# Electric Resistance. Combinations of Resistors | iCalculator™

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15.2Electric Resistance. Combinations of Resistors

In this Physics tutorial, you will learn:

• The definition of resistance
• The unit of resistance
• The factors affecting the resistance of a conductor
• Methods of combining resistors in a circuit (series & parallel)
• How to calculate the equivalent resistance is a series setup of resistors
• How to calculate the equivalent resistance is a parallel setup of resistors
• How to calculate the equivalent resistance is a complex setup of resistors

## Introduction

Does electricity flow freely through a conductor? Do you think all electrical energy produced by the source converts into the desired form? Why?

What is the main purpose of an electric heater? What kind of energy transformations take place when we turn it on? Is there any undesired form of energy produced during the operation of an electric heater?

Now, answer the above questions considering a light bulb. What conclusions do you draw regarding the undesired form of energy in this case? Is it the same as for electric heater? Why?

In this tutorial, you will learn about a phenomenon known as electrical resistance. We encounter it everywhere, every time we turn on an electrical appliance. Therefore, this tutorial has a great importance on the practical aspect.

## Definition of Resistance. Resistance of a Conducting Wire

When electrons flow through a conductor, they collide with the atoms of conductor. As a result, the electrons slow down. We have found in the previous tutorial that the drift velocity of electrons in a conducting wire is very low due to a large number of collisions they experience during their motion through the conductor, as shown in the figure. The electron in the figure experiences a number of collisions with the atoms of conductor during its movement from position 1 to position 2. As a result, the electron loses some of its initial energy, which converts into other forms (mainly thermal, because the atoms start vibrate faster around their equilibrium positions when hit by the electron). As a result, the conductor gets hotter and the flow of electrons in such conditions becomes more difficult. We say the conductor makes resistance to the flow of electrons.

By definition, the electrical resistance R, is a quantity related to the opposition that electrons encounter during their flow through a conductor.

Resistance is measured in Ohms [Ω], in honour to the German scientist Georg Simon Ohm.

All electrical appliances - some more and some less - make resistance to the electrons flow. Conductors in general provide low resistance while insulators a very high resistance to the flow of charges.

## Factors Affecting the Resistance of Materials

Let's explain this point by considering the analogy between the electrons' flow through a conducting wire and a trip made by car. The road symbolically represents the conducting wire, and the drivers - including their cars - represent the moving electrons. Obviously, the physical state of a driver is highly affected by the trip conditions. If the trip is tedious, this is analogue to the resistance encountered by electrons during their flow through the conductor, i.e. more difficult the trip becomes for the driver, higher the resistance encountered by electrons during their flow through the circuit, based on the above analogy.

Let's consider the following scenarios:

1. If the trip is long, the car has to travel many kilometres. This takes a lot of time to the driver and as a result, he gets tired and loses concentration. There is an increased risk for accidents to occur if the driver doesn't stop to take a rest. This means longer the path, more difficult the trip becomes.
Similarly, an electron flowing through a long wire has more probability to collide with other particles in the conductor than if the wire was shorter. This means the resistance of a conducting wire (Rw) increases with the increase in its length (L), i.e. resistance of wire is proportional to the length of conductor. We write
Rw ~ L
2. If the road is new and well maintained, it is easier for the driver to travel along, but if the road is unpaved and full of holes etc., it becomes more difficult for the driver to travel on it. As a result, the trip takes more time and the driver gets tired faster.
Similarly, the flow of electrons through a good conductor is easier than when they flow through a poorer conductor. This means a good conductor makes less resistance to electrons flow than a poorer conductor. This phenomenon is represented in formulae by a quantity known as resistivity, ρ, which shows how many ohms of resistance does a cube of 1 m side length makes to the flow of electrons through it (i.e. we put all materials in the same conditions to compare their conductive ability). Resistivity of a conductor (measured in Ω ∙ m) is an intrinsic property of material (i.e. it is given in tables, like density etc.) and is proportional to the resistance, i.e.
Rw ~ ρ
3. If the road is wider (when the car in traveling on highways for example), the travel becomes easier than when the road is narrow. This is because the driver does not have to care about other cars; he can shift with his car through any lane he considers as more appropriate to avoid the traffic. Hence, the trip becomes easier when the road is wider.
Similarly, electrons encounter less resistance when they flow through a wider cable than when they flow through a narrow wire. This means the resistance of a conducting wire is inversely proportional to its thickness (cross-sectional area, A). We write
Rw ~ 1/A
Combining all the three above factors in the same formula, we obtain for the resistance of conductor ### Example 1

What is the resistance of a 20 m long and 6 mm2 thick copper wire id the resistivity of copper is 1.69 × 10-8 Ω ∙ m? Take the figure above as a reference.

