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Chapter 3
Resistance variation At the end of this chapter you should be able to: • recognise three common methods of resistor construction • appreciate that electrical resistance depends on four factors ρl • appreciate that resistance R = , where ρ is the resistivity a • recognize typical values of resistivity and its unit ρl • perform calculations using R = a • define the temperature coefficient of resistance, α • recognize typical values for α • perform calculations using Rθ = R0 (1 +αθ) • determine the resistance and tolerance of a fixed resistor from its colour code • determine the resistance and tolerance of a fixed resistor from its letter and digit code
3.1
Resistor construction
There is a wide range of resistor types. Three of the most common methods of construction are:
(i) Wire wound resistors A length of wire such as nichrome or manganin, whose resistive value per unit length is known, is cut to the desired value and wound around a ceramic former prior to being lacquered for protection. This type of resistor has a large physical size, which is a disadvantage; however, they can be made with a high degree of accuracy, and can have a high power rating. Wire wound resistors are used in power circuits and motor starters.
(ii) Metal oxide resistors With a metal oxide resistor a thin coating of platinum is deposited on a glass plate; it is then fired and a thin track etched out. It is then totally enclosed in an outer tube. DOI: 10.1016/B978-1-85617-770-2.00003-3
Metal oxide resistors are used in electronic equipment.
(iii) Carbon resistors This type of resistor is made from a mixture of carbon black resin binder and a refractory powder that is pressed into shape and heated in a kiln to form a solid rod of standard length and width. The resistive value is predetermined by the ratio of the mixture. Metal end connections are crimped onto the rod to act as connecting points for electrical circuitry. This type of resistor is small and mass-produced cheaply; it has limited accuracy and a low power rating. Carbon resistors are used in electronic equipment.
3.2
Resistance and resistivity
The resistance of an electrical conductor depends on 4 factors, these being: (a) the length of the conductor, (b) the cross-sectional area of the conductor, (c) the type of material and (d) the temperature of the material.
Part 1
18 Electrical Circuit Theory and Technology Resistance, R, is directly proportional to length, l, of a conductor, i.e. R ∝ l. Thus, for example, if the length of a piece of wire is doubled, then the resistance is doubled. Resistance, R, is inversely proportional to crosssectional area, a, of a conductor, i.e. R ∝ 1/a. Thus, for example, if the cross-sectional area of a piece of wire is doubled then the resistance is halved. Since R ∝ l and R ∝ 1/a then R ∝ l/a. By inserting a constant of proportionality into this relationship the type of material used may be taken into account. The constant of proportionality is known as the resistivity of the material and is given the symbol ρ (Greek rho). Thus, resistance
R=
ρl ohms a
ρ is measured in ohm metres (m). The value of the resistivity is that resistance of a unit cube of the material measured between opposite faces of the cube. Resistivity varies with temperature and some typical values of resistivities measured at about room temperature are given below: Copper
1.7 × 10−8 m (or 0.017 µm)
Aluminium
2.6 × 10−8 m (or 0.026 µm)
Carbon (graphite) 10 × 10−8 m
(or 0.10 µm)
Glass
1 × 1010 m
104 µm)
(or
Mica
1 × 1013 m
(or 107 µm)
Note that good conductors of electricity have a low value of resistivity and good insulators have a high value of resistivity. Problem 1. The resistance of a 5 m length of wire is 600 . Determine (a) the resistance of an 8 m length of the same wire, and (b) the length of the same wire when the resistance is 420 . (a)
Resistance, R, is directly proportional to length, l, i.e. R ∝ l Hence, 600 ∝ 5 m or 600 =(k)(5), where k is the coefficient of proportionality. Hence, k=
600 = 120 5
When the length l is 8 m, then resistance R = kl =(120)(8) =960
(b) When the resistance is 420 , 420 = kl, from which, length l =
420 420 = = 3.5 m k 120
Problem 2. A piece of wire of cross-sectional area 2 mm2 has a resistance of 300 . Find (a) the resistance of a wire of the same length and material if the cross-sectional area is 5 mm2 , (b) the cross-sectional area of a wire of the same length and material of resistance 750 . Resistance R is inversely proportional to cross-sectional 1 area, a, i.e. R ∝ a 1 1 or 300 = (k) , Hence 300 ∝ 2 mm2 2 from which, the coefficient of proportionality, k = 300 ×2 = 600 (a)
When the cross-sectional area a = 5 mm2 then R = (k)
1 1 = (600) = 120 5 5
(Note that resistance has decreased as the crosssectional area is increased.) (b) When the resistance is 750 then 750 =(k)(1/a), from which cross-sectional area, a=
600 k = = 0.8 mm2 750 750
Problem 3. A wire of length 8 m and cross-sectional area 3 mm2 has a resistance of 0.16 . If the wire is drawn out until its cross-sectional area is 1 mm2 , determine the resistance of the wire. Resistance R is directly proportional to length l, and inversely proportional to the cross-sectional area, a, i.e. l l R ∝ or R = k , where k is the coefficient of a a proportionality. 8 Since R = 0.16, l = 8 and a = 3, then 0.16 =(k) , 3 from which 3 k = 0.16 × = 0.06 8
Resistance variation
Problem 4. Calculate the resistance of a 2 km length of aluminium overhead power cable if the cross-sectional area of the cable is 100 mm2 . Take the resistivity of aluminium to be 0.03 ×10−6 m.
