4.3.9  Cable rating calculation
The Regulations indicate
the following symbols for use when selecting cables:
Iz

is the current carrying
capacity of the cable in the situation where it
is installed 
It

is the tabulated current
for a single circuit at an ambient temperature of 30°C 
Ib

is the design current,
the actual current to be carried by the cable 
In

is the rating of the
protecting fuse or circuit breaker 
I2

is the operating current for the fuse or circuit
breaker (the current at which the fuse blows or the
circuit breaker opens) 
Ca

is the correction factor
for ambient temperature 
Cg

is the correction factor for grouping 
Ci

is the correction factor
for thermal insulation. 
The correction factor for protection by
a semienclosed (rewirable) fuse is not given a symbol but
has a fixed value of 0.725.
Under all circumstances, the cable current
carrying capacity must be equal to or greater than the circuit
design current and the rating of the fuse or circuit breaker
must be at least as big as the circuit design current. These
requirements are common sense, because otherwise the cable
would be overloaded or the fuse would blow when the load
is switched an.
To
ensure correct protection from overload, it is important
that the protective device operating current (I2) is not
bigger than 1.45 times the current carrying capacity of
the cable (Iz). Additionally, the rating of the fuse or
circuit breaker (In) must not be greater than the the cable
current carrying capacity (Iz) It is important to appreciate
that the operating current of a protective device is always
larger than its rated value. In the case of a backup fuse,
which is not intended to provide overload protection, neither
of these requirements applies.
To select a cable for a particular application,
take the following steps: (note that to save time it may
be better first to ensure that the expected cable for the
required length of circuit will] not result in the maximum
permitted volt drop being exceeded {4.3.11}).
1.
 Calculate the expected (design) current in the
circuit (Ib)
2.
 Choose the type and rating of protective device
(fuse or circuit breaker) to be used (In)
3.
 Divide the protective device rated current by the
ambient temperature
 correction factor (Ca)
if ambient temperature
differs from 30°C
4.
 Further divide by the grouping correction factor
(Cg)
5.
 Divide again by the thermal insulation correction
factor (CI)
6.
 Divide by the semienclosed fuse factor of 0.725
where applicable
7.
 The result is the rated current of the cable required,
which must be chosen
 from the appropriate
tables {4.6
to 4.9}.
Observe that one should divide by the correction
factors, whilst in the previous subsection we were multiplying
them. The difference is that here we start with the design
current of the circuit and adjust it to take account of
factors which will derate the cable. Thus, the current carrying
capacity of the cable will be equal to or greater than the
design current. In {4.3.7}
we were calculating by how much the current carrying capacity
was reduced due to application of correction factors.
{Tables
4.6 to 4.9} give current ratings and volt
drops for some of the more commonly used cables and sizes.
The Tables assume that the conductors and the insulation
are operating at their maximum rated temperatures. They
are extracted from the Regulations Tables shown in square
brackets e.g. [4D1A]
The examples below will illustrate the
calculations, but do not take account of volt drop requirements
(see
{4.3.11}).
Example 4.1
An immersion heater rated at 240 V, 3 kW
is to be installed using twin with protective conductor
p.v.c. insulated and sheathed cable. The circuit will be
fed from a 15 A miniature circuit breaker type 2, and will
be run for much of its 14 m length in a roof space which
is thermally insulated with glass fibre. The roof space
temperature is expected to rise to 50°C in summer, and where
it leaves the consumer unit and passes through a 50 mm insulationfilled
cavity, the cable will be bunched with seven others. Calculate
the crosssectional area of the required cable.
First calculate the design current Ib
Ib= 
P

= 3000A =

12.5A 

U

240


The ambient temperature correction factor
is found from {Table
4.3} to be 0.71. The group correction
factor is found from {Table
4.4} as 0.52. (The circuit in question is bunched
with seven others, making eight in all).
The thermal insulation correction factor
is already taken into account in the current rating table
(4D2A ref. method 4] and need not be further considered.
This is because we can assume that the cable in the roof
space is in contact with the glass fibre but not enclosed
by it. What we must consider is the point where the bunched
cables pass through the insulated cavity. From {Table
4.5} we have a factor of 0.89.
The correction factors must now be considered
to see if more than one of them applies to the same part
of the cable. The only place where this happens is in the
insulated cavity behind the consumer unit. Factors of 0.52
(Cg) and 0.89 (CI) apply. The combined value of these (0.463),
which is lower than the ambient temperature correction factor
of 0.71, and will thus be the figure to be applied. Hence
the required current rating is calculated:
Table 4.6  Current
ratings and volt drops for unsheathed single core p.v.c.
insulated cables 
Cross sectional area

In conduit in thermal insulation

In conduit in thermal insulation

In conduit on wall

In conduit on wall

Clipped direct

Clipped direct

Volt drop

Volt drop

(mm²)

(A)

