First, let me clarify the question:
You want to use a 16V, 4.5 amp power supply to operate a device that uses 16V, 1.5 amps.
That's not a problem, the larger power supply simply has the capacity of 4.5 amps, meaning that you can use anything UP TO 4.5 amps.
On the other hand...
if you want to try using a device that pulls 4.5 amps, using a 1.5 amp power supply... sorry, won't work out well. That would overheat the power supply and it would fail.
Yes, you can use a 16V 4.5A power supply for a device that requires 16V 1.5A. The device will only draw the current it needs, so having a higher amperage rating on the power supply is not a problem. Just ensure the voltage matches and polarity is correct.
Yes, you can.
yes you can
Multiplying 400 volts by 45 amps would give you a power output of 18000 watts, which is equivalent to 18 kilowatts (kW).
To calculate the amperage per phase, you can use the formula: Current (A) = Power (W) / (Square Root(3) x Voltage (V)). In this case, for a 45 kW unit at 380 volts, the amperage per phase would be approximately 70 amps.
To calculate the amperage load, use the formula: Amperage = Power (in watts) / (Voltage (in volts) * Power Factor). For a 45 kVA load at 208-120 volts, assuming a power factor of 1, the amperage would be approximately 173A.
To determine the running amps of a 45 kW motor, you would need to know the voltage at which it operates. You can use the formula: Amps = (kW x 1000) / (Volts x power factor). Once you have the voltage and power factor information, you can plug them into the formula to calculate the running amps.
A #8 copper conductor with an insulation factor of 60 degrees C is rated at 40 amps, insulation factor of 75 or 90 degrees C is rated at 45 amps.If the breakers rating is 90 degrees C then the amperage can be boosted for 60 degree wire to 40 amps, 75 degree C wire to 50 amps and 90 degree C wire to55 amps.
The current is 1.4 amps, as already stated. The voltage is 45 x 1.4 volts.
Each phase supplies 15 kVA. The primary has a line-to-neutral voltage of 277 v so the line current is 15,000 / 277 or 54 amps. The secondary has a line-to-neutral voltage of 120v so the current is 15,000/120 or 125 amps.
Multiplying 400 volts by 45 amps would give you a power output of 18000 watts, which is equivalent to 18 kilowatts (kW).
To calculate the amperage per phase, you can use the formula: Current (A) = Power (W) / (Square Root(3) x Voltage (V)). In this case, for a 45 kW unit at 380 volts, the amperage per phase would be approximately 70 amps.
Just add the amps (3.2 amps).
The answer to the LCM of 45 and 15 is 45.
To calculate the amperage load, use the formula: Amperage = Power (in watts) / (Voltage (in volts) * Power Factor). For a 45 kVA load at 208-120 volts, assuming a power factor of 1, the amperage would be approximately 173A.
A walmart ever last wiil be fine at 350 c-amps, if you use your mower at temps below 45-f, or have more than 20 hp then go up to 420 c-amps.
3 * 15 = 45 or 15 = 45 / 3 or 3 = 45 / 15
To determine the running amps of a 45 kW motor, you would need to know the voltage at which it operates. You can use the formula: Amps = (kW x 1000) / (Volts x power factor). Once you have the voltage and power factor information, you can plug them into the formula to calculate the running amps.
45 - 15 = 3030 - 15 = 15
45 - 15 - 15 = 15