3 ESO The electricity Unit

2,419 views
2,364 views

Published on

Published in: Education, Business, Technology
0 Comments
3 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
2,419
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
Downloads
173
Comments
0
Likes
3
Embeds 0
No embeds

No notes for slide

3 ESO The electricity Unit

  1. 1. Electricity Unit 3 1
  2. 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. 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. 4. 3.1 IntroductionWhat would happen if we didn’t have electricity? ? 4
  5. 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. 6. 3.1 IntroductionWhat are we going to learn today?The electric current and its physic principles 6
  7. 7. 3.1 Introduction So, what is electricity?The concept of electricity includes all the phenomena related to electric charges 7
  8. 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. 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. 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. 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. 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. 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. 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. 15. 3.2.2 Electricity generationWhat are we going to learn next?Explain how the electricity can be created. 15
  16. 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. 17. 3.2.2 Electricity generationLong time ago… Hans Christian Oersted discovered that a electric current can disturb a compass 17
  18. 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. 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. 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. 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. 22. 3.2.2 Electricity generation This is the diagram of an industrial electric generator Artificial magnetClosed circuit 22
  23. 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. 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. 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. 26. 3.2.2 Electricity generationWhat are we going to learn next?Difference between the direct and the alternate current 26
  27. 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. 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. 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. 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. 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. 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. 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. 34. 3.3.1 CIRCUITSWhat are we going to learn next?Analysis of the direct current circuits. 34
  35. 35. 3.3.1 Exercise 3.3.1:Electricity magnitudesExercise 3.3.1:Define Voltage, Intensity and Resistance and its units Solution 35
  36. 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. 37. 3.3.1 Electricity magnitudesIn order to understand electricity, we have to first know the main electric magnitudes: VOLTAGE INTENSITYResistance 37
  38. 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. 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. 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. 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. 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. 43. 3.3.1 Electricity magnitudesExercise 3Which one of these lamps will give light? 43
  44. 44. 3.3.1 Electricity magnitudesExercise 4Will this lamp turn on? 44
  45. 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. 46. 3.3.2 Ohm’s law. Electric PowerWhat are we going to learn next?Magnitudes analysis using the Ohm’s Law. 46
  47. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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=100WCalculate the Kwh that the electricity meter will show after 20 hours.P=100W=0,1kWPKWh=0,1x20h=2Kwh 59
  60. 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. 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. 62. 3.3.3 Elements and SymbolsWhat are we going to learn next?Electric components and their applications. 62
  63. 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. 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. 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. 66. 3.3.3 Elements and symbolsControl elements: Change Direction diverter switch: We use this element to invert the direction of the electric current through a component. 66
  67. 67. 3.3.3 Elements and symbolsControl 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. 68. 3.3.3 Elements and symbolsSafityIt is used to control high electrical volt devices, because we don’t touch the main circuit 68
  69. 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 engineResistance: we use it to decrease the intensity of a circuit 69
  70. 70. 3.3.3 Elements and symbolsConductor: 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. 71. 3.3.3 Elements and symbolsProtection 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. 72. 3.3.3 Elements and symbolsElectric symbols are used to represent electric circuits with drawings that replace the real circuit elements. 72
  73. 73. 3.3.3 Elements and symbolsExercise: Draw the following circuit using electric symbols 73
  74. 74. 3.3.3 Elements and symbolsExercise: Draw the following circuit using electric symbols 74
  75. 75. 3.3.3 Elements and symbolsExercise: Draw the following circuit using electric symbols 75
  76. 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. 77. 3.3.3 Elements and symbols Control elements Push button Switch Diverter switch 77
  78. 78. 3.3.3 Elements and symbols Protection elements Fuse Receptors: Lamp Resistances: they have two symbols Engines 78
  79. 79. 3.3.4 Circuit associationsThe behaviour of electrical elements depends on how they connect to each other.There are three possible configurations :SeriesParallelMixed 79
  80. 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. 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. 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. 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. 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. 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. 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. 87. 3.3.4 Circuit associationsIf 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. 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. 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. 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. 91. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt 91 Solution
  92. 92. 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt 92
  93. 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. 94. 3.3.4 Circuit associations. Calculations VT I1 ET RT V1 IT I2 I3 V2 V3 94
  95. 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. 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. 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. 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. 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. 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. 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. 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. 103. 3.3.4 Circuit associations VT=V123=V1= V2=V3 IT=I123=I1+ I2+I3 103
  104. 104. 3.3.4 Circuit associations PARALLEL association INTENSITY= DIVED IT=I1+I2+I3 I1 I2 IT I3 104
  105. 105. 3.3.4 Circuit associations PARALLEL association INTENSITY= DIVED IT=I1+I2+I3 I1 I2 IT I3 105
  106. 106. 3.3.4 Circuit associations PARALLEL association VOLTAGE: IQUAL VT=V1= V2=V3 V1 V2 V3 VT 106
  107. 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. 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. 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. 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. 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. 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. 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. 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. 115. 3.3.4 Circuit associations MIXED associationIT=I123=I4 series associationI123=I1+ I2+I3 parallel association 115
  116. 116. 3.3.4 Circuit associations MIXED associationIT=I123=I4 series associationI123=I1+ I2+I3 parallel association I1 I2 I1 116
  117. 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. 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. 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. 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. 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. 122. 3.3.4 Circuit associations. Calculations 1º RT= 6 Ω IT= VT= 20V 122
  123. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 137. 3.3.4 Circuit associationsExercise:Which of these lamps have more lightness? 137
  138. 138. 3.3.4 Circuit associationsExercise:Which of these assemblies generates electricity? 138
  139. 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. 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. 141. 3.3.4 Circuit associations Why are these assemblies no correct? 141
  142. 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. 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. 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. 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. 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. 147. 3.3.6 Electric Measurements. PolimeterAmperimeter: we chose the position accordingto the scale of the current. 4K7 Ω 147
  148. 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. 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. 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. 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. 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. 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. 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. 155. 3. Draw the diagram of a mobile phone power supply. DC AC Back 155
  156. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 174. 3.3.4 Circuit associations. Calculations 174
  175. 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. 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. 177. 3.3.4 Circuit associations. Calculations 177
  178. 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. 179. 3.3.4 Circuit associations. Calculations 179
  180. 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. 181. 3.3.4 Circuit associations. CalculationsRT1 1 1 = +RT R1−5 R61 1 1 = +RT 160 401 1 =RT 32RT = 32Ω 181
  182. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 195. 3.3.4 Circuit associations. Calculations 4Ω 4Ω 12 Ω 220V 8Ω 4Ω 8/3Ω 12 Ω220V 56/3 Ω220V 195
  196. 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. 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. 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. 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. 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. 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. 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. 203
  204. 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. 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. 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. 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. 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. 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. 210
  211. 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. 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. 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. 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. 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. 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. 217
  218. 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. 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. 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. 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. 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. 223
  224. 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. 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. 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. 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. 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. 229
  230. 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

×