(A)

(A)

(A)

(A)

(A)

(mV/A/m)

(mV/A/m)



2 cables

3 or 4 cables

2 cables

3 or 4 cables

2 cables

3 or 4 cables

2 cables

3 or 4 cables

1.0

11.0

10.5

13.5

12.0

15.5

14.0

44.0

38.0

1.5

14.5

13.5

17.5

15.5

20.0

18.0

29.0

25.0

2.5

19.5

18.0

24.0

21.0

27.0

25.0

18..0

15.0

4.0

26.0

24.0

32.0

28.0

37.0

33.0

11.0

9.5

6.0

34.0

31.0

41.0

36.0

47.0

43.0

7.3

6.4

10.0

46.0

42.0

57.0

50.0

65.0

59.0

4.4

3.8

16.0

61.0

56.0

76.0

68.0

87.0

79.0

2.8

2.4

Table 4.7  Current
ratings and volt drops for sheathed multicore p.v.c.insulated
cables 
Cross sectional area

In conduit in thermal insulation

In conduit in thermal insulation

In conduit on wall

In conduit on wall

Clipped direct

Clipped direct

Volt drop

Volt drop

(mm²)

(A)

(A)

(A)

(A)

(A)

(A)

(mV/A/m)

(mV/A/m)



2 core

3 or 4 core

2 core

3 or 4 core

2 core

3 or 4 core

2 core

3 or 4 core

1.0

11.0

10.0

13.0

11.5

15.0

13.5

44.0

38.0

1.5

14.0

13.0

16.5

15.0

19.5

17.5

29.0

25.0

2.5

18.5

17.5

23.0

20.0

27.0

24.0

18.0

15.0

4.0

25.0

23.0

30.0

27.0

36.0

32.0

11.0

9.5

6.0

32.0

29.0

38.0

34.0

46.0

41.0

7.3

6.4

10.0

43.0

39.0

52.0

46.0

63.0

57.0

4.4

3.8

16.0

57.0

52.0

69.0

62.0

85.0

76.0

2.8

2.4

Iz =
in = 15
A
= 32.4 A
Cg x Ci 0.52 x 0.89
From {Table 4.7}, 6 mm² p.v.c. twin with
protective conductor has a current rating of 32 A. This
is not quite large enough, so 10 mm²with a current rating
of 43 A is indicated. Not only would this add considerably
to the costs, but would also result in difficulties due
to terminating such a large cable in the accessories.
A more sensible option would be to look
for a method of reducing the required cable size. For example,
if the eight cables left the consumer unit in two bunches
of four, this would result in a grouping factor of 0.65
(from
{Table 4.4}). Before applying this, we must check
that the combined grouping and thermal insulation factors
(0.65 x 0.89 = .0.58) are still less than the ambient temperature
factor of 0.71, which is the case.
Table 4.8  Current
ratings of mineral insulated cables clipped direct 
Crosssectional area

Volt

p.v.c. sheath
2 x single or twin

p.v.c. Sheath 3 core

p.v.c. Sheath 3 x single
or twin

Bare sheath 2 x single

Bare sheath 3 x single

(mm²)


(A)

(A)

(A)

(A)

(A)

1.0

500v

18.5

16.5

16.5

22.0

21.0

1.5

500v

24.0

21.0

21.0

28.0

27.0

2.5

500v

31.0

28.0

28.0

38.0

36.0

4.0

500v

42.0

37.0

37.0

51.0

47.0

1.0

750v

20.0

17.5

17.5

24.0

24.0

1.5

750v

25.0

22.0

22.0

31.0

30.0

2.5

750v

34.0

30.0

30.0

42.0

41.0

4.0

750v

45.0

40.0

40.0

55.0

53.0

6.0

750v

57.0

51.0

51.0

70.0

67.0

10.0

750v

78.0

69.0

69.0

96.0

91.0

16.0

750v

104.0

92.0

92.0

127.0

119.0

Note that in (Tables
4.8 and 4.9) 'P.V.C. Sheath means bare and exposed to touch
or having an overall covering of p.v.c. or LSF and 'Bare'
means bare and neither exposed to touch nor in contact with
combustible materials.
Table 4.9  Volt drops
for mineral insulated cables 
Crosssectional area

Singlephase p.v.c. Sheath

Singlephase bare

Threephase p.v.c. Sheath

Threephase bare

(mm²)

(mV/A/m)

(mV/A/m)

(mV/A/m)

(mV/A/m)

