V1 - V3 V2 = 31,52V V3 = 31,52VV1− 2 = 220 − 47,12 −141,36 = 31,52V Two ways to calculate V1 and V2 200
- 201. RT= 18,67Ω R1 = 4 Ω R2= 4 Ω R3= 8 Ω R4= 12 ΩIT= 11,78 A I1= 11,78A I2=7,88A I3=3,94A I4= 11,78 AVT= 220V V1= 47,12V V2=31,52 V3=31,52 V4= 141,36V V V 5º I2 = V2 31,52 = = 7,88A ⇒ I2 = 7,88A 6º R2 4 V3 31,52 I3 = = = 3,94 ⇒ I3 = 3,94 A R3 8 back 201
- 202. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 202
- 203. 203
- 204. RT=20 Ω R1= 7 Ω R2= 6 Ω R3= 3 Ω R4= 11 ΩIT= I1= I 2= I3= I4= 1,5AVT= 30V V1= V 2= V 3= V4=R T = R 1 + R 2 −3 + R 4 R T = R 1 + R 2 −3 + R 4 1 1 1 = + R T = 7 + 2 + 11R 2 −3 R 2 R 3 1 1 1 = + R T = 20ΩR 2 −3 6 3 1 3 =R 2 −3 6R2−3 = 2Ω 204
- 205. RT=20 Ω R1= 7 Ω R2= 6 Ω R3= 3 Ω R4= 11 ΩIT= 1,5A I1= 1,5A I 2= I3= I4= 1,5AVT= 30V V1= V 2= V 3= V4= VT 30 IT = = = 1,5 A R T 20 I T = 1,5A Series association : I T = I1 = I 2-3 = I 4 = 1,5A 205
- 206. RT=20 Ω R1= 7 Ω R2= 6 Ω R3= 3 Ω R4= 11 ΩIT= 1,5A I1= 1,5A I2= I3= I4= 1,5AVT= 30V V1= 10,5V V2=3V V3=3V V4= 16,5VV1 = I1R1 = 1,5×7V1 = 10,5VV4 = I4 R4 = 1,5×11V1 = 16,5VIn series ⇒ VT = V1 + V2−3 + V430 = 10,5 + V2−3 +16,5V2−3 = V2 = V3 = 3V 206
- 207. RT=20 Ω R1= 7 Ω R2= 6 Ω R3= 3 Ω R4= 11 ΩIT= 1,5A I1= 1,5A I2= 0,5A I3= 1A I4=1,5AVT= 30V V1= 10,5V V2=3V V3=3V V4= 16,5V V2 3 I2 = = R2 6 I 2 = 0,5 A V3 3 I3 = = R3 3 I 3 = 1A 207
- 208. RT=20 Ω R1= 7 Ω R2= 6 Ω R3= 3 Ω R5= 11 ΩIT= 1,5A I1= 1,5A I2= 0,5A I3= 1A I5=1,5AVT= 30V V1= 10,5V V2=3V V3=3V V5= 16,5VPT= 45V P1=15,75W P2=4,5W P3=3W P5= 24,75W P1 = V1I1 = 15,75W P2 = V2 I 2 = 4,5W P3 = V3 I 3 = 3W P4 = V4 I 4 = 24,75W 208
- 209. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 209
- 210. 210
- 211. RT=19 Ω R1= 8 Ω R2= 6 Ω R3= 10 Ω R4= 5 Ω R4= 5 ΩIT = I1 = I2= I3 = I4= I4=VT= 19V V1= V2= V3= V4= V4=PT= P1= P2= P3= P4= P4=R T = R 1 + R 2 + R 3− 4 -5 1 1 1 = +R 3− 4 -5 R 3 R 4 -5R 4 -5 = R 4 + R 5R 4−5 = 5 + 5 = 10ΩR 4−5 = 10Ω 1 1 1 = +R 3− 4 -5 R 3 R 4 −5 R T = R 1 + R 2 + R 3− 4 -5 1 1 1 1 RT = 8 + 6 + 5 = + =R 3− 4 -5 10 10 5 R T = 19Ω 211R 3− 4-5 = 5Ω
- 212. RT=19 Ω R1= 8 Ω R2= 6 Ω R3= 10 Ω R4= 5 Ω R5= 5 ΩIT=1A I1=1A I2=1A I3 = I4= I5=VT= 19V V1= 8V V2=6V V3=5V V4=2,5V V5=2,5VPT= P1= P2= P3= P4= P5= V3-4-5 = V3 = V4−5 = V4 + V5 VT 19 V1 = I1R 1 = 1 ⋅ 8IT = = R T 19 V1 = 8VI T = 1A V2 = I 2 R 2 = 1⋅ 6I T = I1 = I 2 = I 3-4-5 = 1A V1 = 6VParallel I T = I 3-4-5 = I 3 + I 4-5 V3-4-5 = I 3-4-5 R 3-4-5 = 1 ⋅ 5 212Series I 4-5 = I 4 = I 5 V3-4-5 = 5V
