Volt amps is the same way of saying watts. This is seen in the formula Watts = Volts x Amps. Using the formula I = W/E we transpose I = 20000/240 = 83.33 amps.
That's a powerful motor, and would need a supply greater than 13 amps at 240 volts, also the conversion might be difficult. I have a feeling that you would be better off finding out how powerful a motor you would need for the task you have, then buying a suitable 240 volt motor. You are heading into territory where you should really consult a qualified electrician.
The formula you are looking for is I = W/E. Amps = Watts/Volts.
On a 1kva you have 1000 watts capacity. To fine the current the formula is I = W/E. The secondary side of the transformer has the capacity of 1000/120 = 8.3 amps. In your question you do not put the amps across the secondary you draw amps from it. Using the transformer to its maximum, without overloading it, the primary will be 4.16 amps at 240 volts and the secondary will be 8.33 at 120 volts. <<>> voltage times amps equals wattage
A three phase 30 HP at 240 volts draws 80 amps. Motor feeders have to be rated at 125% of the motors full load amps. 80 x 125% = 100 amps. A # 3 copper conductor with an insulation factor of 75 or 90 degrees C are rated at 100 and 105 amps respectively. Non-time delay fusing is 225 amps, time delay fusing is 125 amps and a circuit breaker is sized by 250% x 80, the full load amps, which equals 200 amps.
24.87amps 1 hp=746 watts P=IxV ... (746x8)/240
It would be at least 250 amps, maybe 300 amps.
To calculate the amp draw for a 3800-watt water heater at 240 volts, use the formula: Amps = Watts / Volts. Thus, the amp draw would be 3800 watts / 240 volts, which equals approximately 15.83 amps. Therefore, the water heater draws about 16 amps.
To find amps if watts and volts are known, use the formula; watts / volts = amps or 5000 / 240 = 20.83 amps
AWG #10 copper on a 30 amp breaker.
Presuming single phase and 1500 rpm. Normal running current would nearly 10 Amps but varies considerably. A real cheap one running at 3000/3600 rpm could exceed 12 amps. At 240 volt, all should be run off 15 Amp line.
In a 240-volt motor, the three leads typically refer to the three phases in a three-phase motor. Each phase will carry current, and the amperage can be measured at each lead. However, the total current draw of the motor will depend on the load and the motor's design, so the readings on each lead may vary. It's important to ensure that all three phases are balanced for optimal motor performance.
no
No, a 31-watt motor operating at 240 volts would draw approximately 0.13 amps. This calculation is done using the formula P = IV, where P is power (31 watts), I is current (amps), and V is voltage (240 volts).
The formula for amps is I = W/E. Amps = 40/240 = .17 primary amperage. For the secondary amperage I = W/E. Amps = 40/24 = 1.7 amps.
Depends on the size of the LED light and the voltage applied. An example is an LED 24 volt globe light that pulls 8 watts which draw 0.333333 amps. Take an LED 120 volt light bulb draws 12 watts and will pull 0.1 amps. The same bulb at 240 volts wil draw 0.05 amps. it really depends on the watts and voltage applied. An average would be about 0.1 amps.
Minimum current would be 10000 divided by 240 but it might be up to 30% more if the load has a poor power factor.