1.0

42.0

47.0

36.0

40.0

1.5

28.0

31.0

24.0

27.0

2.5

17.0

19.0

14.0

16.0

4.0

10.0

12.0

9.1

10.0

6.0

7.0

7.8

6.0

6.8

10.0

4.2

4.7

3.6

4.1

16.0

2.6

3.0

2.3

2.6

This leads to a cable current rating of 15 A = 25.9 A
0.65
x 0.89
This is well below the rating for 6
mm² of 32 A, so a cable of this size could be selected.
Example 4.2
The same installation as in Example 4.1 is proposed.
To attempt to make the cable size smaller, the run in the
roof space is to be kept clear of the glass fibre insulation.
Does this make any difference to the selected cable size?
There is no correction factor for the presence
of the glass fibre, so the calculation of Iz will be exactly
the same as Example 4.1 at 32.4 A.
This time reference method I (clipped direct)
will apply to the current rating {Table
4.7}. For a two core cable, 4.0 mm², two core has a
rating of 36 A, so this Will be the selected size.
It is of interest to notice how quite a
minor change in the method of installation, in this case
clipping the cable to the joists or battens clear of the
glass fibre, has reduced the acceptable cable size.
Example 4.3
Assume that the immersion heater indicated in the
two previous examples is to be installed, but this time
with the protection of a 15 A rewirable (semienclosed)
fuse. Calculate the correct cable size for each of the alternatives,
that is where firstly the cable is in contact with glass
fibre insulation, and secondly where it is held clear of
it.
This time the value of the acceptable current
carrying capacity Iz will be different because of the need
to include a factor for the rewirable fuse as well as the
new ambient temperature and grouping factors for the rewirable
fuse from {Tables
4.3 and 4.4}.
Iz = in = 15 A = 45.7 A
Cg x Ca x 0.725 0.52
x 0.87 x 0.725
In this case, the cable is in contact with
the glass fibre, so the first column of {Table
4.7} of current ratings will apply. The acceptable cable
size is 16 mm² which has a current rating of 57 A.
This cable size is not acceptable on the
grounds of high cost and because the conductors are likely
to be too large to enter the connection tunnels of the immersion
heater and its associated switch. If the cables leaving
the consumer unit are rearranged in two groups of four,
this will reduce the grouping factor to 0.65, so that the
newly calculated value of Iz is 36.6 A. This means using
10 mm² cable with a current rating of 43A (from
{Table 4.7}), since 6 mm² cable is shown to have a current
rating in these circumstances of only 32 A. By further rearranging
the cables leaving the consumer unit to be part of a group
of only two, Cg is increased to 0.8, which reduces Iz to
29.7 A which enables selection of a 6 mm² cable.
Should it be possible to bring the immersion
heater cable out of the consumer unit on its own, no grouping
factor would apply and Iz would fall to 23.8 A, allowing
a 4 mm² cable to be selected.
Where the cable is not in contact with
glass fibre there will be no need to repeat the calculation
of Iz, which still has a value of 29.7 A provided that it
is possible to group the immersion heater cable with only
one other where it leaves the consumer unit. This time we
use the 'Reference Method 1 (clipped direct)' column of
the current rating {Table
4.7}, which shows that 4 mm² cable with a current rating
of 36 A will be satisfactory.
Examples
4.1, 4.2 and 4.3 show clearly how forward planning will
enable a more economical and practical cable size to be
used than would appear necessary at first. It is, of course,
important that the design calculations are recorded and
retained in the installation manual.
Example 4.4
A 415 V 50 Hz threephase motor with an output of
7,5 kW, power factor 0.8 and efficiency 85% is the be wired
using 500 V light duty threecore mineral insulated p.v.c.
sheathed cable. The length of run from the HBC protecting
fuses is 20 m, and for about half this run the cable is
clipped to wall surfaces. For the remainder it shares a
cable tray, touching two similar cables across the top of
a boiler room where the ambient temperature is 50°C. Calculate
the rating and size of the correct cable.
The first step is to calculate the line
current of the motor.
Input =
output = 7.5 x 100 kW = 8.82 kW
efficiency 85
Line current Ib = P
=
8.82 x 103 A
= 15.3 A
 Ö3 x UL x
cosÆ Ö3
x 415 x 0.8
We must now select a suitable fuse. {Fig
3.15} for BS 88 fuses shows the 16 A size to be the
most suitable. Part of the run is subject to an ambient
temperature of 50°C, where the cable is also part of a group
of three, so the appropriate correction factors must be
applied from {Tables
4.3 and 4.4}.
Iz =
In =
16A
= 34.2A
Cg x Ca  0.70x 0.67
Note that the grouping factor of 0.70 has
been selected because where the cable is grouped it is clipped
to a metallic cable tray, and not to a nonmetallic surface.
Next the cable must be chosen from {Table
4.8}. Whilst the current rating would be 15.3 A if all
of the cable run were clipped to the wall, part of the run
is subject to the two correction factors, so a rating of
34.2 A must be used. For the clipped section of the cable
(15.3 A), reference method I could be used which gives a
size of 1.0 mm² (current rating 16.5 A). However, since
part of the cable is on the tray (method 3) the correct
size for 34.2 A will be 4.0 mm², with a rating of 37 A.