3 eso electricity unit light

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Electricity

Unit 3

8.1 Electricity3.1 Introduction

3.2 Electricity fundamentals

2.1 Particles and laws

2.2 Electricity generation

2.3 Electricity magnitudes

3.3 Circuits

3.1 Ohm’s law. Electric Power

3.2 Elements and Symbols

3.3 Circuit associations. Calculations

3.4 Basic circuits

3.5 Electric Measurements

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


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

The electrons and protons

Atom

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: repulsion

Different charges: attraction

attraction repulsion repulsion

2


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


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

3

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

solution



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


4

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


In order to obtain a direct current we use

powers supplies that include different elements that transforms AC into DC.

AC DC


These are the elements of a Power supply

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

regulator.




regulator.




regulator.

3.3.1 CIRCUITS


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

Solution

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:

VOLTAGEINTENSITY

Resistance

5


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


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


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


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

Lower Resistance

Higher Resistance

3.3.1 Electricity magnitudes Exercise 3

Which one of these lamps will give light?


Exercise 4

Will 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

3.3.2 Ohm’s law. Electric Power


Magnitudes analysis using the Ohm’s Law.

6

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

Solution


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

R

VI =

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


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 =


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

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 (Ω) I (A)

2 14

2 2

20 2

12 6

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 Kw

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.

Extra data: house electrical supply 230V, 50Hz

Solution


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

t

WP=


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=

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

=

7


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.

P =100W P =VI

V = 220V I =P

V=100

220= 0,45A



Power of the lamp.

As is says in the wording it is 100W

P=100W

Calculate the Kwh that the electricity

meter will show after 20 hours.

P=100W=0,1kW

PKWh=0,1x20h=2Kwh

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

is 0,90€/Kwh

Cost = Energy consumed × Price

Energy

Cost = Kwh ×€

Kwh

Cost = 2Kwh ×0, 9€

Kwh Cost = 1, 8€


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

hPmedPowerConsu •=

3.3.3 Elements and Symbols


Electric components and their applications.

3.3.3 Elements and symbols

An electric circuit is a group of elements that allows us to control electric current through some sort of material


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

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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 symbolsSafity

It 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.


Electric symbols are used to represent

electric circuits with drawings that replace the real circuit elements.


Exercise: Draw the following circuit using electric symbols



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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

Fuse

Receptors:

Lamp

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 :

Series

Parallel

Mixed

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 A

2 & 3 are connected only by B

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.

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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

CUT

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

cut

3.3.4 Circuit associations3.3.4 Circuit associations

Exercises

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

3.3.4 Circuit associationsExercises

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

3.3.4 Circuit associationsExercises

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


Calculate in each case the resistance equivalent Rt

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SERIES circuit

In this association we find that:

VT=V1+ V2+V3 RT=R1+R2+R3IT=I1= I2=I3 ET=E1+ E2+E3

RTET

IT

VT

ET=VT


RTET

IT

VTI1

I2 I3

V1

V2 V3


SERIES circuit

INTENSITY= EQUAL

IT=I1=I2=I3

RTET

IT

VT

ET=VT

I1

I2 I3IT IT


SERIES circuit

VOLTAGE= DIVED

VT=V1+ V2+V3

RTET

IT

VT

ET=VTIT IT

V1

V2 V3

VT

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



3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It

and Vt. Calculate the I and V in each Resistance

Ω=

++=

++=

8

5

63,6

2

1

321

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=

Ω=

=

=

5,27

8

220

T

T

T

TT

I

I

R

VI



Series⇒I t= I1 = I2 = I3 = 27,5A

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=



Series⇒V t=V1 +V2 +V3 = 220V

V1 = I1R1 = 0,5 • 27,5 =13,75V

V2 = I2R2 = 6,3• 27,5 =173,25V

V3 = I3R3 =1,2 • 27,5 = 33V

Series⇒V t=13,75 +173,25 + 33 = 220V

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

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VT=V1= V2=V3

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

321

1111

RRRRT++=


VT=V123=V1= V2=V3

IT=I123=I1+ I2+I3


INTENSITY= DIVEDIT=I1+I2+I3

I3

I2

I1

IT


INTENSITY= DIVEDIT=I1+I2+I3

I3

I2

I1

IT


VOLTAGE: IQUALVT=V1= V2=V3

V1

V2

V3VT


V1

V2

V3VT

RT= R1= 6 Ω R2= 18 Ω R3= 4 Ω

IT= I1= I2= I3=

VT= 20V V1= V2= V3=

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=

Parallel⇒1

RT=

1

R1+

1

R2

+1

R3

1

RT=1

6+1

18+1

4=

6

36+

2

36+

9

36

1

RT=17

36⇒ RT =

36

17= 2,11Ω


V1

V2

V3VT

RT=2,11Ω R1= 6 Ω R2= 18 Ω R3= 4 Ω

IT= I1= I2= I3=

VT= 20V V1= V2= V3=

Parallel⇒1

RT=

1

R1+

1

R2

+1

R3

1

RT=1

6+1

18+1

4=

6

36+

2

36+

9

36

1

RT=17

36⇒ RT =

36

17= 2,11Ω


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=

IT =VT

RT

IT =20

2,11

IT = 9,48A

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3.3.4 Circuit associationsVOLTAGE: 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


