5.4.5  Protective conductor crosssection calculation
The c.s.a. of the circuit protective conductor
(c.p.c.) is of great importance since
the level of possible shock in the event
of a fault depends on it (as
seen in {5.4.4}).
Safety could always be assured if we assessed
the size using {Table
5.7} as a basis.
However, this would result in a more expensive
installation than necessary because we would often use protective
conductors which are larger than those found to be acceptable
by calculation. For example. twin with cpc insulated and
sheathed cables larger than 1 mm² would be ruled out because
in all other sizes the CPC is smaller than required by {Table
5.7}.
In very many cases, calculation of the
CPC size will show that a smaller size than that detailed
in {5.4.4}
is perfectly adequate. The formula to be used is:
S =

Ö(Ia²t)



k


where

S
is the minimum protective conductor crosssectional
area (mm^{2}) 

Ia
is the fault current (A) 

t
is the opening time of the protective device
(s) 

k
is a factor depending on the conductor material
and insulation, and the initial and maximum insulation
temperatures. 
This is the same formula as in {3.7.3},
the adiabatic equation, but with a change in the subject.
To use it, we need to have three pieces of information,
Ia, t and k.
1) To find Ia 


Since Ia =

Uo

we need values for Ia
= uo and Zs 

Zs






Uo

is simply the supply
voltage, which in most cases will be 240V. 

Zs

is the earthfault loop
impedance assuming that the fault has zero impedance. 
Since we must assume that we
are at the design stage, we cannot measure the loop impedance
and must calculate it by adding the loop impedance external
to the installation (Ze) to the resistance of the conductors
to the furthest point in the circuit concerned. This technique
was used in
{5.3.6}.
Thus, Zs = Ze + R1 + R2 where R1 and R2
are the resistances of the phase and protective conductors
respectively from {Table
5.5}.
2) To find t
We can find t from the time/current characteristics
of {Figs
3.13 to 3.19} using the value of Ia already calculated
above. For example, if the protective device is a 20 A miniature
circuit breaker type I and the fault current is 1000 A,
we shall need to consult {Fig
3.16}, when we can read off that operation will be in
0.01 s (10 ms). (It is of interest here to notice that if
the fault current had been 80 A the opening time could have
been anything from 0.04 s to 20 s,so the circuit would not
have complied with the required opening times).
3) To find k
k is a constant, which we cannot calculate but must
obtain from a suitable table of values. Some values of k
for typical protective conductors are given in
{Table 5.8}.
It is worth pointing out here that correctly
installed steel conduit and trunking will always meet the
requirements of the Regulations in terms of protective conductor
impedance.
Although appearing a little complicated,
calculation of acceptable protective conductor size is worth
the trouble because it often allows smaller sizes than those
shown in {Table
5.7}.
Table 5.8  Values
of k for protective conductors 
Nature of protective conductor

Initial temp.
(°C)

Final temp
(°C)

Conductor
material

K

p.v.c. insulated, not in cable
or bunched

30

160

Copper

143



30

160

Aluminium

95



30

160

Steel

52



p.v.c. insulated, in cable
or bunched

70

160

Copper

115



70

160

Aluminium

76



Steel conduit or trunking

50

160

Steel

47



Bare conductor

30

200

Copper

159



30

200

Aluminium

105



30

200

Steel

58

Example 5.2
A load takes 30 A from a 240 V single phase supply
and is protected by a 32 A HBC fuse to BS 88. The wiring
consists of 4 mm² single core p.v.c. insulated cables run
in trunking, the length of run being 18 m. The earthfault
loop impedance external to the installation is assessed
as 0.7 Ohms. Calculate the crosssectional area of a suitable
p.v.c. sheathed protective conductor.
This is one of those cases where we need
to make an assumption of the answer to the problem before
we can solve it. Assume that a 2.5 mm² protective conductor
will be acceptable and calculate the combined resistance
of the phase and protective conductors from the origin of
the installation to the end of the circuit. From {Table
5.5}, 2.5 mm² cable has a resistance of 7.4 mohms/m
and 4 mm² a resistance of 4.6 mOhms/m. Both values must
be multiplied by 1.2 to allow for increased resistance as
temperature rises due to fault current.
Thus, R1 + R2 = 
(7.4 + 4.6) x 1.2 x 18

Ohms = 
12.0 x 1.2 x 18

= 0.26 Ohms 

1000


1000


This conductor resistance must be added
to external loop impedance to give the total earthfault
loop impedance.
Zs = Ze + Rl + R2 = 0.7+ 0.26
Ohms = 0.96 Ohms 
We can now calculate the fault current:
la = 
Uo

= 
240

= 250A 

Zs


0.96


Next we need to find the operating
time for a 32 A BS 88 fuse carrying 250 A. Examination of
{Fig
3.15} shows that operation will take place after 0.2
s.
Finally, we need a value for k. From {Table
5.8} we can read this off as 115, because the protective
conductor will be bunched with others in the trunking.
We now have values for Ia, t and k so we
can calculate conductor size.
S = 
Ö(Ia²t)

