You need a breaker rated for 10 amps and 250 volts. The breaker must also fit properly in your panel.
For a 10 amp, 250 volt circuit, you should install a circuit breaker rated for at least 10 amps and 250 volts. A common option would be a single-pole, 10 amp circuit breaker designed for 240-250 volt applications. Be sure to check the specifications and compatibility with your specific electrical panel.
Assuming each 250-watt fixture draws about 2.1 amps, you could install up to 9 fixtures on a twenty-amp breaker. This calculation is based on the formula for determining amperage (Watts / Volts = Amps) and the maximum continuous load of an electrical circuit being 80% of the circuit's rated capacity.
The horsepower rating of a 250 amp circuit breaker can be determined by using the formula: HP = (V x A x Eff) / 746, where V is the voltage (in volts), A is the amperage (in amps), and Eff is the efficiency factor (usually 0.8 for typical motors). Plug in the values and calculate to find the horsepower rating of the circuit breaker.
No, the electrical insulation rating of a 15 amp duplex receptacle is only rated at 130 volts. The 15 amp 240 volt receptacle is rated at 250 volts. The 240 volt receptacle also has a tandem pin configuration instead of the parallel configuration of the 120 volt device. The different pin configuration is so that a 120 volt electrical device can not be plugged into a 240 voltage receptacle.
The circuit breaker is used to protect the conductor that is connected to it from an overload condition. On a motor load the breaker has to be sized 250% of the motors full load amperage. Some circuits can be only loaded to 80% capacity and this is governed by the sizing of the breaker. Without knowing what the load amperage is, this question can not be answered.
A 250 volt outlet with 30 amps can handle up to 7500 watts (250V x 30A = 7500W) of power. This calculation is based on the formula Watts = Volts x Amps.
My calculation shows 15 such fixtures on that circuit.
The circuit breaker is sized to the full load amps of the motor times 250%.
The breaker should be sized to 250% of the motor's full load amperage.
Assuming each 250-watt fixture draws about 2.1 amps, you could install up to 9 fixtures on a twenty-amp breaker. This calculation is based on the formula for determining amperage (Watts / Volts = Amps) and the maximum continuous load of an electrical circuit being 80% of the circuit's rated capacity.
Tripping is caused by excessive current, which is usually the result of a fault in cable or equipment. The circuit breaker is there for safety and it trips to prevent a possible fire starting.
The horsepower rating of a 250 amp circuit breaker can be determined by using the formula: HP = (V x A x Eff) / 746, where V is the voltage (in volts), A is the amperage (in amps), and Eff is the efficiency factor (usually 0.8 for typical motors). Plug in the values and calculate to find the horsepower rating of the circuit breaker.
No, the electrical insulation rating of a 15 amp duplex receptacle is only rated at 130 volts. The 15 amp 240 volt receptacle is rated at 250 volts. The 240 volt receptacle also has a tandem pin configuration instead of the parallel configuration of the 120 volt device. The different pin configuration is so that a 120 volt electrical device can not be plugged into a 240 voltage receptacle.
In this situation, to calibrate a transmitter you need a power circuit and communicator circuit. The Hart communicator used in the calibration process is connected to the power source circuit in parallel. The power source circuit is the one that has ammeter, 250 Ohm resistor, and power source all connected in series. As the transmitter sends output mA, it creates volt drop across the 250 Ohm resister. Let's say the volt drop across the resistor was 1 Volt. Now, back to the Hart communicator. It is a load, meaning there will be a volt drop across the Hart communicator. Since it is in parallel with the power circuit, it is also parallel with the resistor. So, the 1 volt drop across the 250 Ohm resistor will also make 1 volt drop across the Hart communicator. Technically speaking, the 1 volt drop across the Hart communicator is only true if its resistor is also 250 Ohm. However, it does NOT matter what voltage drop is in the Hart communcator. It only sees the "relative" voltage drop changes to measure the changes in transmitter outputs.
The circuit breaker is used to protect the conductor that is connected to it from an overload condition. On a motor load the breaker has to be sized 250% of the motors full load amperage. Some circuits can be only loaded to 80% capacity and this is governed by the sizing of the breaker. Without knowing what the load amperage is, this question can not be answered.
A 250 volt outlet with 30 amps can handle up to 7500 watts (250V x 30A = 7500W) of power. This calculation is based on the formula Watts = Volts x Amps.
The formula you are looking for is W = I x E, Watts = Amps x Volts.
Answer for USA, Canada and countries running a 60 Hz supply service.If they fit, which I doubt, yes, you can use 600 volt fuses in place of 250 volt fuses so long as the amp rating is correct.YOU CANNOT USE 250 VOLT FUSES IN PLACE OF 600 VOLT FUSES !!!You can use a fuse with a higher volt rating but not a smaller volt rating than the voltage applied to the circuit.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 JOBSAFELY AND COMPETENTLYREFER THIS WORK TO QUALIFIED PROFESSIONALS.