Chapter 3

Resistance variation
At the end of this chapter you should be able to:
• appreciate that electrical resistance depends on four factors
• appreciate that resistance R = ρl/a, where ρ is the resistivity
• recognize typical values of resistivity and its unit
• perform calculations using R = ρl/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 Resistance and resistivity Resistivity varies with temperature and some typ-
ical values of resistivities measured at about room
The resistance of an electrical conductor depends on temperature are given below:
four factors, these being: (a) the length of the conductor,
(b) the cross-sectional area of the conductor, (c) the Copper 1.7 × 10−8  m (or 0.017 μ m)
type of material and (d) the temperature of the material. Aluminium 2.6 × 10−8  m (or 0.026 μ m)
Resistance, R, is directly proportional to length, l, of a
Carbon (graphite) 10 × 10−8  m (0.10 μ m)
conductor, i.e. R ∝ l. Thus, for example, if the length of
a piece of wire is doubled, then the resistance is doubled. Glass 1 × 1010  m (or 104 μ m)
Resistance, R, is inversely proportional to cross- Mica 1 × 1013  m (or 107 μ m)
sectional area, a, of a conductor, i.e. R ∝ 1/a. Thus,
for example, if the cross-sectional area of a piece of Note that good conductors of electricity have a low value
wire is doubled then the resistance is halved. of resistivity and good insulators have a high value of
Since R ∝ l and R ∝ 1/a then R ∝ l/a. By inserting resistivity.
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 Problem 1. The resistance of a 5 m length of wire
the material and is given the symbol ρ (Greek rho). Thus, is 600 . Determine (a) the resistance of an 8 m
length of the same wire, and (b) the length of the
ρl
resistance R= ohms same wire when the resistance is 420 .
a
ρ is measured in ohm metres ( m). The value of the
resistivity is that resistance of a unit cube of the material (a) Resistance, R, is directly proportional to length, l,
measured between opposite faces of the cube. i.e. R ∝ l. Hence, 600  ∝ 5 m or 600 = (k)(5),
 Resistance variation 21

where k is the coefficient of proportionality. Since R = 0.16, l = 8 and a = 3, then 0.16 = (k)(8/3),

Section 1
600 from which k = 0.16 × 3/8 = 0.06
Hence, k = = 120 If the cross-sectional area is reduced to 1/3 of its
5
original area then the length must be tripled to 3 × 8,
When the length l is 8 m, then resistance i.e. 24 m
R = kl = (120)(8) = 960     
l 24
(b) When the resistance is 420 , 420 = kl, from New resistance R = k = 0.06 = 1.44 
which, a 1

420 420 Problem 4. Calculate the resistance of a 2 km

length l = = = 3.5 m
k 120 length of aluminium overhead power cable if the
cross-sectional area of the cable is 100 mm2 . Take
Problem 2. A piece of wire of cross-sectional the resistivity of aluminium to be 0.03 × 10−6  m.
area 2 mm2 has a resistance of 300 . Find (a) the
resistance of a wire of the same length and material
Length l = 2 km = 2000 m,
if the cross-sectional area is 5 mm2 , (b) the
area a = 100 mm2 = 100 × 10−6 m2
cross-sectional area of a wire of the same length
and resistivity ρ = 0.03 × 10−6  m.
and material of resistance 750 .
ρl
Resistance R =
Resistance R is inversely proportional to cross-sectional a
area, a, i.e. R ∝ l/a (0.03 × 10−6  m)(2000 m)
=
Hence 300  ∝ 21 mm2 or 300 = (k) ( 21 ) (100 × 10−6 m2 )
from which, the coefficient of proportionality, 0.03 × 2000
=  = 0.6 
k = 300 × 2 = 600 100

