3 eso the electricity unit

Electricity

Unit 3

1

8.1 Electricity3.1 Introduction3.2 Electricity fundamentals

2.1 Particles and laws2.2 Electricity generation2.3 Electricity magnitudes

3.3 Circuits3.1 Ohm’s law. Electric Power3.2 Elements and Symbols3.3 Circuit associations. Calculations3.4 Basic circuits 3.5 Electric Measurements

2

Indicador Prueba Valor

Comprensión del concepto de corriente eléctrica y su base física. Diferenciación entre corriente continua y alterna.

Ex El 2

Interpretación de esquemas de circuitos sencillos de corriente continua

Poster 2

Cálculos en circuitos de corriente continua basados en la ley de Ohm

ex el 4

Conocimiento de la función de cada componente eléctrico y de su simbología

Ex El 2

Formulación y resolución de problemas Ex El 2

3

3.1 Introduction

What would happen if we didn’t have electricity?

?4

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

3.1 Introduction

What are we going to learn today?

The electric current and its physic principles

6

3.1 Introduction

So, what is electricity?The concept of electricity includes all

the phenomena related to electric charges

7

3.2.1 Particles and laws

Matter can be electrically charged when the charge distribution is upset

For 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


Matter is formed by atoms, which contain smaller particles inside with electric charges:

The electrons and protons

Atom

Electron

Proton

9

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: repulsionDifferent charges: attraction

attraction repulsion repulsion 10


The use of electricity needs a flow of electrons, the electric current

c

The electric current is the displacement of the electrical charges through the matter

c

11


In solid metals such as wires, the positive charge carriers are immobile, and only the negatively charged electrons can flow.

12

All the flowing charges are assumed to have positive polarity, and this flow is called conventional current.


13


Because the electron carries negatives charges, the electron motion in a metal is in the direction opposite to that of conventional (or electric) current.

14

3.2.2 Electricity generation

What are we going to learn next?Explain how the electricity can be created.

15

Exercise 3.2.2a :

Explain how the electricity can be created. Describe all elements needed in a common electric generator.


Solution

16

3.2.2 Electricity generationLong time ago… Hans Christian Oersted

discovered that a electric current can disturb a compass

17


Since the compass only is disturbed with a magnet, Volta concluded that an electric current creates a artificial magnetic

Natural magnet

Artificialmagnet

18


The 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

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

In conclusion, if we want to create an artificial electric current, we need:

Mechanical energy

to move the circuit or the

magnet


Closed circuit

Artificial magnet

21

This is the diagram of an industrial electric generator


Artificial magnet

Closed circuit

22

The electricity is created when the mechanical energy moves the closed circuit inside the artificial magnet


Electric engine video Electric generator and Engine video

23

When 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

Exercise 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.

3.2.2 Exercise 3.2.2b Electricity generation

solution25


What are we going to learn next?Difference between the direct and the alternate current

26

The 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

But, most electric at devices like an iPhone, need a Direct Current that keeps its magnitude and direction constant


28

If 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


In order to obtain a direct current we use powers supplies that include different elements that transforms AC into DC.

AC DC

30

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



33

3.3.1 CIRCUITS

What are we going to learn next?

Analysis of the direct current circuits.

34

3.3.1 Exercise 3.3.1:Electricity magnitudes

Exercise 3.3.1:Define Voltage, Intensity and

Resistance and its units

Solution35

3.3.1 Electricity magnitudes

To 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


In order to understand electricity, we have to first know the main electric magnitudes:

VOLTAGE

INTENSITYResistance

37


The electrons need energy to be able to move through a material. This is the Voltage

We define the Voltage as the energy per charge unit that makes them flow through a material. This magnitude is measured in Volts (V).

Energy

38


The higher the Voltage is, the more energy the electric charges will have to keep on moving.

Lower voltage

Higher voltage

39


Intensity is the amount of charges that goes through a conductor per time unit. It is measured in Ampere (A)

Lower intensity

Higher intensity

40


We 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

Less water pressure

More water pressure

41


Electric resistance is the opposition to the movement of the charges through a conductor. It is measured in Ohms ()

Lower Resistance

Higher Resistance

42

3.3.1 Electricity magnitudesExercise 3Which one of these lamps will give light?

43


Exercise 4Will this lamp turn on?

44


The electric current always goes through the route with less resistance, like water

Here we have the resistance needed to create light

No resistance 45

3.3.2 Ohm’s law. Electric Power


Magnitudes analysis using the Ohm’s Law.

