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  • 1. 8.1 Electricity 3.1 Introduction 3.1 Introduction 3.2 Electricity fundamentals What would happen if we didn’t have 2.1 Particles and laws electricity? 2.2 Electricity generation Electricity 2.3 Electricity magnitudes 3.3 Circuits Unit 3 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 ?3.1 Introduction 3.1 Introduction 3.1 Introduction However, we have to remember that we can What are we going to learn today? So, what is electricity? decrease the amount of energy that we waste everyday, helping our sustainable The concept of electricity includes all the development. The electric current and its physic phenomena related to electric charges principles3.2.1 Particles and laws 3.2.1 Particles and laws 3.2.1 Particles and laws Electrons and protons have negative andMatter can be electrically charged when positive charges respectively . the charge distribution is upset Matter is formed by atoms, which contain smaller particles inside with electric These charges create forces between charges: them that can be attraction or repulsion Proton forces according to the value of the The electrons and protons charge: Electron Equal charges: repulsionFor example, you can change the charge Different charges: attraction distribution of a pen by rubbing it against your hair. Then you can attract some pieces of paper Atom attraction repulsion repulsion 1
  • 2. 3.2.1 Particles and laws 3.2.1 Particles and laws 3.2.1 Particles and laws The use of electricity needs a flow of In solid metals such as All the flowing electrons, the electric current wires, the positive charge charges are carriers are immobile, and assumed to have only the negatively positive polarity, charged electrons can c flow. and this flow is called conventional The electric current is the displacement of current. the electrical charges through the matter 3.2.1 Particles and laws 3.2.2 Electricity generation 3.2.2 Electricity generationBecause the electron What are we going to learn next? Exercise 3.2.2a : carries negatives charges, the electron Explain how the electricity can be Explain how the electricity can be created. motion in a metal is in created. Describe all elements needed in a the direction opposite to common electric generator. that of conventional (or electric) current. Solution 3.2.2 Electricity generation 3.2.2 Electricity generation 3.2.2 Electricity generation Long time ago… Hans Christian Oersted discovered that a electric current can The magnet is stronger if we use a spiral disturb a compass instead of a simple circuit. Since the compass only is disturbed with a magnet, Volta concluded that an electric current creates a artificial magnetic Artificial magnet Therefore electric magnets are formed by Natural several spirals connected to a battery magnet 2
  • 3. 3.2.2 Electricity generation 3.2.2 Electricity generation 3.2.2 Electricity generation Using Volta experiment, Mr In conclusion, if we want to create an This is the diagram of an industrial electric Michael Faraday discovered artificial electric current, we need: generator that an electric current can be Mechanical energy created when a magnet is Artificial magnet to move the circuit or the moving close to an electric Closed circuit circuit or vice versa magnet Artificial magnet Closed circuit3.2.2 Electricity generation 3.2.2 Electricity generation 3.2.2 Exercise 3.2.2b Electricity generation When we have all elements together, weThe electricity is created when the find that we create an alternate electric Exercise 3.2.2b : mechanical energy moves the closed current due to the movement of the circuit 1. Explain the difference between the AC and the DC. circuit inside the artificial magnet inside the magnetic field. 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. Electric engine video Electric generator and solution Engine video3.2.2 Electricity generation 3.2.2 Electricity generation 3.2.2 Electricity generation The difference between AC and DC is :What are we going to learn next? AC is an alternating current (the amount of But, most electric at devices like an iPhone, electrons) that flows in both directions and DC is need a Direct Current that keeps itsDifference between the direct and direct current that flows in only one direction magnitude and direction constant the alternate current AC changes its Voltage value from +A to –A (+ and – represent the direction) and the DC has a constant Voltage value 3