= 
Ö(250²
x 0.02)

mm² = 0.97 mm² 

k


115


This result suggests that a 1.0 mm² protective
conductor will suffice. However, it may he dangerous to
make this assumption because the whole calculation has been
based on the resistance of a 2.5 mm² conductor. Let us start
again assuming a 1.5 mm² protective conductor and work the
whole thing through again.
The new size protective conductor has a
resistance of 18.1 mOhms/m, see {Table
5.5}, and with the 4 mm² phase conductor gives a total
conductor resistance, allowing for increased temperature,
of 0.491 Ohms. When added to external loop impedance this
gives a total earthfault loop impedance of 1.191 Ohms and
a fault current at 240 V of 202 A. From {Fig
3.15} operating time will be 0.6 s. The value of k will
be unchanged at 115.
S = 
Ö(Ia²t)

= 
Ö(202²
x 0.6)

mm² = 1.36 mm² 

k


115


Thus, a 1.5 mm² protective conductor
can be used in this case. Note that if the size had been
assessed rather than calculated, the required size would
be 4 mm², two sizes larger. A point to notice here is that
the disconnection time with a 1.5mm² protective conductor
is 0.6 s, which is too long for socket outlet circuits (0.4
s max.).
Example 5.3
A 240 V, 30 A ring circuit for socket outlets is 45
m long and is to be wired in 2.5 mm² flat twin p.v.c. insulated
and sheathed cable incorporating a 1.5 mm² cpc. The circuit
is to be protected by a semienclosed (rewirable) fuse to
BS 3036, and the earthfault loop impedance external to
the installation has been ascertained to be 0.3 Ohms. Verify
that the 1.5 mm² cpc enclosed in the sheath is adequate.
First use {Table
5.5} to find the resistance of the phase and cpc conductors.
These are 7.4 mOhms/m and 12.1 mOhms/m respectively, so
for a 45 m length and allowing for the resistance increase
with temperature factor of 1.2.
R1 + R2 = 
(7.4 + 12.1) x 1.2 x 45

Ohms = 
19.5 x 1.2 x 45

Ohms = 1.05 Ohms 

1000


1000


Zs = Ze + 
R1+ R2

= 0.3 + 
1.05

Ohms = 0.3 + 0.263 Ohms
= 0.563 Ohms 

4


4


The division by 4 is to allow for the ring
nature of the circuit.
Ia = 
Uo

= 
240

A = 426A 

Zs


0.563


We must then use the time/current
characteristic of {Fig
3.13} to ascertain an operating time of 0.10 s.
From {Table
5.8} the value of k is 115.
Then 
S = 
Ö(Ia²t)

= 
Ö(426²
x 0.10)

mm² = 1.17 mm² 


k


115


Since this value is smaller than the intended
value of 1.5 mm², this latter value will be satisfactory.
Example 5.4
A 240 V singlephase circuit is to be wired in p.v.c.
insulated single core cables enclosed in plastic conduit.
The circuit length is 45 m and the live conductors are 16
mm² in crosssectional area. The circuit will supply fixed
equipment, and is to be protected by a 63 A HBC fuse to
BS 88. The earthfault loop impedance external to the installation
has been ascertained to be 0.58 Ohms. Calculate a suitable
size for the circuit protective conductor.
With the information given this time the
approach is somewhat different. We know that the maximum
disconnection time for fixed equipment is 5 s, so from the
time/current characteristic for the 63 A fuse {Fig
3.15} we can see that the fault current for disconnection
will have a minimum value of 280 A.
Thus, Zs = 
Uo

= 
240

Ohms
= 0.857 Ohms 

Ia


280


If we deduct the external loop
impedance, we come to the resistance of phase and protective
conductors.
R1 +R2 = ZsZe = 0.8570.58Ohms
= 0.277 Ohms 
Converting this resistance to
the combined value of R1 and R2 per metre,
(R1 + R2) per metre = 
0.277 x 1000

mOhms/m = 5.13 mOhms/m 

45 x 1.2


Consulting
{Table 5.5} we find that the resistance of 16 mm² copper
conductor is 1.15 mOhms/m, whilst 10 mm² and 6 mm² are 1.83
and 3.08 mOhms/m respectively. Since 1.15 and 3.08 add to
4.23, which is less than 5.13, it would seem that a 6 mm²
protective conductor will be large enough. However, to be
sure we must check with the adiabatic equation.
R1 + R2 = 
(1.15+3.08) x 1.2 x 45

Ohms = 0.228 Ohms 

1000


Zs = Ze+(R1 +R2) = 0.58+0.228
Ohms = 0.808 Ohms 
Ia = 
Uo

= 
240

A = 297 A 

Zs


0.808


From {Fig
3.15} the disconnection time for a 63 A fuse carrying
297 A is found to he 3.8s.
From {Table
5.8} the value of k is 115.
Then S = 
Ö(Ia²t)

= 
Ö(297²
x 3.8)

mm² = 5.03 mm² 

k


115


Since 5.03 is less than 6 then a 6 mm²
protective conductor will be large enough to satisfy the
requirements