46

3.3.2 Exercise 3.3.2a Ohm’s law. Electric Power

Exercise 3.3.2a:Justify how the Intensity will be if: We have a low voltage V We have a low Resistance R

Solution47


Ohm’s law links the three electric magnitudes as is shown:

R

VI

V= Voltage (voltsV)I= Intensity (ampereA)R= Resistance (ohm )

48


Intensity 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

R

VI

49


Intensity is inversely proportional to Resistance

If there is a high Resistance, the intensity will be low.

R

VI The charges will cross the

material slowly

50

3.3.2 Exercise 3.3.2.b Ohm’s law. Electric Power

Exercise 3.3.2.b:Calculate the value

of the intensity in these cases:

R

VI

V (V) R () I (A)

2 14

2 2

20 2

12 6

252 64

1000 20

Solution

51

3.3.2 c Exercise Ohm’s law. Electric Power

We 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

3.3.2 D Exercise Ohm’s law. Electric Power

A) 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

Solution53


Power is defined as the work consumed per time unit, and it’s measured in watt W

t

WP

54


If we apply the Ohm’s law, we obtain a new formula that explains the power of an electrical device

VIt

VQ

t

WP

t

QI

VQWQ

WV

VIP 55

3.3.2 Ohm’s law. Electric PowerFrom this expression we can derivate

in other two using the Ohm’s Law

R

VI

IRV

VIP

RIP 2

R

VP

2

56


There 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



Intensity throw the lamp.

58



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

3.3.2 Ohm’s law. Electric PowerIndicate the price of the electricity consumed if the

price is 0,90€/Kwh

60


Electrical companies measure our electrical consume in Kwh with a electricity meter situated in the basement of a building.

hPmedPowerConsu 61

3.3.3 Elements and Symbols


Electric components and their applications.

62

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

3.3.3 Elements and symbols

The 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

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

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

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

3.3.3 Elements and symbolsSafityIt is used to control high electrical volt

devices, because we don’t touch the main circuit

68

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


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


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 reload them. 71


Electric symbols are used to represent electric circuits with drawings that replace the real circuit elements.

72


Exercise: Draw the following circuit using electric symbols

73



74



75


Generator

Battery

Battery association

Conductors:

When two conductors are crossed without any contact we indicate it with a curve

76


Control elements

Push button

Switch

Diverter switch

77


Protection elements

Fuse

Receptors:

Lamp

Resistances: they have two symbols

Engines

78

3.3.4 Circuit associations

The behaviour of electrical elements depends on how they connect to each other.

There are three possible configurations :

SeriesParallelMixed

79

3.3.4 Circuit associationsSERIES circuit

The 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 A2 & 3 are connected only by B 80

3.3.4 Circuit associationsPARALLEL association

In this association, all the elements are connected by two points.

So, 1, 2 & 3 are connected by A and B

81

3.3.4 Circuit associationsMIXED association

Mixed 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


But what happens to receptors when they are connected in series or parallel associations?

Series and parallel association change the value of intensity and voltage through receptors.

83

3.3.4 Circuit associationsVoltage

Series Parallel

Voltage is distributed between elements, that is the reason why they have less energy for each lamp, so the light is lower in each lamp.

Voltage is the same in all elements, so all the lamps have the same energy and the light is higher.

84

Intensity

Series Parallel

All the lamps are in line, so they create a high resistance, that is the reason why the intensity is lower but the battery will have a longer life.

All lamps are separated so they create a low resistance, that is the reason why the intensity through the lamps is higher, but the battery will have a shorter life

85

Circuit

Series Parallel

If there is any cut along the conductor, the electric current will not be able to go from the positive to the negative pole.

If there is any cut along the conductor, the electric current can go through any other way

CUT

86

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

3.3.4 Circuit associationsExercises

Which of these elements are associated with series, parallel or mixed circuits? Name the connection points with letters.

88


Which of these elements are associated with series, parallel or mixed circuits. Name the connection points with letters

89


Which of these elements are associated with series, parallel or mixed circuits?. Name the connection points with letters.