(a) When the cross-sectional area a = 5 mm2 then Problem 5. Calculate the cross-sectional area, in
mm2 , of a piece of copper wire, 40 m in length and
R = (k)( 15 ) = (600)( 15 ) = 120  having a resistance of 0.25 . Take the resistivity of
copper as 0.02 × 10−6  m.
(Note that resistance has decreased as the cross-
sectional is increased.) Resistance R = ρl/a hence cross-sectional area
(b) When the resistance is 750  then
  ρl (0.02 × 10−6  m)(40 m)
1 a= =
750 = (k) R 0.25 
a = 3.2 × 10−6 m2
from which = (3.2 × 10−6 ) × 106 mm2 = 3.2 mm2
k 600
cross-sectional area, a = =
750 750 Problem 6. The resistance of 1.5 km of wire of
= 0.8 mm 2 cross-sectional area 0.17 mm2 is 150 . Determine
the resistivity of the wire.
Problem 3. A wire of length 8 m and
cross-sectional area 3 mm2 has a resistance of Resistance, R = ρl/a hence
0.16 . If the wire is drawn out until its Ra
cross-sectional area is 1 mm2 , determine the resistivity ρ =
l
resistance of the wire. (150 )(0.17 × 10−6 m2 )
=
(1500 m)
Resistance R is directly proportional to length l, and
inversely proportional to the cross-sectional area, a, i.e. = 0.017 × 10−6  m
R ∝ l/a or R = k(l/a), where k is the coefficient of or 0.017 μ m
proportionality.
 22 Electrical and Electronic Principles and Technology

Problem 7. Determine the resistance of 1200 m

Section 1

4. Find the resistance of 800 m of copper cable of

of copper cable having a diameter of 12 mm if the cross-sectional area 20 mm2 . Take the resistiv-
resistivity of copper is 1.7 × 10−8  m. ity of copper as 0.02 μ m. [0.8 ]

5. Calculate the cross-sectional area, in mm2 , of

Cross-sectional area of cable, a piece of aluminium wire 100 m long and hav-
 2 ing a resistance of 2 . Take the resistivity of
a = πr 2 = π
12 aluminium as 0.03 × 10−6  m.
2 [1.5 mm2 ]

= 36π mm2 = 36π × 10−6 m2 6. The resistance of 500 m of wire of cross-

sectional area 2.6 mm2 is 5 . Determine the
ρl resistivity of the wire in μ m.
Resistance R = [0.026 μ m]
a

(1.7 × 10−8  m)(1200 m) 7. Find the resistance of 1 km of copper cable

= having a diameter of 10 mm if the resistivity
(36π × 10−6 m2 )
of copper is 0.017 × 10−6  m.
1.7 × 1200 × 106 [0.216 ]
= 
108 × 36π

1.7 × 12 3.2 Temperature coefficient of

=  = 0.180 
36π resistance

In general, as the temperature of a material increases,

Now try the following exercise
most conductors increase in resistance, insulators
decrease in resistance, whilst the resistance of some
Exercise 11 Further problems on resistance special alloys remain almost constant.
and resistivity The temperature coefficient of resistance of a mate-
rial is the increase in the resistance of a 1  resistor
1. The resistance of a 2 m length of cable is of that material when it is subjected to a rise of tem-
2.5 . Determine (a) the resistance of a 7 m perature of 1◦ C. The symbol used for the temperature
length of the same cable and (b) the length coefficient of resistance is α (Greek alpha). Thus, if
of the same wire when the resistance is some copper wire of resistance 1  is heated through
6.25 . [(a) 8.75  (b) 5 m] 1◦ C and its resistance is then measured as 1.0043 
then α = 0.0043 / ◦ C for copper. The units are usu-
2. Some wire of cross-sectional area 1 mm2 has ally expressed only as ‘per ◦ C’, i.e. α = 0.0043/◦ C
a resistance of 20 . for copper. If the 1  resistor of copper is heated
Determine (a) the resistance of a wire of the through 100◦ C then the resistance at 100◦ C would
same length and material if the cross-sectional be 1 + 100 × 0.0043 = 1.43 . Some typical values of
area is 4 mm2 , and (b) the cross-sectional area temperature coefficient of resistance measured at 0◦ C
of a wire of the same length and material if the are given below:
resistance is 32 .
[(a) 5  (b) 0.625 mm2 ] Copper 0.0043/◦ C
Nickel 0.0062/◦ C
Constantan 0
3. Some wire of length 5 m and cross-sectional Aluminium 0.0038/◦ C
area 2 mm2 has a resistance of 0.08 . If the Carbon −0.00048/◦ C
wire is drawn out until its cross-sectional area Eureka 0.00001/◦ C
is 1 mm2 , determine the resistance of the wire.
[0.32 ] (Note that the negative sign for carbon indicates that
its resistance falls with increase of temperature.)
 Resistance variation 23

If the resistance of a material at 0◦ C is known the i.e.