Cross-sectional area of cable, a = πr 2 = π
(0.03 × 10−6 m)(2000 m) ρl = a (100 × 10−6 m2 ) 0.03 × 2000 = 100 = 0.6
= 36π × 10−6 m2 Resistance R = =
=
Exercise 7 Further problems on resistance and resistivity 1.
The resistance of a 2 m length of cable is 2.5 . Determine (a) the resistance of a 7 m length of the same cable and (b) the length of the same wire when the resistance is 6.25 . [(a) 8.75 (b) 5 m]
2.
Some wire of cross-sectional area 1 mm2 has a resistance of 20 . Determine (a) the resistance of a wire of the same length and material if the cross-sectional area is 4 mm2 , and (b) the cross-sectional area of a wire of the same length and material if the resistance is 32 . [(a) 5 (b) 0.625 mm2 ]
3.
Some wire of length 5 m and cross-sectional area 2 mm2 has a resistance of 0.08 . If the wire is drawn out until its cross-sectional area is 1 mm2 , determine the resistance of the wire. [0.32 ]
4.
Find the resistance of 800 m of copper cable of cross-sectional area 20 mm2 . Take the resistivity of copper as 0.02 µm. [0.8 ]
5.
Calculate the cross-sectional area, in mm2 , of a piece of aluminium wire 100 m long and having a resistance of 2 . Take the resistivity of [1.5 mm2 ] aluminium as 0.03 ×10−6 m.
6.
(a) What does the resistivity of a material mean? (b) The resistance of 500 m of wire of crosssectional area 2.6 mm2 is 5 . Determine the resistivity of the wire in µm. [0.026 µm]
(0.02 × 10−6 m)(40 m) 0.25
= (3.2 × 10−6) × 106 mm 2 = 3.2 mm2 Problem 6. The resistance of 1.5 km of wire of cross-sectional area 0.17 mm2 is 150 . Determine the resistivity of the wire. Resistance, R =
ρl a
hence, resistivity ρ =
(150 )(0.17 × 10−6 m2 ) Ra = l (1500 m)
= 0.017×10−6 m or 0.017 µm Problem 7. Determine the resistance of 1200 m of copper cable having a diameter of 12 mm if the resistivity of copper is 1.7 × 10−8 m.
1.7 × 12 1.7 × 1200 × 106 = 108 × 36π 36π
Now try the following exercise
ρl ρl hence cross-sectional area a = a R
= 3.2 × 10−6 m2
ρl (1.7 × 10−8 m)(1200 m) = a (36π × 10−6 m2 )
= 0.180
Problem 5. Calculate the cross-sectional area, in mm2 , of a piece of copper wire, 40 m in length and having a resistance of 0.25 . Take the resistivity of copper as 0.02 × 10−6 m. Resistance R =
2
= 36π mm 2
Length l = 2 km = 2000 m; area, a = 100 mm2 = 100 × 1 0 −6 m2 ; resistivity ρ = 0.03 ×10−6 m Resistance R =
12 2
Part 1
If the cross-sectional area is reduced to 13 of its original area then the length must be tripled to 3 ×8, i.e. 24 m l 24 New resistance R = k = 0.06 = 1.44 a 1
19
Part 1
20 Electrical Circuit Theory and Technology 7. Find the resistance of 1 km of copper cable having a diameter of 10 mm if the resistivity [0.216 ] of copper is 0.017 ×10−6 m.
