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Collection by: Scott Starks. E-mail the author

Module by: Scott Starks. E-mail the author

Summary: This module is part of a collection of modules written for students enrolled in a Pre Calculus Course for Pre Engineers (MATH 1508) at the University of Texas at El Paso. The module introduces how fractions can be added through the use of the lowest common denominator method. To illustrate the principle, an introduction to Ohm's Law is presented. In addition, the rules that govern the determination of the equivalent resistance of series and parallel connections of resistors are introduced. Finally, exercises involving the addition of fractions in the context of parallel connections of resistors are included to emphasize the importance of the lowest common denominator method in engineering calculations.

### Introduction

In order to enjoy success as an engineer, it is important to learn how to add fractions. In this module, you will learn to add fractions using the lowest common denominator (LCD) method. Also, you will learn the role that the addition of fractions plays in determining the equivalent resistance of resistors connected in parallel.

### Lowest Common Denominator (LCD) Method

In the course of working algebraic problems, one often encounters situations that require the addition of fractions with unequal denominators. For example, let us consider the following

3 10 + 1 12 3 10 + 1 12 size 12{ { {3} over {"10"} } + { {1} over {"12"} } } {}
(1)

In order to add the two fractions, it is important to rewrite each fraction with the same denominator. In order to accomplish this, we begin by expressing the denominator of the first fraction in terms of a product of its factors

10 = 2 × 5 10 = 2 × 5 size 12{"10"=2 times 5} {}
(2)

We do the same with the denominator of the second fraction

12 = 2 × 2 × 3 = 2 2 × 5 12 = 2 × 2 × 3 = 2 2 × 5 size 12{"12"=2 times 2 times 3=2 rSup { size 8{2} } times 5} {}
(3)

We can express the lowest common denominator as

LCD = 2 2 × 3 × 5 = 60 LCD = 2 2 × 3 × 5 = 60 size 12{ ital "LCD"=2 rSup { size 8{2} } times 3 times 5="60"} {}
(4)

We proceed to express the sum of fractions using the lowest common denominator just found

3 10 + 1 12 = 3 × 6 10 × 6 + 1 × 5 12 × 5 = 18 60 + 5 60 = 23 60 3 10 + 1 12 = 3 × 6 10 × 6 + 1 × 5 12 × 5 = 18 60 + 5 60 = 23 60 size 12{ { {3} over {"10"} } + { {1} over {"12"} } = { {3 times 6} over {"10" times 6} } + { {1 times 5} over {"12" times 5} } = { {"18"} over {"60"} } + { {5} over {"60"} } = { {"23"} over {"60"} } } {}
(5)

Thus we obtain the result of the addition as 23/60.

Let us consider another example in which three fractions are to be added

2 3 + 3 2 + 5 7 2 3 + 3 2 + 5 7 size 12{ { {2} over {3} } + { {3} over {2} } + { {5} over {7} } } {}
(6)

The denominator of each fraction is a prime number, so the lowest common denominator is their product

LCD = 3 × 2 × 7 = 42 LCD = 3 × 2 × 7 = 42 size 12{ ital "LCD"=3 times 2 times 7="42"} {}
(7)

We must rewrite each fraction as an equivalent fraction with a denominator of 42

2 3 = 2 × 2 × 7 3 × 2 × 7 = 28 42 2 3 = 2 × 2 × 7 3 × 2 × 7 = 28 42 size 12{ { {2} over {3} } = { {2 times 2 times 7} over {3 times 2 times 7} } = { {"28"} over {"42"} } } {}
(8)

size 12{~} {}

3 2 = 3 × 3 × 7 3 × 2 × 7 = 63 42 3 2 = 3 × 3 × 7 3 × 2 × 7 = 63 42 size 12{ { {3} over {2} } = { {3 times 3 times 7} over {3 times 2 times 7} } = { {"63"} over {"42"} } } {}
(9)
5 7 = 5 × 3 × 2 3 × 2 × 7 = 30 42 5 7 = 5 × 3 × 2 3 × 2 × 7 = 30 42 size 12{ { {5} over {7} } = { {5 times 3 times 2} over {3 times 2 times 7} } = { {"30"} over {"42"} } } {}
(10)

Next, we express the sum of fractions using those equivalent fractions just determined

2 3 + 3 2 + 5 7 = 28 42 + 63 42 + 30 42 = 28 + 63 + 30 42 = 121 42 2 3 + 3 2 + 5 7 = 28 42 + 63 42 + 30 42 = 28 + 63 + 30 42 = 121 42 size 12{ { {2} over {3} } + { {3} over {2} } + { {5} over {7} } = { {"28"} over {"42"} } + { {"63"} over {"42"} } + { {"30"} over {"42"} } = { {"28"+"63"+"30"} over {"42"} } = { {"121"} over {"42"} } } {}
(11)

So 121/42 is the desired result.

