Use AWG # 3 copper.
10 gauge
A # 14 copper conductor will be fine to carry 8 amps at 120 volts. This size conductor is rated at 15 amps.
To answer this question a voltage needs to be stated. Wire is sized by the amount of amperage the load takes. W = Amps x Volts. Amps = 650/ volts.
The V stands for volts and A is amps. If for example you have a 12kVA device and are running off a voltage of 120 volts then Amps = 12000/120 = 100. You then use the calculated amps in a wire size table to get the correct size.
Current is inversely proportional to resistance. If you double the resistance, you halve the current. Ohm's Law: Volts = Amps * Ohms Solve for Amps: Amps = Volts / Ohms
Fuses are rated in Amps. Although the physical size of a fuse is to do with volts; the further the terminals are apart the less likelihood there is of 'sparkover' between them.
Assuming 120 volts, you would need at least 4 AWG, which would give you a 5.6 percent voltage drop at 56.25 amps (i.e., full-time 45 amps derated for 80 percent design rule). At 240 volts you would only need 6 AWG, giving you a 4.5 percent voltage drop at 45 amps.
10 guage
resisters actually slow down the amount of volts and or amps going thru them.If you have say 12 volts and you want only 6 volts you use a formula to determine the size opf resisters you need to accomplish this.
#8 copper
Depending on size of Fridge. But AVERAGE is 12 volts for fridge, circuit necessity 15 amps 15 amps X 120 Volts=1800 watts minimum...I'm LEARNING myself
The formula you are looking for is Watts = Amps x Volts. Amps = Watts/Volts. This comes to 4 amps load. Minimum size fuse would be 5 amps.