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Anonymous Posted on Apr 06, 2014

Fuse for co lemme pmp400 inverter

Fuse

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cwradio73

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  • Master 1,162 Answers
  • Posted on Jul 19, 2017
cwradio73
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1 Related Answer

Anonymous

  • 577 Answers
  • Posted on Aug 27, 2011

SOURCE: I have a coleman powermate

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 :

f9ae53a99f2b2b6a74146fb04fb3ff73.png


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

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I have a coleman powermate pmp400-t inverter 400w/800w will it power a sump pump? I tried it and it beeped and the fault light came on

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 :

f9ae53a99f2b2b6a74146fb04fb3ff73.png


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
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