### Application: Combining Resistors in Parallel

Figure 1 depicts a physical device known as a resistor. A resistor is often used in an electrical circuit to control the amount of current that flows throughout the circuit. The relationship between voltage, current and resistance in an electric circuit is governed by a fundamental law of Physics known as Ohm’s Law. Stated in words, Ohm’s Law tells us that the potential difference (V) measured in Volts across a resistor is directly proportional the current (I) measured in Amps that flows through the resistor. Additionally, the constant of proportionality is the value of the resistance (R), measured in Ohms. Ohm’s Law can be stated mathematically as

V = I R V = I R size 12{V=IR} {}
(12)

Instead of a single resistor, electric circuits often employ multiple resistors. Let us consider the case where we have two resistors that are denoted as R1 and R2.

The two resistors can be connected in an end-to-end manner as shown in Figure 2 (a).

Resistors connected in this manner are said to be connected in series. We may replace a series connection of two resistors by a single equivalent resistance, Req. From an electrical standpoint, the single equivalent resistance will behave exactly the same as the combination of the two resistors connected in series. The equivalent resistance of two resistors connected in series can be calculated by summing the resistance value of each of the two resistors.

R eq = R 1 + R 2 R eq = R 1 + R 2 size 12{R rSub { size 8{ ital "eq"} } =R rSub { size 8{1} } +R rSub { size 8{2} } } {}
(13)

Let us now consider the case where the two resistors are placed side-by-side and then connected at both ends. This situation is depicted in Figure 2 (b) and is called a parallel combination of resistors. Whenever resistors are connected in this manner, they are said to be connected in parallel. The equivalent resistance of two resistors connected in parallel obeys the following relationship.

1 R eq = 1 R 1 + 1 R 2 1 R eq = 1 R 1 + 1 R 2 size 12{ { {1} over {R rSub { size 8{ ital "eq"} } } } = { {1} over {R rSub { size 8{1} } } } + { {1} over {R rSub { size 8{2} } } } } {}
(14)

Once this quantity is calculated, one may easily take the reciprocal of the result to obtain the value of Req.

The rule governing the determination of the equivalent resistance for the series connection of more than two resistors can be expanded to accommodate any number (n) of resistors. For n resistors connected in series, the equivalent resistance is equal to the sum of the n resistance values.

In addition, the rule governing the determination of the equivalent resistance for the parallel connection of more than two resistors can be expanded. For n resistors connected in parallel,

1 R eq = 1 R 1 + 1 R 2 + 1 R n 1 R eq = 1 R 1 + 1 R 2 + 1 R n size 12{ { {1} over {R rSub { size 8{ ital "eq"} } } } = { {1} over {R rSub { size 8{1} } } } + { {1} over {R rSub { size 8{2} } } } + dotsaxis { {1} over {R rSub { size 8{n} } } } } {}
(15)

It is clear to see that one’s ability to determine the equivalent resistance of parallel resistors depends upon one’s ability to add fractions. The following exercises are included to reinforce this idea.

Example 1: Two resistors are connected in series. The value of resistance for the first resistor is 5 Ω, while that of the second is 9 Ω. Find the equivalent resistance of the series combination.

R eq = 5 Ω + 9 Ω = 14 Ω R eq = 5 Ω + 9 Ω = 14 Ω size 12{R rSub { size 8{ ital "eq"} } =5 %OMEGA +9 %OMEGA ="14" %OMEGA } {}
(16)

Example 2: Consider the two resistors presented in Example 1. Let the two resistors now be connected in parallel. Find the equivalent resistance of the parallel combination.

1 R eq = 1 5 Ω + 1 9 Ω 1 R eq = 1 5 Ω + 1 9 Ω size 12{ { {1} over {R rSub { size 8{ ital "eq"} } } } = { {1} over {5 %OMEGA } } + { {1} over {9 %OMEGA } } } {}
(17)

The lowest common denominator is 45 Ω. So we incorporate it into our analysis.