90

3.3.4 Circuit associations. Calculations

Calculate in each case the resistance equivalent Rt

Solution91


Calculate in each case the resistance equivalent Rt

92


SERIES circuit In this association we find that:

VT=V1+ V2+V3

RT=R1+R2+R3

IT=I1= I2=I3 ET=E1+ E2+E3RTET

IT

VT

ET=VT

93


RTET

IT

VTI1

I2 I3

V1

V2 V394


SERIES circuit

INTENSITY= EQUAL IT=I1=I2=I3

RTET

IT

VT

ET=VT

I1

I2 I3

IT IT

95


SERIES circuit

VOLTAGE= DIVED VT=V1+ V2+V3

RTET

IT

VT

ET=VT

IT IT

V1

V2V3

VT

96

3.3.4 Series Exercise Circuit associations. Calculations

Calculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance

97



RT= 1ºR1= 0,5 Ω R2= 6,3 Ω R3= 6/5 Ω

IT= 2º I1= 2º I2= 2º I3= 2º

VT= 220VV1= 3º V2= 4º V3=5º

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

85

63,6

2

1321

t

t

t

R

R

RRRR

RT= 8 Ω R1= 0,5 Ω R2= 6,3 Ω R3= 6/5 Ω

IT= 27,5 A I1= I2= I3=

VT= 220V V1= V2= V3=

5,278

220

T

T

T

TT

I

I

R

VI

1º

2º

99



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= V2= V3=

100



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

101


VT=V1= V2=V3

IT=I1+ I2+I3 ET=E1+ E2+E3

321

1111

RRRRT

102


VT=V123=V1= V2=V3

IT=I123=I1+ I2+I3103


INTENSITY= DIVEDIT=I1+I2+I3

I3

I2

I1

IT104


INTENSITY= DIVEDIT=I1+I2+I3

I3

I2

I1

IT105


VOLTAGE: IQUALVT=V1= V2=V3

V1

V2

V3VT 106


V1

V2

V3 VT

RT= R1= 6 Ω R2= 18 Ω R3= 4 Ω

IT= I1= I2= I3=

VT= 20V V1= V2= V3=

RT= R1= 6 Ω R2= 18 Ω R3= 4 Ω

IT= I1= I2= I3=

VT= 220V V1= V2= V3=

107


V1

V2

V3 VT

RT= R1= 6 Ω R2= 18 Ω R3= 4 Ω

IT= I1= I2= I3=

VT= 20V V1= V2= V3=

RT= 1ºR1= 6 Ω R2= 18 Ω R3= 4 Ω

IT= 2º I1= 3º I2= 4º I3= 5º

VT= 220VV1= 20V V2= 20V V3=20V

108

3.3.4 Circuit associationsPARALLEL EXERCISE

V1

V2

V3VT

RT=2,11Ω R1= 6 Ω R2= 18 Ω R3= 4 Ω

IT= I1= I2= I3=

VT= 20V V1= V2= V3=

109


V1

V2

V3VT

RT=2,11Ω R1= 6 Ω R2= 18 Ω R3= 4 Ω

IT= I1= I2= I3=

VT= 20V V1= V2= V3=

110


V1

V2

V3VT

RT=2,11Ω R1= 6 Ω R2= 18 Ω R3= 4 Ω

IT=9,48A I1= I2= I3=

VT= 20V V1= V2= V3=

111

3.3.4 Circuit associations VOLTAGE: IQUAL

VT=V1= V2=V3

VT=20

V1

V2

V3VT

RT=2,11Ω R1= 6 Ω R2= 18 Ω R3= 4 Ω

IT=9,48A I1= I2= I3=

VT= 20V V1=20V V2=20V V3=20V

112


V1

V2

V3VT

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

113


We have to convert the parallel associations into series in order to obtain Vt, It and Rt. T

1ºVt, It and Rt2º

V4, I4 and V123

3ºI1, I2 , I3 and

V123

114


IT=I123=I4 series association

I123=I1+ I2+I3 parallel association

115




I1

I2

I1116


I123=I1+ I2+I3, but they don’t have to be equal, it depends on the R in each path.I1

