1. Chapter 3 Resistance
ECET 1010 Fundamentals

2. 3.1 Resistance
Opposing force – due to collisions between electrons and between electrons and other atoms
Converts electrical energy into another form of energy such as heat

3. Resistance Determined by the following four factors R = ρ l / A
Material
Length
Cross-sectional area
Temperature
R = ρ l / A

4. 3.2 Resistance: Circular Wires
The higher the resistivity, the more the resistance
The longer the length of the conductor, the more the resistance
The smaller the area of the conductor, the more the resistance
The higher the temperature of the conductor, the more the resistance

5. Area of Circular Wires
The area of the conductor is measured in circular mils (CM)
1 mil is one one-thousandth of an inch
A wire with a diameter of 1 mil has an area of 1 circular mil (CM)
Area of Circle = π r2 = π d2 / 4

6. Area of Circular Wires
Relationship between circular mils (CM) and square mils
1 CM = π / 4 sq. mils
Relationship between area in circular mils (CM) and diameter in mils
ACM = (dmils)2

7. Example 3.1
What is the resistance of a 100-ft length of copper wire with a diameter of in. at 20°C?
l = 100 ft
ρ = CM-Ω/ft (Table 3.1)
d = in = 20 mils
A = d2 CM = 400 CM
R = * 100 / 400 = 2.59 Ω

8. Example 3.2
You have been given a carton of wire where an undetermined number of feet of the wire has been used. Find the length of the remaining copper wire if it has a diameter of 1/16 in. and a resistance of 0.5 Ω.
ρ = CM-Ω/ft
d = 1/16 in = 62.5 mils
A = (62.5)2 CM = CM
R = 0.5 Ω
l = * 0.5 / = ft

9. Example 3.3
What is the resistance of a copper bus-bar (as used in the power distribution panel of a high-rise office building) 3 ft long, 5 in. wide, and 0.5 in. thick?
l = 3 ft
ρ = CM-Ω/ft
A = ½ in * 5 in = 500 mil * 5,000 mil = 2.5 * 106 mil2
ACM = 2.5 * 106 * (4/ π) = * 106 CM
R = * 3 / * 106 = * 10-6 Ω

10. 3.3 Wire Tables
Standardize the size of wire produced by manufacturers throughout the United States
See Table 3.2 – American Wire Gage (AWG) sizes
AWG number
Area in circular mils
Resistance in Ω per 1,000 feet at 20°C
Maximum allowable current

11. Example 3.4
Find the resistance of 650 ft of #8 copper wire (T = 20°C).
from Table 3.2
Ω / 1,000 ft
R = 650 ft * ( Ω / 1,000 ft)
= Ω

12. Example 3.5 What is the diameter, in inches, of #12 copper wire?
from Table 3.2
ACM = 6529 CM = (dmils)2
d = 80.8 mil = in ~ 1/12 in

13. Example 3.6
For the following system, the total resistance of each power line cannot be more than Ω, and the maximum current to be drawn by the load is 95 A. What gage wire should be used?
Need a picture of an input connected to a load by two 100 ft lengths of solid round copper.

14. Example 3.6 (continued)
From Table 3.2 we choose #3 wire since its maximum allowable current is 100 A and this is greater than 95 A
We need to check and make sure 100 ft of #3 wire is NOT more than Ω
R = 100 ft * ( Ω / 1,000 ft)
= Ω ‹ Ω

15. 3.5 Temperature Effects
Temperature has a significant effect on the resistance of
Conductors
Semiconductors
Insulators

16. Conductors
For good conductors, an increase in temperature will result in an increase in the resistance level (due to an increase in the random motion of the particles in the material). Consequently, conductors have a positive temperature coefficient.

17. Semiconductors
For semiconductor materials, an increase in temperature will result in a decrease in the resistance level (due to an increase in free carriers). Consequently, semiconductors have a negative temperature coefficient.

18. Insulators
As with semiconductors, an increase in temperature will result in a decrease in the resistance of an insulator. The result is a negative temperature coefficient.