- 213. RT=19 Ω R1= 8 Ω R2= 6 Ω R3= 10 Ω R4= 5 Ω R5= 5 ΩIT=1A I1=1A I2=1A I3 = I4= I5=VT= 19V V1= 8V V2=6V V3=5V V4= V5=PT= P1= P2= P3= P4= P5=Voltaje in each R V3-4-5 = V3 = V4−5 = V4 + V5V1 = I1R 1 = 1 ⋅ 8 V3-4-5 = V3 = 5VV1 = 8VV2 = I 2 R 2 = 1⋅ 6V1 = 6VV3-4-5 = I 3-4-5 R 3-4-5 = 1 ⋅ 5V3-4-5 = 5V 213
- 214. RT=19 Ω R1= 8 Ω R2= 6 Ω R3= 10 Ω R4= 5 Ω R5= 5 ΩIT=1A I1=1A I2=1A I3=0,5A I4=0,5A I5=0,5AVT= 19V V1= 8V V2=6V V3=5V V4=2,5V V5=2,5VPT= P1= P2= P3= P4= P5=Let s calculate the I in the V4 = I 4 R 4 = 0,5 ⋅ 5 R in parallel V4 = 2,5V V 5I3 = 3 = V5 = I 5 R 5 = 0,5 ⋅ 5 R 3 10 V5 = 2,5VI 3 = 0,5A V4-5 5I 4 −5 = = R 4-5 10I 4−5 = 0,5A = I 4 = I 5 214
- 215. RT=19 Ω R1= 8 Ω R2= 6 Ω R3= 10 Ω R4= 5 Ω R5= 5 ΩIT=1A I1=1A I2=1A I3=0,5A I4=0,5A I5=0,5AVT= 19V V1= 8V V2=6V V3=5V V4=2,5V V5=2,5VPT=19W P1=8W P2=6W P3=2,5W P4=1,25W P5=1,25W P1 = V1I1 = 8W P2 = V2 I 2 = 6W P3 = V3 I 3 = 2,5W P4 = V4 I 4 = 1,25W P4 = V4 I 4 = 1,25W 215
- 216. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 216
- 217. 217
- 218. RT=14 Ω R1= 1 Ω R2= 6 Ω R3= 3 Ω R4= 3 Ω R5= 16 Ω R6= 16 ΩIT= I1= I2= I3= I4= I5= I6=VT=28V V1= V2= V3 = V4 = V5= V6=PT = P1= P2= P3 = P4 = P5= P6= R T = R 1 + R 2 −3 + R 4 + R 5−6 1 1 1 = + R 2 −3 R 2 R 3 R T = R 1 + R 2−3 + R 4 + R 5−6 1 1 1 1 = + = RT = 1+ 2 + 3 + 8 R 2 −3 6 3 2 R 2−3 = 2Ω R T = 14Ω 1 1 1 = + R 4 -5 R 4 R 5 1 1 1 1 = + = R 4-5 16 16 8 218 R 4-5 = 8Ω
- 219. RT=14 Ω R1= 1 Ω R2= 6 Ω R3= 3 Ω R4= 3 Ω R5= 16 Ω R6= 16 ΩIT=2A I1=2A I2= I3= I4=2A I5= I6=VT=28V V1=2V V2=4V V3= 4V V4=6V V5= 16V V6= 16V VT 28IT = = R 2 14 V1 = I1R 1 = 1 ⋅ 2I T = 2A V1 = 2VI T = I1 = I 2-3 = I 4 = I 5-6 V2-3 = V2 = V3 V2-3 = I 2-3 R 2-3 = 2 ⋅ 2I T = I1 = 2 A V2 = 4V V2-3 = 4VIT = I4 = 2 A V3 = 4V V4 = I 4 R 4 = 2 ⋅ 3 V5-6 = V5 = V6 V4 = 6V V5 = 16V V5-6 = I 5-6 R 5-6 = 2 ⋅ 8 V6 = 16V 219 V5-6 = 16V
- 220. RT=14 Ω R1= 1 Ω R2= 6 Ω R3= 3 Ω R4= 3 Ω R5= 16 Ω R6= 16 ΩIT=2A I1=2A I2=2/3A I3=4/3A I4=2A I5=1A I6=1AVT=28V V1=2V V2=4V V3=4V V4=6V V5=16V V6=16VPT = P1= P2= P3 = P4 = P5= P6= V2 4I2 = = R2 6 V5 16 I5 = = 2 R 5 16I2 = A I 5 = 1A 3 V3 4 V6 16I3 = = I6 = = R3 3 R 6 16 4 I 2 = 1AI2 = A 3 220
- 221. RT=14 Ω R1= 1 Ω R2= 6 Ω R3= 3 Ω R4= 3 Ω R5= 16 Ω R6= 16 ΩIT=2A I1=2A I2=2/3A I3=4/3A I4=2A I5=1A I6=1AVT=28V V1=2V V2=4V V3=4V V4=6V V5=16V V6=16VPT=56W P1=4W P2=8/3W P3=16/3W P4=12W P5=16W P6=16W PT = VT IT = 56W P1 = V1I1 = 4W P2 = V2 I 2 = 8 / 3W P3 = V3 I 3 = 16 / 3W P4 = V4 I 4 = 12W P5 = V5 I 5 = 16W P6 = V6 I 6 = 16W 221
