how many volts are in a AA-cell battery?
604652
All round household alkaline batteries are 1.5 volts, the ni-cad or rechargable ones are slightly higher, but they don't last as long.  It's due to the makeup of materials used to create a charge.
1095215
The above answers about using a voltage regulator sound ok.  Just make sure to check out the regulator's data sheet so that it can safely regulate what DC voltage you are giving it (though with this 12V case you shouldn't have a problem).\n\nAlso, when a battery is rated at 12V, 30A the battery doesn't necessarily output 30A at all times; that is just a full load rating given for safety reasons (don't want to overload the battery).  The battery is a constant voltage source so you can certainly have a load that pulls much less than 30A.\n\nWidth of wire?  You shouldn't be too concerned with this at these low voltages.  Just use breadboard type wire (around 22 AWG I believe).\n\nAlso, one final comment from me... why use a car battery when you can use a more convenient source of DC voltage?  If you want something that can conveniently power your IC's try one of those nifty AC/DC plugs that charge cell phones and the like.  They are rated in various output voltages and can surely supply all the current your IC's will need.  Just cut off the DC plug end and strip the wire.  Boom, now you have DC voltage for your breadboard supply terminals or whatever.
31337
It is DC voltage.\n\nhttp://www.ecs.umass.edu/ece/allerton/papers1999/36/
604716
Need to know a need or application. AC voltage by definition is bipolar. What are you trying to do?
591652
Batteries use the different potential that certain chemicals have. Battery consists of 2 chemicals which potential different are significantly enough to generate voltage difference. Reaction happening in a battery is redox (reduction-oxidation) which allows electron transfer. Positive terminal of the cell has the higher potential.\n\nAs the reactant are all used up, reaction reaches equilibrium. This is when we found out that the battery is 'dead'.
502455
In order to get 12V, you would need 8 AAA's in series (1.5V per).  Now, how many of those series combinations of batteries needs to be attached in parallel to safely provide the 500A?\n\n-AAA's have a rating of about 1000 mA-hrs = 1 A-hr (see Wikipedia source).\n\n-CCAs is defined as the battery's current draw for 30secs and not fall below a certain voltage (Battery FAQ).\n\nThe bare minimum number of parallel branches would be something like this: (A-hr ratings add in parallel)\n\n(500A)x(30sec/3600)= branches x 1 A-hr\nbranches = 4.17 => 5\n\nBut that would almost drain your battery pack within only 30 seconds, subject each branch to 100A (500amp/5), and probably force the voltage below 12V at some point.  I'm not sure what the safe current rating of AAA batteries is, but likely a little bit below 100A, haha.\n\nI would say well above 5 parallel branches to be re-usable/safe.  Maybe 500 branches would be safe and usable (though impractical).  Then you only have 1A through each branch and could start the engine about 120 times (30secs per start => 3600sec => 1hr).\n\nbatteries = 500 branches x 8 per branch = 4000... yikes
264809
AAA could be a battery size, or it could be the American Automobile Association. The category you placed this question in makes it difficult to decide just what you're asking.
420687
Without applying a potential, nothing.  The net voltage of the cell is less than zero (silver's oxidation is -0.80V, while aluminum's reduction is -1.68V).  Total voltage would be -2.48V, which is non-spontaneous.
652086
1. volts\n2. (b) negatively charged.\n3.  short.  Could also be "conductor," but the lack of control seems to indicate an unwanted short.\n4.  watt-hours.\n5.  \n6.  (a) The formula is P=V*V/R.  Where V is the voltage and R is the resistance.  So P=45*45/15=135 watts.  Actually the object is consumming the power, not producing it.\n    (b)  135*10 = 1350 watt-hours, or 1.35 kw-hr.\n    (c) 0.08*1.35 = $1.08\n7.  (a)  V=IR, so if the voltage is increased while the resistance is kept constant, then the current will also increase. (I=V/R)\n    (b)  If the resistance is increased and the voltage is kept constant, then the current has to decrease. (I=V/R)
1127734
Dunno for sure but dont be too hasty on blaming the pute. Go to radioshack and invest in a cheap volt meter, put it on dc volts and measure the voltage at the output of the charger....the charger will tell you what it should be for voltage....sounds more like the charger took a dump to me....it may also be the battery.
913090
google it??  I dont think there are any, a 9v battery is not strong enough to be fatal by itself.  actually, maybe i was wrong thinking that, if the darwin awards can be believed.
934641
If a voltage spec (v volts) is given on the resistor, then the power\nrating, p = v^2/r  ( r = resistance in ohm, you already know).\n\nIf v is not given then the only way is to connect the resistor to the poles of a battery cell via an Ammeter, and see the electric current measurement in ampere (say, i amperes).\n\nThen p = i^2 x r  (p, in watts).
209317
The straight calculation is this:\nVolts X Amps = Watts(power)\n\n4.5 watts would be the answer.
951550
The storage capacity of small batteries is generally rated in milliampere hours (mAh).  For example, an "AAA" size nickel metal hydride (NiMH) battery is rated at about 850 mAh, whereas a "D" size NiMH battery is rated at 2500 mAh.  What this means is that the voltage produced by the battery remains relatively constant for one hour when the battery is supplying the specified current in milliamperes.  This is considered the useful life of the battery, before it must be recharged (if it is a type that is rechargeable).  The battery voltage starts to drop rapidly after the mAh capacity has been consumed, until the battery is drained.\n\nIf the load on the battery is, say, 50% of the rated load, the useful life of the battery is approximately two hours.  Conversely, if the load is 200% of the rated load, the useful life of the battery is approximately a half hour.  In actuality a reduced load will result in a battery life somewhat longer than would be expected by the mAh rating, whereas an increased load would will result in a shorter life than expected.\n\nThere are other factors that come into play, such as the internal resistance of the battery, but these are beyond the scope of your question.\n\nHowever, you should know that most electronic devices draw a current that is much less than the mAh rating of the recommended battery.  (Who would want to change or recharge batteries every hour?)  The actual current usually varies, depending on the user settings or activity.  For example, a higher radio volume requires more battery current.\n\nTo answer your second question, you can substitute a battery with a higher mAh rating as long as the voltage is the same.  This will give you a correspondingly longer useful battery life.  On the other hand, substituting a battery with a lower mAh rating may not work well because the higher internal resistance mentioned above will give a somewhat reduced battery voltage.
631579
2 electrodes and a battery
275873
The Heath AC voltmeter is illustrated in the link.  Click on the diagram to enlarge it.
225854
12.5 is fine. the battery is 12 volts.