I2

I1117


VT=V123+V4 series association

V123=V1= V2=V3 parallel association

V1

VT=V123+V4 V2

V3118






119


1ºVt, It and Rt

2ºV4, I4 and

V123

3º I1, I2 , I3 and

V1





120

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

RT= 1ºR1= 6 Ω R2= 18 Ω R3= 3/2 Ω

IT= 2º I1= 5º I2= 6º I3=2º

VT= 20VV1= 4º V2= 4º V3=3º

121


RT= 6 Ω

IT=

VT= 20V

1º

122

RT= 6 Ω R1= 6 Ω R2= 18 Ω R3= 3/2 Ω

IT= 10/3 A I1= I2= I3=

VT= 20V V1= V2= V2= 2º

123

RT= 6 Ω R1= 6 Ω R2= 18 Ω R3= 3/2 Ω

IT= 10/3 A I1= I2= I3= 10/3 A

VT= 20V V1= V2= V3=

I1

I2

I3I1-2

I3

2º

124

RT= 6 Ω R1= 6 Ω R2= 18 Ω R3= 3/2 Ω

IT= 10/3 A I1= I2= I3= 10/3 A

VT= 20V V1= V2= V3= 5V

3º

125

RT= 6 Ω R1= 6 Ω R2= 18 Ω R3= 3/2 Ω

IT= 10/3 A I1= I2= I3= 10/3 A

VT= 20V V1= 15V V2= 15V

V3= 5V

Two ways to calculate V1 and V2

4º

126

RT= 6 Ω R1= 6 Ω R2= 18 Ω R3= 3/2 Ω

IT= 10/3 A I1= 2,5 A I2= 5/6 A I3= 10/3 A

VT= 20V V1= 15v V2= 15V V2= 5 V

AIR

VI

AIR

VI

6

5

6

5

18

15

5,22

5

6

15

12

22

11

11

5º6º

127


1ºVt, It and Rt

2ºV4, I4 and

V123

3º I1, I2 , I3 and

V1





128

3.3.4 Ex 2 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I


RT= R1= 4 Ω R2= 4 Ω R3= 8 Ω R4= 12 Ω

IT= I1= I2= I3= I4=

VT= 220V V1= V2=

V3 V4=

4 Ω

4 Ω

8 Ω12 Ω

220V

sol

129


E represents the electrometric force that is the Total Voltage that provides a electric source like a battery or a power supply,

130


Exercise 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

3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I


SolSol132



133



134



135



136

3.3.4 Circuit associationsExercise:

Which of these lamps have more lightness?

137


Which of these assemblies generates electricity?

138


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


Draw the electric circuit of a car that has :

1.2 front lamps

2.2 Back lamps

3.A forward and backward engine that stops automatically when it crash with a object

140


Why are these assemblies no correct?

141

3.3.6 Electric Measurements. Polimeter

Ohmmeter: It measures the resistance and the continuity of a circuit or component.

Resistance control

If there is no resistance that means that there is a shortcircuit If there is a maximum resistance that means that the cricuit is open

142

3.3.6 Electric Measurement. Polimeter

4K7

• In order to get a precise lecture we have to choose the correct range of measurement.

143

3.3.6 Electric Measurements. PolimeterVoltimeterVoltimeter

AplicationAplication: If we want to measure the voltage present in a component we have to conect the electrodes in parallel

V

Voltage in a component 144

If we want to measure the voltage of battery we have to place the control in the Direct Current and in the proper scale


145

Amperimeter: In order to be able to measure the current we have to place the electrodes BETWEEN the circuitr or in sereies

A

CONTROL DE CONSUMO


146

4K7

Amperimeter: we chose the position according to the scale of the current.


147


Exercise:

According to this assembly, chose the correct answer:

The I through the lamp is 2,3 AThe voltage between the lamp is

2,3 VThe resistance of the lamp is 2,3

Ω

148

A closed

B open

A open

B closed

A closed

B closed

Engine

lamp 1

lamp 2

Point out in the table if the engine and lamps works for the following situations


149

Exercise 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.


Back150

1. 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

3.2.2 Sol Exercise 3.2.2b Electricity generation

151

Exercise 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 is what fuels our homes. The wires outside of our house are connected at two ends to AC 152

Exercise 3.2.2b :



DC is found in batteries and solar cells and it used for most of our electric devices. We need to convert the AC into DC in most of electric devices using power suplies

153

Exercise 3.2.2b :



AC DCTV TORCHFridge PhoneBedroom lamp

154

AC DC

3. Draw the diagram of a mobile phone power supply.

Back 155

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

3.3.1 Exercise 3.3.1:SolutionElectricity magnitudes

Exercise 3.3.1:Define Voltage, Intensity and Resistance and

its unitsSimbol Name Definition Unit

V VoltageVoltage Is the energy per

charge unit that makes them flow through a material

Volts (V)

I IntensityIntensity is the amount of

charges that goes through a conductor per time unit

Ampere (A)

R Resistance

Electric resistance is the opposition to the movement of

the charges through a conductor.

ohm (Ω)

BAck

157

3.3.2 a Exercise Ohm’s law. Electric Power

Intensity 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

R

VI

158

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.

