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I think I might be losing power through my 2000 watt inverter. It won't run 130 watts all night long. I have 2 6volt Trojan t105re batteries rated at 225ah each wired together. They connect to the 2000 watt inverter which connects to a power strip with a fan and a 75 watt light. They don't make it through the night when the batteries are fully charged. Shouldn't they?
And got the following answer:
You DO realize that even when idling, an inverter draws a substantial amount of power, yes? OK, so how much power is lost to overhead in your inverter? From my experience, inverters run from about 50% efficient to as much as 90% efficient depending on the load. The more load, the greater the efficiency. I have a small inverter, only 500 watts that I run off a car battery as backup power for my computers. Without any load, it draws about 1 amp at 12 volts, for an overhead cost of 12 watts. If I put on a 12 watt load, it transfers across and the inverter now draws 24 watts, or 2 amps from the battery. With the bulb, you've added 75 watts. On my inverter, that would correspond to about 7.5 amps, plus overhead for a total drain of 8.7 amps. Since the overhead cost in current is relatively constant for my tiny inverter, my efficiency went from 50% with a 12 watt load, to about 87% with a 75 watt load. 75 watts out plus 12 watts overhead, means 87 watts from the battery. You say the load is 130 watts. If so, on my system, that would correspond to 142 watts from the battery, for 14.2 amps. That is equivalent to running the headlights on a typical car. My 80AH car battery would run that much load for about 4 to 5 hours in an ideal world, but in reality the inverter would stop on a low voltage situation after about 2 to 3 hours because a car battery is not designed for that kind of use. I really should have a deep cycle RV type of battery. Anyway, do you see the picture here? What you need to do is measure the current drain from the batteries, under no load and various known loads and figure out your overhead cost to use the inverter. If you don't have one, you need to get a current shunt. A current shunt is simply an accurately calibrated low value resistor. They come in various ranges. For example, a 50 amp shunt might be calibrated to have a voltage drop of 500 millivolts at 50 amps. Then, you put the shunt in the circuit in series with the battery and use a millivolt meter to measure the voltage drop across the shunt. Every current meter works in the same manner, voltage drop across a known resistance. My 50 amp shunt is accurate to 1/10th of an amp in a DC circuit since it reads 10 millivolts per amp of current. 130 watts of load would drain my 80AH battery in a few hours and I would expect to see a 14.2 amp current drain. Your batteries have 3x the capacity of mine (assuming your batteries are in series, for 12 volts 225AH), so I would expect maybe 9 hours to my 3 hours. There is another factor to consider, wave form. That fan is an induction motor. An induction motor works best with a sine wave. If your inverter is putting out square waves, the efficiency of the motor goes down drastically, which means it draws more current, as much as 30% less efficient, which means if rated at 55 watts, 130 watts - 75 watts, it really is drawing about 70 watts worth though rated at 55 watts because of the less efficient use of square waves instead of sine waves. Unless your inverter says it puts out sine waves, it is putting out square waves or modified square waves.