It depends on the voltage. A load of 32 amps at 120 volts will be 3.84 kW. A load of 32 amps at 240 volts will be 7.68 kW. For any other voltage, multiply the voltage by 0.032. All these calculations assume a resistive (non-reactive) load.
32amp IN kielowat
There are zero kW in 32 amps. Watts are the product of amps times volts. Once you find the voltage of the system multiply it times 32 amps and then divide that answer by 1000. This will give you the answer in kW.
The code book states that the motor will draw 1.8 amps. <<>> 1 amp
It depends on what kind of amp it is. Is it an audio amp or an rf amp . . A 3 kW audio amp would draw about 300 watts on average at most, so with a good reservoir capacitor the supply current would be 0.2 amps. A 3 kW rf amp for AM radio would draw about 4500 watts so the supply current would be 3 amps.
The equation that you are looking for is kW = Amps x Volts/1000. My average Sony audio amp at home uses about 20 watts or .02 kW at a low audio output level.
The amount of 4 kW is equal to zero amps. To find the amperage the following equation should be used. Amps = Watts/Volts. So you can see that it depends on the voltage of the circuit if the amperage is to be found.
No; drawing more than the rated amperage from a transformer will cause it to overheat.
There are zero kW in 32 amps. Watts are the product of amps times volts. Once you find the voltage of the system multiply it times 32 amps and then divide that answer by 1000. This will give you the answer in kW.
What is the KW of 4 50 amp rectifiers at a 48 volt DC system
no
Look at the motor nameplate and it shoud have the amp draw on it. If the nameplate is missing, then the amp draw depends on what type of motor it is. The basic calculation to get you in the ball park would be as follows: 1 HP = 0.75 KW 7.5 HP = 5.63 KW Assume the efficiency of the motor is 80%, then the power supplied will need to be 5.63/0.8 = 7.04 KW amp draw = 7040/220 = 32 amps <<>> For calculation purposes the electrical code book states that a 7.5 HP motor draws 40 amps.
The most you could use would be 48 kW. P = E x I = 240 x 200 = 48,000 = 48 kW
A 32 amp fixture can not be fed from a 20 amp breaker as the breaker will trip every time.
The code book states that the motor will draw 1.8 amps. <<>> 1 amp
It depends on what kind of amp it is. Is it an audio amp or an rf amp . . A 3 kW audio amp would draw about 300 watts on average at most, so with a good reservoir capacitor the supply current would be 0.2 amps. A 3 kW rf amp for AM radio would draw about 4500 watts so the supply current would be 3 amps.
4 pole, 5 wire
P= (1/1000) UI (cos φ ) √3where P is in kW; I in A; U in V ;hence I= 1000 P / U (cos φ ) √3
The equation that you are looking for is kW = Amps x Volts/1000. My average Sony audio amp at home uses about 20 watts or .02 kW at a low audio output level.