2007-07-01 19:12:45 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Engineering
Since the load is100W, the current draw @ 12 Volts would ideally be 100/12 = 8.33 Amps IF the inverter were 100% efficient.
Low-power inverters are notoriously in-efficient, around 50 - 60%, so you will be draining almost twice that 8.333 Amps, or about 16 amps.
16 amps is approximately 1/10th the rated capacity, which is a good thing -- the battery will probably last for its rated capacity. 150 AH / 16 A = 9.3 hours. But the inverter will probably provide useful power past that time, perhaps 10 or 11 hours.
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2007-07-02 04:10:18 · answer #1 · answered by tlbs101 7 · 0⤊ 0⤋
Since watts = volts x amps, multiply the 12 volts by the 150 amp-hours to get 1800 watt-hours. Divide that by the wattage you are going to use (100) and you should get 18 hours out of the battery, regardless of the voltage you use. That's the theoretical maximum, but you should be able to get 16 hours at least.
2007-07-01 19:23:02 · answer #2 · answered by TitoBob 7 · 0⤊ 0⤋
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2017-01-23 08:14:31 · answer #3 · answered by ? 3 · 0⤊ 0⤋
the power supplied by the battery=100w
so current supplied =100/12
since the battery is 150 AH,
150=current *time
time=18hrs
i am a student so dont use this ans for professional purpose
2007-07-01 19:26:57 · answer #4 · answered by tejas p 1 · 1⤊ 0⤋
Its Simple, Work It out
2007-07-01 19:21:11 · answer #5 · answered by Anonymous · 0⤊ 1⤋