This depends on what rating your over-current device is. For a 48 amp load, the minimum breaker size would normally be 60A. If your breaker or fuse is 60A, then your wire size would be #4 copper, or IF both wire and connection points were rated for 75 degrees C or higher you could use #6 copper. See NEC Article 310 table 310.16.
Your situation may be different due to the type of appliance you are connecting and your local codes. Best to check with a licensed electrician to be sure.
Keep in mind that the electrical code only allows the conductor to be loaded to 80% for loads classed as continuous. If the 48 amps is true load current then a #6 copper wire with a insulation rating of 90 degrees C would be used. The amp rating of a #6 R90 copper wire is 65 amps x 80% = 52 amps for a continuous load, and 65 amps for a noncontinuous load. See 2005 NEC 215.2(A)(1) including the exception.
<><><> As always, if you are in doubt about what to do, the best advice anyone should give you is to call a licensed electrician to advise what work is needed. Before you do any work yourself,
on electrical circuits, equipment or appliances,
always use a test meter to ensure the circuit is, in fact, de-energized. IF YOU ARE NOT ALREADY SURE YOU CAN DO THIS JOB
SAFELY AND COMPETENTLY
REFER THIS WORK TO QUALIFIED PROFESSIONALS.
The ampacity of a wire has nothing to do with the voltage of the conductor. The voltage of a conductor is an insulation factor. The common voltage ratings are 300, 600 and 1000 volts. The current rating of a #12 wire with an insulation rating of 60, 75 or 90 degrees C is 20 amps. A derated by 80%, 20 amp conductor for a continuous load is, 20 x .8 = 16 amps. The distance of 23 feet does not enter into the equation.
Answer for USA, Canada and countries running a 60 Hz supply service.
The simple answer is #8 AWG.
But this assumes you are using THHN or similarly insulated wire. Generally, THHN is what you get when you go to the hardware store, and it DOES MATTER!
<><><>
As always, if you are in doubt about what to do, the best advice anyone should give you is to call a licensed electrician to advise what work is needed.
Before you do any work yourself,
on electrical circuits, equipment or appliances,
always use a test meter to ensure the circuit is, in fact, de-energized.
IF YOU ARE NOT ALREADY SURE YOU CAN DO THIS JOB
SAFELY AND COMPETENTLY
REFER THIS WORK TO QUALIFIED PROFESSIONALS.
The length of wire needed will dictate wire size as much as amps and volts.
6 gauge
10 AWG in copper.
Amps * Volts = Watts So, Watts / Volts = Amps 2000 / 240 = 8.333 Amps You should run the circuit on a two pole 15 Amp breaker, using 14 AWG, 2 conductor (plus ground) wire, just so you have a little safety factor in the circuit size.
Wire is sized by the amperage applied to the wire. To answer this question a voltage needs to be stated. I = W/E. Amps = Watts/Volts. A #14 copper conductor is rated at 15 amps.
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.
Cable sizing is based on load amperage. The formula you need to use is Amps = Watts/Volts. Amps = 1000/220 = 4.55 amps. A #14 copper wire with a insulation factor of 60, 75 and 90 degrees C is rated at 15, 15 and 15 amps respectively.
A # 14 copper conductor will be fine to carry 8 amps at 120 volts. This size conductor is rated at 15 amps.
3/0 wire 3/0 wire
#8 copper
10 gauge
Use AWG # 3 copper.
10 AWG in copper.
You physically can but sit will be unsafe, and will not protect the circuit because it will not blow when it should. Only replace a fuse with the same size fuse.
4/0 (4 ought) in copper will handle 250 amps. The voltage drop at 175 ft. is about 2.14 volts which should be okay.
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.
Amps * Volts = Watts So, Watts / Volts = Amps 2000 / 240 = 8.333 Amps You should run the circuit on a two pole 15 Amp breaker, using 14 AWG, 2 conductor (plus ground) wire, just so you have a little safety factor in the circuit size.
Wire is sized by the amperage applied to the wire. To answer this question a voltage needs to be stated. I = W/E. Amps = Watts/Volts. A #14 copper conductor is rated at 15 amps.
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