(I'm guessing you're not thinking about attempting to charge the battery to 40 V. if that is the case, the battery will fail, possibly even explosively, which could be counted as dramatically reducing the amp-hours!)
So I'll assume you're planing on the doing the voltage boost by some kind of voltage converter. Even if the converter were 100% efficient (hypothetical) you just can't get away from...
Electrical power = current * voltage
So I could say
Pbatt=Abatt*Vbatt
Pcircuit=Acurcuit * Vcircuit
and assuming an idea conversion
Pbatt=Pcircuit
You've fixed Vbatt and Vcircuit, so all that can be done is increase amp hours is to reduce Pbatt via Pcircuit via Acurcuit.
Scenerio you describe might not be quite as daft as it first seems. With some output circuits (those that draw the power, connected after your converter) you MIGHT just draw proportionally less current, hence power, when they are run at higher voltage. You'd need a current lower than 12/40 of the original output current, to gain though. That's the really difficult part.
2007-12-09 13:02:59
·
answer #1
·
answered by Steve C 6
·
0⤊
0⤋
A lot of the previous answers are missing the point I think.
At the point you have 12V, you have the rated amp*hours. If you increase the voltage to 40V through a converter, you still have rated amp*hours on the 12V side but only 12/40 as much on the output 40V side (not counting efficiency loss). If you increase the voltage to 40V by adding more batteries in series, assuming the added batteries also have the same amp*hour rating, then you have the rated amp*hours at the output..
2007-12-09 16:16:19
·
answer #2
·
answered by semdot 4
·
0⤊
0⤋
Sorry the question doesn't make sense. if you have a twelve volt battery. You have a twelve volt battery period. That will supply 1 amp for 1000 hours. (Damn big batery) How you supposed to increase the voltage output. If you could you wouldn't have a 12 v battery would would have a 40V battery. If you have a 40V battery but have it set to only put out 12V at 1amp then it's still going to depend on the load
2007-12-09 12:43:56
·
answer #3
·
answered by Anonymous
·
0⤊
0⤋
For a given battery, you cannot increase the voltage from 12v to 40v. You would have to have a different battery and then the ampere-hours would be a function of the batteries construction.
I really don't understand what you are after with your question. The first answer is correct with regard to transmission of ac utility power, but is not really germane to your question.
2007-12-09 12:46:11
·
answer #4
·
answered by Anonymous
·
0⤊
0⤋
You have a certain amount of energy or power. In this case, watts: 12x1K=12K/hour. 12K/40V=300A. How you suddenly get 40V from a 12V battery is a mystery to me; 2000amps for a 1/2 hour from a 1000amp/hour battery makes more sense.
2007-12-09 12:41:04
·
answer #5
·
answered by Anonymous
·
0⤊
0⤋
I suppose you're talking about using an inverter of some sort. And yes, it will greatly reduce the number of amp hours you can get out of it. The total amount of energy is supposed to remain constant, and that is the product of the voltage, the number of amps, and the duration of the load. Good night!
2007-12-09 12:38:16
·
answer #6
·
answered by anonymous 7
·
2⤊
0⤋
Amp hors times volts equals watt hours. That is 12000 watt hours. Divide that by 40 volts, and you get 300 amp hours.
2007-12-09 15:52:36
·
answer #7
·
answered by Anonymous
·
0⤊
0⤋
thats usualy the case , you cant have it both ways, higher voltage lower amps, thats why transmission line are high voltage so that the amperes are lower
2007-12-09 12:36:08
·
answer #8
·
answered by grd_jck(AU) 4
·
0⤊
0⤋
upload man or woman KVA load = KA * V assumed capability element used interior the generating unit say PF = 0.80 the definitely capability intake KW is KW = KVA * PF KW = KVA * 0.80 please be conscious which you may't vectorial upload KW yet you may upload KVA. because of the fact each load has diverse capability element (PF).
2016-11-14 06:06:01
·
answer #9
·
answered by Anonymous
·
0⤊
0⤋