100v at 1A is 100 watts, 240 v 5A is 1200 watts. The other numbers give intermediate amounts of watts.
Yeah all you need is a completely friction less system.
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.
such a motor would be theoretically 82.8888% efficient
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.
Watts and horsepower are both units of power.
40 watts
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.
Yeah all you need is a completely friction less system.
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.
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.
The absolute power gain is the output power divided by the input power, in this case 20 / 0.1 or 200. In decibels this is 10 log 200 which is 23 dBs.
1 kp=746w1.5 hp=1119w ( 746+(746/2) )AnswerIt depends upon whether you are referring to the machine's input or output power. In North America, horsepower is still used to define a motor's output power (elsewhere in the world, the watt is used). To convert this into watts, you need to multiply by 746. However, if you want to know the input power, you must take the efficiency of the motor into account, because the input power is always higher than the output power.
If you want to work in watts, convert 25dB to a scalling factor: 3dB = 2 x input 10dB = 10 x input 20dB = 100 x input ...25dB = 10 ^ (25/10) = 316.2 x input So the output is 15 Watts x 316.2 = 4.7kW If you want to work in dB, then convert 15 watts to dB: 10 * log |P| = dB = 10*log |15| = 11.76dB so the output is 11.76 + 25 = 36.76dB
the unit which are measured for power of lights are watts....