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

I1 =V1R1

=20

6= 3,33A

I2 =V2

R2

=20

18=1,11A

I3 =V3

R3

=20

4= 5A


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


IT=I123=I4 series association

I123=I1+ I2+I3 parallel association


IT=I123=I4 series association


I1

I2

I1


VT=V123+V4 series associationV123=V1= V2=V3 parallel association

V1

VT=V123+V4V2

V3


VT=V123+V4 series associationIT=I123=I4 series associationV123=V1= V2=V3 parallel association



VT=V1234 IT=I1234

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

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R T = R1 − 2 + R 3 R T = R 1 − 2 + R 3

1

R 1 − 2

=1

R 1

+1

R 2

R T =9

2+

3

2

1

R1 − 2

=1

6+

1

18=

3

18+

1

18R T =

12

2

1

R1 − 2

=2

9R T = 6Ω

R1 − 2 =9

2Ω

RT= 6 Ω

IT=

VT= 20V

IT =VT

RT=20

6

IT =10

3A

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=

Mixed

IT = I1 + I2 = I1−2 = I3

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

Mixed

10 /3A = I1 + I2 = I1−2 =10/3

Mixed

VT =V1−2 +V3

V1−2 =V1 =V2

I3 = IT =10

3A ⇒ V3 = I3R3 =

10

3•3

2= 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= V2= V3= 5V

One way

Mixed :

VT = V1−2 +V3

V1−2 = VT1 −V3

V1−2 = 20 − 5 =15V

Other way

V1−2 = I1−2R1−2 =10

3•9

2=15V

V1−2 = V1 = V2 =15V

V1 =15V

V2 =15V

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

AIR

VI

AIR

VI

6

5

6

5

18

15

5,22

5

6

15

1

2

22

1

1

11

=⇒===

=⇒===


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


Another example


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.

15









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

16


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.

ΩΩΩΩ

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

4K7ΩΩΩΩ

• 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

V

Voltage in a component

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

A

CONTROL DE CONSUMO


4K7ΩΩΩΩ

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 A

The voltage between the lamp is

2,3 V

The resistance of the lamp is 2,3 Ω

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


17

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.


Back

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

Exercise 3.2.2b :

2.-Indicate what type of current do we use in this electrical

devices: Torch, Bedroom lamp, TV, Mobil phone, Fridge.


DC is found in batteries 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 :

2.-Indicate what type of current do we use in this electrical devices: Torch, Bedroom lamp, TV, Mobil phone, Fridge.


AC DC

TV TORCH

Fridge Phone

Bedroom

lamp

AC DC

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

Back

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 magnitudesExercise 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 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 =

18

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

V (V) R (Ω) 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 ===

V (V) R (Ω) I (A)

2 4

10 5

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

8VV

24V

IRV

=

⋅=

=

Ω=

=

=

2R

5

10R

I

VR

V (V) R (Ω) 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

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;

I= ? ; P=?

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

I =V

R⇒ R =

V

I=230

0,52= 442,3Ω

R = 442,3Ω

3.3.2 c Solution Ohm’s law. Electric PowerWe have a fan connected to the domestic electric supply. The

amperimeter indicate a 0,52A on the circuit. Extra data: house electrical supply 230V, 50Hz

1º Text analysis:

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

I= ? ; P=?

2º- What’s the power measured in Kw

P =VI

P = 230 • 0,52 = 0,12kw

P = 0,12kw

or you could aslo do

P = V⋅V

R=V 2

R=

2302

442,3= 0,12kw

3.3.2 c Solution Ohm’s law. Electric PowerCalculate the Intensity that flaws in a electrical engine of 1,2 Kw installed

in my washing machine.

Extra data: house electrical supply 230V, 50Hz

Text Analysis:

R= ?; V=230V; I= ?; P=1,2kW

P =VI

P =VI⇒ I =P

V=1200

230= 5,22A

I = 5,22A

3.3.2 c Solution Ohm’s law. Electric PowerCalculate the Intensity that flaws in a electrical engine of 1,2 Kw installed

in my washing machine. What would be the Intensity and the power if we have 380V at home.

