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UNIT 1
Electric Circuits
AGORA International School
Technology
3 ESO
1. INTRODUCTION
Electricity has an important place in
modern society. It is a controllable and
convenient form of energy for a variety
of uses in homes, schools, hospitals,
industries and so on.
SECTION 1
Basic Concepts
1. Introduction
2. Current
3. Voltage
4. Resistance
5. Ohm’s Law
6. Power and Energy
INTRODUCTION TO ELECTRICITY
▪ What constitutes electricity?
▪ How does it flow in an electric
circuit?
▪ What are the factors that
control or regulate the current
through an electric circuit?
NOTE TAKING
In this section we will talk about the
main magnitudes of electricity.
During the explanations, identify the
following information and complete
it on the table.
SECTION 1
Basic Concepts
1. Introduction
2. Current
3. Voltage
4. Resistance
5. Ohm’s Law
6. Power and Energy
INTRODUCTION TO ELECTRICITY
Definition
Hydraulic
analogy
Symbol and
Units
Current
Voltage
Resistance
2. CURRENT
You are familiar with air current and water current.
SECTION 1
Basic Concepts
1. Introduction
2. Current
3. Voltage
4. Resistance
5. Ohm’s Law
6. Power and Energy
INTRODUCTION TO ELECTRICITY
Similarly, if electrons flow through a
conductor (for example a metallic wire),
we say that there is an electric current
in the conductor.
Units: Electric current is expressed by -the amount of charge flowing
through a particular area in unit time-. The unit is called ampere (A), named
after the French scientist, Andre-Marie Ampere (1775–1836).
In river current, there is a flow of water
molecules.
Symbol: current is represented with an “I” (intensity)
3. VOLTAGE (Potential Difference)
What makes the water molecules (electrons) to
flow in a river (copper wire)?
SECTION 1
Basic Concepts
1. Introduction
2. Current
3. Voltage
4. Resistance
5. Ohm’s Law
6. Power and Energy
INTRODUCTION TO ELECTRICITY
Charges do not flow in a copper wire by
themselves, just as water in a perfectly horizontal
tube does not flow.
For flow of charges in a conducting metallic wire, the
electrons move only if there is a difference of electric
pressure –potential difference – along the conductor.
If one end of the tube is connected to a tank of water
kept at a higher level, such that there is a pressure
difference between the two ends of the tube, water
flows out of the other end of the tube.
Potential
difference
Potential
difference
Units: It is measured in volts, named after the Italian
physicist Alessandro Volta who invented the first
chemical battery.
Symbol: Voltage is represented in equations by the letter "V".
4. RESISTANCE
Resistance, as the name suggests, always tries
to stop the current from flowing. In the case of
the flow of water through pipes, the resistance
would be the frictional effects between water
and the pipe surfaces as well as the resistance
offered by obstacles that are present in its path
(like the constriction of a water pipe or a hose). It
is this resistance that hinders the water flow and
reduces both its flow rate and its drift speed.
SECTION 1
Basic Concepts
1. Introduction
2. Current
3. Voltage
4. Resistance
5. Ohm’s Law
6. Power and Energy
INTRODUCTION TO ELECTRICITY
Units: The SI unit for resistance is ohm, represented by the Greek letter Ω.
Symbol: Resistance is represented in equations by the letter “R".
In electricity, resistance is the property of a conductor to resist the flow of
charges. The circuit with the higher resistance will allow less charge to flow,
meaning the circuit with higher resistance has less current flowing through
it.
4. FACTORS ON WHICH THE RESISTANCE OF A CONDUCTOR DEPENDS
The resistance of the conductor depends on its length, on its cross-sectional
area, and on the nature of its material as follow:
SECTION 1
Basic Concepts
1. Introduction
2. Current
3. Voltage
4. Resistance
5. Ohm’s Law
6. Power and Energy
INTRODUCTION TO ELECTRICITY
Length: The longer the wire, the
more resistance that there will be.
After all, if resistance occurs as the
result of collisions between charge
carriers and the atoms of the wire,
then there is likely to be more
collisions in a longer wire. More
collisions mean more resistance.
𝑅 ∝ 𝑙
Cross-sectional area:
Water will flow through with lower
amount of resistance in a wider pipe.
In the same manner, the wider the
wire, the less resistance that there
will be to the flow of electric charge.
Resistance and area are inversely
proportional.
