introduction to DC circuits

Our objective in this experiment was to determine the max cable resistance required, the max distance the battery and load can be separated with awg#30 cable, the distribution efficiency, and the time it takes for the battery to discharge

we determine the theoretical value for R(load) to be 1000Ω

Vload=11.01V
Ibatt=11.6mA
Rcabletot=73Ω

Time to discharge
.8Ahr/.0116A=68.97 hours

power to the load=.1309 W
power to the cable=.0098W

the efficiency is (.1309/(.1309+.0098))100=93%

we are not exceeding the power capability of the resistor box because we are well below .13w

the max distance of the wire is 73/.3451=105.77m


e. (optional)
the ohm per foot ratio in AWG#28 wire is 0.06490Ω
V=IR
5=(.020)R
R=250Ω
250/.06490=3,852 ft.

however we will probably not be getting ideal values, therefore done with a  voltage of .2
.2=(.020)R
R=10Ω
10/.06490=154 ft.

only being able to drop 12V I calculated R=1.2ohms dividing that by 150 ft. I get .024
therefore,
if the sub is approximately 150 ft. away then the minimum cable gauge would be #22