6 gauge if it is a very short distance.
You will need a 40 amp breaker using AWG # 8 wire running it a distance of no more than 50 feet.
50 amps.
the size of the wire that you need to run depends on the the amperage of the device/appliance you'll be hooking up to that line. Use #12 wire for 20 amp, #10 for 30 amp, #8 wire for 40 amps and #6 for 50 amps. Hope that helps.
A #6 copper conductor is rated at 65 amps. To connect to a 50 amp RV outlet you will need a four conductor #6 SOW 600 volt cabtire. The black and red wires go on terminals X and Y. The white wire goes on terminal W. The green wire goes on terminal G.
6 gauge if it is a very short distance.
You will need a 40 amp breaker using AWG # 8 wire running it a distance of no more than 50 feet.
50 amps.
the size of the wire that you need to run depends on the the amperage of the device/appliance you'll be hooking up to that line. Use #12 wire for 20 amp, #10 for 30 amp, #8 wire for 40 amps and #6 for 50 amps. Hope that helps.
14 AWG is fine for this application.
Wire size depends on amps. Use #10 for 30a. #8 for 40 or 50a. #6 for 60a.
Depends, there are fuses ranging from 5 amps up to 50 amps in use on the Neon.
10 awg wire can have a breaker size of 30amps. It can actually carry more but as for codes the wire is usually allowed to carry 80% of its max capability which puts the breaker at a max size of 30 amps.
For the short distance of 50 feet no voltage drop calculations are needed. A #10 copper conductor with a insulation factor of 60, 75 or 90 degrees C is rated at 30 amps respectively. This conductor can legally be loaded to 80% capacity. This allows 30 x .8 = 24 amps. If the device draws more that 24 amps then go to the next wire size which is a #8. A #8 copper conductor with an insulation factor of 75 and 90 degrees C is rated at 45 amps respectively. This legally allows 45 x .8 = 36 amps to flow on the wire. Of course with this size wire only 30 amps will be allow to flow before the breaker will trip.
Yes, there will be enough capacity to run a 500 watt sensor light. Assuming by the wire size, the question is from a 50 Hz country where the operating voltage is 240 volts. Amps = Watts/Volts = 500/240 = 2.08 amps. The ampacity of a 1.5 mil conductor is 15 amps. This size wire is equal to a #14 wire AWG.
You need to mention the current or connected load to define the size of the wire
To answer this question, wire size is rated in the amount of amperage that it can legally carry. The formula to find amperage when the HP is known is I = HP x 746/1.73 x E x %eff x pf. A standard motor's efficiency between 5 to 100 HP is .84 to .91. A standard motor's power factor between 10 to 100 HP is .86 to .92. Amps = 10 x 746 = 7460 = 7460/1.73 x 220 x .84 x .86 = 7460/275 = 27.1 amps. The electrical code states that a motor conductor has to be rated at 125% of the motors full load amperage. 27.1 x 125% = 33.8 A #8 copper conductor with an insulation factor of 60, 75 or 90 degrees C is rated at 40, 50 and 55 amps respectively.