1
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
3.2.1 Particles and laws
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
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3.2.1 Particles and laws
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
3.2.1 Particles and laws
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.
3.2.1 Particles and laws
3.2.1 Particles and laws
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.
3.2.2 Electricity generation
Solution
3.2.2 Electricity generationLong time ago… Hans Christian Oersted discovered that a electric current can disturb a compass
3.2.2 Electricity generation
Since the compass only is disturbed with a magnet, Volta concluded that an electric current creates a artificial magnetic
Natural magnet
Artificialmagnet
3.2.2 Electricity generation
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
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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
3.2.2 Electricity generation
In conclusion, if we want to create an
artificial electric current, we need:
Mechanical energy
to move the circuit or the
magnet
3.2.2 Electricity generation
Closed circuit
Artificial magnet
This is the diagram of an industrial electric
generator
3.2.2 Electricity generation
Artificial magnet
Closed circuit
The electricity is created when the
mechanical energy moves the closed circuit inside the artificial magnet
3.2.2 Electricity generation
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.
3.2.2 Electricity generation
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
3.2.2 Electricity generation
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
3.2.2 Electricity generation
But, most electric at devices like an iPhone,
need a Direct Current that keeps its magnitude and direction constant
3.2.2 Electricity generation
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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:
3.2.2 Electricity generation
At home, the pulse of
the AC supply is so fast that the light
pulse cannot be detected by the
human eyes
3.2.2 Electricity generation
In order to obtain a direct current we use
powers supplies that include different elements that transforms AC into DC.
AC DC
3.2.2 Electricity generation
These are the elements of a Power supply
that that transforms AC into DC: transformer, rectifier, smoothing and
regulator.
3.2.2 Electricity generation
These are the elements of a Power supply
that that transforms AC into DC: transformer, rectifier, smoothing and
regulator.
3.2.2 Electricity generation
These 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
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
3.3.1 Electricity magnitudes
In order to understand electricity, we have to first know the main electric magnitudes:
VOLTAGEINTENSITY
Resistance
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3.3.1 Electricity magnitudes
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
3.3.1 Electricity magnitudes
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
3.3.1 Electricity magnitudes
The higher the Voltage is, the more energy the electric charges will have to keep on moving.
Lower voltage
Higher voltage
3.3.1 Electricity magnitudes
Intensity is the amount of charges that goes through a conductor per time unit. It is measured in Ampere (A)
Lower intensity
Higher intensity
3.3.1 Electricity magnitudes
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?
3.3.1 Electricity magnitudes
Exercise 4
Will this lamp turn on?
3.3.1 Electricity magnitudes
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
What are we going to learn next?
Magnitudes analysis using the Ohm’s Law.
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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
3.3.2 Ohm’s law. Electric Power
Ohm’s law links the three electric magnitudes as is shown:
R
VI =
V= Voltage (voltsV)I= Intensity (ampereA)R= Resistance (ohm ΩΩΩΩ)
3.3.2 Ohm’s law. Electric Power
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 =
3.3.2 Ohm’s law. Electric Power
Intensity is inversely proportional to Resistance
If there is a high Resistance, the intensity will be low.
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
3.3.2 Ohm’s law. Electric Power
Power is defined as the work consumed per time unit, and it’s measured in watt W
t
WP=
3.3.2 Ohm’s law. Electric Power
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
=
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3.3.2 Ohm’s law. Electric Power
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
3.3.2 Ohm’s law. Electric Power
There is a bulb lamp of 100w connected to 220V electric grid. Calculate:
Intensity throw the lamp.
P =100W P =VI
V = 220V I =P
V=100
220= 0,45A
3.3.2 Ohm’s law. Electric Power
There is a bulb lamp of 100w connected to 220V electric grid. Calculate:
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€
3.3.2 Ohm’s law. Electric Power
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
What are we going to learn next?
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
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
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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
3.3.3 Elements and symbols
Conductor: all the elements have to be connected to a material that transmits the electric charges.
The circuit has to be CLOSED in order to allow the electricity to circulate around it from the positive to the negative pole.
3.3.3 Elements and symbols
Protection elements: they keep all the circuit elements safe from high voltage rises that can destroy the receptors.Fuse: the first one will blow, cutting the circuit in case of a voltage rise. They are easily replaced
Circuit breaker: they are used in new electrical installations, at home or at factories. If there is a voltage rise, you don’t have to replace them, only reload them.
3.3.3 Elements and symbols
Electric symbols are used to represent
electric circuits with drawings that replace the real circuit elements.