### Solution 2

First, we must convert the units into the standard ones. We have

L = 20 m = 2 × 101 m
A = 6 mm2 = 6 × 10-6 m2
ρ = 1.69 × 10-8 Ω ∙ m
Rw = ?

Applying the equation for the resistance of conductor, we obtain

Rw = ρ ∙ L/A
= (1.69 × 10-8 Ω ∙ m) ∙ (2 × 101 m)/(6 × 10-6 m2 )
= 5.63 × 10-2 Ω

Remark! There is a fourth factor - the temperature of conductor - which affects the conducting ability of materials. As we have seen in the previous chapter, an increase in temperature brings an increase in the vibrational activity of material molecules around their equilibrium position. This brings an increase in the resistance of material as in such conditions it becomes more difficult for electrons to flow through the material without colliding. Imagine yourself walking through a hall in which people are running around. You have difficulty to walk, as besides carrying out the walking process you must take care to avoid collisions with other people. As a result, it takes you more time to reach the opposite wall. Therefore, we say temperature of material is proportional to the resistance. However, this factor is not represented in the above formula (it is intended only for cases in which the temperature of conductor is assumed as constant).

## Resistors

Resistors are electrical appliances that are used to convert electrical energy into heat. This is achieved by winding a conducting wire in a spring-like form, in order to confine a long wire in a narrow space. As discussed in the previous paragraph, longer the conductor, higher the resistance it displays to the current flow. Therefore, resistors usually have a high resistance.

Practically, all electric appliances installed in a circuit have a certain resistance, so they can be considered as resistors. However, typically we refer to resistors when the main purpose of an electrical appliance is to convert the electrical energy into heat.

The symbol of resistor in an electric circuit is ## Combination of Resistors

As you remember in capacitors, we explained that it is possible change the capacitance of a system by combining two or more capacitors in series or in parallel. This is because capacitors have fixed values but the circuit may require a different capacitance.

Similarly, we can combine two or more resistors in series or in parallel to change the resistance of a circuit. Let's explain what happens during these combinations.

### a. Series combination of resistors

If two or more resistors are connected in series, they are permeated by the same current. As a result, electrons encounter resistance in two successive positions. This causes an accumulative effect in the total (otherwise known as "equivalent") resistance, Req of the circuit, i.e.

Req = R1 + R2 + ⋯

Therefore, we can replace the two above resistors connected in series by a single resistor whose resistance is the mathematical sum of each single resistor. ### Example 2

Five resistors of resistance 2Ω, 8Ω, 6Ω, 12Ω and 10Ω are connected in series, as shown in the figure. What is the equivalent resistance of this branch of resistors?

### Solution 2

We just have to add all individual resistors to calculate the resistance of this series branch. Thus,

Req = R1 + R2 + R3 + R4 + R5
= 2Ω + 8Ω + 6Ω + 12Ω + 10Ω
= 38Ω

This is the equivalent figure: As you see, the equivalent resistance of a series combination of resistors gives a higher resistance than that of each single resistor. Therefore, a series setup is used when we want to increase the resistance of a circuit and the resistors available are small.

The equation used for a series combination of resistors is analogue to the equations used for parallel combination of springs or capacitors.

### b. Parallel combination of resistors

A parallel setup is when there are at least two branches containing one or more resistors each, as shown in the figure below. In a parallel setup the current comes as a whole from the source, then it divides into two smaller currents I1 and I2 flowing through the resistors R1 and R2 respectively. As a result, the equivalent resistance of a parallel setup is smaller than the resistance of each single resistor, as electric charges flow easier through two resistors than if there it was a single resistor. The concept of equivalent resistance is used to ease the study of such situations as in this case we assume it was only a single resistor in a single branch, as shown below. This situation (of parallel setup of resistors) is analogue to when there are two bridges built across the same river. The traffic in each bridge is less intense than if there was a single bridge.

The equation for the equivalent resistance of a parallel setup is

1/Req =1/R1 + 1/R2 + ⋯

(we will show the proof of this equation in the next tutorial). This equation is analogue to the equation used for series combination of springs or capacitors.

