3 eso the electricity unit

Electricity

Unit 3

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

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

3.1 Introduction

What would happen if we didn’t have electricity?

3.1 Introduction

However, we have to remember that we can decrease the amount of energy that we waste everyday, helping our sustainable

development.

3.1 Introduction

What are we going to learn today?

The electric current and its physic principles

3.1 Introduction

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

the phenomena related to electric charges

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

Electron

Proton

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

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

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

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

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

3.2.2 Electricity generation

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

Exercise 3.2.2a :

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

Solution

3.2.2 Electricity generationLong time ago… Hans Christian Oersted

discovered that a electric current can disturb a compass

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

Natural magnet

Artificialmagnet

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

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

This is the diagram of an industrial electric generator

Artificial magnet

Closed circuit

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

Electric engine video Electric generator and Engine video

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.

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

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

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

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.

3.2.2 Electricity generationThese are the elements of a Power supply that

that transforms AC into DC: transformer, rectifier, smoothing and regulator.

3.3.1 CIRCUITS

What are we going to learn next?

Analysis of the direct current circuits.

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

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

VOLTAGE

INTENSITYResistance

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

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

Lower voltage

Higher voltage

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

Lower intensity

Higher intensity

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

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

Lower Resistance

Higher Resistance

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

Exercise 4Will this lamp turn on?

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.

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:

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

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

Intensity is inversely proportional to Resistance

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

VI The charges will cross the

material slowly

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:

V (V) R () I (A)

252 64

1000 20

Solution

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

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

VIP 55

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

in other two using the Ohm’s Law

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

Intensity throw the lamp.

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

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 with a electricity meter situated in the basement of a building.

hPmedPowerConsu 61

3.3.3 Elements and Symbols

Electric components and their applications.

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

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.

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

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.

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

3.3.3 Elements and symbolsSafityIt is used to control high electrical volt

devices, because we don’t touch the main circuit

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

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.

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.

Exercise: Draw the following circuit using electric symbols

Generator

Battery

Battery association

Conductors:

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

Control elements

Push button

Switch

Diverter switch

Protection elements

Receptors:

Resistances: they have two symbols

Engines

3.3.4 Circuit associations

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

There are three possible configurations :

SeriesParallelMixed

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

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

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.

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.

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

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

If there is any cut in series association, the water can’t go further. But in parallel association there will be no problem.

3.3.4 Circuit associationsExercises

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

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

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

3.3.4 Circuit associations. Calculations

Calculate in each case the resistance equivalent Rt

Solution91

Calculate in each case the resistance equivalent Rt

SERIES circuit In this association we find that:

VT=V1+ V2+V3

RT=R1+R2+R3

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

V2 V394

SERIES circuit

INTENSITY= EQUAL IT=I1=I2=I3

SERIES circuit

VOLTAGE= DIVED VT=V1+ V2+V3

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

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

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= I2= I3=

VT= 220V V1= V2= V3=

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=

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

VT=V1= V2=V3

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

VT=V123=V1= V2=V3

IT=I123=I1+ I2+I3103

INTENSITY= DIVEDIT=I1+I2+I3

VOLTAGE: IQUALVT=V1= V2=V3

V3VT 106

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=

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

3.3.4 Circuit associationsPARALLEL EXERCISE

RT=2,11Ω R1= 6 Ω R2= 18 Ω R3= 4 Ω

IT= I1= I2= I3=

VT= 20V V1= V2= V3=

RT=2,11Ω R1= 6 Ω R2= 18 Ω R3= 4 Ω

IT= I1= I2= I3=

VT= 20V V1= V2= V3=

RT=2,11Ω R1= 6 Ω R2= 18 Ω R3= 4 Ω

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

VT= 20V V1= V2= V3=

3.3.4 Circuit associations VOLTAGE: IQUAL

VT=V1= V2=V3

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

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

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

IT=I123=I4 series association

I123=I1+ I2+I3 parallel association

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

VT=V123+V4 series association

V123=V1= V2=V3 parallel association

VT=V123+V4 V2

1ºVt, It and Rt

2ºV4, I4 and

3º I1, I2 , I3 and

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

RT= 6 Ω

VT= 20V

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

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=

I3I1-2

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

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

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

5º6º

1ºVt, It and Rt

2ºV4, I4 and

3º I1, I2 , I3 and

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=

8 Ω12 Ω

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

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.

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

SolSol132

3.3.4 Circuit associationsExercise:

Which of these lamps have more lightness?

Which of these assemblies generates electricity?

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

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

Why are these assemblies no correct?

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

3.3.6 Electric Measurement. Polimeter

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

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

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

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

CONTROL DE CONSUMO

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

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

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

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

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

Exercise 3.2.2b :

AC DCTV TORCHFridge PhoneBedroom lamp

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

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)

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

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.

VI The charges will cross the

material slowly

exercise

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:

V (V) R () I (A)

2 14 0,14A

20 2 10A

72 V 12 6

252 64 3,94 A

1000 20 50A

exercise

V (V) R () I (A)

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

A5,0I 4

V (V) R () I (A)

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

V (V) R () I (A)

20 1000

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

A 0,5I

V00002V

200001V

exercise

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?

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

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

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

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

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

321321

RRRRRR

1311213

111111

321321

RRRRRR

5001000

543 543

15001500

543215_1

20001000

8768_6

8_65_1

432432

54325_2

552521

RT= 80 Ω R1= 25 Ω R2= 55 Ω

IT= 2,75 A I1= I2=

VT= 220V V1= V2=

75,280

VVVVSeries

AIIISeries

25,15175,255

75,6875,225

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

back185

1213RRR

RT= 200/13 Ω

VT= 20V

1,3AI13200120

1320020

32121T

circuitMixed

RT= 200/13 Ω

IT= 1,3 A

VT= 20V

8,0III

212121

32121T

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

128,0RIV

138,0RIV

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

AIIIInassociatioSeries

151023

RT= R1= 3 Ω R2= 2 Ω R3= 10 Ω

IT= I1= I2=

VT= V1= V2= V3=

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

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

8 Ω12 Ω

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

8 Ω12 Ω

4 Ω 8/3Ω 12 Ω 220V

56/3 Ω 220V

RT= 6 Ω

VT= 220V

8 Ω12 Ω

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=

131/8 Ω 220V

8 Ω12 Ω

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=

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

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

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

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

IT= I1= I2=

I4= 1,5A

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

20ΩR3

1127RR

RRRRRRRR

4321T4321T

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

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

I4= 1,5A

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

A5,1IIII

43-21T

nassociatioSeries

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

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

I4= 1,5A

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

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

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

75,24IVP

5,4IVP

75,15IVP

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

IT= I1= I2=

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

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

10Ω55R

5435-43

5-435-43

5-4321T

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

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

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

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

5-435-4-3T

5-4-321T

III Series

IIII Parallel

1AIIII

545435-4-3 VVVVV

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

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

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

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

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

Reach in Voltaje

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

35-4-3

545435-4-3

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=

II0,5AI

parallelin R

in the I thecalculate sLet'

55,0RIV

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

25,1IVP

5,2IVP

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=

654321T

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

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

6-543-21T

6-56-56-5

3-23-23-2

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=

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

3/16IVP

3/8IVP

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

5435-4-3

5-4-321T

:elements series In the

4,93AI

5-4-321T

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

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

5435-4-3

09,1867,393,4RIV

16,591293,4RIV

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

5435-4-3

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

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=

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

85,22IVP

23,15IVP

57,19IVP

18,89IVP

66,291IVP

77,438IVP

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

5435-4-3

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