3.3.3 Elements and symbols
Exercise: Draw the following circuit using electric symbols
3.3.3 Elements and symbols
Exercise: Draw the following circuit using electric symbols
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3.3.3 Elements and symbols
Exercise: Draw the following circuit using electric symbols
3.3.3 Elements and symbols
Generator
Battery
Battery association
Conductors:
When two conductors are crossed without any contact we indicate it with a
curve
3.3.3 Elements and symbols
Control elements
Push button
Switch
Diverter switch
3.3.3 Elements and symbols
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
3.3.4 Circuit associations
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
3.3.4 Circuit associations. Calculations
Calculate in each case the resistance equivalent Rt
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3.3.4 Circuit associations. Calculations
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
3.3.4 Circuit associations. Calculations
RTET
IT
VTI1
I2 I3
V1
V2 V3
3.3.4 Circuit associations. Calculations
SERIES circuit
INTENSITY= EQUAL
IT=I1=I2=I3
RTET
IT
VT
ET=VT
I1
I2 I3IT IT
3.3.4 Circuit associations. Calculations
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. 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
3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It
and Vt. Calculate the I and V in each Resistance
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=
3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It
and Vt. Calculate the I and V in each Resistance
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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3.3.4 Circuit associationsPARALLEL association
VT=V1= V2=V3
IT=I1+ I2+I3 ET=E1+ E2+E3
321
1111
RRRRT++=
3.3.4 Circuit associations
VT=V123=V1= V2=V3
IT=I123=I1+ I2+I3
3.3.4 Circuit associationsPARALLEL association
INTENSITY= DIVEDIT=I1+I2+I3
I3
I2
I1
IT
3.3.4 Circuit associationsPARALLEL association
INTENSITY= DIVEDIT=I1+I2+I3
I3
I2
I1
IT
3.3.4 Circuit associationsPARALLEL association
VOLTAGE: IQUALVT=V1= V2=V3
V1
V2
V3VT
3.3.4 Circuit associationsPARALLEL association
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
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
3.3.4 Circuit associationsPARALLEL EXERCISE
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
3.3.4 Circuit associationsPARALLEL EXERCISE
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
3.3.4 Circuit associationsMIXED association
We have to convert the parallel associations into series in order to obtain Vt, It and Rt. T
3.3.4 Circuit associationsMIXED association
IT=I123=I4 series association
I123=I1+ I2+I3 parallel association
3.3.4 Circuit associationsMIXED association
IT=I123=I4 series association
I123=I1+ I2+I3 parallel association
I1
I2
I1
3.3.4 Circuit associations
VT=V123+V4 series associationV123=V1= V2=V3 parallel association
V1
VT=V123+V4V2
V3
3.3.4 Circuit associations
VT=V123+V4 series associationIT=I123=I4 series associationV123=V1= V2=V3 parallel association
I123=I1+ I2+I3 parallel association
3.3.4 Circuit associations
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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3.3.4 Circuit associations. Calculations
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
=⇒===
=⇒===
3.3.4 Circuit associations. Calculations
E represents the electrometric force that is the Total Voltage that provides a electric source like a battery or a power supply,
3.3.4 Circuit associations. Calculations
Another example
3.3.4 Circuit associations
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.
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3.3.4 Circuit associations. CalculationsCalculate 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
3.3.4 Circuit associations. Calculations
Calculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance
3.3.4 Circuit associations. Calculations
Calculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance
3.3.4 Circuit associations. Calculations
Calculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance
3.3.4 Circuit associationsExercise:
Which of these lamps have more
lightness?
3.3.4 Circuit associationsExercise:
Which of these assemblies generates electricity?
3.3.4 Circuit associationsExercise:
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
3.3.4 Circuit associations
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
3.3.4 Circuit associations
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
3.3.6 Electric Measurements. Polimeter
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
3.3.6 Electric Measurements. Polimeter
4K7ΩΩΩΩ
Amperimeter: we chose the position according to the scale of the current.
3.3.6 Electric Measurements. Polimeter
3.3.4 Circuit associations
Exercise:
According to this assembly, chose the correct answer:
The I through the lamp is 2,3 A
The voltage between the lamp is
2,3 V
The resistance of the lamp is 2,3 Ω
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
3.3.4 Circuit associations
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.
3.2.2 Electricity generation
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.
3.2.2 Electricity generation
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.
3.2.2 Electricity generation
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.
3.2.2 Electricity generation
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.
3.2.2 Electricity generation
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
3.3.4 Circuit associations. Calculations
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
3.3.4 Circuit associations. Calculations
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
3.3.4 Circuit associations. Calculations
Ω=Ω=
=+=
+=+=
+=+=
−
−
−
−−
13
20025
200
1311213
40
1
25
11
111111
21
21
2121
321321
T
t
t
tt
RR
RR
RRRR
RRRRRR
3.3.4 Circuit associations. Calculations
3.3.4 Circuit associations. Calculations
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
3.3.4 Circuit associations. Calculations
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
3.3.4 Circuit associations. Calculations
20
3.3.4 Circuit associations. Calculations
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
3.3.4 Circuit associations. Calculations
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. 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=
+=
+=
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
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
3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance
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
=⋅==
=⋅==
=⋅==
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=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
==
==
==
==
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=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
==
==
==
==
==
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=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
RT=14 Ω R1= 1 Ω R2= 6 Ω R3= 3 Ω R4= 3 Ω R5= 16 Ω R6= 16 Ω
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
=
=
==
=
=
==
RT=14 Ω R1= 1 Ω R2= 6 Ω R3= 3 Ω R4= 3 Ω R5= 16 Ω R6= 16 Ω
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
=
==
=
==
RT=14 Ω R1= 1 Ω R2= 6 Ω R3= 3 Ω R4= 3 Ω R5= 16 Ω R6= 16 Ω
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
==
==
==
==
==
==
==
3.3.4 Circuit associations. CalculationsCalculate in each case the resistance equivalent Rt, It and Vt. Calculate the I and V in each Resistance
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
=
++=
=
=++=
++=
+++=
−
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