Section 1
resistance at any other temperature can be determined
Rθ = 1000[1 + (−0.0005)(80)]
from:
= 1000[1 − 0.040] = 1000(0.96) = 960 
Rθ = R0 (1 + α0 θ)

where R0 = resistance at 0◦ C If the resistance of a material at room temperature

(approximately 20◦ C), R20 , and the temperature coef-
Rθ = resistance at temperature θ ◦ C ficient of resistance at 20◦ C, α20 , are known then the
α0 = temperature coefficient of resistance at 0◦ C resistance Rθ at temperature θ ◦ C is given by:
Rθ = R20 [1 + α20 (θ − 20)]

Problem 8. A coil of copper wire has a resistance Problem 11. A coil of copper wire has a
of 100  when its temperature is 0◦ C. Determine its resistance of 10  at 20◦ C. If the temperature
resistance at 70◦ C if the temperature coefficient of coefficient of resistance of copper at 20◦ C is
resistance of copper at 0◦ C is 0.0043/◦ C. 0.004/◦ C determine the resistance of the coil when
the temperature rises to 100◦ C.
Resistance Rθ = R0 (1 + α0 θ). Hence resistance at
100◦ C, Resistance at θ ◦ C,
Rθ = R20 [1 + α20 (θ − 20)]
R100 = 100[1 + (0.0043)(70)]
Hence resistance at 100◦ C,
= 100[1 + 0.301]
= 100(1.301) = 130.1  R100 = 10[1 + (0.004)(100 − 20)]
= 10[1 + (0.004)(80)]

Problem 9. An aluminium cable has a resistance = 10[1 + 0.32]

of 27  at a temperature of 35◦ C. Determine its = 10(1.32) = 13.2 
resistance at 0◦ C. Take the temperature coefficient
of resistance at 0◦ C to be 0.0038/◦ C.
Problem 12. The resistance of a coil of
aluminium wire at 18◦ C is 200 . The temperature
Resistance at θ ◦ C, Rθ = R0 (1 + α0 θ). Hence resistance of the wire is increased and the resistance rises to
at 0◦ C, 240 . If the temperature coefficient of resistance
of aluminium is 0.0039/◦ C at 18◦ C determine the
Rθ 27 temperature to which the coil has risen.
R0 = =
(1 + α0 θ) [1 + (0.0038)(35)]
27 Let the temperature rise to θ ◦ C. Resistance at θ ◦ C,
=
1 + 0.133
27 Rθ = R18 [1 + α18 (θ − 18)]
= = 23.83 
1.133 i.e.
240 = 200[1 + (0.0039)(θ − 18)]
240 = 200 + (200)(0.0039)(θ − 18)
Problem 10. A carbon resistor has a resistance of
1 k at 0◦ C. Determine its resistance at 80◦ C. 240 − 200 = 0.78(θ − 18)
Assume that the temperature coefficient of 40 = 0.78(θ − 18)
resistance for carbon at 0◦ C is −0.0005/◦ C.
40
= θ − 18
0.78
Resistance at temperature θ ◦ C, 51.28 = θ − 18, from which,
Rθ = R0 (1 + α0 θ) θ = 51.28 + 18 = 69.28◦ C
 24 Electrical and Electronic Principles and Technology

Hence the temperature of the coil increases to

69.28◦ C its resistance at 100◦ C if the temperature coef-
Section 1

ficient of resistance of aluminium at 0◦ C is

0.0038/◦ C [69 ]
If the resistance at 0◦ C is not known, but is known
at some other temperature θ1 , then the resistance at any 2. A copper cable has a resistance of 30  at
temperature can be found as follows: a temperature of 50◦ C. Determine its resis-
tance at 0◦ C. Take the temperature coefficient
R1 = R0 (1 + α0 θ1 ) of resistance of copper at 0◦ C as 0.0043/◦ C
and R2 = R0 (1 + α0 θ2 ) [24.69 ]