𝑅 ∝
1
𝐴
Combining both equations we get that:
𝑅 ∝
𝑙
𝐴
𝑜𝑟, 𝑅 ∝ 𝜌
𝑙
𝐴
where ρ (rho) is a constant of proportionality and is called the electrical
resistivity of the material of the conductor.
4. FACTORS ON WHICH THE RESISTANCE OF A CONDUCTOR DEPENDS
The resistance of the conductor depends on its length, on its cross-sectional
area, and on the nature of its material as follow:
SECTION 1
Basic Concepts
1. Introduction
2. Current
3. Voltage
4. Resistance
5. Ohm’s Law
6. Power and Energy
INTRODUCTION TO ELECTRICITY
Resistivity of the material:
The resistivity (Ω m) is a
characteristic property of the
material. The metals and alloys
have very low resistivity in the
range of 10–8 Ω m to 10–6 Ω m.
They are good conductors of
electricity. Insulators like rubber
and glass have resistivity of the
order of 1012 to 1017 Ω m.
Alloys do not oxidise (burn)
readily at high temperatures. For
this reason, they are commonly
used in electrical heating
devices, like electric iron,
toasters etc. Copper and
aluminium are generally used for
electrical transmission lines.
SAMPLE PROBLEM : Determine the resistance
of a 20-mile length of 12-gauge copper wire.
Given:
 1 mi = 1609 meters
 Diameter of 12-gauge wire = 0.2117 cm.
INTRODUCTION TO ELECTRICITY
SECTION 1
Basic Concepts
1. Introduction
2. Current
3. Voltage
4. Resistance
5. Ohm’s Law
6. Power and Energy
STEP 1: Identify data:
 Lenght → 𝑙 = 20 mi ( 1 mile = 1609 meters)
 Diameter → ∅ = 0.2117 cm
 Resistivity → 𝜌 = 1.62 ∙ 10−8
Ω ∙ 𝑚
 Resistance → R?
SAMPLE PROBLEM : Determine the resistance
of a 20-mile length of 12-gauge copper wire.
Given:
 1 mi = 1609 meters
 Diameter of 12-gauge wire = 0.2117 cm.
INTRODUCTION TO ELECTRICITY
SECTION 1
Basic Concepts
1. Introduction
2. Current
3. Voltage
4. Resistance
5. Ohm’s Law
6. Power and Energy
STEP 1: Identify data
 𝑙 = 20 mi
 ∅ = 0.2117 cm
 𝜌 = 1.62 ∙ 10−8
Ω ∙ 𝑚
 R?
STEP 2: Define Steps
1. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 of the wire → 𝒓 =
𝒅
𝟐
2. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 sectional area of the wire →
𝑨 = 𝝅𝒓 𝟐
2. Find resistance → 𝑹 =
𝝆 ∙ 𝒍
𝑨
INTRODUCTION TO ELECTRICITY
SECTION 1
Basic Concepts
1. Introduction
2. Current
3. Voltage
4. Resistance
5. Ohm’s Law
6. Power and Energy
 𝑙 = 20 mi
 ∅ = 0.2117 cm
 𝜌 = 1.62 ∙ 10−8
Ω ∙ 𝑚
 R?
1. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 sectional area of the wire
𝐴 = 𝜋𝑟2
2. Find resistance 𝑅 =
𝜌 ∙ 𝑙
𝐴
STEP 3: Mathematical Procedure
1. Find the radius of the wire
1.1 First we have to pass centimeters to meters:
Factor 1m = 100 cm ⟶ 0.2117 𝑐𝑚 ∙
1𝑚
100 𝑐𝑚
= 0,002117 m = 2.117 ∙ 10−3
𝑚
1.2 With the diameter, we find the radius:
𝑑 = 2𝑟 ⟶ 𝑟 =
𝑑
2
=
2.117 ∙ 10−3 𝑚
2
= 1.089 ∙ 10−3
𝑚
2. We find the area:
𝐴 = 𝜋𝑟2
= 𝜋 ∙ (1.089 ∙ 10−3
𝑚 )2
= 3.520 ∙ 10−6
𝑚2
3. Find resistance
3.1 First we have to pass from miles to meters:
Factor 1 mi = 1609 m ⟶ 20 𝑚𝑖 ∙
1609 𝑚
1 𝑚𝑖
= 32180 m = 3.218 ∙ 104
𝑚
3.2 First we have to pass from miles to meters:
𝑅 =
𝜌 ∙ 𝑙
𝐴
=
1.62 ∙ 10−8 Ω∙𝑚 ∙ (3.218 ∙ 104 𝑚 )
(3.520 ∙ 10−6 𝑚2)
= 148,10 𝜴

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Electric Circuits Explained

  • 1. UNIT 1 Electric Circuits AGORA International School Technology 3 ESO
  • 2. 1. INTRODUCTION Electricity has an important place in modern society. It is a controllable and convenient form of energy for a variety of uses in homes, schools, hospitals, industries and so on. SECTION 1 Basic Concepts 1. Introduction 2. Current 3. Voltage 4. Resistance 5. Ohm’s Law 6. Power and Energy INTRODUCTION TO ELECTRICITY ▪ What constitutes electricity? ▪ How does it flow in an electric circuit? ▪ What are the factors that control or regulate the current through an electric circuit?