  • 4. 3.2.2 Electricity generation 3.2.2 Electricity generation 3.2.2 Electricity generation These are the elements of a Power supply If we apply an AC to an electrical device like In order to obtain a direct current we use that that transforms AC into DC: a light bulb, we obtain a light pulse as you powers supplies that include different transformer, rectifier, smoothing and see in a bike dynamo: elements that transforms AC into DC. regulator. At home, the pulse of the AC supply is so fast that the light DC pulse cannot be AC detected by the human eyes 3.2.2 Electricity generation 3.2.2 Electricity generationThese are the elements of a Power supply These are the elements of a Power supply that that transforms AC into DC: that that transforms AC into DC: 3.3.1 CIRCUITS transformer, rectifier, smoothing and transformer, rectifier, smoothing and regulator. regulator. What are we going to learn next? Analysis of the direct current circuits.3.3.1 Exercise 3.3.1: 3.3.1 Electricity magnitudes 3.3.1 Electricity magnitudesElectricity magnitudesExercise 3.3.1: To better understand the concept of the In order to understand electricity, we have to electric current, we can think of it as a first know the main electric magnitudes:Define Voltage, Intensity and stream where the drops are the electric Resistance and its units charges VOLTAGE We use the water INTENSITY power from the Resistance drops’ movement to create energy Solution 4
  • 5. 3.3.1 Electricity magnitudes 3.3.1 Electricity magnitudes 3.3.1 Electricity magnitudes The electrons need energy to be able to We can see that the stream will have The higher the Voltage is, the more move through a material. This is the more strength if there is more water in energy the electric charges will have to Voltage the tank. It’s similar to what happened keep on moving. Energy to the electricity More Higher water voltage pressure We define the Voltage as the energy per charge unit that makes them flow Less Lower through a material. This magnitude is water voltage measured in Volts (V). pressure 3.3.1 Electricity magnitudes 3.3.1 Electricity magnitudes 3.3.1 Electricity magnitudes Exercise 3Intensity is the amount of charges that Electric resistance is the opposition to Which one of these lamps will give light? goes through a conductor per time unit. the movement of the charges through a It is measured in Ampere (A) conductor. It is measured in Ohms (Ω) Higher intensity Higher Resistance Lower Lower Resistance intensity 3.3.1 Electricity magnitudes 3.3.1 Electricity magnitudes Exercise 4 The electric current always goes through 3.3.2 Ohm’s law. Electric Power Will this lamp turn on? the route with less resistance, like water What are we going to learn next? Here we have the Magnitudes analysis using the resistance needed to Ohm’s Law. create light No resistance 5
  • 6. 3.3.2 Exercise 3.3.2a Ohm’s law. Electric Power 3.3.2 Ohm’s law. Electric Power 3.3.2 Ohm’s law. Electric Power Exercise 3.3.2a: Ohm’s law links the three electric Intensity is directly proportional to Justify how the Intensity will be if: magnitudes as is shown: voltage: We have a low voltage V If the voltage is high, the charges will have a lot of energy. Therefore the We have a low Resistance R V Intensity will be high too V I= V= Voltage (voltsV) I= Solution R I= Intensity (ampereA) R= Resistance (ohm Ω ) R Solution 3.3.2 Exercise 3.3.2.b 3.3.2 Ohm’s law. Electric Power 3.3.2 c Exercise Ohm’s law. Electric Power Ohm’s law. Electric Power We have a fan connected to the domestic electric Intensity is inversely proportional to Exercise 3.3.2.b: V (V) R (Ω) I (A) supply. The amperimeter indicate a 0,52A on the Resistance Calculate the value of circuit. If there is a high Resistance, the intensity will the intensity in 1. What’s is the resistance of the fan? 2 14 be low. these cases: 2.What’s the power measured in Kw 2 2 Calculate the Intensity that flaws in a electrical engine of 1,2 Kw installed in my washing machine. V V 20 2 What would be the Intensity and the power if weThe charges will cross thematerial slowly I= I= 12 6 have 380V at home. R R 252 64 Extra data: house electrical supply 230V, 50Hz 1000 20 Solution 3.3.2 Ohm’s law. Electric Power 3.3.2 Ohm’s law. Electric Power 3.3.2 Ohm’s law. Electric Power From this expression we can derivate in Power is defined as the work W W VQ other two using the Ohm’s Law V= ⇒W =VQ consumed per time unit, and Q P= = =VI it’s measured in watt W t t P =VI I= Q P = I 2R t V = IR W If we apply the Ohm’s law, we obtain a P= new formula that explains the power of V V2 t an electrical device I= P= P=VI R R 6
  • 7. 3.3.2 Ohm’s law. Electric Power 3.3.2 Ohm’s law. Electric Power 3.3.2 Ohm’s law. Electric PowerThere is a bulb lamp of 100w connected to There is a bulb lamp of 100w connected to There is a bulb lamp of 100w connected to 220V electric grid. Calculate: 220V electric grid. Calculate: 220V electric grid. Calculate: Intensity throw the lamp. Intensity throw the lamp. Power of the lamp. Power of the lamp. As is says in the wording it is 100W Calculate the Kwh that the electricity P = 100W P = VI P=100W meter will show after 20 hours. Calculate the Kwh that the electricity P 100 Indicate the price of the electricity V = 220V I= = = 0,45A meter will show after 20 hours. consumed if the price is 0,90€/Kwh V 220 P=100W=0,1kW PKWh=0,1x20h=2Kwh 3.3.2 Ohm’s law. Electric Power 3.3.2 Ohm’s law. Electric PowerIndicate the price of the electricity consumed if the price is 0,90€/Kwh Electrical companies measure our electrical consume in Kwh 3.3.3 Elements and Symbols Price with a electricity meter What are we going to learn next? Cost = Energy consumed × situated in the basement of a Energy € building. Electric components and their Cost = Kwh × Kwh applications. 