When considering cable ratings care needs to taken as this is not as straight forward as it may seem. For example, if you check the manufacturers current rating for 1.5 twin and earth cable (suitable for most lighting circuits in a domestic environment) it will state that it is around 20 amps, but this value would only apply if the cable was in free air (as it can cool easily). If you put the same cable above a plasterboard ceiling covered by thermal insulation 100mm thick (17th edition regs reference method 100) its current carrying capacity is reduced to 16 amps (because it can no long cool as easily).
Consider the table below and the reference methods below that to see how the current carrying capacity is affected by installing the cable in different situations.
Conductor area (mm²) 
Current carrying capacity (AMPS) 

A 
B 
C 
100 
101 
102 
103 

1 
11.5 
13 
16 
13 
10.5 
13 
8 
1.5 
14.5 
16.5 
20 
16 
13 
16 
10 
2.5 
20 
23 
27 
21 
17 
21 
13.5 
4 
26 
30 
37 
27 
22 
27 
17.5 
6 
32 
38 
47 
34 
27 
35 
23.5 
10 
44 
52 
64 
45 
36 
47 
32 
16 
57 
69 
85 
57 
46 
63 
42.5 
Examples of installation method
BS7671 17th Editon Reference Method 
Example of installation method 
A  In conduit in an insulated wall with the conduit close to or touching the inner skin. 
B  Enclosed in conduit or trunking on or in a wall. 
C  Clipped direct, or sheathed cables embedded directly in masonry, brickwork, concrete, plaster or the like (other than thermally insulating materials) 
100  Above a plasterboard ceiling covered by thermal insulation, insulation thickness less than 100 mm. 
101  Above a plasterboard ceiling covered by thermal insulation, insulation thickness greater than 100 mm. 
102  In a stud wall with thermal insulation with the cable touching the inner wall surface. 
103  In a stud wall with thermal insulation with the cable NOT touching the inner wall surface. 
Cable sizing example
So lets say we wish to rewire a lighting circuit for the upstairs of our house. We have 3 bedrooms, a bathroom and a landing light. So lets assume we will be powering 5 100w lamps which gives us 500 watts in total.
We plan to use our standard 1.5mm twin and earth cable that the manufacturer rates at 20 amps, but the cables are run under thick (more than 100mm) insulation in the loft (installation method 101) so our cables current carrying capacity is reduced to 13 amps (from top table).
So if we now take our total wattage and divide it by the voltage of the circuit we will see how much current (amps) our lights will draw:
500(watts) / 240(Volts) = 2.08 amps
This is well within the 13 amp current carrying capacity of our cable so the 1.5 twin and earth cable will be fine for the job.
Note
This is quite a crude calculation, circuit designers will take additional factors into consideration when sizing cables, for example, in large installations the length of cable (and the resulting volt drop along it, due to its resistance) will need to be considered. Also my power calculation is not entirely accurate because we would need to consider the power factor of the circuit; inductive loads (like the chokes in fluorescent lamps and low voltage lighting transformers) have the effect of putting the current and the voltage in the circuit ‘out of phase’ which will affect the current drawn.
In summary, if you have an average size 3 4 bedroom house with runofthemill lamps and the odd fluorescent lamp, and after doing the calculation above you are well within the current carrying capacity of the cable (after allowing for the installation method as described) and you protect the circuit with a 6 amp circuit breaker you can’t go far wrong.
In a large dwelling (with potentially long cable runs) and a mix of exotic lighting with low voltage lighting transformers etc the basic calculations above may not be sufficient and I would recommend consulting an expert.