A domestic consumer unit on a TN-C-S system uses a 15A Type 2 mcb, to BS3036 to protect a radial circuit wired in 2.5mm^2 two-core pvc insulated cable with a 1.5mm^2 protective conductor clipped directly to a surface. If this circuit supplies socket outlets for hand held equipment, the maximum length of run is:
The answer from the paper gives a correct answer of 35m.
My assumptions are:
Best way to tackle this problem is to rearrange the formula for voltage drop i.e.
V_drop = [(mV/A/m)*(I_b)*L]/1000
L = [1000*(V_drop)] /[(mV/A/m)*(I_b)]
Problem is the question doesn't give me a value for the design current: I_b.
So I've used a hypothetical value of 14A, since 14A < 15A satisfying I_b < I_n.
Using table 4D5 from the regs I get a voltage drop of 18V (mV/A/m) for installation ref method C clipped direct for 2.5 twin + earth cable.
Using these values in the above formula gives:
L_max = (1000*18)/(18*14) = 71.42m
Some questions then,
Am I missing something?
Am I ok to use a design current of 14A?