19. Inferred Absolute Temperature
Effect of temperature on the resistance of copper
R1,t1 and R2,t2 °C
Similar triangles yield
( t1) / R1 = ( t2) / R2
Adapting to any material   ( |T| + t1) / R1 = ( |T| + t2) / R2

20. Inferred Absolute Temperatures
Material °C
α20
Silver
-243
0.0038
Copper
-234.5
Gold
-274
0.0034
Aluminum
-236
Tungsten
-204
0.005
Nickel
-147
0.006
Iron
-162
0.0055
Nichrome
-2,250
Constantan
-125,000

21. Example 3.9
If the resistance of copper wire is 50 Ω at 20° C, what is its resistance at 100° C (boiling point of water)?
( ) / 50 Ω = ( ) / R
R = Ω

22. Example 3.10
If the resistance of copper wire at freezing (0° C) is 30 Ω, what is its resistance at -40° C?
( ) / 30 Ω = ( ) / R
R = Ω

23. Example 3.11
If the resistance of an aluminum wire at room temperature (20° C) is 100 mΩ (measured by a milliohmeter), at what temperature will its resistance increase to 120 mΩ?
( ) / 100 mΩ = (236 + t) / 120 mΩ
t = 71.2° C

24. Temperature Coefficient of Resistance
Definition of temperature coefficient of resistance at 20° C.
α20 = 1 / (|T| + 20° C)
For copper, α20 = Ω/° C/Ω
The higher the temperature coefficient of resistance for a material, the more sensitive the resistance level to changes in temperature.

25. Temperature Coefficient of Resistance
Resistance R at a temperature t given by:
R = R20 [1 + α20(t - 20° C)]
Which can be written as:
R = ρ (l/A) [1 + α20(ΔT)]

26. Example
If the nominal resistance of a copper wire is 5 Ω, what will its resistance be at 30 ° C?
R = 5 Ω [ (30° C - 20° C)]
R = 5 Ω (1.0393) = Ω

27. PPM/°C Parts per million per degree Celsius For resistors
5000-PPM is high
20-PPM is low
1000-PPM/°C says a 1° C change in temperature gives a change in resistance equal to 1000 parts per million or 1,000/1,000,1000 or 1/1,000 of its value.
ΔR = (Rnominal/106) (PPM) (ΔT)

28. Example 3.12
For a 1-kΩ carbon composition resistor with a PPM of 2500, determine the resistance at 60°C.
Rnominal = 1,000 Ω
PPM = 2,500
ΔT = t – 20° C = 60° C – 20° C = 40° C
ΔR = (1,000/106) (2,500) (40) = 100 Ω
R = Rnominal + ΔR = 1,100 Ω

29. 3.8 Color Coding and Standard Resistor Values
See Table 3.7
Way to identify
Resistance
Tolerance
Reliability

30. Color Code Table Color Band 1 Band 2 Band 3 Band 4 Band 5 Silver 10-2
10%
Gold
10-1 5%
Black
100
Brown 1
101 1%
Red 2
102 2%
0.1%
Orange 3
103 3%
0.01%
Yellow 4
104 4%
0.001%
Green 5
105
Blue 6
106
Violet 7
107
Gray 8
108
White 9
109
None
20%

31. Bands First and second band give first two digits
Third band gives power-of-ten multiplier
Fourth band gives tolerance, ±percent
Fifth band gives reliability, failures per 1,000 hours use
See Table 3.8 for standard resistor values

32. Standard Resistor Values
08/16/99
Standard Resistor Values 6

33. Example 3.13a
Find the range in which a resistor having the following color bands must exist to satisfy the manufacturer’s tolerance:
Gray, Red, Black, Gold, Brown
% 1%
82 * 100 ± 5% Ω
77.9 – 86.1 Ω

34. Example 3.13b
Find the range in which a resistor having the following color bands must exist to satisfy the manufacturer’s tolerance:
Orange, White, Gold, Silver, No Color %
39 * 10-1 ± 10% Ω
3.51 – 4.29 Ω

35. Conductance Reciprocal of resistance G = 1 / R (Siemens, S)
As a function of area, length, and resistivity
G = A / (ρl)

36. Example 3.14
What is the relative increase or decrease in conductivity of a conductor if the area is reduced by 30% and the length is increased by 40%? (The resistivity is fixed.)
Gi = Ai / (ρili)
Gr = Ar / (ρrlr) = 0.7 Ai / (ρi (1.4) li)
= (0.7/1.4) Ai / (ρili) = 0.5 Gi

37. Fixed Composition Resistors of Different Wattage Ratings
08/16/99
Fixed Composition Resistors of Different Wattage Ratings 4

38. Measuring the Resistance of a Single Element
08/16/99
Measuring the Resistance of a Single Element 7