for example i have "AAA"battery 1.5 volts,i want to check the initial voltage of this battery and what equipment i will use and how to use it.
211417
You'll need, as the name may suggest, a voltmeter.  Also applicable is a multimeter - which you'd then set to voltmeter mode.\n\nAt which point you'd just put one contact on one one end of the battery, and one on the other, and see what the equipment reads.\n\nMy money's on 1.5V :)
542151
Check the owners manual.
474107
http://froogle.google.com/froogle?q=220+volt+,+20Kw+generator&hl=en&lr=&safe=off&sa=X&oi=froogle&ct=title
561189
you'll need an adapter...different types of electricity
100111
Hi! You haven't mentioned which of the resistances are in series or parallel. Too bad we can't provide images in Yahoo! Answers.\n:(\n\nDo this: Remove V2 and V3 from the circuit, and replace them by a simple wire. Then find the equivalent resistance of the circuit taking V1 as the battery source. Then find the voltage drops across each of the resistances R1, R2, R3, R4 (you can do this by finding current through them using simple Ohm's Law and then multiplying by the R value).\n\nTAKE CARE that you note the DIRECTIONS of the voltage drops. The direction of voltage drop across R is the direction of current through R. So you must see if its positive or negative.\n\nOnce you have done this, repeat the procedure by removing V1, V3 and taking only V2 in circuit. This time, again specifically note the direction (+ or -). Then do it with V3.\n\nFinally, add the drops across each R, and this is when you use the signs (directions) of the voltage drops. For instance (just an EXAMPLE. You will get diff. values), if you get VR1 with V1 battery as -6V, with V2 as 4V, and with V3 as -2V, then VR1 will be:\n\nVR1 = (-6)+(4)+(-2) volts = -4 volts\n\nYou can use the same procedure to compute IR1: find current each time with V1, V2, V3 resp.\n\nHope this helps!
90043
.01uF / 1210/ 50v/ avx /digikey.com/\nMIKE
464521
First, what is the input voltage and the desired output voltage?  Is the output voltage fixed, or do you need a range of adjustment (taps)?  Do you need isolation between the input and the output?  If not, an autotransformer (single winding) might be acceptable.  What is the frequency, and is it single phase, three phase Y, or three phase delta?\n\nNext, what is the power rating?  Small transformers are usually rated in output current, whereas larger transformers are rated in KVA.  This is the maximum load;  a smaller load would not be a problem.\n\nFinally, what are the mechanical factors?  Does it need to be hermetic or can it be open frame?  Do you want terminals or wire leads?  What shape and size?  What is the mounting?\n\nIt's not really that hard.  Usually you can browse through a catalog and find the transformer you need in minutes.
360901
1 volt theoretically, actuaully closer to 1.3
480080
I don't know what methods you are using in class, but here is the way I'd start:\nProblem 7.63:\nAt t<0, you have a current source feeding a resistor network.  You can calculate the initial voltage on the capacitor.\nAt t=0, the circuit is now a capacitor (with an initial voltage) discharging through 40 k-ohms in parallel with 10 k-ohms.  You can calculate the voltage after 400 micro-sec.\nAt t=400 micro-sec, the capacitor is now discharging through 40 k-ohms.  You know the voltage at the start of this stage, so you can calculate the voltage 1 ms later.\nAt t=1.400 ms, the capacitor is again discharging through a parallel resistor network.  Knowing the voltage on the capacitor at  the begining of this stage, you can calculate the voltage on the capacitor 0.2 ms later.\nKnowing the voltage on the capacitor at t = 1.600 ms, Vo can be found.  The 5 k-ohm and the 35 k-ohm resistors form a divider network.\n\nThere you have it, just look at the circuit at the various stages and calculate the voltage on the energy storage element.
81644
OUCH!  A bit over $200. I suggest you recode the one you have and live with it. Or go to the junk yard and buy 1 or two, then recode the set. You will have to recode both when you change a battery anyway.\nIf you recode one from the junk yard, replace the battery's first.....Might want to change battery's anyway to get max life from them. \nP/S The recode proceedure is in your glove box manual. By the way, the stolen one wont work after the recode.
601076
Perhaps the link will help.  At a minimum, you risk voiding your warrant if you use other that the voltage and frequency required on the unit name plates.  At worse you risk injury and a possible fire if you really don't know what you are doing.  Good luck.
196639
sure. go to a science store or look up schematics on the web.
621136
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.
204147
http://en.wikipedia.org/wiki/Lithium_ion_battery
431730
The voltage across each resistor will be 12 volts as each is effectively directly connected to the battery.\nIt is the current flow through the resistors which will be different.\nVolts/Ohms =Amps - that is\n12/7 amps or 12/5 amps now get your calculator out!\nRoyS
350051
The amp-hour rating is the current flowing out of the battery in Amps multiplied by the time in hours it is flowing until the battery is dead.\n\nYou could measure this by putting a resistor across the battery and an ammeter in series with it - measure the current - take readings periodically - until the battery dies. Average the readings then multiply the average by the number of hours it took to die.
478522
A long time ago I found a voltage regulator for an automotive alternator that would allow output at "line" voltage (for an application similar to yours), but unfortunately I lost the information on the source.  Check at an RV (Recreational Vehicle) dealer in your area for something like this - those folks have information on stuff like that..\n\nA standard automotive alternator isn't built for the sustained high power output you're describing.  You might be able to get a higher capacity unit better suited for your project from a heavy duty truck (or an RV) out of a junkyard and get it rebuilt (complete with the new voltage regulator) for a fraction of the price of a new one.\n\nWarning:  alternators typically output full capacity at frequencies far above 50 or 60 Hertz.  Your power diodes may not work well at higher line frequencies.\n\nAlso, have you checked into getting a gasoline engine powered 120 volt power supply, like what are typically sold for emergencies power outages and campers?  You can find them for a low few hundred dollars for 1000 watts, and would a lot cheaper in the long run than risking trashing car alternators....