100v at 1A is 100 watts, 240 v 5A is 1200 watts. The other numbers give intermediate amounts of watts.
No, you cannot always get 250 watts of power out of a mechanical system when 250 watts of power is input. Some energy may be lost due to friction, heat, or other inefficiencies in the system, resulting in a lower power output.
The efficiency of the motor is calculated by dividing the output power by the input power and multiplying by 100%. In this case, the efficiency would be (373/450)*100 = 82.9%. This means that the motor is operating at an efficiency of approximately 83%.
You need the EER rating, that will let you calculate the power used. EER is a measure of efficiency, power input versus power output. The Energy Efficiency Ratio (EER) of a particular cooling device is the ratio of output cooling (in Btu/hr) to input electrical power (in Watts) at a given operating point (indoor and outdoor temperature and humidity conditions). one ton HVAC capacity = 12,000 BTU/hour = 3500 watts so if the EER is 10, then input power is 12000/10 = 1200 watts Current used, if this is a 240 volt system, is 1200/240 = 5 amps Another similar rating is COP (Coefficient of power) which is the output power in watts (instead of BTU/h) divided by input power, so with a COD = 3, for example, input power is 3500 /3 = 1200 watts.
1 hp equals 746 Watts. You do the math. Update: There are a few different conversion rates for horsepower (metric, electric, etc), but approximately 1 watt = 0.0134 horsepower. Therefore 2500W = approx 3.35 horsepower, or 3.4 using metric horsepower. For future conversion questions use http://www.onlineconversion.com/ (not a plug, actually a useful site)
You can measure the current and power of a 'power supply', using an ammeter and a wattmeter. With the power supply connected to its load, the ammeter must be connected in series with the power supply's input. The wattmeter's current coil must also be connected in series with the power supply's input, and its voltage coil must be connected in parallel with the supply, taking the instrument's polarity markings into account.
40 watts
No, you cannot always get 250 watts of power out of a mechanical system when 250 watts of power is input. Some energy may be lost due to friction, heat, or other inefficiencies in the system, resulting in a lower power output.
You need to divide the output power by the input power. If you want to express that as a percentage, you would also multiply this result by 100.
Electrical efficiency is calculated by dividing the useful output power (in watts) by the input power (in watts) and multiplying by 100 to get a percentage. The formula is: Efficiency = (Useful output power / Input power) * 100. The higher the percentage of efficiency, the more effective the electrical system is at converting input power into useful output power.
The efficiency of a DC generator can be calculated using the formula: Efficiency (%) = (Output Power / Input Power) × 100. In this case, the input power is 200 watts and the output power is 160 watts. Thus, the efficiency is (160 W / 200 W) × 100 = 80%. This means the generator operates at 80% efficiency.
It is very critical to never exceed the input voltage of an IC to begin with. To calculate the power dissipation ( in watts) VxI (Voltage x Current) will give you power in watts.
Anything over 3400 watts is ok for a converter to power the product. The converter's rating is a maximum value so it can power anything up to 3400 watts.
Usually the power (in watts) is close to the product of volts x amperes.
1 HP = 746 watts therefore, 3 HP = 3 x 746 watts ... get out your calculator or "guesstimate" like me and say ~2250 watts or 2.3 Kilowatts. The above answer will give you the output, but for input power you have to divide by the efficiency factor (generally in the range of 0.7 to 0.9), so the input power required will be closer to 3.0 Kilowatts.
The amperage input of a transformer on a 208V three-phase system can be calculated using the formula: Amperage (A) = Power (Watts) / (Voltage (V) x √3). To determine the amperage, you need to know the power rating of the transformer in watts (or kilowatts). Once you have that, you can plug the values into the formula to find the input amperage.
The efficiency of the cyclist's body is 10% in this scenario. This is calculated by dividing the useful power output (100 watts) by the total power input (1000 watts) and multiplying by 100 to get the percentage.
The efficiency of the motor is calculated by dividing the output power by the input power and multiplying by 100%. In this case, the efficiency would be (373/450)*100 = 82.9%. This means that the motor is operating at an efficiency of approximately 83%.