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.
yes you can
400 Volts X 45 Amps = 18,000 Watts
The equation that you are looking for is; amps when kilowatts are known. kW x 1000/1.73 x Volts x pf.45 x 1000 = 45,000/208 x 1.73 x pf. Power factor to use will be .9. 45000/ 324 = 139 amps.
I am assuming that you are talking single phase. 45 kva is k = 1000, v = volts, a = amps. 45 kva is 45000 volt / amps. Input 45000 divided by 208 volts = 216 amps. Output 45000 divided by 120 volts = 375 amps. There are other losses in the transformer but as a general rule of thumb this is the calculation that you would use.
The formula you are looking for is I = W/E. Amps = Watts/Volts.
A #8 copper wire with an insulation rating of 90 degrees C is rated at 45 amps.
The current is 1.4 amps, as already stated. The voltage is 45 x 1.4 volts.
400 Volts X 45 Amps = 18,000 Watts
The equation that you are looking for is; amps when kilowatts are known. kW x 1000/1.73 x Volts x pf.45 x 1000 = 45,000/208 x 1.73 x pf. Power factor to use will be .9. 45000/ 324 = 139 amps.
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.
I am assuming that you are talking single phase. 45 kva is k = 1000, v = volts, a = amps. 45 kva is 45000 volt / amps. Input 45000 divided by 208 volts = 216 amps. Output 45000 divided by 120 volts = 375 amps. There are other losses in the transformer but as a general rule of thumb this is the calculation that you would use.
The formula you are looking for is I = W/E. Amps = Watts/Volts.
Just add the amps (3.2 amps).
The answer to the LCM of 45 and 15 is 45.
A #8 copper wire with an insulation rating of 90 degrees C is rated at 45 amps.
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
45 - 15 = 3030 - 15 = 15