R

VI The charges will cross the

material slowly

exercise

159

3.3.2 Exercise 3.3.2.b Ohm’s law. Electric Power

Exercise 3.3.2.b:Calculate the value

of the intensity in these cases:

R

VI

V (V) R () I (A)

2 14 0,14A

2 1 2

20 2 10A

72 V 12 6

252 64 3,94 A

1000 20 50A

exercise

160

V (V) R () I (A)

2 2

2 4

3.32 b Solution Ohm’s law Calculations with Ohm’s law 5º Exercise Solution

1AI 2

2I

R

VI

A5,0I 4

2I

R

VI

161

V (V) R () I (A)

2 4

10 5

6.4 Ohm’s law Calculations with Ohm’s law Solution

8VV

24V

IRV

2R5

10R

I

VR

162

V (V) R () I (A)

5 10

20 1000

6.4 Ohm’s law Calculations with Ohm’s law Solution

A 0,5I

10

5I

R

VI

20kvV

V00002V

200001V

IRV

exercise

163

3.3.2 c Solution Ohm’s law. Electric Power

We have a fan connected to the domestic electric supply. The amperimeter indicate a 0,52A on the circuit. Extra data: house electrical supply 230V, 50Hz

1º Text analysis:

I= 0,52A ; V=230V;

R= ? ; P=?

1.What’s is the resistance of the fan?

164


We have a fan connected to the domestic electric supply. The amperimeter indicate a 0,52A on the circuit. Extra data: house electrical supply 230V, 50Hz

1º Text analysis:

I= 0,52A ; V=230V;

I= ? ; P=?

2º- What’s the power measured in Kw

165


Calculate 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=1,2kW

166


Calculate 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=1,2kW

Back 167

3.3.4 Circuit associationsSolutions

Parallel: 1, 2 and 3 are joined together by A and B

Series: 1 and 2 are joined by A; 2 and 3 are joined by B

168

3.3.4 Circuit associationsSolutions

Mixed: Series: 1 and 2 are joined to ASeries: 2 and the 4 & 3 association are joined by B Parallel: 4 and 3 are joined by B and C

169

3.3.4 Circuit associationsSolutions: red-series Blue: parallel

SERIES: 1 and 2 are joined by A. 2 & 3 are joined by B3 & 4 are joined by C

MIXED: Series:1 is joined to 2 and 3 by A Parallel: 2 and 3 are joined by A and B Series: 2 and 3 are joined to 4 by C

