Power Inverter Upgrade 15+
05-01-2017, 01:16 PM
#11
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Location: In way too hot southern Georgia
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[QUOTE=
Also, will the truck's inverter charge a laptop? I haven't tied it yet but I thought I remember somebody saying it would.[/QUOTE]
It will run a laptop, charge camera batteries and other low draw items.
Also, will the truck's inverter charge a laptop? I haven't tied it yet but I thought I remember somebody saying it would.[/QUOTE]
It will run a laptop, charge camera batteries and other low draw items.
05-01-2017, 03:35 PM
#12
Texas A&M Aggie
Thread Starter
I've used it to charge my computer, but couldn't run a circular saw. I've heard several people have issues with charging batteries on it though for power tools.
09-30-2017, 12:48 PM
#13
No issues for me charging my Ryobi 18 volt battery packs or running a flat screen TV while tailgating but I wish it had more power for things like running a shop vac when I'm cleaning it out or the occasional circular saw use when I'm doing projects on a job site. Would love to hear a success story on slightly upgrading it.
08-14-2020, 03:34 PM
#15
09-30-2020, 08:48 AM
#16
Reviving an old thread. Anyone upgrade?
12-12-2020, 10:56 PM
#17
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12-13-2020, 04:13 AM
#18
5000W / 13.8VDC = 362.3188A (assuming 100% efficiency)
Multiplying this by 0.9, as you've indicated, would reduce the current consumption and push efficiency beyond 100%. Since we're all fans of the laws of thermodynamics, we'll avoid that.
Dividing this by 0.9 would increase the current consumption, which would be accurate for reduced efficiency. Thus, 5KW / 13.8VDC / 0.9 = 402.576A.
But, good job reviving a dead thread and being a one post wonder to be pedantically incorrect about basic math.
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F175 (10-12-2021)
12-13-2020, 12:34 PM
#19
Other than the fact I accidentally stated 500W rather than 5000W (although the proper value is clear from the context), my math is quite correct.
5000W / 13.8VDC = 362.3188A (assuming 100% efficiency)
Multiplying this by 0.9, as you've indicated, would reduce the current consumption and push efficiency beyond 100%. Since we're all fans of the laws of thermodynamics, we'll avoid that.
Dividing this by 0.9 would increase the current consumption, which would be accurate for reduced efficiency. Thus, 5KW / 13.8VDC / 0.9 = 402.576A.
But, good job reviving a dead thread and being a one post wonder to be pedantically incorrect about basic math.
5000W / 13.8VDC = 362.3188A (assuming 100% efficiency)
Multiplying this by 0.9, as you've indicated, would reduce the current consumption and push efficiency beyond 100%. Since we're all fans of the laws of thermodynamics, we'll avoid that.
Dividing this by 0.9 would increase the current consumption, which would be accurate for reduced efficiency. Thus, 5KW / 13.8VDC / 0.9 = 402.576A.
But, good job reviving a dead thread and being a one post wonder to be pedantically incorrect about basic math.
https://www.edn.com/efficiency-calcu...er-converters/
The biggest problem with the stupid converters is they use a modified sine wave rather than a true sine wave. This is what causes many devices not to work.
https://www.prostarsolar.net/wp-cont...-Sine-Wave.jpg
12-13-2020, 04:41 PM
#20
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Other than the fact I accidentally stated 500W rather than 5000W (although the proper value is clear from the context), my math is quite correct.
5000W / 13.8VDC = 362.3188A (assuming 100% efficiency)
Multiplying this by 0.9, as you've indicated, would reduce the current consumption and push efficiency beyond 100%. Since we're all fans of the laws of thermodynamics, we'll avoid that.
Dividing this by 0.9 would increase the current consumption, which would be accurate for reduced efficiency. Thus, 5KW / 13.8VDC / 0.9 = 402.576A.
But, good job reviving a dead thread and being a one post wonder to be pedantically incorrect about basic math.
5000W / 13.8VDC = 362.3188A (assuming 100% efficiency)
Multiplying this by 0.9, as you've indicated, would reduce the current consumption and push efficiency beyond 100%. Since we're all fans of the laws of thermodynamics, we'll avoid that.
Dividing this by 0.9 would increase the current consumption, which would be accurate for reduced efficiency. Thus, 5KW / 13.8VDC / 0.9 = 402.576A.
But, good job reviving a dead thread and being a one post wonder to be pedantically incorrect about basic math.
100/.9
And
100* .9
just saying.....