Hello, the answer is no. This inverter will work with a pump that below or about
400 watts power for operation. I notice it says that the inverter 400w/800w. The
800 watts is the surge required to get an electric motor to run. On a single
phase motor (120 AC) and it take 4 amperes to operator this motor. But to start
the it will 12 amperes to start it. This is one reason why on a lot of
electrical motors there will be a big tube or can, this is the starting
capacitor, it stores energy in form of current (amperes) to drop across the
windings of the motor when it starts. If some electric motors didn't have this
capacitor. The most likely scenario would be tripping a breaker because it
drawing to much current.
Now, for you problem. Example of Power and the unit of the formula is V
(volt), R (resistance) I (current) P (power) Ohm's states that :
Now Power would be stated: P = V times(x) I. Therefore in your case, if the
pump at normal operation is drawing 4 amps of current for operation pumping
water. The power factor would be 120 V AC (house voltage) times (X) I (current)
which 4 amps. So here is what you have: P = 120 X 4 equal= 480 watts.
Therefore, your inverter will not work because it operating at 80 watts more
than the power inverter can deliver which is 400 Watts.
If you need to purchase a new power inverter, remember always take into
account the Amperes required for operation. Just by the simple Ohm's Formula.
Best of Luck. GB....stewbison
son