170


85

63,6

2

1321

t

t

t

R

R

RRRR

5,32

7

7

21

28

1

28

7

28

1

4

11

111

21

T

T

T

T

T

R

R

R

R

RRR

171


2

9

69

21

2

12

18

1

18

3

18

1

6

11

2

3

2

9111

21

21

21

2121

321321

R

RR

RR

RRRR

RRRRRR

T

T

T

TT

172


13

20025

200

1311213

40

1

25

11

111111

21

21

2121

321321

T

t

t

tt

RR

RR

RRRR

RRRRRR

173


174


1500

1500

11

6000

1

3000

1

6000

11

1111

1500

5001000

543

543

543

543 543

543

21

21

2121

21

R

R

R

RRRR

R

R

R

RRR

R

1000

1000

11

2000

1

2000

11

111

76

76

76

76 76

76

R

R

R

RRR

R

175


3000

15001500

5_1

5_1

543215_1

5_1

R

R

RRR

R

3000

20001000

8_6

8_6

8768_6

8_6

R

R

RRR

R

1500

1500

11

3000

1

3000

11

111

8_65_1

T

T

T

T

T

R

R

R

RRR

R

176


177


20

20

11

100

1

25

11

111

43

43

43

4343

43

R

R

R

RRR

R

120

10020

432

432

432432

432

R

R

RRR

R

178


179


60

60

11

120

1

120

11

111

5_2

5_2

5_2

54325_2

5_2

R

R

R

RRR

R

180


32

32

11

40

1

160

11

111

651

T

T

T

T

T

R

R

R

RRR

R

181



182



80

552521

t

t

t

R

R

RRR

RT= 80 Ω R1= 25 Ω R2= 55 Ω

IT= 2,75 A I1= I2=

VT= 220V V1= V2=

AI

I

R

VI

T

T

T

TT

75,280

220

183



VVVVSeries

VRIV

VRIV

AIIISeries

t

t

220

25,15175,255

75,6875,225

75,2

21

222

111

21

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

184



back185



13

200R

200

13

40

1

25

1

R

1

R

1

R

1

25ΩR

1213RRR

R

1

R

1

R

1

T

321

21

2121

321

T

T

RT= 200/13 Ω

IT=

VT= 20V

186



1,3AI13200120

1320020

R

VI

T

T

TT

32121T

2121

321T

VVVVV

III

III

circuitMixed

RT= 200/13 Ω

IT= 1,3 A

VT= 20V

187

A

A

VT

5,0I

40

20

R

VI

8,0III

25

20

R

VIII

20VVV

3

3

33

2121

21

212121

321

32121T

2121

321T

VVVVV

III

III

circuitMixed

RT=200/13 R1= 13 Ω R2= 12 Ω R3= 40 Ω

IT=1,3A I1= I2= I3=0,5A

VT= 20V V1= V2= V3=20V

188

V

V

6,9V

128,0RIV

4,10V

138,0RIV

2

222

1

111

RT= 200/13 Ω R1= 13 Ω R2= 12 Ω R3= 40 Ω

IT= 1,3 A I1= 0,8 A I2= 0,8 A I3= 0,5 A

VT= 20V V1= 10,4 V V2= 9,6 V V3= 20 V

189



190

AIIIInassociatioSeries

AR

VI

VEEEV

RR

RRRR

T

T

TT

TT

TT

T

1

115

15

15205

151023

321

21

321

RT= R1= 3 Ω R2= 2 Ω R3= 10 Ω

IT= I1= I2=

I3=

VT= V1= V2= V3=

191

VRIV

VRIV

VRIV

10101

221

331

333

222

111

RT= Ω R1= 3 Ω R2= 2 Ω R3= 10 Ω

IT= 1A I1=1A I2= 1A

I3= 1A

VT= 15V V1= 3V V2= 2V V3=10V

PT= 15W P1= 3W P2= 2W P3=10W

WIVP

WIVP

WIVP

10110

212

313

333

222

111

192

3.3.4 sol Ex 2 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I


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= 220V

V1= =3º V2= 5º V3=5º V4=4º

4 Ω

4 Ω

8 Ω12 Ω

220V

193

3.3.4 sol Ex 2 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I


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= 220V

V1= =3º V2= 5º V3=5º V4=4º

4 Ω

4 Ω

8 Ω12 Ω

220V

194


4 Ω

4 Ω

8 Ω12 Ω

220V

4 Ω 8/3Ω 12 Ω 220V

56/3 Ω 220V

195


RT= 6 Ω

IT=

VT= 220V

1º

220V

4 Ω

4 Ω

8 Ω12 Ω

196

RT= 18,67Ω R1= 4 Ω R2= 4 Ω R3= 8 Ω R4= 12 Ω

IT= 11,78 A I1= I2= I3=

VT= 220V V1= V2= V3= V4=

2º

131/8 Ω 220V

197

2º

4 Ω

4 Ω

8 Ω12 Ω

220V

RT= 18,67Ω R1= 4 Ω R2= 4 Ω R3= 8 Ω R4= 12 Ω

IT= 11,78 A I1= 11,78A

I2= I3= I4= 11,78 A

VT= 220V V1= V2= V2=

198

3º

RT= 18,67Ω R1= 4 Ω R2= 4 Ω R3= 8 Ω R4= 12 Ω

IT= 11,78 A I1= 11,78A

I2= I3= I4= 11,78 A

VT= 220V V1= 47,12V V2= V3= V4= 141,36V

199

Two ways to calculate V1 and V2

4ºRT= 18,67Ω R1= 4 Ω R2= 4 Ω R3= 8 Ω R4= 12 Ω

IT= 11,78 A I1= 11,78A

I2= I3= I4= 11,78 A

VT= 220V V1= 47,12V V2=31,52 V

V3=31,52 V

V4= 141,36V

200

5º6º

RT= 18,67Ω R1= 4 Ω R2= 4 Ω R3= 8 Ω R4= 12 Ω

IT= 11,78 A I1= 11,78A

I2=7,88A I3=3,94A I4= 11,78 A

VT= 220V V1= 47,12V V2=31,52 V

V3=31,52 V

V4= 141,36V

back201



202

RT=20 Ω R1= 7 Ω R2= 6 Ω R3= 3 Ω R4= 11 Ω

IT= I1= I2=

I3=

I4= 1,5A

VT= 30V V1= V2= V3= V4=

2

6

3

R

1

20ΩR3

1

6

1

R

1

1127RR

1

R

1

R

1

RRRRRRRR

32

32

T32

T3232

4321T4321T

R204

RT=20 Ω R1= 7 Ω R2= 6 Ω R3= 3 Ω R4= 11 Ω

IT= 1,5A I1= 1,5A I2=

I3=

I4= 1,5A

VT= 30V V1= V2= V3= V4=

A5,1IIII

:

