What seems to be missing out of this discussion is Power. Every electrical item the designers install use it.
As a review: Power (P) = Current (I) times Voltage (E) or P=IE.
That can then also be reworked to P/E = I, which is where the wire size comes into play.
For example,
If a landing light consumes 250 Watts, a typical value:
@ 14 V (AC or DC, it does not matter) that is 17 Amps (round values)
@ 28 V that is 9 Amps
@ 115 V that is 2 Amps
Now, let us look at the wire size to carry that. From the circuit breaker in the cockpit to the wing root on Boeing 707 (think back when this was first being done, and probably more like a DC 4) that wire run would be about 80 feet.
To carry 14 Volts @ 9 Amps and only have 1 volt drop (allowed for intermittent loads like landing lights) an 8 AWG wire is required, which weights in around 5 lbs and is 0.17 in diameter.
28 V at 9 Amps uses a 12 AWG wire and 2 volt drop which weights in around 2 lbs, and is .09 in diameter
115 V at 2 Amps is smaller than 24 AWG and an 8 volt drop. Back then about the smallest that was used was 20 AWG which weights in around 1/2 lbs at 0.04 in diameter.
As noted, designs have changed in time and we can now generate higher voltage DC systems. Back in the 707 day the only way was AC systems. In addition, the AC system was 3 phase, which helps efficiency in things like motors and power conversion systems. DC switching systems now change all of that.