Dividing one equation by the other gives: 3. The temperature coefficient of resistance for
R1 1 + α0 θ 1 carbon at 0◦ C is −0.00048/◦ C. What is the
= significance of the minus sign? A carbon resis-
R2 1 + α0 θ 2
tor has a resistance of 500  at 0◦ C. Determine
where R2 = resistance at temperature θ2 its resistance at 50◦ C. [488 ]

Problem 13. Some copper wire has a resistance 4. A coil of copper wire has a resistance of 20 
of 200  at 20◦ C. A current is passed through the at 18◦ C. If the temperature coefficient of resis-
wire and the temperature rises to 90◦ C. Determine tance of copper at 18◦ C is 0.004/◦ C, determine
the resistance of the wire at 90◦ C, correct to the the resistance of the coil when the temperature
nearest ohm, assuming that the temperature rises to 98◦ C [26.4 ]
coefficient of resistance is 0.004/◦ C at 0◦ C.
5. The resistance of a coil of nickel wire at
20◦ C is 100 . The temperature of the wire
R20 = 200 , α0 = 0.004/◦ C is increased and the resistance rises to 130 .
R20 [1 + α0 (20)] If the temperature coefficient of resistance of
and = nickel is 0.006/◦ C at 20◦ C, determine the
R90 [1 + α0 (90)]
temperature to which the coil has risen.
Hence [70◦ C]
R20 [1 + 90α0 ]
R90 =
[1 + 20α0 ] 6. Some aluminium wire has a resistance of 50 
200[1 + 90(0.004)] at 20◦ C. The wire is heated to a temperature
= of 100◦ C. Determine the resistance of the
[1 + 20(0.004)]
wire at 100◦ C, assuming that the temperature
200[1 + 0.36] coefficient of resistance at 0◦ C is 0.004/◦ C.
=
[1 + 0.08] [64.8 ]
200(1.36) 7. A copper cable is 1.2 km long and has a
= = 251.85 
(1.08) cross-sectional area of 5 mm2 . Find its
i.e. the resistance of the wire at 90◦ C is 252 , correct resistance at 80◦ C if at 20◦ C the resistivity of
to the nearest ohm copper is 0.02 × 10−6  m and its temperature
coefficient of resistance is 0.004/◦ C.
[5.95 ]
Now try the following exercise

Exercise 12 Further problems on the

temperature coefficient of 3.3 Resistor colour coding and
resistance ohmic values
1. A coil of aluminium wire has a resistance of
(a) Colour code for fixed resistors
50  when its temperature is 0◦ C. Determine
The colour code for fixed resistors is given in Table 3.1
 Resistance variation 25

Table 3.1 The fourth band, i.e. brown, indicates a toler-

Section 1
ance of ±1% from Table 3.1. Hence a colour coding
Colour Significant Multiplier Tolerance
of orange-orange-silver-brown represents a resistor of
Figures
value 0.33  with a tolerance of ±1%
Silver – 10−2 ±10%
Problem 15. Determine the value and tolerance
Gold – 10−1 ±5%
of a resistor having a colour coding of:
Black 0 1 – brown-black-brown.

Brown 1 10 ±1%
The first two bands, i.e. brown-black, give 10 from
Red 2 102 ±2% Table 3.1.
The third band, brown, indicates a multiplier of 10
Orange 3 103 – from Table 3.1, which means that the value of the
resistor is 10 × 10 = 100 
Yellow 4 104 – There is no fourth band colour in this case; hence,
from Table 3.1, the tolerance is ±20%. Hence a colour
Green 5 105 ±0.5%
coding of brown-black-brown represents a resistor of
Blue 6 106 ±0.25% value 100  with a tolerance of ±20%

Violet 7 107 ±0.1%

Problem 16. Between what two values should a
Grey 8 108 – resistor with colour coding
brown-black-brown-silver lie?
White 9 109 –

