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1

Chapter Four

Resistors

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Objectives

Notion of Resistor and Resistance

2. The characteristic curve of a resistor

Ohm’s Law

Types of Resistors

Joule’s Law

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1. Notion of Resistor and Resistance

Resistor is an electrical component to create resistance in the flow of electric current.

Resistors are used for delimit electric current, heat generation,...

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1. Notion of Resistor and Resistance

The representation of a resistor is:

The SI unit of the resistance is: Ohm (Ω)

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1. Notion of Resistor and Resistance

The resistance of a resistance is measured by Ohmmeter.

We can measure the resistance with Multi-meter used as Ohmmeter.

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2. The characteristic curve of a resistor

Set up the circuit formed of: an electric source, resistor, and an Ammeter all connected in series. Also connect a voltmeter in parallel with the resistor.

Give different values of current; then read in each case the value of the voltage.

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2. The characteristic curve of a resistor

Circuit Representation

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2. The characteristic curve of a resistor

The graph of potential difference (V) as a function of the Current (I) is a straight line passing through the origin.

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2. The characteristic curve of a resistor

The equation of this line in the form of VAB =kI with k > 0.

The slope k is a characteristic of the load; we call it electric resistance (R or r).

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2. The characteristic curve of a resistor

Example:

A student designed an experiment in order to measure the current through a resistor when manipulating the potential difference. The following data was recorded:

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2. The characteristic curve of a resistor

1. Draw the set-up of the electric circuit.
2. Draw a graph of U as a function of I.
3. Determine the equation of the St. Line.
4. Deduce the value of the resistance of the resistor.

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2. The characteristic curve of a resistor

Solution of the Example:

1. See the figure.

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2. The characteristic curve of a resistor

Solution of the Example:

2. See the graph.

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2. The characteristic curve of a resistor

Solution of the Example:

3. The graph is a St. Line passing through origin; and its equation is U= kI; where k is the slope.

Slope = k= (1.2V-0.9V)-(150×10^-3 -100× 10^-3)→ k=1.8Ω.

The equation of the St. Line is: U=1.8I.

4. The slope k = the resistance R =1.8Ω.

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3. Ohm’s law

Statement of Ohm’s Law:

The potential difference (VAB) between the terminals of a resistor is proportional to the current (I) traverses it.

 V_AB = V_A – V_B = RI

or

V_BA = V_B – V_A = - RI

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3. Ohm’s law

Example: applying ohm’s Law

Calculate the voltage across the terminals of a resistor of 20kΩ resistance when a current of 20mA traverses it.

Solution of Example:

Ohm’s Law:  V = RI → V = (20×10³)× (20× 10^-3 )

→ V = 400V.

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4. Types of Resistors

1. The coiled metallic wires:

They are used in laboratories (rheostats). They are used in domestic appliances (water-heaters, electric radiators, stoves,...).

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4. Types of Resistors

2. Color Coding Resistor:

Are used in electronic circuits (radios, televisions, computers, ...).

They are made of a thin layer of carbon powder covering an insulating rod.

The resistance of carbon resistors is indicated by rings colored according to international code.

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4. Types of Resistors

Fig. How to read color stripes on carbon resistors for R in ohms.

• Resistor Color Code

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4. Types of Resistors

Example: calculate the resistance of a resistor using the given table.

To calculate the color code of the above 4 band resistor:

The resistance R =

8

2

×10³

5%

R = 82KΩ 5%

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4. Types of Resistors

3. Resistance of wire resistor:

The resistance of a homogeneous and cylindrical wire resistor is given by:

Where ρ is a constant called resistivity depends on the nature of the substance. The constant ρ expressed in Ω.m

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4. Types of Resistors

Example:

calculate the resistance of a cylindrical wire of diameter 0.32mm and length 5m.

The resistivity of the wire is 5×10^-7Ωm.

Solution of the Example:

R= ρ L/A → R = ρ L/ ∏r² → R= 5×10^-7Ωm × (5m/∏ 0.16× 10^-3).

→ R = 49761.1×10^-7Ω

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5. Joule’s Law

The electric energy W consumed by a load when traversed by a current I, under a potential difference V during a time t is given by:

W = VIt

In a resistor this energy is completely transomed into heat: W = VIt

And V = RI; then W = RI (It)

W = RI²t

Joule’s Law

The SI unit of W is joules (J)

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1. Joule’s Law

The electric power P consumed by the resistor is:

P = W/t → P= VIt/t

P= VI

The SI unit of power is watt (w).