0, 9€ Cost = 2Kwh × Cost = 1, 8€ Kwh PowerConsu med = P • h 3.3.3 Elements and symbols 3.3.3 Elements and symbols 3.3.3 Elements and symbols The essential elements in a circuit are: An electric circuit is a group of Control elements: we can manipulate the The essential elements in a circuit are: elements that allows us to control electricity through the circuit. Generator: it creates an electric current electric current through some Switch: they keep the ON or OFF positions. supplying voltage to the circuit. They can be: sort of material Batteries: they supply electric current but only for a For example, the switch lights at the bathroom short time. Power supplies: they give a constant and continuous Push button: The On position only works while electric current. 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 7
  • 8. 3.3.3 Elements and symbols 3.3.3 Elements and symbols 3.3.3 Elements and symbolsControl elements: Control elements: Safity Relay: It is an electrically operated switch. Change Direction diverter switch: Thanks to that we can control a circuit from a It is used to control high electrical volt We use this element to invert the remote control applying just electricity devices, because we don’t touch the main direction of the electric current circuit through a component. An electrical magnet attracts the switch that closes the circuit3.3.3 Elements and symbols 3.3.3 Elements and symbols 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 Protection elements: they keep all Conductor: all the elements have to be example: connected to a material that transmits the circuit elements safe from Incandescent lights: when the electric the electric charges. high voltage rises that can destroy current goes through the lamp filament it the receptors. gets really hot and starts emitting light. Fuse: the first one will blow, cutting the circuit in case of a voltage rise. Engine: the electricity creates a magnetic They are easily replaced field that moves the metal elements of The circuit has to be CLOSED in order to the engine Circuit breaker: they are used in new allow the electricity to circulate around it electrical installations, at home or at from the positive to the negative pole. factories. If there is a voltage rise, Resistance: we use it to decrease the you don’t have to replace them, only intensity of a circuit reload them.3.3.3 Elements and symbols 3.3.3 Elements and symbols 3.3.3 Elements and symbols Electric symbols are used to represent Exercise: Draw the following circuit using Exercise: Draw the following circuit using electric circuits with drawings that replace electric symbols electric symbols the real circuit elements. 8
  • 9. 3.3.3 Elements and symbols 3.3.3 Elements and symbols 3.3.3 Elements and symbols Generator Control elements Exercise: Draw the following circuit using Battery electric symbols Push button Battery association Switch Conductors: Diverter switch When two conductors are crossed without any contact we indicate it with a curve3.3.3 Elements and symbols 3.3.4 Circuit associations 3.3.4 Circuit associations Protection elements SERIES circuit The behaviour of electrical elements The series circuit connects the electrical Fuse depends on how they connect to each elements one behind the other. other. Receptors: There are three possible configurations : Lamp Series Resistances: they have two symbols Parallel In this way, there is only one connection Mixed point between elements Engines 1 & 2 are connected only by A 2 & 3 are connected only by B3.3.4 Circuit associations 3.3.4 Circuit associations 3.3.4 Circuit associations PARALLEL association MIXED associationIn this association, all the elements are Mixed association has elements But what happens to receptors when they connected by two points. associated in parallel and in series. are connected in series or parallel associations?So, 1, 2 & 3 are connected by A and B 1, 2 & 3 are in parallel and all of them are Series and parallel association change the value in series with 4 of intensity and voltage through receptors. 9
