00:01
So we need to identify the power supply to the house, which is p equals v i.
00:06
So the rate at which electrical energy is dissipated in the wires is i squared of r, where r equals pl divided by a.
00:18
So just a little side note here, copper would be p equals 1 .72 times 10 to the negative 8.
00:29
So for part a here, the line voltage current to be drawn and wire diameter are what must be considered in household writing or wiring.
00:42
So let me write that out.
00:44
The line voltage current to be drawn and wire diameter are what must be considered in household.
01:16
Wiring.
01:21
For part b, we have p equals vi, which gives i equals p over v, which is 3 ,900 divided by 230.
01:46
That leaves you with 35 as.
01:49
So the 8 -gauge wire is necessary, since it can carry up to for a's.
02:02
So for part c, we have p equals i squared of r, which equals i squared pl over a, which is 35 a's squared times 1 .72 times 10 to the negative 8.
02:24
That's the number i gave you at the very beginning, times 42 .0 meters divided by pi over 4 times 0 .00326 meters squared...