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.
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.
For a 50 feet distance with 120 volts and 12 amps, you should use a 14-gauge wire to ensure electrical safety and efficiency. A 14-gauge wire is rated to handle up to 15 amps for household wiring applications.
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.
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
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.