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V = I R size 12{V=I`R} {}

Photograph of a resistor.

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 R 1 and R 2 .

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

Series and parallel connections of resistors.

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, R eq . 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 size 12{R rSub { size 8{ ital "eq"} } =R rSub { size 8{1} } +R rSub { size 8{2} } } {}

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

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

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 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} } } } } {}

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 size 12{R rSub { size 8{ ital "eq"} } =5` %OMEGA +9` %OMEGA ="14"` %OMEGA } {}

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 size 12{ { {1} over {R rSub { size 8{ ital "eq"} } } } = { {1} over {5` %OMEGA } } + { {1} over {9` %OMEGA } } } {}

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

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 } } } {}
1 R eq = 14 45 Ω size 12{ { {1} over {R rSub { size 8{ ital "eq"} } } } = { {"14"} over {"45"` %OMEGA } } } {}
R eq = 45 Ω 14 = 3 . 21 Ω size 12{R rSub { size 8{ ital "eq"} } = { {"45"` %OMEGA } over {"14"} } =3 "." "21"` %OMEGA } {}

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 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 } } } {}

The lowest common denominator for the fractional terms is

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 } {}

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 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 } } } {}
1 R eq = 31 30 , 000 Ω size 12{ { {1} over {R rSub { size 8{ ital "eq"} } } } = { {"31"} over {"30","000"` %OMEGA } } } {}

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?

Questions & Answers

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The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
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1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
Pawel
2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
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x*x=2
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×/×+9+6/1
Debbie
Q2 x+(x+2)+(x+4)=60 3x+6=60 3x+6-6=60-6 3x=54 3x/3=54/3 x=18 :. The numbers are 18,20 and 22
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Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
mariel Reply
Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
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x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
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Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
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(61/11,41/11,−4/11)
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Need help solving this problem (2/7)^-2
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x+2y-z=7
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-1
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A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
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Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Source:  OpenStax, Math 1508 (laboratory) engineering applications of precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11337/1.3
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