A5,1I

5,120

30

R

VI

43-21T

T

T

TT

nassociatioSeries

A

205

RT=20 Ω R1= 7 Ω R2= 6 Ω R3= 3 Ω R4= 11 Ω

IT= 1,5A I1= 1,5A I2=

I3=

I4= 1,5A

VT= 30V V1= 10,5V V2=3V V3=3V V4= 16,5V

206

RT=20 Ω R1= 7 Ω R2= 6 Ω R3= 3 Ω R4= 11 Ω

IT= 1,5A I1= 1,5A I2= 0,5A

I3= 1A

I4=1,5A

VT= 30V V1= 10,5V V2=3V V3=3V V4= 16,5V

A

A

1I

3

3

R

VI

5,0I

6

3

R

VI

3

3

33

2

2

22

207

RT=20 Ω R1= 7 Ω R2= 6 Ω R3= 3 Ω R5= 11 Ω

IT= 1,5A I1= 1,5A I2= 0,5A

I3= 1A

I5=1,5A

VT= 30V V1= 10,5V V2=3V V3=3V V5= 16,5V

PT= 45V P1=15,75W P2=4,5W P3=3W P5= 24,75W

W

W

W

W

75,24IVP

3IVP

5,4IVP

75,15IVP

444

333

222

111

208



209

RT=19 Ω R1= 8 Ω R2= 6 Ω R3= 10 Ω R4= 5 Ω R4= 5 Ω

IT= I1= I2=

I3=

I4=

I4=

VT= 19V V1= V2= V3= V4= V4=

PT= P1= P2= P3= P4= P4=

5ΩR

5

1

10

1

10

1

R

1

R

1

R

1

R

1

10ΩR

10Ω55R

RRR

R

1

R

1

R

1

RRRR

5-43

5-43

5435-43

54

54

545-4

5-435-43

5-4321T

19ΩR

568R

RRRR

T

T

5-4321T

211

RT=19 Ω R1= 8 Ω R2= 6 Ω R3= 10 Ω R4= 5 Ω R5= 5 Ω

IT=1A I1=1A I2=1A

I3=

I4=

I5=

VT= 19V V1= 8V V2=6V V3=5V V4=2,5V V5=2,5V

PT= P1= P2= P3= P4= P5=

545-4

5-435-4-3T

5-4-321T

T

T

TT

III Series

IIII Parallel

1AIIII

1AI

91

91

R

VI

545435-4-3 VVVVV

5VV

51RIV

6VV

61RIV

8VV

81RIV

5-4-3

5-4-35-4-35-4-3

1

222

1

111

212

RT=19 Ω R1= 8 Ω R2= 6 Ω R3= 10 Ω R4= 5 Ω R5= 5 Ω

IT=1A I1=1A I2=1A

I3=

I4=

I5=

VT= 19V V1= 8V V2=6V V3=5V V4= V5=

PT= P1= P2= P3= P4= P5=

5VV

51RIV

6VV

61RIV

8VV

81RIV

Reach in Voltaje

5-4-3

5-4-35-4-35-4-3

1

222

1

111

V5VV

VVVVV

35-4-3

545435-4-3

213

RT=19 Ω R1= 8 Ω R2= 6 Ω R3= 10 Ω R4= 5 Ω R5= 5 Ω

IT=1A I1=1A I2=1A

I3=0,5A

I4=0,5A

I5=0,5A

VT= 19V V1= 8V V2=6V V3=5V V4=2,5V V5=2,5V

PT= P1= P2= P3= P4= P5=

5454

5-4

5-454

3

3

33

II0,5AI

10

5

R

VI

0,5AI

10

5

R

VI

parallelin R

in the I thecalculate sLet'