  • 10. 3.3.4 Circuit associations Intensity Circuit Voltage Series Parallel Series Parallel Series Parallel All the lamps are in line, All lamps are separated so If there is any cut along If there is any cut along so they create a high they create a lowVoltage is distributed Voltage is the same in all the conductor, the the conductor, the resistance, that is the resistance, that is thebetween elements, that is elements, so all the electric current will not electric current can go reason why the intensity reason why the intensitythe reason why they have lamps have the same be able to go from the through any other way is lower but the battery through the lamps isless energy for each energy and the light is positive to the negative will have a longer life. higher, but the battery willlamp, so the light is higher. pole. have a shorter lifelower in each lamp. CUT3.3.4 Circuit associations If there is any cut in series association, 3.3.4 Circuit associations 3.3.4 Circuit associations the water can’t go further. But in Exercises Exercises parallel association there will be no Which of these elements are associated Which of these elements are associated problem. with series, parallel or mixed circuits? with series, parallel or mixed circuits. Name the connection points with Name the connection points with cut letters. letters3.3.4 Circuit associations 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations ExercisesWhich of these elements are associated with Calculate in each case the resistance Calculate in each case the resistance series, parallel or mixed circuits?. Name the equivalent Rt equivalent Rt connection points with letters. 10
  • 11. 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations SERIES circuit SERIES circuit VT In this association we find that: I1 ET INTENSITY= EQUAL RT VT=V1+ V2+V3 RT=R1+R2+R3 V1 IT=I1=I2=I3 IT=I1= I2=I3 ET=E1+ E2+E3 IT I1 ET RT VT I2 I3 ET RT VT ET=VT ET=VT IT IT IT I V2 V3 I2 I3 T 3.3.4 Series Exercise Circuit associations. 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations Calculations Calculate in each case the resistance equivalent Rt, It SERIES circuit Calculate in each case the resistance and Vt. Calculate the I and V in each Resistance VOLTAGE= DIVED equivalent Rt, It and Vt. Calculate the I and V in each Resistance VT=V1+ V2+V3 V1 ET RT VT VT ET=VT IT IT I V2 V3 T 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It Calculate in each case the resistance equivalent Rt, It Calculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance and Vt. Calculate the I and V in each Resistance and Vt. Calculate the I and V in each Resistance R T= 8 R1= 0,5 R2= 6,3 R3= 6/5 RT = 8 R1= 0,5 R2= 6,3 R3= 6/5 RT = 8 R1= 0,5 R2= 6,3 R3= 6/5 IT= 27,5 A I1= I2= I3= IT = 27,5 A I1= 27,5 A I2= 27,5 A I3= 27,5 A IT = 27,5 A I1= 27,5 A I2= 27,5 A I3= 27,5 A VT = 220V V1= V2= V3= VT = 220V V1= V2= V3= VT = 220V V1= 13,75V V2= 173,25V V3= 33V R t = R1 + R2 + R3 VT Series ⇒I t = I1 = I2 = I3 = 27,5A Series ⇒V t= V1 + V2 + V3 = 220V IT = 1 6 RT R t = + 6,3 + V1 = I1R1 = 0,5 • 27,5 = 13,75V 2 5 220 IT = V2 = I2 R2 = 6,3 • 27,5 = 173,25V R t = 8Ω 8 V3 = I3 R3 = 1,2 • 27,5 = 33V I T = 27,5Ω Series ⇒V t= 13,75 +173,25 + 33 = 220V 11
  • 12. 3.3.4 Circuit associations 3.3.4 Circuit associations 3.3.4 Circuit associations PARALLEL association PARALLEL associationVT=V1= V2=V3 1 1 1 1 INTENSITY= DIVED = + + RT R1 R2 R3 IT=I1+I2+I3IT=I1+ I2+I3 ET=E1+ E2+E3 I1 I2 VT=V123=V1= V2=V3 IT=I123=I1+ I2+I3 IT I33.3.4 Circuit associations 3.3.4 Circuit associations 3.3.4 Circuit associations PARALLEL association PARALLEL association PARALLEL association INTENSITY= DIVED VOLTAGE: IQUAL RT= R1= 6 R2= 18 R3= 4 IT = I1= I2= I3= IT=I1+I2+I3 VT=V1= V2=V3 I1 VT= 20V V1= V2= V3= V1 V1 I2 V2 V2 IT V3 V3 I3 VT VT3.3.4 Circuit associations 3.3.4 Circuit associations 3.3.4 Circuit associations PARALLEL EXERCISE PARALLEL EXERCISE PARALLEL EXERCISE RT=2,11 R1= 6 R2= 18 R3= 4 RT=2,11 R1= 6 R2= 18 R3= 4 RT=2,11 R1= 6 R2= 18 R3= 4 IT = I1= I2= I3= IT = I1= I2= I3= IT =9,48A I1= I2= I3= VT= 20V V1= V2= V3= VT= 20V V1= V2= V3= VT= 20V V1= V2= V3= V1 1 1 1 1 V1 1 1 1 1 V1 Parallel ⇒ = + + Parallel ⇒ = + + VT RT R1 R2 R3 RT R1 R2 R3 IT = RT V2 1 1 1 1 6 2 9 V2 1 1 1 1 6 2 9 V2 = + + = + + = + + = + + 20 RT 6 18 4 36 36 36 RT 6 18 4 36 36 36 IT = V3 V3 V3 2,11 1 17 36 1 17 36 IT = 9,48A VT = ⇒ RT = = 2,11Ω VT = ⇒ RT = = 2,11Ω VT RT 36 17 RT 36 17 12
  • 13. 3.3.4 Circuit associations 3.3.4 Circuit associations 3.3.4 Circuit associations VOLTAGE: IQUAL PARALLEL EXERCISE MIXED association VT=V1= V2=V3 RT=2,11 R1= 6 R2= 18 R3= 4 We have to convert the parallel associations VT=20 IT=9,48A I1=3,33A I2=1,11A I3=5A into series in order to obtain Vt, It and RT=2,11 R1= 6 R2= 18 R3= 4 VT= 20V V1=20V V2=20V V3=20V Rt. T IT =9,48A I1= I2= I3= V1 V1 V 20 VT= 20V V1=20V V2=20V V3=20V I1 = 1 = = 3,33A R1 6 V2 V2 V2 20 I2 = = = 1,11A R2 18 V3 V3 V3 20 VT VT I3 = = = 5A R3 43.3.4 Circuit associations 3.3.4 Circuit associations 3.3.4 Circuit associations MIXED association MIXED association VT=V123+V4 series association IT=I123=I4 series association IT=I123=I4 series association V123=V1= V2=V3 parallel associationI123=I1+ I2+I3 parallel association I123=I1+ I2+I3 parallel association I1 V1 VT=V123+V4 I2 V2 I1 V33.3.4 Circuit associations 3.3.4 Circuit associations 3.3.4 Circuit associations. Calculations Calculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance VT=V123+V4 series association IT=I123=I4 series association VT=V1234 IT=I1234 V123=V1= V2=V3 parallel association I123=I1+ I2+I3 parallel association 13