Resistance Rθ = R0 (1 +α0 θ) Hence resistance at 70◦ C, R70 = 100[1 +(0.0043)(70)] = 100[1 + 0.301] = 100(1.301) = 130.1
3.3 Temperature coefficient of resistance In general, as the temperature of a material increases, most conductors increase in resistance, insulators decrease in resistance, whilst the resistance of some special alloys remain almost constant. The temperature coefficient of resistance of a material is the increase in the resistance of a 1 resistor of that material when it is subjected to a rise of temperature of 1◦ C. The symbol used for the temperature coefficient of resistance is α (Greek alpha). Thus, if some copper wire of resistance 1 is heated through 1◦C and its resistance is then measured as 1.0043 then α = 0.0043 /◦ C for copper. The units are usually expressed only as ‘per ◦ C’, i.e. α = 0.0043/◦C for copper. If the 1 resistor of copper is heated through 100◦C then the resistance at 100◦C would be 1 +100 ×0.0043 =1.43 . Some typical values of temperature coefficient of resistance measured at 0◦C are given below: Copper
0.0043/◦C Aluminium
Nickel
0.0062/◦C Carbon
Constantan
0
Eureka
0.0038/◦C −0.000 48/◦C 0.000 01/◦C
(Note that the negative sign for carbon indicates that its resistance falls with increase of temperature.) If the resistance of a material at 0◦C is known the resistance at any other temperature can be determined from: Rθ = R0(1 + α 0 θ) where R0 = resistance at 0◦ C Rθ = resistance at temperature θ ◦ C α0 = temperature coefficient of resistance at
Problem 9. An aluminium cable has a resistance of 27 at a temperature of 35◦C. Determine its resistance at 0◦ C. Take the temperature coefficient of resistance at 0◦ C to be 0.0038/◦C. Resistance at θ ◦ C, Rθ = R0 (1 + α0θ) Hence resistance at 0◦ C, R0 =
Rθ (1 + α0 θ)
=
27 [1 + (0.0038)(35)]
=
27 27 = 1 + 0.133 1.133
= 23.83 Problem 10. A carbon resistor has a resistance of 1 k at 0◦ C. Determine its resistance at 80◦ C. Assume that the temperature coefficient of resistance for carbon at 0◦ C is −0.0005/◦C. Resistance at temperature θ ◦ C, Rθ = R0 (1 + α0 θ) i.e. Rθ = 1000[1 + (−0.0005)(80)] = 1000[1 − 0.040] = 1000(0.96) = 960 If the resistance of a material at room temperature (approximately 20◦ C), R20 , and the temperature coefficient of resistance at 20◦C, α20, are known then the resistance Rθ at temperature θ ◦ C is given by: Rθ = R20 [1 + α 20(θ − 20)]
0◦ C
Problem 8. A coil of copper wire has a resistance of 100 when its temperature is 0◦C. Determine its resistance at 70◦C if the temperature coefficient of resistance of copper at 0◦ C is 0.0043/◦C.
Problem 11. A coil of copper wire has a resistance of 10 at 20◦C. If the temperature coefficient of resistance of copper at 20◦C is 0.004/◦C determine the resistance of the coil when the temperature rises to 100◦C.
Resistance at θ ◦ C, R = R20 [1 +α20 (θ − 20)] Hence resistance at 100◦C, R100 = 10[1 +(0.004)(100 − 20)] = 10[1 +(0.004)(80)] = 10[1 +0.32]
R20 = 200 , α0 = 0.004/◦ C R20 [1 + α0 (20)] = R90 [1 + α0 (90)] R20 [1 + 90α0 ] Hence R90 = [1 + 20α0]
= 10(1.32)
=
200[1 + 90(0.004)] [1 + 20(0.004)]
=
200[1 + 0.36] [1 + 0.08]
=
200(1.36) = 251.85 (1.08)
= 13.2 Problem 12. The resistance of a coil of aluminium wire at 18◦C is 200 . The temperature of the wire is increased and the resistance rises to 240 . If the temperature coefficient of resistance of aluminium is 0.0039/◦C at 18◦ C determine the temperature to which the coil has risen. Let the temperature rise to θ ◦ Resistance at θ ◦ C, Rθ = R18 [1 +α18(θ − 18)] i.e.