- 222. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 222
- 223. 223
- 224. RT=18,04 Ω R1= 12 Ω R2= 3,67Ω R3= 7 Ω R4= 9 Ω R5= 6 ΩIT=4,93A I1=4,93A I2=4,93A I3= I4= I5=VT=89V V1= V2= V3= V4= V5=PT = P1= P2= P3= P4= P5= R T = R 1 + R 2 + R 3-4-5 1 1 1 1 VT 89 = + + IT = = R 3-4-5 R3 R4 R5 R T 18,04 1 1 1 1 53 I T = 4,93A = + + = R 3-4-5 7 9 6 126 In the series elements : 126 I T = I1 = I 2 = I 3-4-5 R 3− 4 − 5 = Ω 53 I1 = 4,93 A 126 I 2 = 4,93 A R T = 12 + 3,67 + 53 R T = 18,04Ω 224
- 225. RT=18,04 Ω R1= 12 Ω R2= 3,67Ω R3= 7 Ω R4= 9 Ω R5= 6 ΩIT=4,93A I1=4,93A I2=4,93A I3= I4= I5=VT=89V V1=59,16V V2=18,09V V3= V4= V5=PT = P1= P2= P3= P4= P5=V1 = I1R 1 = 4,93 ⋅12 = 59,16 AV2 = I 2 R 2 = 4,93 ⋅ 3,67 = 18,09 AV3-4-5 = V3 = V4 = V5 V3-4-5 = I3-4-5 R 3-4-5 126 V3-4-5 = 4,93 ⋅ 53 V3-4-5 = 11,72V 225
- 226. RT=18,04 Ω R1= 12 Ω R2= 3,67Ω R3= 7 Ω R4= 9 Ω R5= 6 ΩIT=4,93A I1=4,93A I2=4,93A I3= I4= I5=VT=89V V1=59,16V V2=18,09V V3=11,72V V4=11,72V V5=11,72VPT = P1= P2= P3= P4= P5=V3-4-5 = I3-4-5 R 3-4-5 126V3-4-5 = 4,93 ⋅ 53V3-4-5 = 11,72VThe parallel elements have the same VV3-4-5 = V3 = V4 = V5 = 11,72V 226
- 227. RT=18,04 Ω R1= 12 Ω R2= 3,67Ω R3= 7 Ω R4= 9 Ω R5= 6 ΩIT=4,93A I1=4,93A I2=4,93A I3=1,67A I4=1,30A I5=1,95AVT=89V V1=59,16V V2=18,09V V3=11,72V V4=11,72V V5=11,72VPT = P1= P2= P3= P4= P5= V3 11,72 I3 = = R3 7 I 3 = 1,67 A V4 11,72 I4 = = R4 9 I 4 = 1,30 A V5 11,72 I5 = = R5 6 I 3 = 1,95 A 227
- 228. RT=18,04 Ω R1= 12 Ω R2= 3,67Ω R3= 7 Ω R4= 9 Ω R5= 6 ΩIT=4,93A I1=4,93A I2=4,93A I3=1,67A I4=1,30A I5=1,95AVT=89V V1=59,16V V2=18,09V V3=11,72V V4=11,72V V5=11,72VPT=438,77W P1=291,66W P2=89,18W P3=19,57W P4=15,23W P5=22,85W PT = VT IT = 438,77W P1 = V1I1 = 291,66W P2 = V2 I 2 = 89,18W P3 = V3 I 3 = 19,57W P4 = V4 I 4 = 15,23W P5 = V5 I 5 = 22,85W 228
- 229. 229
- 230. RT=18,04 Ω R1= 12 Ω R2= 3,67Ω R3= 7 Ω R4= 9 Ω R5= 6 ΩIT=4,93A I1=4,93A I2=4,93A I3=1,67A I4=1,30A I5=1,95AVT=89V V1=59,16V V2=18,09V V3=11,72V V4=11,72V V5=11,72VPT = P1= P2= P3= P4= P5= R T = R1 + R 2 + R 3 + R 4 1 1 1 1 = + + R 3-4-5 R3 R4 R5 1 1 1 1 53 = + + = R 3-4-5 7 9 6 126 126 R 2 −3 = Ω 53 126 R T = 12 + 3,67 + 53 R T = 18,04Ω 230

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