  • 14. RT = 6 R1= 6 R2= 18 R3= 3/2 RT = 6 R1= 6 R2= 18 R3= 3/2 3.3.4 Circuit associations. Calculations IT= 10/3 A I1= I2= I3= IT= 10/3 A I1= I2= I3= 10/3 A VT = 20V V1= V2= V2= VT = 20V V1= V2= V3= R T= 6 I T= Mixed Mixed VT= 20V IT = I1 + I2 = I1−2 = I3 10 / 3A = I1 + I2 = I1− 2 = 10 / 3 RT = R1 − 2 + R 3 R = R 1− 2 + R T 3 VT 20 1 1 1 9 3 IT = = I2 = + R T = + RT 6 R 1− 2 R1 R2 2 2 1 1 1 3 1 12 10 = + = + R = IT = A R 1− 2 6 18 18 18 T 2 3 I3 1 = 2 RT = 6Ω I1 I1-2 I3 R1 − 2 9 9 R1 − 2 = Ω 2RT = 6 R1= 6 R2= 18 R3= 3/2 RT = 6 R1= 6 R2= 18 R3= 3/2 RT = 6 R1= 6 R2= 18 R3= 3/2IT= 10/3 A I1= I2= I3= 10/3 A IT = 10/3 A I1= I2= I3= 10/3 A IT = 10/3 A I1= 2,5 A I2= 5/6 A I3= 10/3 AVT = 20V V1= V2= V3= 5V VT = 20V V1= 15V V2= 15V V3= 5V VT = 20V V1= 15v V2= 15V V2= 5 V Two ways to calculate V1 and V2 Other way Mixed 10 9 One way V1−2 = I1−2R1−2 = • = 15V VT = V1−2 + V3 3 2 Mixed : V1−2 = V1 = V2 = 15V V1− 2 = V1 = V2 VT = V1−2 + V3 V1 = 15V V2 = 15V V1 15 5 10 10 3 V1− 2 = VT1 − V3 I1 = = = ⇒ I1 = 2,5 A I3 = IT = A ⇒ V3 = I 3 R3 = • = 5V R1 6 2 3 3 2 V1− 2 = 20 − 5 = 15V V2 15 5 5 I2 = = = ⇒ I1 = A R2 18 6 6 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations E represents the electrometric force that is Another example the Total Voltage that provides a electric Exercise 3.3.4. Try to make a cheat sheet source like a battery or a power supply, with all the formulas needed to solve electric circuits. It has to be no bigger than 25cm2. 14
  • 15. 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. CalculationsCalculate in each case the resistance Calculate in each case the resistance equivalent Calculate in each case the resistance equivalent equivalent Rt, It and Vt. Calculate the I Rt, It and Vt. Calculate the I and V in each Rt, It and Vt. Calculate the I and V in each and V in each Resistance Resistance Resistance 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations Calculate in each case the resistance Exercise: Calculate in each case the resistance equivalent equivalent Rt, It and Vt. Calculate the I Rt, It and Vt. Calculate the I and V in each Which of these lamps have more and V in each Resistance Resistance lightness? 3.3.4 Circuit associations 3.3.4 Circuit associations 3.3.4 Circuit associations Exercise: Exercise: Which of these assemblies generates What happen in this Draw the electric circuit of a car that has : exercise if we electricity? activate: 1.2 front lamps 1. 1 and 7 2.2 Back lamps 2. 1 and 8 3. 1,2 and 3 3.A forward and backward engine that stops 4. 1,2,3,4,5,6 and 7 automatically when it crash with a object 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 15
  • 16. 3.3.6 Electric Measurements. Polimeter 3.3.6 Electric Measurement. Polimeter3.3.4 Circuit associations Ohmmeter: It measures the resistance and the continuity of a circuit or component. • In order to get a precise lecture we have to If there is no resistance that means that there is a shortcircuit Why are these assemblies no correct? If there is a maximum resistance that means that the cricuit is open choose the correct range of measurement. Ω 4K7Ω Ω Resistance control 3.3.6 Electric Measurements. Polimeter 3.3.6 Electric Measurements. Polimeter 3.3.6 Electric Measurements. Polimeter Voltimeter If we want to measure the voltage of battery Amperimeter: In order to be able to measure the we have to place the control in the Direct current we have to place the electrodes BETWEEN Aplication: If we want to measure the voltage present in a component we have to conect the electrodes in parallel Current and in the proper scale the circuitr or in sereies V A Voltage in a component CONTROL DE CONSUMO 3.3.6 Electric Measurements. Polimeter 3.3.4 Circuit associations 3.3.4 Circuit associations Point out in the table if the engine andAmperimeter: we chose the position lamps works for the following situationsaccording to the scale of the current. Exercise: 4K7 According to this assembly, Ω chose the correct answer: The I through the lamp is 2,3 A A closed A open A closed The voltage between the lamp is B open B closed B closed 2,3 V Engine The resistance of the lamp is 2,3 lamp 1 lamp 2 16