V

V

5,2V

55,0RIV

5,2V

55,0RIV

5

555

4

444

214

RT=19 Ω R1= 8 Ω R2= 6 Ω R3= 10 Ω R4= 5 Ω R5= 5 Ω

IT=1A I1=1A I2=1A

I3=0,5A

I4=0,5A

I5=0,5A

VT= 19V V1= 8V V2=6V V3=5V V4=2,5V V5=2,5V

PT=19W P1=8W P2=6W P3=2,5W P4=1,25W P5=1,25W

W

W

W

W

W

25,1IVP

25,1IVP

5,2IVP

6IVP

8IVP

444

444

333

222

111

215



216

RT=14 Ω R1= 1 Ω R2= 6 Ω R3= 3 Ω R4= 3 Ω R5= 16 Ω R6= 16 Ω

IT= I1= I2= I3= I4= I5= I6=

VT=28V V1= V2= V3= V4= V5= V6=

PT= P1= P2= P3= P4= P5= P6=

Ω8R

8

1

16

1

16

1

R

1

R

1

R

1

R

1

Ω2R

2

1

3

1

6

1

R

1

R

1

R

1

R

1

RRRRR

5-4

5-4

545-4

32

32

3232

654321T

14R

8321R

RRRRR

T

T

654321T

218


IT=2A I1=2A I2= I3= I4=2A I5= I6=

VT=28V V1=2V V2=4V V3= 4V V4=6V V5= 16V V6= 16V

A

A

2II

2II

IIIII

2AI

14

28

R

VI

4T

1T

6-543-21T

T

2

TT

16VV

82RIV

6VV

32RIV

4VV

22RIV

2VV

21RIV

6-5

6-56-56-5

4

444

3-2

3-23-23-2

1

111

V16V

6V1V

VVV

4VV

4VV

VVV

6

5

656-5

3

2

323-2

219


IT=2A I1=2A I2=2/3A I3=4/3A I4=2A I5=1A I6=1A

VT=28V V1=2V V2=4V V3=4V V4=6V V5=16V V6=16V

PT= P1= P2= P3= P4= P5= P6=

A

A

3

4I

3

4

R

VI

3

2I

6

4

R

VI

2

3

33

2

2

22

A

A

1I

16

16

R

VI

1I

16

16

R

VI

2

6

66

5

5

55

220


IT=2A I1=2A I2=2/3A I3=4/3A I4=2A I5=1A I6=1A

VT=28V V1=2V V2=4V V3=4V V4=6V V5=16V V6=16V

PT=56W P1=4W P2=8/3W P3=16/3W P4=12W P5=16W P6=16W

W

W

W

W

W

W

WTTT

16IVP

16IVP

12IVP

3/16IVP

3/8IVP

4IVP

56IVP

666

555

444

333

222

111

221



222

RT=18,04 Ω R1= 12 Ω R2= 3,67Ω R3= 7 Ω R4= 9 Ω R5= 6 Ω

IT=4,93A I1=4,93A I2=4,93A I3= I4= I5=

VT=89V V1= V2= V3= V4= V5=

PT= P1= P2= P3= P4= P5=

Ω04,18R53

12667,312R

Ω53

126R

126

53

6

1

9

1

7

1

R

1

R

1

R

1

R

1

R

1

RRRR

T

T

543

5-4-3

5435-4-3

5-4-321T

A

A

93,4I

93,4I

IIII

:elements series In the

4,93AI

18,04

89

R

VI

2

1

5-4-321T

T

T

TT

224

RT=18,04 Ω R1= 12 Ω R2= 3,67Ω R3= 7 Ω R4= 9 Ω R5= 6 Ω

IT=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=

V72,11V53

12693,4V

RIV

5-4-3

5-4-3

5-4-35-4-35-4-3

5435-4-3

222

111

VVVV

09,1867,393,4RIV

16,591293,4RIV

A

A

225

RT=18,04 Ω R1= 12 Ω R2= 3,67Ω R3= 7 Ω R4= 9 Ω R5= 6 Ω

IT=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=11,72V

PT= P1= P2= P3= P4= P5=

11,72VVVVV

V same thehave elements parallel The

11,72VV53

1264,93V

RIV

5435-4-3

5-4-3

5-4-3

5-4-35-4-35-4-3

226

RT=18,04 Ω R1= 12 Ω R2= 3,67Ω R3= 7 Ω R4= 9 Ω R5= 6 Ω

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=

A

A

A

95,1I

6

11,72

R

VI

30,1I

9

11,72

R

VI

67,1I

7

11,72

R

VI

3

5

55

4

4

44

3

3

33

227

RT=18,04 Ω R1= 12 Ω R2= 3,67Ω R3= 7 Ω R4= 9 Ω R5= 6 Ω

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=438,77W P1=291,66W P2=89,18W P3=19,57W P4=15,23W P5=22,85W

W

W

W

W

W

WTTT

85,22IVP

23,15IVP

57,19IVP

18,89IVP

66,291IVP

77,438IVP

555

444

333

222

111

228

RT=18,04 Ω R1= 12 Ω R2= 3,67Ω R3= 7 Ω R4= 9 Ω R5= 6 Ω

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=

Ω04,18R53

12667,312R

Ω53

126R

126

53

6

1

9

1

7

1

R

1

R

1

R

1

R

1

R

1

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T

T

32

5-4-3

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230

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electricity generationwhat

electricity magnitudes3

electricity fundamentals

electricity generationthis

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