#### Example 3

Two resistors R1 = 24 Ω and R2 = 36 Ω are connected in parallel, as shown in the figure. #### Solution 3

Using the equation of the parallel combination of resistors, we obtain

1/Req =1/R1 + 1/R2
1/Req = 1/24 + 1/36
1/Req = 3/72 + 2/72
1/Req = 5/72
Req = 72/5 Ω
Req = 14.4 Ω

### c. Complex combination of resistors

This kind of setup is composite; it takes place when resistors are combined both in series and in parallel in the same circuit. In these cases, the equivalent resistance is calculated in steps (stages) and we start calculating it from the innermost part of the setup. Let's consider an example to clarify this point.

#### Example 4

Four resistors, R1 = 12 Ω, R2 = 18 Ω, R3 = 10 Ω and R4 = 7 Ω are placed in a circuit as shown in the figure below. Calculate the equivalent resistance of the entire setup.

#### Solution 4

The innermost part of the setup is the series combination of R1 and R2 as we cannot continue with the rest without finding the equivalent resistance of this part first. Thus, we have

Rs = R1 + R2
= 12Ω + 18Ω
= 30Ω

Thus, we obtain the following simplified equivalent figure: Now, we continue with the parallel part of the setup. We have:

1/Rp = 1/Rs + 1/R3
1/Rp = 1/30 + 1/10
1/Rp = 1/30 + 3/30
1/Rp = 4/30
Rp = 30/4 Ω
Rp=7.5Ω

Again, we can draw the simplified equivalent setup to have a better understanding of the situation. As you see, now there is only a series setup to calculate. Hence, we obtain for the equivalent resistance of the entire circuit:

Req = Rp + R4
= 7.5Ω + 7Ω
= 14.5Ω

## Appendix: Dependence of Resistance from Temperature of Material

As we explained earlier, resistance of a conductor depends on temperature but this quantity does not appear in the formula of the resistance of a conductor as temperature was assumed as constant. However, we explained that electricity flow is accompanied by a heating effect in the conductor because electrons collide with the atoms of conductor during their flow and as a result, they make these atoms vibrate faster around their equilibrium positions. This brings an increase in the internal energy of atoms, which is nothing else but the average kinetic energy of all particles of a material.

All values of resistance we discussed earlier, are taken at "normal" temperature, i.e. at T = 20°C. This temperature is known as the "reference temperature", Tref and the corresponding resistance of a certain conductor at this temperature is known as the "conductor resistance at reference temperature", Rref. The equation for the resistance R of a conductor at a certain temperature T therefore is

R = Rref ∙ [1 + α ∙ (T - Tref )]

where α is the temperature coefficient of resistance for a given conducting material. It is an intrinsic property of the material itself and is given in tables. (Do not confuse this quantity with the coefficient of linear expansion α we have discussed in the tutorial 13.2 "Thermal Expansion"; they are different concepts). The unit of temperature coefficient of resistance is [K-1] or [1/K].

### Example 5

What is the resistance of a 12 m copper wire at 50°C if its thickness is 4mm2? Take the resistivity of copper as 1.69 × 10-8 Ω ∙ m and the temperature coefficient of resistance for copper as 0.00393 K-1.

### Solution 5

Clues:

L = 12 m
T = 50°C
A = 4 mm2 = 4 × 10-6 m2
ρ = 1.69 × 10-8 Ω ∙ m
α = 0.00393 K-1 = 3.93 × 10-3 K-1
Tref = 20°C
R = ?

First, we must find the reference temperature for this copper wire. It is calculated using the formula for the resistance of conductor. We have

Rref = ρ ∙ L/A
= (1.69 × 10-8 Ω ∙ m) ∙ (12m)/4 × 10-6 m2
= 5.07 × 10-2 Ω

Now, we calculate the resistance of the wire at the given temperature. We have:

R=Rref ∙ [1 + α ∙ (T - Tref )]
= (5.07 × 10-2 Ω) ∙ [1 + 3.93 × 10-3 K-1 ∙ (50-20)0 C]
= (5.07 × 10-2 Ω) ∙ [1 + 3.93 × 10-3 K-1 ∙ 30 K]
= (5.07 × 10-2 Ω) ∙ (1 + 0.1179)
= (5.07 × 10-2 Ω) ∙ 1.1179
= 5.668 × 10-2 Ω

As you see, the temperature of the copper wire at 50°C is slightly higher than the temperature of the same wire at 20°C.

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