  • 17. 3.2.2 Sol Exercise 3.2.2b Electricity 3.2.2 Electricity generation 3.2.2 Electricity generation Exercise 3.2.2a :Explain how the electricity can be created. Describe all generation Exercise 3.2.2b : 2.-Indicate what type of current do we use in this electrical devices: Torch, Bedroom elements needed in a common electric generator. lamp, TV, Mobil phone, Fridge. Thanks to the discoveries made by Faraday and Oesterd, we know that when a 1. Explain the difference between the AC and the DC. closed circuit moves inside a magnetic field an electric current is created The difference between AC and DC is that AC is an inside the circuit. Therefore, nowadays we use an artificial magnet (made with a electric current alternating current (the amount of electrons) that flows in that flows through a spiral that has a iron bar inside), closed circuit (spirals) both directions and DC is direct current that flows in only AC power is and mechanical energy to create electricity. The closed circuit is moved one direction; Also, AC changes its Voltage value from what fuels 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 +A to –A (+ and – represent the direction) and the DC our homes. circuit start to move creating the electric current. has a constant Voltage value The wires outside of our house are connected at two ends Back to AC 3. Draw the diagram of a mobile phone power supply. 3.2.2 Electricity generation 3.2.2 Electricity generationExercise 3.2.2b :2.-Indicate what type of current do we use in this electrical Exercise 3.2.2b : devices: Torch, Bedroom lamp, TV, Mobil phone, Fridge. DC 2.-Indicate what type of current do we use in this electrical devices: Torch, Bedroom lamp, TV, Mobil ACDC is found in batteries andsolar cells and it used for most phone, Fridge.of our electric devices. We AC DCneed to convert the AC into TV TORCHDC in most of electric devices Fridge Phoneusing power suplies Bedroom lamp Back 3.3.1 Exercise 3.3.1:Solution BAck 3.2.2 Electricity generation 3.3.2 a Exercise Ohm’s law. Electric Power Exercise 3.2.2b : Electricity magnitudes Exercise 3.3.1: 4.Find the input and output current in your Intensity is directly proportional to Define Voltage, Intensity and Resistance and its mobile phone power supply. units voltage: Simbol Name Definition Unit If the voltage is Low, the charges willInput; 100-240 V Voltage Is the energy per have less energy to move. ThereforeOutput: 5,3 V V Voltage charge unit that makes them Volts (V) the Intensity will be low too flow through a material Intensity is the amount of V I Intensity charges that goes through a conductor per time unit Ampere (A) I= R Resistance Electric resistance is the opposition to the movement of the charges through a ohm ( ) R conductor. 17
  • 18. exercise 3.32 b Solution Ohm’s law 6.4 Ohm’s law 3.3.2 a Exercise Ohm’s law. Electric Power Calculations with Ohm’s law 5º Exercise Solution Calculations with Ohm’s law Solution Intensity is inversely proportional to V (V) R (Ω) I (A) V (V) R (Ω) I (A) Resistance If there is Low Resistance, the intensity will V 2 V = IR 2 2 I= I= I = 1A be High because there is less opostion to R 2 V = 4⋅2 2 4 the movement of electrons. V 2 I= I= I = 0,5A V = 8V R 4 V 2 4 R = V The charges will cross the material slowly I= I 10 R 10 R = 5 5 R = 2Ω exercise 6.4 Ohm’s law 3.3.2 c Solution Ohm’s law. Electric Power 3.3.2 c Solution Ohm’s law. Electric Power Calculations with Ohm’s law Solution We have a fan connected to the domestic electric We have a fan connected to the domestic electric supply. The V (V) R (Ω) I (A) amperimeter indicate a 0,52A on the circuit. Extra data: house V supply. The amperimeter indicate a 0,52A on the electrical supply 230V, 50Hz I= circuit. Extra data: house electrical supply 230V, 1º Text analysis: R 50Hz I= 0,52A ; V=230V; 5 10 5 I= ? ; P=? I= 1º Text analysis: 2º- What’s the power measured in Kw 10 I= 0,52A ; V=230V; P = VI I = 0,5 A I= ? ; P=? P = 230 • 0,52 = 0,12kw V = IR 1. What’s is the resistance of the fan? P = 0,12kw V = 1000 ⋅ 20 V V 230 20 1000 or you could aslo do V = 20000V I= ⇒R= = = 442,3Ω R I 0,52 V V 2 230 2 V = 20kv P = V⋅ = = = 0,12kw R = 442,3Ω R R 442,3 3.3.2 c Solution Ohm’s law. Electric Power 3.3.2 c Solution Ohm’s law. Electric Power 3.3.4 Circuit associationsCalculate the Intensity that flaws in a electrical engine of 1,2 Kw installed Calculate the Intensity that flaws in a electrical engine of 1,2 Kw installed in my washing machine. in my washing machine. What would be the Intensity and the power if we SolutionsExtra data: house electrical supply 230V, 50Hz have 380V at home.Text Analysis: Text Analysis: Parallel: 1, 2 and 3 Series: 1 and 2 areR= ?; V=230V; I= ?; P=1,2kW R= ?; V=230V; I= ?; P=1,2kW are joined together joined by A; 2 and 3 by A and B are joined by BP = VI P 1200P = VI ⇒ I = = = 5,22A V 230I = 5,22A Back 18
  • 19. 3.3.4 Circuit associations 3.3.4 Circuit associations 3.3.4 Circuit associations. Calculations Solutions Solutions: red-series Blue: parallel Mixed: SERIES: MIXED: 1 1 1 1 and 2 are joined by A. Series:1 is joined to 2 and 3 by A = + Series: 1 and 2 are joined to A RT R1 R2 2 & 3 are joined by B Parallel: 2 and 3 are joined by A and Series: 2 and the 4 & 3 association are joined by B 1 1 1 7 1 3 & 4 are joined by C B R t = R1 + R2 + R3 = + = + Parallel: 4 and 3 are joined by B and C Series: 2 and 3 are joined to 4 by C RT 4 28 28 28 1 6 1 2 Rt = + 6,3 + = 2 5 RT 7 Rt = 8 RT = 7 2 RT = 3,5Ω 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. CalculationsRT = R1− 2 + R3 RT = R1− 2 + R3 1 1 1 9 3 1 1 1 1 1 1 = + RT = + = + = +R1− 2 R1 R2 2 2 Rt R1−2 R3 Rt R1−2 R3 1 1 1 1 1 1 3 1 12 R1− 2 = R1 + R2 = + = + = + RT = Rt 25 40R1− 2 6 18 18 18 2 1 13 1 2 R1− 2 = 13 + 12 = = RT = 6Ω Rt 200R1− 2 9 200 9 R1− 2 = 25Ω RT = ΩR1− 2 = Ω 13 2 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations R1− 2 R1 _ 5 R1− 2 = R1 + R 2 RT R1− 2 = 1000 + 500 R1 _ 5 = R1− 2 + R3 − 4 − 5 1 1 1 R1− 2 = 1500 Ω R1 _ 5 = 1500 + 1500 = + R3 − 4 − 5 RT R1 _ 5 R 6 _ 8 1 1 1 1 R6 − 7 R1 _ 5 = 3000 Ω = + + 1 1 1 R 3 − 4 −5 R3 R 4 R5 1 = 1 + 1 R6 _ 8 = + R 6 −7 R6 R7 RT 3000 3000 1 1 1 1 R 6 _ 8 = R6 − 7 + R8 = + + 1 1 1 1 1 R 3 − 4 −5 6000 3000 6000 = + = R 6−7 2000 2000 R 6 _ 8 = 1000 + 2000 RT 1500 1 1 = 1 1 R 3 − 4 −5 1500 = R 6 _ 8 = 3000 Ω RT = 1500 Ω R 6−7 1000 R 3−4−5 = 1500 Ω R 6−7 = 1000 Ω 19
