Strategy for powering surveillance cameras
The closest table of conductor resistances I have to hand is for 50Hz AC resistances of metric wire. For the purpose of this exercise, assume that the DC resistance is approximately the same as the AC resistance at 50Hz.

Suppose each camera requires 6 watts at 5 volts. This is 1.2 amperes.


The a.c. resistance of 2.5mm² conductor (common household mains wiring size) is 9.01 Ω/km at 75° C.
You need 50 metres of it (25 metres each way), so the circuit resistance is 9.01 Ω/km ×  0.050 km = 0.45 Ω.
By Ohm's Law, 1.2 amps × 0.45 Ω = 0.54 V of voltage drop.
Typically, a voltage supply is expected to be within 10% of the nominal voltage. In this case, 10% of 5V is 0.50V, so the voltage drop of 0.54 V is probably too much.


If you go up to 4mm² conductor, with 5.61 Ω/km resistance, the voltage drop is only 0.33 V which is probably tolerable.

Personally, I would be inclined to run 4mm² conductors at 5 VDC. Low voltage is less of an electrocution risk, and you can install it yourself. You should (must?) mitigate the risk of fire (from faulted circuit) by installing a 4 amp fuse on each circuit.

Running 240 VAC to each camera means that you would need an electrician to do the work, you need to buy 16 individual power supplies, and those power supplies will be in very inconvenient places when you need to replace them.