Text Analysis:

R= ?; V=230V; I= ?; P=1,2kW

Back

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

19

3.3.4 Circuit associationsSolutions

Mixed:

Series: 1 and 2 are joined to A

Series: 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


8

5

63,6

2

1

321

=

++=

++=

t

t

t

R

R

RRRR

Ω=

=

=

+=+=

+=

5,3

2

7

7

21

28

1

28

7

28

1

4

11

111

21

T

T

T

T

T

R

R

R

R

RRR


Ω=

Ω==

=+=+=

+=+=

+=+=

−

−

−

−

−−

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


Ω=Ω=

=+=

+=+=

+=+=

−

−

−

−−

13

20025

200

1311213

40

1

25

11

111111

21

21

2121

321321

T

t

t

tt

RR

RR

RRRR

RRRRRR



Ω=

=

++=

++=

Ω=

+=

+=

−−

−−

−−

−−

−−

−

−

−

−

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

Ω=

=

+=

+=

−

−

−

−

−

1000

1000

11

2000

1

2000

11

111

76

76

76

76 76

76

R

R

R

RRR

R


Ω=

+=

+=−−−

3000

15001500

5_1

5_1

543215_1

5_1

R

R

RRR

R

Ω=

+=

+=−

3000

20001000

8_6

8_6

8768_6

8_6

R

R

RRR

R

Ω=

=

+=

+=

1500

1500

11

3000

1

3000

11

111

8_65_1

T

T

T

T

T

R

R

R

RRR

R


20


Ω=

=

+=

+=

−

−

−

−

−

20

20

11

100

1

25

11

111

43

43

43

4343

43

R

R

R

RRR

R

Ω=

+=

+=

−−

−−

−−−

−−

120

10020

432

432

432432

432

R

R

RRR

R

3.3.4 Circuit associations. Calculations 3.3.4 Circuit associations. Calculations

Ω=

=

+=

+=

−−

60

60

11

120

1

120

11

111

5_2

5_2

5_2

54325_2

5_2

R

R

R

RRR

R

Ω=

+=

+=

160

10060

5_1

5_1

55_25_1

5_1

R

R

RRR

R


Ω=

=

+=

+=

−

32

32

11

40

1

160

11

111

651

T

T

T

T

T

R

R

R

RRR

R



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

in each Resistance

Ω=

+=

+=

80

5525

21

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

75,2

80

220

=

=

=

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

in each Resistance

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


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

I and V in each Resistance

Ω=

=+=+=

=

+=+=

+=

−

−

−

−

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

21

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

I and V in each Resistance

1,3AI

13

2001

20

13

200

20

R

VI

T

T

TT

=

===

32121T

2121

321T

VVVVV

III

III

circuitMixed

=+==

==

+=

−

−

−

RT= 200/13 Ω

IT= 1,3 A

VT= 20V

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

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


AIIIInassociatioSeries

AR

VI

VEEEV

RR

RRRR

T

T

TT

TT

TT

T

1

115

15

15205

151023

321

21

321

====⇒

===

−=−=+==

Ω=⇒++=

++=

RT= R1= 3 Ω R2= 2 Ω R3= 10 Ω

IT= 1A I1=1A I2= 1A I3= 1A

VT= 15V V1= V2= V3=

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

=⋅==

=⋅==

=⋅==


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=

Ω=

=

=+=

++=+=

++=++=

−

−

−

−

−−

2

6

3

R

1

20ΩR3

1

6

1

R

1

1127RR

1

R

1

R

1

RRRRRRRR

32

32

T

32

T

3232

4321T4321T

R

22

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

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

V1 = I1R1 =1,5⋅ 7

V1 =10,5V

V4 = I4R4 =1,5⋅ 11

V1 =16,5V

In series⇒ VT =V1 +V2−3 +V4

30 =10,5 +V2−3 +16,5

V2−3 =V2 =V3 = 3V

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

=

==

=

==

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

==

==

==

==


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

=

++=

++=−

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

=

⋅==

=

⋅==

=

⋅==

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

==

+===−

23

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

=

⋅==

=

⋅==

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

==

==

==

==

==


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

=

=+=

+=

=

=+=

+=

+++=

−

−

−

−−

Ω=

+++=

+++=−−

14R

8321R

RRRRR

T

T

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

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

=

=

==

=

=

==


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

=

==

=

==


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

==

==

==

==

==

==

==


24

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=

Ω04,18R

53

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

=

=

===

=

==

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=

V72,11V

53

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

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

11,72VV

53

1264,93V

RIV

5435-4-3

5-4-3

5-4-3

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=

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

=

==

=

==

=

==

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

==

==

==

==

==

==

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=

Ω04,18R

53

12667,312R

Ω53

126R

126

53

6

1

9

1

7

1

R

1

R

1

R

1

R

1

R

1

RRRRR

T

T

32

5-4-3

5435-4-3

4321T

=

++=

=

=++=

++=

+++=

−

3 eso electricity unit light

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