  • 20. 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations R3 − 4 1 1 1 R2 _ 5 = + R3 − 4 R3 R 4 R 2 − 3− 4 1 1 1 = + R1 _ 5 1 1 1 R 2 − 3− 4 = R 2 + R3 − 4 R 2 _ 5 R 2 − 3 − 4 R5 = + R1 _ 5 = R 2 _ 5 + R5 R3− 4 25 100 R 2 − 3− 4 = 20 + 100 1 1 1 = + 1 1 R 2 − 3− 4 = 120 Ω R2 _ 5 120 120 R1 _ 5 = 60 + 100 = R3− 4 20 1 1 R1 _ 5 = 160 Ω = R3 − 4 = 20 Ω R2 _ 5 60 R 2 _ 5 = 60 Ω 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations Calculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V Calculate in each case the resistance in each Resistance RT equivalent Rt, It and Vt. Calculate the I RT = 80 R1= 25 R2= 55 1 1 1 and V in each Resistance IT = 2,75 A I1= I2= = + RT R1− 5 R6 VT = 220V V1= V2= 1 1 1 = + VT RT 160 40 IT = R t = R1 + R2 RT 1 1 = R t = 25 + 55 220 RT 32 IT = RT = 32 Ω R t = 80Ω 80 I T = 2,75 A 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations Calculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each ResistanceCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance Calculate in each case the resistance RT= 200/13 RT = 80 R1= 25 R2= 55 equivalent Rt, It and Vt. Calculate the I IT = IT = 2,75 A I1= 2,75 A I2= 2,75 A and V in each Resistance VT = 20V VT = 220V V1= 68,75 V V2= 151,25V 1 1 1 = + RT R 1− 2 R 3 Series ⇒ I t= I1 = I 2 = 2,75 A R 1− 2 = R 1 + R 2 = 13 + 12 V1 = I1 R1 = 25 • 2,75 = 68,75V R 1− 2 = 25 1 1 1 1 1 13 V2 = I 2 R2 = 55 • 2,75 = 151,25V RT = + = + = R 1− 2 R 3 25 40 200 Series ⇒V t= V1 + V2 = 220V RT = 200 Ω 13 20
  • 21. RT =200/13 R1= 13 R2= 12 R3= 40 RT = 200/13 R1= 13 R2= 12 R3= 40 3.3.4 Circuit associations. Calculations IT =1,3A I1= I2= I3=0,5A IT = 1,3 A I1= 0,8 A I2= 0,8 A I3= 0,5 ACalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance VT = 20V V1= V2= V3=20V VT = 20V V1= 10,4 V V2= 9,6 V V3= 20 V RT = 200/13 IT= 1,3 A VT = 20V V1 = I1R 1 = 0,8 • 13 Mixed circuit V1 = 10 , 4V 20 V 20 I T = I1 − 2 + I 3 VT = V1− 2 = V3 = 20V IT = T = = 1 V1− 2 20 V2 = I 2 R 2 = 0,8 • 12 RT 200 200 Mixed circuit I1 − 2 = I 1 = I 2 I1− 2 = I1 = I 2 = = 13 13 R 1− 2 25 V2 = 9, 6V I T = I1− 2 + I 3 VT = V1− 2 = V1 + V2 = V3 I T = 1,3A I1− 2 = I1 = I 2 = 0 ,8 A I1− 2 = I1 = I 2 V 20 VT = V1− 2 = V1 + V2 = V3 I3 = 3 = R 3 40 I 3 = 0 ,5 A RT = R1= 3 R2= 2 R3= 10 RT = R1= 3 R2= 2 R3= 10 IT= 1A I1=1A I2= 1A I3= 1A IT = 1A I1=1A I2= 1A I3= 1A 3.3.4 Circuit associations. Calculations VT= 15V V1= V2= V3= VT = 15V V1= 3V V2= 2V V3=10VCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I PT = 15W P1= 3W P2= 2W P3=10W and V in each Resistance V1 = I1 R1 = 1⋅ 3 = 3V P = V1 I1 = 3 ⋅1 = 3W 1 RT = R1 + R2 + R3 P2 = V2 I 2 = 2 ⋅1 = 2W V2 = I 2 R2 = 1⋅ 2 = 2V RT = 3 + 2 + 10 ⇒ RT = 15Ω P3 = V3 I 3 = 10 ⋅1 = 10W V3 = I 3 R3 = 1⋅10 = 10V VT = ET = E1 + E2 = 5 − 20 = −15V VT 15 IT = = = 1A RT 15 Series association ⇒ I T = I1 = I 2 = I 3 = 1A RT=20 R1= 7 R2= 6 R3= 3 R4= 11 IT = I1= I2= I3= I4= 1,5A 3.3.4 Circuit associations. Calculations VT = 30V V1= V2= V3= V4=Calculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance 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 + 11 R 2−3 R 2 R 3 1 1 1 = + R T = 20 R 2−3 6 3 1 3 = R 2−3 6 R2 −3 = 2Ω 21
  • 22. RT =20 R1= 7 R2= 6 R3= 3 R4= 11 RT =20 R1= 7 R2= 6 R3= 3 R4= 11 RT =20 R1= 7 R2= 6 R3= 3 R4= 11IT= 1,5A I1= 1,5A I2= I3= I4= 1,5A IT = 1,5A I1= 1,5A I2= I3= I4= 1,5A IT = 1,5A I1= 1,5A I2= 0,5A I3= 1A I4=1,5AVT = 30V V1= V2= V3= V4= VT = 30V V1= 10,5V V2=3V V3=3V V4= 16,5V VT = 30V V1= 10,5V V2=3V V3=3V V4= 16,5V VT 30 V1 = I1R1 = 1,5⋅ 7 V2 3 IT = = = 1,5 A I2 = = R T 20 V1 =10,5V R2 6 I T = 1,5A V4 = I 4 R4 = 1,5⋅ 11 I 2 = 0,5 A Series associatio n : V1 =16,5V I T = I1 = I 2 -3 = I 4 = 1,5A V3 3 In series ⇒ VT = V1 + V2−3 + V4 I3 = = R3 3 30 =10,5 + V2−3 +16,5 I 3 = 1A V2− 3 = V2 = V3 = 3VRT =20 R1= 7 R2= 6 R3= 3 R5= 11IT = 1,5A I1= 1,5A I2= 0,5A I3= 1A I5=1,5A 3.3.4 Circuit associations. CalculationsVT = 30V V1= 10,5V V2=3V V3=3V V5= 16,5V Calculate in each case the resistancePT = 45V P1=15,75W P2=4,5W P3=3W P5= 24,75W equivalent Rt, It and Vt. Calculate the I and V in each Resistance P1 = V1I1 = 15 ,75W P2 = V2 I 2 = 4,5W P3 = V3 I 3 = 3W P4 = V4 I 4 = 24 ,75WRT=19 R1= 8 R2= 6 R3= 10 R4= 5 R4= 5 RT=19 R1= 8 R2= 6 R3= 10 R4= 5 R5= 5 RT=19 R1= 8 R2= 6 R3= 10 R4= 5 R5= 5IT= I1= I2= I3= I4= I4= IT=1A I1=1A I2=1A I3= I4= I5= IT=1A I1=1A