240 = 200[1 +(0.0039)(θ − 18)] 240 = 200 +(200)(0.0039)(θ − 18) 240 − 200 =0.78(θ − 18) 40 = 0.78(θ − 18) 40 = θ − 18 0.78 51.28 = θ − 18, from which, θ = 51.28 +18 =69.28◦ C
i.e. the resistance of the wire at 90◦ C is 252 . Now try the following exercise Exercise 8 Further problems on temperature coefficient of resistance 1.
A coil of aluminium wire has a resistance of 50 when its temperature is 0◦ C. Determine its resistance at 100◦C if the temperature coefficient of resistance of aluminium at 0◦ C is [69 ] 0.0038/◦C.
2.
A copper cable has a resistance of 30 at a temperature of 50◦C. Determine its resistance at 0◦C. Take the temperature coefficient of resistance of copper at 0◦ C as 0.0043/◦C. [24.69 ]
3.
The temperature coefficient of resistance for carbon at 0◦C is −0.00048/◦C. What is the significance of the minus sign? A carbon resistor has a resistance of 500 at 0◦ C. Determine its [488 ] resistance at 50◦ C.
4.
A coil of copper wire has a resistance of 20 at 18◦C. If the temperature coefficient of resistance of copper at 18◦C is 0.004/◦C, determine the resistance of the coil when the temperature [26.4 ] rises to 98◦C.
5.
The resistance of a coil of nickel wire at 20◦C is 100 . The temperature of the wire is increased and the resistance rises to 130 . If the temperature coefficient of resistance of nickel is 0.006/◦C at 20◦C, determine the temperature to which the coil has risen. [70◦C]
Hence the temperature of the coil increases to 69.28◦C. If the resistance at 0◦C is not known, but is known at some other temperature θ1 , then the resistance at any temperature can be found as follows: R1 = R0 (1 + α0 θ1 ) and R2 = R0 (1 + α0 θ2 ) Dividing one equation by the other gives: R1 1 + α 0 θ 1 = R2 1 + α 0 θ 2 where R2 = resistance at temperature θ2 . Problem 13. Some copper wire has a resistance of 200 at 20◦ C. A current is passed through the wire and the temperature rises to 90◦ C. Determine the resistance of the wire at 90◦C, correct to the nearest ohm, assuming that the temperature coefficient of resistance is 0.004/◦C at 0◦C.
21
Part 1
Resistance variation
Part 1
22 Electrical Circuit Theory and Technology 6. Some aluminium wire has a resistance of 50 at 20◦C. The wire is heated to a temperature of 100◦ C. Determine the resistance of the wire at 100◦C, assuming that the temperature coefficient of resistance at 0◦C is 0.004/◦C. [64.8 ] 7. A copper cable is 1.2 km long and has a cross-sectional area of 5 mm2 . Find its resistance at 80◦C if at 20◦ C the resistivity of copper is 0.02 ×10−6 m and its temperature coefficient of resistance is 0.004/◦C. [5.952 ]
3.4 Resistor colour coding and ohmic values (a) Colour code for fixed resistors The colour code for fixed resistors is given in Table 3.1 Table 3.1 Colour
Significant Figures
Multiplier
Tolerance
Silver
–
10−2
±10%
Gold
–
10−1
±5%
Black
0
1
–
Brown
1
10
±1%
Red
2
102
±2%
Orange
3
103
–
Yellow
4
104
–
Green
5
105
±0.5%
Blue
6
106
±0.25%
Violet
7
107
±0.1%
Grey
8
108
–
White
9
109
–
None
–
–
±20%
(i) For a four-band fixed resistor (i.e. resistance values with two significant figures): yellow-violet-orange-red indicates 47 k with a tolerance of ±2% (Note that the first band is the one nearest the end of the resistor) (ii) For a five-band fixed resistor (i.e. resistance values with three significant figures): red-yellowwhite-orange-brown indicates 249 k with a tolerance of ±1% (Note that the fifth band is 1.5 to 2 times wider than the other bands) Problem 14. Determine the value and tolerance of a resistor having a colour coding of: orange-orange-silver-brown