→ P= RI²

But V =RI and P = VI → P= (RI )I →P = RI²

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6. Grouping of Resistors

1. Grouping in series:

The equivalent resistance of many resistors placed in series is equal to the sum of their resistances.

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6. Grouping of Resistors

I = I₁ = I₂ = I₃ (resistors in series)

And V = V₁ + V₂ + V₃

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But V = IR (Ohm’s Law)

I R_total = I₁R₁ + I₂R₂ + I₃R₃

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I R_total = IR₁ + IR₂ + IR₃

R_total = R₁ + R₂ + R₃ + … R_n

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6. Grouping of Resistors

Example:

The resistors R₁, R₂ and R₃ are all connected together in series between points A and B with a common current, I flowing through them. Calculate the equivalent resistance.

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6. Grouping of Resistors

Solution of the Example:

The three resistors are connected in series then:

R_eq = R₁ + R₂ + R₃

R_eq = 1kΩ +2kΩ + 6kΩ

→ R_eq = 9kΩ

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6. Grouping of Resistors

V = V₁ = V₂ = V₃

2. Grouping in parallel:

And I = I₁ + I₂ + I₃ (resistors in Parallel)

But V = IR (Ohm’s Law); then: I = V/R

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6. Grouping of Resistors

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6. Grouping of Resistors

Example:

Find the equivalent resistance, R_eq of the following resistors connected in a parallel network.

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6. Grouping of Resistors

Solution of the Example:

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Grade 10 Solving Problems 2019 – 2020

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Exercise 1: Experimental study of a resistor

to determine the resistance R of a resistor, we connect it in series with a rheostat across the terminals of a battery. A voltmeter and an ammeter are connected in the circuit to measure the voltage U across the terminals of R and the intensity of the current I traversing it. The values of U and I are summarized in the table below:

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Grade 10 Solving Problems 2019 – 2020

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1. Draw a figure of the circuit.
2. Represent the graph of U as function of I using the scale: x – Axis: 1cm → 0.01A

y – Axis: 1cm → 2V

3. Determine the equation of the characteristic curve (U-I). Which law is thus verified?
4. Deduce the value of the resistance R.

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Grade 10 Solving Problems 2019 – 2020

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Solution of Exercise 1:

1. See the figure.

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Grade 10 Solving Problems 2019 – 2020

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2. See the figure.

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Grade 10 Solving Problems 2019 – 2020

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3. Referring to part (2); we have a St. line passing through the origin of equation U= kI, where k is the slope. k=(9-6)/(0.09-0.06) → slope =k=100V/A.

Then the equation of the st. Line is: U=100I.

The equation of the st. Line verifies Ohm’s law.

4. R= slope of the st. Line; then R=100Ω.

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Grade 10 Solving Problems 2019 – 2020

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Exercise 2:

A color resistor R is coded as shown below.

Calculate the value of the resistance of R using the given table.

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Grade 10 Solving Problems 2019 – 2020

The resistance is:

R =

4

7

×10¹

5%

R = 470Ω 5%

Solution of Exercise 2:

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Grade 10 Solving Problems 2019 – 2020

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The circuit diagram shows the combination of four resistors that are connected to a 24V battery.

Given that: R₁=R₃=R₄=20Ω, and R₂=10Ω.

Exercise 3:

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Grade 10 Solving Problems 2019 – 2020

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1. Show that the resistance of the equivalent resistor to these four resistors is Req=10Ω.
2. Calculate the intensity of the current I.
3. Calculate the current traversing each resistor.
4. Calculate V_AC, V_AB and V_BC.
5. Calculate the power and energy dissipated by this circuit during one minute.

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Grade 10 Solving Problems 2019 – 2020

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Solution of Exercise 3:

1. R3 and R4 are in parallel then:

1/R3,4 = 1/R3 +1/R4 → 1/R3,4 = 1/20 + 1/20

→ 1/R3,4 = (1+1)/20 → R3,4 = 10Ω.

R2 & R3,4 in series then: R2,3,4 = 10Ω + 10Ω → R2,3,4 = 20Ω.

R2,3,4 and R1 in parallel then:

1/Req = 1/R2,3,4 + 1/R1 →1/Req = 1/20 +1/20

→1/Req = (1+1)/20 →1/Req = 2/20 → Req = 10Ω.