I2=1A I3= I4= I5=VT= 19V V1= V2= V3= V4= V4= VT= 19V V1= 8V V2=6V V3=5V V4=2,5V V5=2,5V VT= 19V V1= 8V V2=6V V3=5V V4= V5=PT= P1= P2= P3= P4= P4= PT= P1= P2= P3= P4= P5= PT= P1= P2= P3= P4= P5=R T = R 1 + R 2 + R 3− 4- 5 1 1 1 V3-4-5 = V3 = V4 −5 = V4 + V5 = +R 3− 4 -5 R 3 R 4-5 Voltaje in each R V3-4-5 = V3 = V4 −5 = V4 + V5R 4- 5 = R 4 + R 5 V1 = I1R 1 = 1⋅ 8 V3-4-5 = V3 = 5VR 4 −5 = 5 + 5 = 10 V1 = 8VR 4 −5 = 10 VT 19 V1 = I1R 1 = 1⋅ 8 IT = = V2 = I 2 R 2 = 1⋅ 6 1 1 1 R T 19 V1 = 8V = + V1 = 6VR 3− 4 -5 R 3 R 4 −5 IT = 1A V2 = I 2 R 2 = 1⋅ 6 R T = R 1 + R 2 + R 3− 4- 5 V3-4-5 = I3-4-5 R 3-4-5 = 1 ⋅ 5 1 1 1 1 IT = I1 = I 2 = I3- 4-5 = 1A V1 = 6V = + = RT = 8 + 6 + 5 V3-4-5 = 5VR 3− 4 -5 10 10 5 Parallel I T = I3-4-5 = I3 + I 4-5 V3-4-5 = I3-4-5 R 3-4-5 = 1 ⋅ 5 R T = 19R 3− 4- 5 = 5 Series I 4-5 = I 4 = I5 V3-4-5 = 5V 22
  • 23. RT=19 R1= 8 R2= 6 R3= 10 R4= 5 R5= 5 RT=19 R1= 8 R2= 6 R3= 10 R4= 5 R5= 5IT=1A I1=1A I2=1A I3=0,5A I4=0,5A I5=0,5A IT=1A I1=1A I2=1A I3=0,5A I4=0,5A I5=0,5A 3.3.4 Circuit associations. CalculationsVT= 19V V1= 8V V2=6V V3=5V V4=2,5V V5=2,5V VT= 19V V1= 8V V2=6V V3=5V V4=2,5V V5=2,5V Calculate in each case the resistancePT= P1= P2= P3= P4= P5= PT=19W P1=8W P2=6W P3=2,5W P4=1,25W P5=1,25W equivalent Rt, It and Vt. Calculate the I and V in each ResistanceLet s calculate the I in the P1 = V1I1 = 8W V4 = I 4 R 4 = 0,5 ⋅ 5R in parallel V4 = 2,5V P2 = V2 I 2 = 6W V3 5I3 = = V5 = I 5 R 5 = 0,5 ⋅ 5 R 3 10 V5 = 2,5V P3 = V3 I 3 = 2,5WI3 = 0,5AI 4− 5 = V4-5 = 5 P4 = V4 I 4 = 1, 25W R 4-5 10I 4− 5 = 0,5A = I 4 = I5 P4 = V4 I 4 = 1, 25W RT=14 R1= 1 R2= 6 R3= 3 R4= 3 R5= 16 R6= 16 RT=14 R1= 1 R2= 6 R3= 3 R4= 3 R5= 16 R6= 16 IT= I1= I2= I3= I4= I5= I6= IT=2A I1=2A I2= I3= I4=2A I5= I6= VT=28V V1= V2= V3= V4= V5= V6= VT=28V V1=2V V2=4V V3= 4V V4=6V V5= 16V V6= 16V PT= P1= P2= P3= P4= P5= P6= R T = R1 + R 2−3 + R 4 + R 5−6 1 1 1 VT 28 = + IT = = R 2−3 R 2 R 3 R 2 14 R T = R 1 + R 2 −3 + R 4 + R 5−6 V1 = I1R 1 = 1 ⋅ 2 1 1 1 1 IT = 2A = + = R T = 1+ 2 + 3 + 8 V1 = 2V R 2−3 6 3 2 IT = I1 = I 2-3 = I 4 = I5-6 V2-3 = V2 = V3 R T = 14Ω V2-3 = I 2-3 R 2-3 = 2 ⋅ 2 R 2−3 = 2 IT = I1 = 2 A V2 = 4V V2-3 = 4V 1 1 1 IT = I 4 = 2 A V3 = 4V = + V4 = I 4 R 4 = 2 ⋅ 3 R 4-5 R 4 R 5 V5-6 = V5 = V6 1 1 1 1 V4 = 6V = + = V5 = 16V R 4-5 16 16 8 V5-6 = I 5-6 R 5-6 = 2 ⋅ 8 V6 = 16V R 4-5 = 8 V5-6 = 16VRT=14 R1= 1 R2= 6 R3= 3 R4= 3 R5= 16 R6= 16 RT=14 R1= 1 R2= 6 R3= 3 R4= 3 R5= 16 R6= 16IT=2A I1=2A I2=2/3A I3=4/3A I4=2A I5=1A I6=1A IT=2A I1=2A I2=2/3A I3=4/3A I4=2A I5=1A I6=1A 3.3.4 Circuit associations. CalculationsVT=28V V1=2V V2=4V V3=4V V4=6V V5=16V V6=16V VT=28V V1=2V V2=4V V3=4V V4=6V V5=16V V6=16V Calculate in each case the resistancePT= P1= P2= P3= P4= P5= P6= PT=56W P1=4W P2=8/3W P3=16/3W P4=12W P5=16W P6=16W equivalent Rt, It and Vt. Calculate the I and V in each Resistance V2 4 PT = VT I T = 56WI2 = = R2 6 V5 16 I5 = = P1 = V1 I1 = 4W 2 R 5 16I2 = A I5 = 1A P2 = V2 I 2 = 8 / 3W 3 V 4 I6 = V6 16 = P3 = V3 I 3 = 16 / 3WI3 = 3 = R3 3 R 6 16 P4 = V4 I 4 = 12W 4 I2 = 1AI2 = A P5 = V5 I 5 = 16W 3 P6 = V6 I 6 = 16W 23
  • 24. RT=18,04 R1= 12 R2= 3,67 R3= 7 R4= 9 R5= 6 RT=18,04 R1= 12 R2= 3,67 R3= 7 R4= 9 R5= 6 IT=4,93A I1=4,93A I2=4,93A I3= I4= I5= IT=4,93A I1=4,93A I2=4,93A I3= I4= I5= VT=89V V1= V2= V3= V4= V5= VT=89V V1=59,16V V2=18,09V V3= V4= V5= PT= P1= P2= P3= P4= P5= PT= P1= P2= P3= P4= P5= R T = R 1 + R 2 + R 3-4-5 V1 = I1R 1 = 4,93 ⋅12 = 59,16 A 1 1 1 1 VT 89 = + + IT = = V2 = I 2 R 2 = 4,93 ⋅ 3,67 = 18,09 A R 3-4-5 R 3 R 4 R 5 R T 18,04 1 1 1 1 53 IT = 4,93A V3-4 -5 = V3 = V4 = V5 = + + = R 3-4-5 7 9 6 126 In the series elements : 126 IT = I1 = I 2 = I3-4-5 V3- 4-5 = I 3- 4-5 R 3- 4-5 R 3− 4 −5 = 53 I1 = 4,93 A 126 126 V3- 4-5 = 4,93 ⋅ R T = 12 + 3,67 + I 2 = 4,93 A 53 53 V3- 4-5 = 11,72V R T = 18,04RT=18,04 R1= 12 R2= 3,67 R3= 7 R4= 9 R5= 6 RT=18,04 R1= 12 R2= 3,67 R3= 7 R4= 9 R5= 6 RT=18,04 R1= 12 R2= 3,67 R3= 7 R4= 9 R5= 6IT=4,93A I1=4,93A I2=4,93A I3= I4= I5= IT=4,93A I1=4,93A I2=4,93A I3=1,67A I4=1,30A I5=1,95A IT=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=11,72V VT=89V V1=59,16V V2=18,09V V3=11,72V V4=11,72V V5=11,72V VT=89V V1=59,16V V2=18,09V V3=11,72V V4=11,72V V5=11,72VPT= P1= P2= P3= P4= P5= PT= P1= P2= P3= P4= P5= PT=438,77W P1=291,66W P2=89,18W P3=19,57W P4=15,23W P5=22,85W V3 11,72V3- 4-5 = I 3-4-5 R 3- 4-5 I3 = = R3 7 126 PT = VT I T = 438 ,77WV3- 4-5 = 4,93 ⋅ I3 = 1,67 A 53 V 11,72 P1 = V1 I1 = 291,66WV3- 4-5 = 11,72V I4 = 4 = R4 9 P2 = V2 I 2 = 89 ,18WThe parallel elements have the same V I 4 = 1,30 A P3 = V3 I 3 = 19 ,57WV3- 4-5 = V3 = V4 = V5 = 11,72V V5 11,72 I5 = = P4 = V4 I 4 = 15 , 23W R5 6 I3 = 1,95 A P5 = V5 I 5 = 22 ,85W RT=18,04 R1= 12 R2= 3,67 R3= 7 R4= 9 R5= 6 IT=4,93A I1=4,93A I2=4,93A I3=1,67A I4=1,30A I5=1,95A VT=89V V1=59,16V V2=18,09V V3=11,72V V4=11,72V V5=11,72V PT= P1= P2= P3= P4= P5= R T = R1 + R 2 + R 3 + R 4 1 1 1 1 = + + R 3-4-5 R 3 R 4 R 5 1 1 1 1 53 = + + = R 3-4-5 7 9 6 126 126 R 2− 3 = 53 126 R T = 12 + 3,67 + 53 R T = 18,04 24