  • 3. NOTE TAKING In this section we will talk about the main magnitudes of electricity. During the explanations, identify the following information and complete it on the table. SECTION 1 Basic Concepts 1. Introduction 2. Current 3. Voltage 4. Resistance 5. Ohm’s Law 6. Power and Energy INTRODUCTION TO ELECTRICITY Definition Hydraulic analogy Symbol and Units Current Voltage Resistance
  • 4. 2. CURRENT You are familiar with air current and water current. SECTION 1 Basic Concepts 1. Introduction 2. Current 3. Voltage 4. Resistance 5. Ohm’s Law 6. Power and Energy INTRODUCTION TO ELECTRICITY Similarly, if electrons flow through a conductor (for example a metallic wire), we say that there is an electric current in the conductor. Units: Electric current is expressed by -the amount of charge flowing through a particular area in unit time-. The unit is called ampere (A), named after the French scientist, Andre-Marie Ampere (1775–1836). In river current, there is a flow of water molecules. Symbol: current is represented with an “I” (intensity)
  • 5. 3. VOLTAGE (Potential Difference) What makes the water molecules (electrons) to flow in a river (copper wire)? SECTION 1 Basic Concepts 1. Introduction 2. Current 3. Voltage 4. Resistance 5. Ohm’s Law 6. Power and Energy INTRODUCTION TO ELECTRICITY Charges do not flow in a copper wire by themselves, just as water in a perfectly horizontal tube does not flow. For flow of charges in a conducting metallic wire, the electrons move only if there is a difference of electric pressure –potential difference – along the conductor. If one end of the tube is connected to a tank of water kept at a higher level, such that there is a pressure difference between the two ends of the tube, water flows out of the other end of the tube. Potential difference Potential difference Units: It is measured in volts, named after the Italian physicist Alessandro Volta who invented the first chemical battery. Symbol: Voltage is represented in equations by the letter "V".
  • 6. 4. RESISTANCE Resistance, as the name suggests, always tries to stop the current from flowing. In the case of the flow of water through pipes, the resistance would be the frictional effects between water and the pipe surfaces as well as the resistance offered by obstacles that are present in its path (like the constriction of a water pipe or a hose). It is this resistance that hinders the water flow and reduces both its flow rate and its drift speed. SECTION 1 Basic Concepts 1. Introduction 2. Current 3. Voltage 4. Resistance 5. Ohm’s Law 6. Power and Energy INTRODUCTION TO ELECTRICITY Units: The SI unit for resistance is ohm, represented by the Greek letter Ω. Symbol: Resistance is represented in equations by the letter “R". In electricity, resistance is the property of a conductor to resist the flow of charges. The circuit with the higher resistance will allow less charge to flow, meaning the circuit with higher resistance has less current flowing through it.
  • 7. 4. FACTORS ON WHICH THE RESISTANCE OF A CONDUCTOR DEPENDS The resistance of the conductor depends on its length, on its cross-sectional area, and on the nature of its material as follow: SECTION 1 Basic Concepts 1. Introduction 2. Current 3. Voltage 4. Resistance 5. Ohm’s Law 6. Power and Energy INTRODUCTION TO ELECTRICITY Length: The longer the wire, the more resistance that there will be. After all, if resistance occurs as the result of collisions between charge carriers and the atoms of the wire, then there is likely to be more collisions in a longer wire. More collisions mean more resistance. 𝑅 ∝ 𝑙 Cross-sectional area: Water will flow through with lower amount of resistance in a wider pipe. In the same manner, the wider the wire, the less resistance that there will be to the flow of electric charge. Resistance and area are inversely proportional. 𝑅 ∝ 1 𝐴 Combining both equations we get that: 𝑅 ∝ 𝑙 𝐴 𝑜𝑟, 𝑅 ∝ 𝜌 𝑙 𝐴 where ρ (rho) is a constant of proportionality and is called the electrical resistivity of the material of the conductor.