1 R eq = 9 45 Ω + 5 45 Ω 1 R eq = 9 45 Ω + 5 45 Ω size 12{ { {1} over {R rSub { size 8{ ital "eq"} } } } = { {9} over {"45" %OMEGA } } + { {5} over {"45" %OMEGA } } } {}
(18)
1 R eq = 14 45 Ω 1 R eq = 14 45 Ω size 12{ { {1} over {R rSub { size 8{ ital "eq"} } } } = { {"14"} over {"45" %OMEGA } } } {}
(19)
R eq = 45 Ω 14 = 3 . 21 Ω R eq = 45 Ω 14 = 3 . 21 Ω size 12{R rSub { size 8{ ital "eq"} } = { {"45" %OMEGA } over {"14"} } =3 "." "21" %OMEGA } {}
(20)

Example 3: Three resistors of values 2 kΩ, 3 kΩ, and 5 kΩ are connected in parallel. Find the equivalent resistance.

1 R eq = 1 2, 000 Ω + 1 3, 000 Ω + 1 5, 000 Ω 1 R eq = 1 2, 000 Ω + 1 3, 000 Ω + 1 5, 000 Ω size 12{ { {1} over {R rSub { size 8{ ital "eq"} } } } = { {1} over {2,"000" %OMEGA } } + { {1} over {3,"000" %OMEGA } } + { {1} over {5,"000" %OMEGA } } } {}
(21)

The lowest common denominator for the fractional terms is

LCD = 2 × 3 × 5 × 1, 000 Ω = 30 , 000 Ω LCD = 2 × 3 × 5 × 1, 000 Ω = 30 , 000 Ω size 12{ ital "LCD"=2 times 3 times 5 times 1,"000" %OMEGA ="30","000" %OMEGA } {}
(22)

We rewrite the original equation to reflect the lowest common denominator

1 R eq = 15 30 , 000 Ω + 10 30 , 000 Ω + 6 30 , 000 Ω 1 R eq = 15 30 , 000 Ω + 10 30 , 000 Ω + 6 30 , 000 Ω size 12{ { {1} over {R rSub { size 8{ ital "eq"} } } } = { {"15"} over {"30","000" %OMEGA } } + { {"10"} over {"30","000" %OMEGA } } + { {6} over {"30","000" %OMEGA } } } {}
(23)
1 R eq = 31 30 , 000 Ω 1 R eq = 31 30 , 000 Ω size 12{ { {1} over {R rSub { size 8{ ital "eq"} } } } = { {"31"} over {"30","000" %OMEGA } } } {}
(24)

R eq = 30 , 000 Ω 31 = 968 Ω R eq = 30 , 000 Ω 31 = 968 Ω size 12{R rSub { size 8{ ital "eq"} } = { {"30","000" %OMEGA } over {"31"} } ="968" %OMEGA } {}

### Summary

This module has presented how to add fractions using the lowest common denominator method. Also presented is the relationship among voltage, current and resistance that is known as Ohm’s Law. Examples illustrating the use of the lowest common denominator method to solve for equivalent resistances of parallel combinations of resistors are also provided.

### Exercises

1. Consider a 10 kΩ and a 20 kΩ resistor. (a) What is the equivalent resistance for their series connection? (b) What is the equivalent resistance for their parallel connection?
2. Consider a parallel connection of three resistors. The resistors have values of 25 Ω, 75 Ω, and 100 Ω. What is the equivalent resistance of the parallel connection?
3. Consider a parallel connection of four resistors. The resistors have values of 100 Ω, 100 Ω, 200 Ω and 200 Ω. What is the equivalent resistance of the parallel connection?
4. Conductance is defined as the reciprocal of resistance. Conductance which is typically denoted by the symbol, G, is measured in the units, Siemens. Suppose that you are presented with two resistors of value 500 Ω and 1 kΩ. What is the conductance of each resistor?
5. The equivalent conductance of a parallel connection of two resistors is equal to the sum of the conductance associated with each resistor. What is the equivalent conductance of the parallel connection of the resistors described in exercise 4?
6. What is the equivalent resistance of the parallel connection of resistors described in exercise 5?
7. Suppose that you are presented with 2 resistors. Each resistor has the same value of resistance (say, R). Derive an expression for the equivalent resistance of their parallel connection.
8. Three resistors with resistance values of 100 kΩ, 50 kΩ, and 100 kΩ are connected in parallel. What is the equivalent resistance? (Hint: You may use the result of Exercise 2 to simplify your work.)
9. Four resistors, each with a value of 10 Ω, are connected in parallel. What is the equivalent conductance of the parallel connection? What is the equivalent resistance of the parallel connection?
10. A 30 Ω resistance is connected in series with a parallel connection of two resistors, each with a value of 40 Ω. What is the equivalent resistance of this series/parallel connection?

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