10 AWG in copper.
14 AWG is fine for this application.
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.
You would need at least 3 AWG at 120 volts, giving you a 4.8 percent voltage drop at the maximum load of 37.5 amps (using 30 Amps at the ordinary 80 percent rated capacity of the circuit). For 240 volts you would only need 6 AWG.
It depends on whether this is residential or commercial.Commercial has to be 12 guage in conduit.
10 AWG.
#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.
14 AWG is fine for this application.
A # 14 copper conductor will be fine to carry 8 amps at 120 volts. This size conductor is rated at 15 amps.
For single phase 30 amps at 120 volts you would need a #8 copper wire with an insulation rating of 90 degrees C.
10 guage
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.
Each baseboard heater will draw a little over 4 amps at 120 volts or 2 Amps at 240 volts. The total number of baseboards on a circuit will draw the sum of these amps. Keep the load under 80% of the amperage rating of the breaker.
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.
You would need at least 3 AWG at 120 volts, giving you a 4.8 percent voltage drop at the maximum load of 37.5 amps (using 30 Amps at the ordinary 80 percent rated capacity of the circuit). For 240 volts you would only need 6 AWG.