  • 8. 4. FACTORS ON WHICH THE RESISTANCE OF A CONDUCTOR DEPENDS The resistance of the conductor depends on its length, on its cross-sectional area, and on the nature of its material as follow: SECTION 1 Basic Concepts 1. Introduction 2. Current 3. Voltage 4. Resistance 5. Ohm’s Law 6. Power and Energy INTRODUCTION TO ELECTRICITY Resistivity of the material: The resistivity (Ω m) is a characteristic property of the material. The metals and alloys have very low resistivity in the range of 10–8 Ω m to 10–6 Ω m. They are good conductors of electricity. Insulators like rubber and glass have resistivity of the order of 1012 to 1017 Ω m. Alloys do not oxidise (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc. Copper and aluminium are generally used for electrical transmission lines.
  • 9. SAMPLE PROBLEM : Determine the resistance of a 20-mile length of 12-gauge copper wire. Given:  1 mi = 1609 meters  Diameter of 12-gauge wire = 0.2117 cm. INTRODUCTION TO ELECTRICITY SECTION 1 Basic Concepts 1. Introduction 2. Current 3. Voltage 4. Resistance 5. Ohm’s Law 6. Power and Energy STEP 1: Identify data:  Lenght → 𝑙 = 20 mi ( 1 mile = 1609 meters)  Diameter → ∅ = 0.2117 cm  Resistivity → 𝜌 = 1.62 ∙ 10−8 Ω ∙ 𝑚  Resistance → R?
  • 10. SAMPLE PROBLEM : Determine the resistance of a 20-mile length of 12-gauge copper wire. Given:  1 mi = 1609 meters  Diameter of 12-gauge wire = 0.2117 cm. INTRODUCTION TO ELECTRICITY SECTION 1 Basic Concepts 1. Introduction 2. Current 3. Voltage 4. Resistance 5. Ohm’s Law 6. Power and Energy STEP 1: Identify data  𝑙 = 20 mi  ∅ = 0.2117 cm  𝜌 = 1.62 ∙ 10−8 Ω ∙ 𝑚  R? STEP 2: Define Steps 1. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 of the wire → 𝒓 = 𝒅 𝟐 2. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 sectional area of the wire → 𝑨 = 𝝅𝒓 𝟐 2. Find resistance → 𝑹 = 𝝆 ∙ 𝒍 𝑨
  • 11. INTRODUCTION TO ELECTRICITY SECTION 1 Basic Concepts 1. Introduction 2. Current 3. Voltage 4. Resistance 5. Ohm’s Law 6. Power and Energy  𝑙 = 20 mi  ∅ = 0.2117 cm  𝜌 = 1.62 ∙ 10−8 Ω ∙ 𝑚  R? 1. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 sectional area of the wire 𝐴 = 𝜋𝑟2 2. Find resistance 𝑅 = 𝜌 ∙ 𝑙 𝐴 STEP 3: Mathematical Procedure 1. Find the radius of the wire 1.1 First we have to pass centimeters to meters: Factor 1m = 100 cm ⟶ 0.2117 𝑐𝑚 ∙ 1𝑚 100 𝑐𝑚 = 0,002117 m = 2.117 ∙ 10−3 𝑚 1.2 With the diameter, we find the radius: 𝑑 = 2𝑟 ⟶ 𝑟 = 𝑑 2 = 2.117 ∙ 10−3 𝑚 2 = 1.089 ∙ 10−3 𝑚 2. We find the area: 𝐴 = 𝜋𝑟2 = 𝜋 ∙ (1.089 ∙ 10−3 𝑚 )2 = 3.520 ∙ 10−6 𝑚2 3. Find resistance 3.1 First we have to pass from miles to meters: Factor 1 mi = 1609 m ⟶ 20 𝑚𝑖 ∙ 1609 𝑚 1 𝑚𝑖 = 32180 m = 3.218 ∙ 104 𝑚 3.2 First we have to pass from miles to meters: 𝑅 = 𝜌 ∙ 𝑙 𝐴 = 1.62 ∙ 10−8 Ω∙𝑚 ∙ (3.218 ∙ 104 𝑚 ) (3.520 ∙ 10−6 𝑚2) = 148,10 𝜴