None – – ±20% From Table 3.1, brown-black-brown-silver indicates

10 × 10, i.e. 100 , with a tolerance of ±10%
This means that the value could lie between
(i) For a four-band fixed resistor (i.e. resistance
values with two significant figures): (100 − 10% of 100) 
yellow-violet-orange-red indicates 47 k with a
tolerance of ±2% and (100 + 10% of 100) 
(Note that the first band is the one nearest the end
of the resistor) i.e. brown-black-brown-silver indicates any value
(ii) For a five-band fixed resistor (i.e. resistance between 90  and 110 
values with three significant figures): red-yellow-
white-orange-brown indicates 249 k with a tol-
erance of ±1% Problem 17. Determine the colour coding for a
(Note that the fifth band is 1.5 to 2 times wider 47 k having a tolerance of ±5%.
than the other bands)
From Table 3.1, 47 k = 47 × 103 has a colour coding
Problem 14. Determine the value and tolerance of yellow-violet-orange. With a tolerance of ±5%, the
of a resistor having a colour coding of: fourth band will be gold.
orange-orange-silver-brown. Hence 47 k ± 5% has a colour coding of: yellow-
violet-orange-gold.
The first two bands, i.e. orange-orange, give 33 from
Table 3.1. Problem 18. Determine the value and tolerance
The third band, silver, indicates a multiplier of 102 of a resistor having a colour coding of:
from Table 3.1, which means that the value of the orange-green-red-yellow-brown.
resistor is 33 × 10−2 = 0.33 
 26 Electrical and Electronic Principles and Technology

orange-green-red-yellow-brown is a five-band fixed From Table 3.2, 4M7M is equivalent to: 4.7 M ± 20%
Section 1

resistor and from Table 3.1, indicates: 352 × 104  with

a tolerance of ±1%
Problem 21. Determine the letter and digit code
352 × 104  = 3.52 × 106 , i.e. 3.52 M
for a resistor having a value of 68 k ± 10%.
Hence orange-green-red-yellow-brown indicates
3.52 M ± 1% From Table 3.2, 68 k ± 10% has a letter and digit code
of: 68 KK

(b) Letter and digit code for resistors

Now try the following exercises
Another way of indicating the value of resistors is the
letter and digit code shown in Table 3.2.
Exercise 13 Further problems on resistor
Table 3.2 colour coding and ohmic values

Resistance Marked as: 1. Determine the value and tolerance of a resis-

tor having a colour coding of: blue-grey-
Value
orange-red [68 k ± 2%]
0.47  R47 2. Determine the value and tolerance of a resis-
tor having a colour coding of: yellow-violet-
1 1R0
gold
4.7  4R7 [4.7  ± 20%]
3. Determine the value and tolerance of a resis-
47  47R tor having a colour coding of: blue-white-
black-black-gold [690  ± 5%]
100  100R
4. Determine the colour coding for a 51 k four-
1 k 1K0 band resistor having a tolerance of ±2%
[green-brown-orange-red]
10 k 10 K
5. Determine the colour coding for a 1 M four-
10 M 10 M band resistor having a tolerance of ±10%
[brown-black-green-silver]
6. Determine the range of values expected for a
Tolerance is indicated as follows: F = ±1%, resistor with colour coding: red-black-green-
G = ±2%, J = ±5%, K = ±10% and M = ±20% silver [1.8 M to 2.2 M]
Thus, for example,
7. Determine the range of values expected for
R33M = 0.33  ± 20% a resistor with colour coding: yellow-black-
orange-brown [39.6 k to 40.4 k]
4R7K = 4.7  ± 10%
8. Determine the value of a resistor marked as
390RJ = 390  ± 5%
(a) R22G (b) 4K7F
[(a) 0.22  ± 2% (b) 4.7 k ± 1%]
Problem 19. Determine the value of a resistor
marked as 6K8F. 9. Determine the letter and digit code for a
resistor having a value of 100 k ± 5%
[100 KJ]
From Table 3.2, 6K8F is equivalent to: 6.8 k ± 1%
10. Determine the letter and digit code for a
resistor having a value of 6.8 M ± 20%
Problem 20. Determine the value of a resistor
[6 M8 M]
marked as 4M7M.