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
#8 copper
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.
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.