The first two bands, i.e. orange-orange, give 33 from Table 3.1. The third band, silver, indicates a multiplier of 102 from Table 3.1, which means that the value of the resistor is 33 ×10−2 = 0.33 The fourth band, i.e. brown, indicates a tolerance of ±1% from Table 3.1. Hence a colour coding of orange-orange-silver-brown represents a resistor of value 0.33 with a tolerance of ±1% Problem 15. Determine the value and tolerance of a resistor having a colour coding of: brown-black-brown. The first two bands, i.e. brown-black, give 10 from Table 3.1. The third band, brown, indicates a multiplier of 10 from Table 3.1, which means that the value of the resistor is 10 ×10 = 100 There is no fourth band colour in this case; hence, from Table 3.1, the tolerance is ±20%. Hence a colour coding of brown-black-brown represents a resistor of value 100 with a tolerance of ±20% Problem 16. Between what two values should a resistor with colour coding brown-black-brown-silver lie? From Table 3.1, brown-black-brown-silver indicates 10 ×10, i.e. 100 , with a tolerance of ±10%
Resistance variation This means that the value could lie between
(100 + 10% of 100)
i.e. brown-black-brown-silver indicates any value between 90 and 110 Problem 17. Determine the colour coding for a 47 k having a tolerance of ±5%. From Table 3.1, 47 k = 47 ×103 has a colour coding of yellow-violet-orange. With a tolerance of ±5%, the fourth band will be gold. Hence 47 k ± 5% has a colour coding of: yellow-violet-orange-gold Problem 18. Determine the value and tolerance of a resistor having a colour coding of: orange-green-red-yellow-brown orange-green-red-yellow-brown is a five-band fixed resistor and from Table 3.1, indicates: 352 ×104 with a tolerance of ±1% 352 × 104 = 3.52 × 106 , i.e. 3.52 M Hence orange-green-red-yellow-brown 3.52 M ± 1%
indicates
Resistance Value
Marked as:
0.47
R47
1
1R0
4.7
4R7
47
47R
100
100R
1 k
1K0
10 k
10 K
10 M
10 M
Problem 20. Determine the value of a resistor marked as 4M7M. From Table 3.2, 4M7M is equivalent to: 4.7 M ± 20% Problem 21. Determine the letter and digit code for a resistor having a value of 68 k ±10%. From Table 3.2, 68 k ±10% has a letter and digit code of: 68 KK Now try the following exercises
(b) Letter and digit code for resistors Another way of indicating the value of resistors is the letter and digit code shown in Table 3.2. Tolerance is indicated as follows: F = ±1%, G = ±2%, J = ±5%, K = ±10% and M = ±20% Thus, for example, R33M = 0.33 ± 20% 4R7K = 4.7 ± 10% 390RJ = 390 ± 5% Problem 19. Determine the value of a resistor marked as 6K8F. From Table 3.2, 6K8F is equivalent to: 6.8 k ± 1%
Exercise 9
Further problems on resistor colour coding and ohmic values
1. Determine the value and tolerance of a resistor having a colour coding of: blue-grey-orangered [68 k ± 2%] 2. Determine the value and tolerance of a resistor having a colour coding of: yellow-violet-gold [4.7 ± 20%] 3. Determine the value and tolerance of a resistor having a colour coding of: blue-white-blackblack-gold [690 ±5%] 4. Determine the colour coding for a 51 k fourband resistor having a tolerance of ±2% [green-brown-orange-red]
Part 1
Table 3.2
(100 − 10% of 100) and
23
Part 1
24 Electrical Circuit Theory and Technology 5. Determine the colour coding for a 1 M fourband resistor having a tolerance of ±10% [brown-black-green-silver]
8. Determine the value of a resistor marked as (a) R22G (b) 4K7F [(a) 0.22 ± 2% (b) 4.7 k ± 1%]
6. Determine the range of values expected for a resistor with colour coding: red-black-greensilver [1.8 M to 2.2 M]
9. Determine the letter and digit code for a resistor having a value of 100 k ±5% [100 KJ]
7. Determine the range of values expected for a resistor with colour coding: yellow-blackorange-brown [39.6 k to 40.4 k]
10. Determine the letter and digit code for a resistor having a value of 6.8 M ± 20% [6 M8 M]