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Grade 10 Solving Problems 2019 – 2020

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Solution of Exercise 3:

2. Using Ohm’s law V= RI → 24V = (10 Ω)(I) → I = 24/10 → I = 2.4A.

3. VG = V2,3,4 = V1 =24V (in parallel)

V2,3,4 = R2,3,4 I1 → I1 = V2,3,4 /R2,3,4 → I1 = 24V/20Ω

→ I1 = 1.2A .

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V1 = R1 I2 → I2 = V1 / R1 → I2 = 24V/20 Ω

→ I2 = 1.2A.

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Grade 10 Solving Problems 2019 – 2020

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3. VAB= R2 I1→ VAB = 10Ω × 1.2A → VAB = 12V.

VAC = VG = 24V (parallel)

VBC = VAC – VAB → VBC =24V – 12V → VBC = 12V.

VBC =VR3 = VR4 = 12V

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VR3 = R3I3 → I3 = VR3/R3 → I3 = 12V/20 Ω

→ I3 = 0.6A

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VR4 = R4I4 → I4 = VR4/R4 → I3 = 12V/20 Ω

→ I4 = 0.6A

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Grade 10 Solving Problems 2019 – 2020

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3. VAB= R2 I1→ VAB = 10Ω × 1.2A → VAB = 12V.

VAC = VG = 24V (parallel)

VBC = VAC – VAB → VBC =24V – 12V → VBC = 12V.

VBC =VR3 = VR4 = 12V

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5. P = VI → P = 24V×2.4A → P = 57.6watt.

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P = w/t → w= P× t → w= 57.6 × (1 × 60) →

w = 3456 J.

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Grade 10 Quiz in Physics Time: 20min

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Note: To solve any combinational resistor circuit.

• we need to take is to identify the simple series and parallel resistor branches and replace them with equivalent resistors.
• This step will allow us to reduce the complexity of the circuit and help us transform a complex combinational resistive circuit into a single equivalent resistance.
• Remembering that series circuits are voltage dividers and parallel circuits are current dividers.

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Grade 10 Quiz in Physics Time: 20min

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Find the equivalent resistance, R_EQ for the following resistor combination circuit. (10 points)

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Grade 10 solution of Quiz Time: 30min

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R₈ & R₁₀ are in series; Then:

R8,10 = R8 + R10 → R8,10 = 10Ω +2 Ω → R8,10 = 12Ω.

R8,10 & R9 are in parallel; let R_A the equivalent resistance of the R_8,10 & R₉ then:

1/RA = 1/R8,10 + 1/R9 → 1/RA = 1/12 + 1/6 → 1/RA =(1+2)/12 → 1/RA = 3/12→ RA =12/3 → RA= 4Ω

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Grade 10 solution of Quiz Time: 30min

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R_A & R₇ are in series; Then:

RA,7 = RA + R7 → RA,7 = 4Ω +8 Ω → RA,7 = 12Ω.

RA,7 & R6 are in parallel; let R_B the equivalent resistance of the R_A,7 & R₆ then:

1/RB = 1/RA,7 + 1/R6 → 1/RB = 1/12 + 1/6 → 1/RB =(1+2)/12 → 1/RB = 3/12→ RB =12/3 → RB= 4Ω

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Grade 10 solution of Quiz Time: 30min

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R_B & R₅ are in series; Then:

RB,5 = RB + R5 → RB,5 = 4Ω +4 Ω → RB,5 = 8Ω.

RB,5 & R4 are in parallel; let R_C the equivalent resistance of the R_B,5 & R₄ then:

1/RC = 1/RB,5 + 1/R4 → 1/RC = 1/8 + 1/8 → 1/RC =(1+1)/8 → 1/RC = 2/8→ RC =8/2 → RC= 4Ω

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Grade 10 solution of Quiz Time: 30min

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R_C & R₃ are in series; Then:

RC,3 = RC + R3 → RC,3 = 4Ω +4 Ω → RC,3 = 8Ω.

RC,3 & R2 are in parallel; let R_D the equivalent resistance of the R_C,3 & R₂ then:

1/RD = 1/RC,3 + 1/R2 → 1/RC = 1/8 + 1/8 → 1/RD =(1+1)/8 → 1/RD = 2/8→ RD =8/2 → RD= 4Ω

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Grade 10 solution of Quiz Time: 30min

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R_D & R₁ are in series; Then:

REQ = RD + R1 → REQ = 4Ω +6 Ω → REQ = 10Ω.

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End of Chapter Four